tag:blogger.com,1999:blog-37414104180967168272024-03-18T08:02:33.594-07:00Unizor - Creative Mind through Art of MathematicsUnizor is a site where students can learn the high school math (and in the future some other subjects) in a thorough and rigorous way. It allows parents to enroll their children in educational programs and to control the learning process.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.comBlogger639125tag:blogger.com,1999:blog-3741410418096716827.post-42334317022474411862024-03-18T08:01:00.000-07:002024-03-18T08:01:43.766-07:00Arithmetic+ 06: UNIZOR.COM - Math+ &Problems - Arithmetic<iframe width="480" height="270" src="https://youtube.com/embed/0216LkX8dU4?si=Y1vJXb5pyX68k5Xz" frameborder="0"></iframe>
<i>Notes to a video lecture on http://www.unizor.com</i><br/>
<br/>
<u> Arithmetic+ 06</u><br />
<br />
<i>Problem A</i><br/>
A positive integer number in decimal notation contains only digits <i><b>1</b></i> and <i><b>0</b></i>.<br/>
The digit <i><b>1</b></i> occurs <i>111</i> times, while the digit <i><b>0</b></i> occurs an unknown number of times.<br/>
Can this number be a square of another integer number?<br/>
<br/>
<i>Hint A</i>: Use the rules of divisibility.<br/>
<br/>
<i>Answer A</i>: No.<br/>
<br/>
<br/>
<i>Problem B</i><br/>
This problem is based on <i>Problems 04(B,C)</i> of this course <i>Math+ & Problems</i>.<br/>
Prove that if the sum of digits of some natural number <i><b>N</b></i> is the same as the sum of digits of the number <i><b>k·N</b></i>, where <i><b>k−1</b></i> <b>is not divisible by <i><b>3</b></i></b>, then number <i><b>N</b></i> is divisible by <i><b>9</b></i>.<br/>
<br/>
<i>Hint B</i>:<br/>
<i>Problem 04(B)</i> stated that a remainder of the division of some natural number by 9 is the same as a remainder of the division by 9 of the sum of this number's digits.<br/>
Therefore, both <i><b>N</b></i> and <i><b>k·N</b></i> have the same remainder if divided by <i><b>9</b></i>.<br/>
<br/>
<br/>
<i>Problem C</i><br/>
Given a number <i>N</i> with a decimal representation <i>999...9</i> that contains <i>k</i> digits <i>9</i>.<br/>
Assume for definitiveness, <i>k</i> is a prime number.<br/>
Find a number whose decimal representation contains only digits <i>1</i> that is divisible by <i>N</i>.<br/>
<br/>
<i>Answer C</i>: <i><b>111...1</b></i> should contain <i><b>9·k</b></i> digits <i>1</i>.<br/>
<br/>
<i>Example C</i>: For <i><b>N=99</b></i> (<i><b>k=2</b></i>) the number <i><b>111...1</b></i> that contains <i>9·2=18</i> digits <i>1</i> is divisible by <i><b>N</b></i>.<br/>
<br/>
<br/>
<i>Problem D</i><br/>
Consider the number<br/>
<i><b>N=(k+1)·(k+2)·...·(2k−1)·(2k)</b></i><br/>
How many <i>2</i>'s, depending on <i><b>k</b></i>, are in the representation of this number as a product of prime numbers?<br/>
<br/>
<i>Hint D</i>: Notice that<br/>
<i><b>N=(2k)!<font size=4>/</font>(k!)</b></i><br/>
Then <i><b>N=(2k)!</b></i> can be represented as a product of only odd numbers by only even numbers.<br/>
<br/>
<i>Answer D</i>: <i><b>N=2<sup>k</sup>·M</b></i><br/>where <i><b>M</b></i> is an odd number,<br/>
so the number of <i>2</i>'s in the representation of number <i><b>N</b></i> as a product of prime numbers is <i><b>k</b></i>.<br/>
<br/>
Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-41506120029249907652024-03-15T09:14:00.000-07:002024-03-15T09:14:59.732-07:00Logic+ 05: UNIZOR.COM - Math+ & Problems - Logic<iframe style="background-image:url(https://i.ytimg.com/vi/lCYxKYxsJnQ/hqdefault.jpg)" width="480" height="270" src="https://youtube.com/embed/lCYxKYxsJnQ?si=BgcIb7kJectlen_K" frameborder="0"></iframe>
<i>Notes to a video lecture on http://www.unizor.com</i><br/>
<br/>
<u> Logic+ 05</u><br />
<br />
<i>Problem A</i><br/>
<br/>
A casino manager, analyzing the results of operations of roulette tables during a period of one day, comes to the following statistics:<br/>
there were <i><b>N<sub>1</sub></b></i> people who won at <b>least</b> <i><b>1</b></i> time;<br/>
there were <i><b>N<sub>2</sub></b></i> people who won at <b>least</b> <i><b>2</b></i> times;<br/>
there were <i><b>N<sub>3</sub></b></i> people who won at <b>least</b> <i><b>3</b></i> times;<br/>
etc.<br/>
there were <i><b>N<sub>n</sub></b></i> people who won at <b>least</b> <i><b>n</b></i> times;<br/>
and nobody won more than <i><b>n</b></i> times.<br/>
<br/>
How many times casino has lost in roulette?<br/>
<br/>
<i>Hint A</i>:<br/>
<i>N<sub>1</sub> ≥ N<sub>2</sub> ≥ N<sub>3</sub> ≥...≥ N<sub>n</sub></i><br/>
<br/>
<i>Answer A</i><br/>
Casino has lost<br/>
<i><b>N<sub>1</sub>+N<sub>2</sub>+N<sub>3</sub>+...+N<sub>n</sub></b></i> games.<br/>
<br/>
<br/>
<i>Problem B</i><br/>
<br/>
How many rooks can be placed on a chessboard such that no two rooks prevent each other to move along chessboard from one edge to another vertically or horizontally?<br/>
<br/>
<i>Answer B</i>: <i><b>8</b></i> rooks.<br/>
<br>
<br/>
<i>Problem C</i><br/>
<br/>
Squares on a chessboard are enumerated as follows:<br/>
1st row: <i><b>1, 2,...,8</b></i><br/>
2nd row: <i><b>9, 10,...,16</b></i><br/>
3rd row: <i><b>17, 18,...,24</b></i><br/>
etc.<br/>
8th row: <i><b>57, 58,...,64</b></i><br/>
<br/>
Eight rooks are placed on a chessboard such that no two rooks prevent each other to move along chessboard from one edge to another vertically or horizontally.<br/>
<br/>
What is the sum of numbers on the squares occupied by these rooks?<br/>
<br/>
<i>Hint</i>:<br/>
Each number <i>N</i> on a square can be represented as<br/>
<i>N=(R−1)·8+C</i>, where<br/>
<i>R</i> is the row number (from 1 to 8) and<br/>
<i>C</i> is the column number (also from 1 to 8).<br/>
<br/>
<i>Answer C</i>: <i><b>260</b></i>.<br/>
<br/>
<br/>
<i>Problem D</i><br/>
<br/>
There is a cup of coffee and a cup of milk. Assume, both cups contain the same amount of liquid.<br/>
A spoon of milk is taken from a milk cup and added to coffee.<br/>
Then, after stirring the coffee, a spoon of coffee with milk is taken and added to a milk cup.<br/>
Which is greater,<br/>
a concentration (in %) of milk in a coffee cup or<br/>
a concentration of coffee in a milk cup?<br/>
<br/>
<i>Answer D</i>: They are the same.<br/>
<br>
<br/>
<i>Problem E</i><br/>
<br/>
Given a table with each its cell containing some number.<br/>
Any number in this table is equal to an arithmetic average of its neighbors - those numbers that this number shares a cell's border with.<br/>
If a number is in the middle of a table, it has 4 neighbors (up, down, left, right).<br/>
If a number is at the edge of a table, it has 3 neighbors (for example, for a number near the top border the neighbors are left, right and down).<br/>
If a number is in the corner of a table, it has only 2 neighbors (for example, for a number in the bottom right corner the neighbors are up and left).<br/>
<br/>
Prove that all numbers in the table are equal to each other.<br/>
<br/>
<i>Hint E</i>: Consider the neighbors of the smallest or the biggest number.<br/>
<br/>
<br/>
Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-60062597267122366812024-03-07T08:22:00.000-08:002024-03-07T08:22:21.584-08:00Geometry+ 06: UNIZOR.COM - Math+ & Problems - Geometry<iframe width="480" height="270" src="https://youtube.com/embed/dJOKF9Y7CEA?si=jLp9k2iezCpqADXy" frameborder="0"></iframe>
<i>Notes to a video lecture on http://www.unizor.com</i><br/>
<br/>
<u> Geometry+ 06</u><br />
<br />
<i>Problem A</i><br/>
<br/>
Given an isosceles triangle Δ<i><b>ABC</b></i> with <i><b>AB=BC</b></i> and ∠<i><b>ABC=20</b></i>°.<br/>
Point <i><b>D</b></i> on side <i><b>BC</b></i> is chosen such that ∠<i><b>CAD=60</b></i>°.<br/>
Point <i><b>E</b></i> on side <i><b>AB</b></i> is chosen such that ∠<i><b>ACE=50</b></i>°.<br/>
Find angle ∠<i><b>ADE</b></i>.<br/>
<br/>
<i>Solution A</i><br/>
<img src='http://www.unizor.com/Pictures/ProblemsGeo_06A.png' style='width:200px;height:380px;'><br/>
<br/>
<br/>
<i>Problem B</i><br/>
<br/>
Given two triangles Δ<i><b>A<sub>1</sub>B<sub>1</sub>C<sub>1</sub></b></i> and Δ<i><b>A<sub>2</sub>B<sub>2</sub>C<sub>2</sub></b></i> with the following properties:<br/>
(a) side <i><b>A<sub>1</sub>B<sub>1</sub></b></i> of the first triangle equals to side <i><b>A<sub>2</sub>B<sub>2</b></i> of the second;<br/>
(b) angles opposite to these sides, ∠<i><b>A<sub>1</sub>C<sub>1</sub>B<sub>1</sub></b></i> and ∠<i><b>A<sub>2</sub>C<sub>2</sub>B<sub>2</sub></b></i>, are equal to each other;<br/>
(c) bisectors of these angles, <i><b>C<sub>1</sub>X<sub>1</sub></b></i> and <i><b>C<sub>2</sub>X<sub>2</sub></b></i>, are also equal to each other.<br/>
<br/>
Prove that these triangles are congruent.<br/>
<br/>
<i>Proof</i><br/>
Put side <i><b>A<sub>1</sub>B<sub>1</sub></b></i> on top of <i><b>A<sub>2</sub>B<sub>2</sub></b></i> and draw a circle around Δ<i><b>A<sub>1</sub>B<sub>1</sub>C<sub>1</sub></b></i>.<br/>
Obviously, point <i><b>C<sub>2</sub></b></i> would lie on this circle because of equality of angles opposite to equal sides.<br/>
Assume that points <i><b>C<sub>1</sub></b></i> and <i><b>C<sub>2</sub></b></i> do not coincide.<br/>
Extend bisectors <i><b>C<sub>1</sub>X<sub>1</sub></b></i> and <i><b>C<sub>2</sub>X<sub>2</sub></b></i> to intersection with a circle at point <i><b>P</b></i>. It must be the same point for both bisectors because each divides the arc ⌒<i><b>A<sub>1</sub>B<sub>1</sub></b></i> in half.<br/>
<img src='http://www.unizor.com/Pictures/ProblemsGeo_06B.png' style='width:200px;height:200px;'><br/>
First, let's prove that triangle Δ<i><b>PC<sub>1</sub>C<sub>2</sub></b></i> is similar to triangle Δ<i><b>PX<sub>1</sub>X<sub>2</sub></b></i><br/>
Our proof will be based on congruency of two angles of one triangle to two corresponding angles of another.<br/>
For brevity, when we use an expression "angle is measured by an arc (or a fraction of it)", we mean that this angle is equal to a central angle subtended (supported) by this arc (or a fraction of it).<br/>
<br/>
We will also use known theorems of Geometry:<br/>
(<i>Theorem 1</i>) An angle inscribed into a circle (that is, formed by two chords with one common point on a circle) is measured by half an arc it cuts from a circle by its two rays.<br/>
(<i>Theorem 2</i>) An angle formed by two chords intersecting inside a circle is measured by half of a sum of two opposite arcs these chords cut.<br/>
These theorems were presented in <i>UNIZOR.COM - Math 4 Teens - Geometry - Circles - Mini Theorems 1</i> and <i>...Mini Theorems 2</i>.<br/>
<br/>
1. ∠<i><b>PC<sub>1</sub>C<sub>2</sub></b></i> is measured by half of sum of arcs ⌒<i><b>C<sub>2</sub>B<sub>2</sub></b></i> and ⌒<i><b>B<sub>2</sub>P</b></i>;<br/>
∠<i><b>PX<sub>2</sub>A<sub>2</sub></b></i> is measured by half of sum of arcs ⌒<i><b>C<sub>2</sub>B<sub>2</sub></b></i> and ⌒<i><b>PA<sub>2</sub></b></i>;<br/>
but arcs ⌒<i><b>B<sub>2</sub>P</b></i> and ⌒<i><b>PA<sub>2</sub></b></i> are equal, from which follows that<br/>
∠<i><b>PC<sub>1</sub>C<sub>2</sub></b></i> equal to ∠<i><b>PX<sub>2</sub>A<sub>2</sub></b></i>.<br/>
<br/>
2. ∠<i><b>PC<sub>2</sub>C<sub>1</sub></b></i> is measured by half of sum of arcs ⌒<i><b>C<sub>1</sub>A<sub>2</sub></b></i> and ⌒<i><b>A<sub>1</sub>P</b></i>;<br/>
∠<i><b>PX<sub>1</sub>B<sub>1</sub></b></i> is measured by half of sum of arcs ⌒<i><b>C<sub>2</sub>B<sub>2</sub></b></i> and ⌒<i><b>PA<sub>2</sub></b></i>;<br/>
but arcs ⌒<i><b>B<sub>2</sub>P</b></i> and ⌒<i><b>PA<sub>2</sub></b></i> are equal, from which follows that<br/>
∠<i><b>PC<sub>1</sub>C<sub>2</sub></b></i> equal to ∠<i><b>PX<sub>2</sub>A<sub>2</sub></b></i>.<br/>
<br/>
Since two angles of Δ<i><b>PC<sub>1</sub>C<sub>2</sub></b></i> are congruent to two angles of Δ<i><b>PX<sub>1</sub>X<sub>2</sub></b></i>, these triangles are similar.<br/>
Consequently, their sides are proportional<br/>
<i><b>PC<sub>1</sub>/PX<sub>2</sub> = PC<sub>2</sub>/PX<sub>1</sub></b></i><br/>
<br/>
If angle bisectors <i><b>C<sub>1</sub>X<sub>1</sub></b></i> and <i><b>C<sub>2</sub>X<sub>2</sub></b></i> are equal, segments <i><b>C<sub>1</sub>P</b></i> and <i><b>C<sub>2</sub>P</b></i> must be equal to preserve the proportionality of corresponding sides of triangles Δ<i><b>PC<sub>1</sub>C<sub>2</sub></b></i> and Δ<i><b>PX<sub>1</sub>X<sub>2</sub></b></i>.<br/>
Indeed, if <i><b>C<sub>1</sub>X<sub>1</sub>=C<sub>2</sub>X<sub>2</sub>=d</b></i>, <i><b>PX<sub>1</sub>=x<sub>1</sub></b></i> and <i><b>PX<sub>2</sub>=x<sub>2</sub></b></i>, then<br/>
<i><b>(d+x<sub>1</sub>)/x<sub>2</sub> = (d+x<sub>2</sub>)/x<sub>1</sub></b></i><br/>
<b>⇒</b> <i><b>d·x<sub>1</sub>+x<sub>1</sub>² = d·x<sub>2</sub>+x<sub>2</sub>²</b></i><br/>
<b>⇒</b> <i><b>d·(x<sub>1</sub>−x<sub>2</sub>) = (x<sub>2</sub>−x<sub>1</sub>)·(x<sub>1</sub>+x<sub>2</sub>)</b></i><br/>
<b>⇒</b> <i><b>x<sub>1</sub> = x<sub>2</sub></b></i>, because otherwise <i><b>d</b></i> would be negative <i><b>−(x<sub>1</sub>+x<sub>2</sub>)</b></i>.<br/>
<br/>
From <i><b>C<sub>1</sub>P = C<sub>2</sub>P</b></i> follows<br/>
<b>⇒</b> ⌒<i><b>C<sub>1</sub>A<sub>1</sub>P = </b></i>⌒<i><b>C<sub>2</sub>B<sub>2</sub>P</b></i><br/>
<b>⇒</b> ⌒<i><b>C<sub>1</sub>A<sub>1</sub> = </b></i>⌒<i><b>C<sub>2</sub>B<sub>2</sub></b></i><br/>
<b>⇒</b> <i><b>C<sub>1</sub>A<sub>1</sub> = C<sub>2</sub>B<sub>2</sub></b></i><br/>
Analogously,<br/>
<b>⇒</b> <i><b>C<sub>1</sub>B<sub>1</sub> = C<sub>2</sub>A<sub>2</sub></b></i><br/>
which proves that triangles Δ<i><b>A<sub>1</sub>B<sub>1</sub>C<sub>1</sub></b></i> and Δ<i><b>A<sub>2</sub>B<sub>2</sub>C<sub>2</sub></b></i> are congruent by three sides.<br/>
<br/>
Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-85794008183982410612024-03-04T08:19:00.000-08:002024-03-04T08:19:18.067-08:00Trigonometry+ 04: UNIZOR.COM - "Math+ & Problems" - "Trigonometry"<iframe width="480" height="270" src="https://youtube.com/embed/ykDpJcpCcTw?si=A2sHSF4hipRQ2Rc-" frameborder="0"></iframe>
<i>Notes to a video lecture on http://www.unizor.com</i><br/>
<br/>
<u> Trigonometry+ 04</u><br />
<br />
<i>Problem A</i><br/>
<br/>
Find the sums<br/>
<font size=5>Σ</font><sub>k∈[0,n−1]</sub><i><b>sin²(x+k·π/n)</b></i><br/>
<br/>
<i>Solution A</i><br/>
<br/>
First, convert <i>sin²(...)</i> into <i>cos(...)</i> by using the identity<br/>
<i>cos(2φ) = cos²(φ) − sin²(φ) =<br/>
= 1 − 2sin²(φ)</i><br/>
from which follows<br/>
<i>sin²(φ) = ½(1 − cos(2φ))</i><br/>
<br/>
Now our sum looks like this<br/>
<font size=5>Σ</font><sub>k∈[0,n−1]</sub><br/>
<i><b>½(1−cos(2x+2k·π/n)) =<br/>
= n/2 −<br/>
− ½</b></i><font size=5>Σ</font><sub>k∈[0,n−1]</sub><i><b>cos(2x+2k·π/n)</b></i><br/>
<br/>
To calculate </b></i><font size=5>Σ</font><sub>k∈[0,n−1]</sub> above, let's use the result of the previous lecture <i>Trigonometry 03</i> that proved the following<br/>
<font size=5>Σ</font><sub>k∈[0,n]</sub><i><b>cos(x+k·y) =<br/>
= </b></i>[<i><b>sin(x+(2n+1)·y/2) −<br/>
− sin(x−y/2)</b></i>]<i><b> <font size=4>/</font><br/>
<font size=4>/</font></b></i> [<i><b>2·sin(y/2)</b></i>]<br/>
<br/>
Since in our sum we have <i>n</i> terms instead of <i>n+1</i>, as in the formula above, to use this formula in our case we have to use <i>n−1</i> instead of <i>n</i>, which will result in<br/>
<font size=5>Σ</font><sub>k∈[0,n−1]</sub><i><b>cos(x+k·y) =<br/>
= </b></i>[<i><b>sin(x+(2n−1)·y/2) −<br/>
− sin(x−y/2)</b></i>]<i><b> <font size=4>/</font><br/>
<font size=4>/</font></b></i> [<i><b>2·sin(y/2)</b></i>]<br/>
<br/>
To use this formula in our case, we have to substitute <i>2x</i> instead of <i>x</i> and <i>2π/n</i> instead of <i>y</i>.<br/>
<br/>
In this case the numerator in the formula above would be<br/>
<i><b>sin(x+(2n−1)·π/n) − sin(x−π/n)</b></i><br/>
<br/>
The first term in the above expression for the numerator equals to<br/>
<i><b>sin(x+(2n−1)·π/n) =<br/>
= sin(x+2n·π/n−π/n) =<br/>
= sin(x−π/n+2π) = sin(x−π/n)</b></i><br/>
(since the value <i>2π</i> is a period of function <i>sin</i>)<br/>
which makes the first term exactly the same as the second term and they will cancel each other resulting in zero in the numerator.<br/>
<br/>
That leaves the sum in our case to be equal to<br/>
<font size=5>Σ</font><sub>k∈[0,n−1]</sub><br/>
<i><b>½(1−cos(2x+2k·π/n)) = n/2</b></i><br/>
<br/>
<i>Answer A</i><br/>
<font size=5>Σ</font><sub>k∈[0,n−1]</sub><i><b>sin²(x+k·π/n) = n/2</b></i><br/>
<br/>
<br/>
<i>Problem B</i><br/>
<br/>
Given a regular <i>n</i>-sided polygon with vertices <i><b>A<sub>1</sub></b></i>, <i><b>A<sub>2</sub></b></i>,...,<i><b>A<sub>n</sub></b></i> inscribed into a circle of radius <i><b>R</b></i> with a center at point <i><b>O</b></i> and any point <i><b>P</b></i> on this circle.<br/>
<br/>
<img src='http://www.unizor.com/Pictures/ProblemsTrigonometry_04B.png' style='width:200px;height:200px;'><br/>
Calculate the sum of squares of distances from point <i><b>P</b></i> to all vertices <i><b>A<sub>k</sub></b></i><br/>
<i><b>S = </b></i><font size=5>Σ</font><sub>k∈[<i>1,n</i>]</sub><i><b>(PA<sub>k</sub>)²</b></i>.<br/>
<br/>
<i>Hint</i>:<br/>
Let ∠<i><b>POA<sub>k</sub> = φ<sub>k</sub></b></i>.<br/>
Then <i><b>φ<sub>k</sub> = φ<sub>1</sub> + (k−1)·2π/n</b></i>.<br/>
Use the results of <i>Problem A</i>.<br/>
<br/>
<i>Answer</i><br/>
<i><b>S = 2nR²</b></i<br/.
<br/>
Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-6712401767370545222024-03-03T09:32:00.000-08:002024-03-03T09:32:45.945-08:00Algebra+ 04: UNIZOR.COM - Math+ & Problems - Algebra<iframe style="background-image:url(https://i.ytimg.com/vi/gb1ygDJpsvg/hqdefault.jpg)" width="480" height="270" src="https://youtube.com/embed/gb1ygDJpsvg?si=Vzkz7hOv33t3iWSb" frameborder="0"></iframe>
<i>Notes to a video lecture on http://www.unizor.com</i><br/>
<br/>
<u> Algebra+ 04</u><br />
<br />
<i>Problem A</i><br/>
<br/>
Prove that<br/>
if <i><b>X+Y+Z=1</b></i><br/>
then <i><b>X²+Y²+Z² ≥ 1/3</b></i>.<br/>
<br/>
<i>Hint A1</i><br/>
<i>X² + Y² ≥ 2·X·Y</i>.<br/>
<br/>
<i>Hint A2</i><br/>
<i>X²+Y²+Z² =<br/>
= (X−a)²+(Y−a)²+(Z−a)²+<br/>
+2·a·(X+Y+Z)−3a² ≥<br/>
≥ 2a −3a²</i><br/>
Quadratic polynomial <i>2a−3a²</i> has roots <i>a=0</i> and <i>a=2/3</i> and maximum at <i>a=1/3</i> with value <i>2·(1/3)−3·(1/3)²=1/3</i>.<br/>
<br/>
<br/>
<i>Problem B</i><br/>
<br/>
Solve an equation<br/>
<i><b>X<sup>8</sup> + X<sup>4</sup> + 1 = 0</b></i><br/>
<br/>
<i>Hint B1</i><br/>
Substitute <i>y=X<sup>4</sup></i>.<br/>
<br/>
<i>Hint B2</i><br/>
Represent the left part as a product of <i>4</i> polynomials of a second degree by adding and subtracting <i>X<sup>4</sup></i>.<br/>
<br/>
<i>Answer B</i></b><br/>
<i><b>X<sub>1,2,3,4</sub> = ±1/2 ± i·√<span style='text-decoration:overline'>3</span>/2</b></i><br/>
<i><b>X<sub>5,6,7,8</sub> = ±√<span style='text-decoration:overline'>3</span>/2 ± i/2</b></i><br/>
<br/>
<br/>
<i>Problem C</i><br/>
<br/>
Find a number from <i>10</i> to <i>99</i> knowing that its square equals to sum of its digits in cube.<br/>
<br/>
<i>Hint C</i><br/>
Actually, we have to solve the equation<br/>
<i>(10·X+Y)² = (X+Y)³</i><br/>
for natural<br/>
<i>1 ≤ X ≤ 9</i> and <i>0 ≤ Y≤ 9</i>.<br/>
<br/>
<i>Answer C</i></b><br/>
The only solution is number <i><b>27</b></i>.<br/>
Check:<br/>
<i>27² = 729</i><br/>
<i>(2+7)³ = 729</i><br/>
<br/>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-71241434155251796352024-03-02T15:18:00.000-08:002024-03-02T15:18:29.141-08:00Arithmetic+ 05: UNIZOR.COM - Math+ & Problems - Arithmetic<iframe style="background-image:url(https://i.ytimg.com/vi/1lf_aq71x8U/hqdefault.jpg)" width="480" height="270" src="https://youtube.com/embed/1lf_aq71x8U?si=2dpAyX6L5YETHc0h" frameborder="0"></iframe>
<i>Notes to a video lecture on http://www.unizor.com</i><br/>
<br/>
<u> Arithmetic+ 05</u><br />
<br />
<i>Problem A</i><br/>
Prove that a remainder of division of any prime number <i><b>P</b></i> by <i><b>24</b></i> is a prime number, assuming <i><b>P</b></i> is greater than <i><b>24</b></i>.<br/>
<br/>
<i>Hint A</i><br/>
If <i>P</i> is prime and <i>R</i> is a remainder of division of <i>P</i> by <i>24</i> then <i>P=24·n+R</i> where <i>R≤23</i>.<br/>
If <i>R</i> is not prime, it should be divisible by a product of, at least, two prime numbers.<br/>
Also, <i>24=2·2·2·3</i>.<br/>
<br/>
<br/>
<i>Problem B</i><br/>
Prove that for any natural number <i>N</i> the number <i>10<sup>N</sup>+18·N−1</i> is divisible by <i>27</i>.<br/>
<br/>
<i>Hint B</i><br/>
First, prove a theorem that any natural number <i>n</i> divided by <i>3</i> (or <i>9</i>) has the same remainder as a sum of its digits divided by <i>3</i> (or <i>9</i>).<br/>
Obviously, the given number is divisible by <i>9</i>.<br/>
Then prove that the number <i>111...111</i> consisting of <i>N</i> unit digits with their sum equal to <i>N</i> plus <i>2·N</i> is divisible by <i>3</i> using a theorem proved above.<br/>
<br/>
<br/>
<i>Problem C</i><br/>
Prove that from any set of <i>100</i> natural numbers <i><b>N<sub>1</sub></b></i>, <i><b>N<sub>2</sub></b></i>,... ,<i><b>N<sub>100</sub></b></i> it is possible to choose such a subset with a sum of its numbers divisible by <i>100</i>.<br/>
<br/>
<i>Hint C</i><br/>
Let <i>S<sub>k</sub>=</i><font size=5>Σ</font><sub>i∈[1,k]</sub><i>N<sub>i</sub></i><br/>
where <i>k=1, 2,...,100</i>.<br/>
If there is one particular number <i>S<sub>k</sub></i> divisible by <i>100</i>, the components of this sum can be chosen as a subgroup and the problem is solved.<br/>
Assume, none of our <i>100</i> sums <i>S<sub>k</sub></i> is divisible by <i>100</i>.<br/>
Let <i>R<sub>k</sub></i> is a remainder of division of <i>S<sub>k</sub></i> by 100.<br/>
Note that there are <i>99</i> possible remainders from <i>1</i> to <i>99</i>.<br/>
<br/>
<br/>
Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-67592092344806966422024-02-25T06:44:00.000-08:002024-02-25T06:44:32.171-08:00Trigonometry+ 02: UNIZOR.COM - Math+ & Problems - Trigonometry<iframe width="480" height="270" src="https://youtube.com/embed/WWdv5J9aO9Q?si=KYHi-i13igUa3AXa" frameborder="0"></iframe>
<i>Notes to a video lecture on http://www.unizor.com</i><br/>
<br/>
<u> Trigonometry+ 02</u><br />
<br />
<br />
<i>Problem A</i><br/>
<br/>
Prove the following trigonometric identities using the Euler formula<br/>
<i>e<sup>i·φ</sup> = cos(φ)+i·sin(φ)</i><br/>
[See lecture on <i><a href="#" onClick="window.open('http://unizor.com', '_blank')">UNIZOR.COM</a> - Math 4 Teens - Trigonometry - Complex Numbers and Trigonometry - Euler's Formula</i>]<br/>
<br/>
<i><b>sin(x+y) =<br/>
= sin(x)·cos(y) + cos(x)·sin(y)</b></i><br/>
<br/>
<i><b>cos(x+y) =<br/>
= cos(x)·cos(y) − sin(x)·sin(y)</b></i><br/>
<br/>
<i>Proof A</i><br/>
<i><b><font color=blue>cos(x+y)</font> + i·<font color=red>sin(x+y)</font> =<br/>
= e<sup>i·(x+y)</sup> = e<sup>i·x</sup>·e<sup>i·y</sup> =<br/>
= </b></i>[<i><b>cos(x) + i·sin(x)</b></i>]<i><b> ·<br/>
· </b></i>[<i><b>cos(y) + i·sin(y)</b></i>]<i><b>=<br/>
=cos(x)·cos(y)+i²sin(x)·sin(y)+<br/>
+i·sin(x)·cos(x)+i·cos(x)·sin(y)=<br/>
=<font color=blue>cos(x)·cos(y)−sin(x)·sin(y)</font>+<br/>
+i·</b></i>[<i><b><font color=red>sin(x)·cos(y)+cos(x)·sin(y)</font></b></i>]<br/>
<br/>
<br/>
<i>Problem B</i><br/>
<br/>
Prove the following trigonometric identities<br/>
<br/>
<table border=0 cellpadding=0 cellspacing=0>
<tr>
<td align=center><i><b>tan(x+y) = </b></i>
</td>
<td align=center>
<table>
<tr><td style="border-bottom: 2px solid black;"><i><b>tan(x)+tan(y)</b></i></b></i></td></tr>
<tr><td><i><b>1−tan(x)·tan(y)</b></i></td></tr>
</table>
</td>
</tr>
</table>
<table border=0 cellpadding=0 cellspacing=0>
<tr>
<td align=center><i><b>cot(x+y) = </b></i>
</td>
<td align=center>
<table>
<tr><td style="border-bottom: 2px solid black;"><i><b>cot(x)·cot(y)−1</b></i></b></i></td></tr>
<tr><td><i><b>cot(x)+cot(y)</b></i></td></tr>
</table>
</td>
</tr>
</table>
<br/>
<i>Proof B</i><br/>
<table border=0 cellpadding=0 cellspacing=0>
<tr>
<td align=center><i><b>tan(x+y) = </b></i>
</td>
<td align=center>
<table>
<tr><td style="border-bottom: 2px solid black;"><i><b>sin(x+y)</b></i></b></i></td></tr>
<tr><td><i><b>cos(x+y)</b></i></td></tr>
</table>
</td>
</tr>
</table>
<table border=0 cellpadding=0 cellspacing=0>
<tr>
<td align=center><i><b>= </b></i>
</td>
<td align=center>
<table>
<tr><td style="border-bottom: 2px solid black;"><i><b>sin(x)·cos(y) + cos(x)·sin(y)</b></i></b></i></td></tr>
<tr><td><i><b>cos(x)·cos(y) − sin(x)·sin(y)</b></i></td></tr>
</table>
</td>
</tr>
</table>
Divide numerator and denominator by <i>cos(x)·cos(y)</i>
<table border=0 cellpadding=0 cellspacing=0>
<tr>
<td align=center><i><b>= </b></i>
</td>
<td align=center>
<table>
<tr><td style="border-bottom: 2px solid black;"><i><b>sin(x)/cos(x) + sin(y)/cos(y)</b></i></b></i></td></tr>
<tr><td><i><b>1−sin(x)·sin(y)/(cos(x)·cos(y))</b></i></td></tr>
</table>
</td>
</tr>
</table>
Cotangent formula is proved analogously.<br/>
<br/>
<br/>
<i>Problem C</i><br/>
<br/>
Prove the following trigonometric identities<br/>
<table border=0 cellpadding=0 cellspacing=0>
<tr>
<td align=center><i><b>sin(x) =</b></i>
</td>
<td align=center>
<table>
<tr><td style="border-bottom: 2px solid black;"><i><b>2·tan(½x)</b></i></td></tr>
<tr><td><i><b>1+tan²(½x)</b></i></td></tr>
</table>
</td>
</tr>
</table>
<table border=0 cellpadding=0 cellspacing=0>
<tr>
<td align=center><i><b>cos(x) =</b></i>
</td>
<td align=center>
<table>
<tr><td style="border-bottom: 2px solid black;"><i><b>1−tan²(½x)</b></i></td></tr>
<tr><td><i><b>1+tan²(½x)</b></i></td></tr>
</table>
</td>
</tr>
</table>
<br/>
<i>Proof C</i><br/>
<i><b>sin(x) = sin(½x + ½x)<br/>
= 2sin(½x)·cos(½x)<br/>
= 2</b></i>[<i><b>sin(½x)/cos(½x)</b></i>]<i><b>·cos²(½x)<br/>
= 2·tan(½x) · cos²(½x)</b></i><br/>
At the same time<br/>
<i><b>1/</b></i>[<i><b>1+tan²(½x)</b></i>]<i><b><br/>
= 1/</b></i>[<i><b>1+sin²(½x)/cos²(½x)</b></i>]<i><b><br/>
= cos²(½x)/</b></i>[<i><b>cos²(½x)+sin²(½x)</b></i>]<i><b><br/>
= cos²(½x)</b></i><br/>
<br/>
<br/>
Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-3103834949420699092024-02-24T09:58:00.000-08:002024-02-24T09:58:03.357-08:00Trigonometry+ 03: UNIZOR.COM - Math+ & Problems - Trigonometry<iframe width="480" height="270" src="https://youtube.com/embed/DdRGCrt6Vxk?si=TxClswPDPhVPA0pk" frameborder="0"></iframe>
<i>Notes to a video lecture on http://www.unizor.com</i><br/>
<br/>
<u> Trigonometry+ 03</u><br />
<br />
<br />
<i>Problem A</i><br/>
<br/>
Find the sums<br/>
<i><b>S</b></i><sub>sin</sub><i><b>(x,y,n)= </b></i><font size=5>Σ</font><sub>k∈[0,n]</sub><i><b>sin(x+k·y)</b></i><br/>
<i><b>S</b></i><sub>cos</sub><i><b>(x,y,n)= </b></i><font size=5>Σ</font><sub>k∈[0,n]</sub><i><b>cos(x+k·y)</b></i><br/>
as algebraic expression with arguments <i><b>x</b></i>, <i><b>y</b></i> and <i><b>n</b></i>.<br/>
<br/>
<br/>
<i>Solution A1</i><br/>
<br/>
Let's start with a sum of sines.<br/>
Recall the formula for a cosine of a sum of two angles<br/>
<i>cos(a+b) =<br/>
= cos(a)·cos(b) − sin(a)·sin(b)</i><br/>
From this formula it's easy to derive another trigonometric identity formula<br/>
<i>2·sin(a)·sin(b) =<br/>
= cos(a−b)−cos(a+b)</i><br/>
<br/>
Let's use this formula to transform our series into only a couple of members.<br/>
Multiply <i><b>S</b></i><sub>sin</sub> (that is, each member of this series) by <i><b>2·sin(y/2)</b></i> and use the trigonometric identity above.<br/>
<br/>
Then after this multiplication by <i><b>2·sin(y/2)</b></i> the <i>k</i><sup>th</sup> member of <i><b>S</b></i><sub>sin</sub> would look like<br/>
<i><b>2·sin(x+k·y)·sin(y/2) =<br/>
= cos(x+k·y−y/2) −<br/>
− cos(x+k·y+y/2) =<br/>
= cos(x+(2k−1)·y/2) −<br/>
− cos(x+(2k+1)·y/2)</b></i><br/>
<br/>
Thus, each member of our series is converted into a pair of cosines with a minus in-between:<br/>
<i><b>2·S</b></i><sub>sin</sub><i><b>(x,y,n)·sin(y/2) =<br/>
= </b></i><font size=5>Σ</font><sub>k∈[0,n]</sub>[<i><b>cos(x+(2k−1)·y/2) −<br/>
− cos(x+(2k+1)·y/2)</b></i>]<i><b> =<br/>
= </b></i>[<i><b>cos(x−y/2)−cos(x+y/2)</b></i>]<i><b> +<br/>
+ </b></i>[<i><b>cos(x+y/2)−cos(x+3y/2)</b></i>]<i><b> +<br/>
+ </b></i>[<i><b>cos(x+3y/2)−cos(x+5y/2)</b></i>]<i><b> +<br/>
</b></i>etc. and the last two members of a series are<i><b><br/>
+ </b></i>[<i><b>cos(x+(2n−3)/2) −<br/>
− cos(x+(2n−1)y/2)</b></i>]<i><b> +<br/>
+ </b></i>[<i><b>cos(x+(2n−1)/2) −<br/>
− cos(x+(2n+1)y/2)</b></i>]<i><b></b></i><br/>
<br/>
It's easy to see that in this series every members, except the first and the last, cancel each other and the result is<br/>
<i><b>2·S</b></i><sub>sin</sub><i><b>(x,y,n)·sin(y/2) =<br/>
= cos(x−y/2) −<br/>
− cos(x+(2n+1)·y/2)</b></i><br/>
<br/>
Therefore,<br/>
<i><b>S</b></i><sub>sin</sub><i><b>(x,y,n) =<br/>
= </b></i>[<i><b>cos(x−y/2) −<br/>
− cos(x+(2n+1)·y/2)</b></i>]<i><b> <font size=4>/</font><br/>
<font size=4>/</font></b></i> [<i><b>2·sin(y/2)</b></i>]<br/>
<br/>
The sum of cosines can be treated analogously.<br/>
Recall the formula for a sine of a sum of two angles<br/>
<i>sin(a+b) =<br/>
= sin(a)·cos(b) + cos(a)·sin(b)</i><br/>
From this formula it's easy to derive another trigonometric identity formula<br/>
<i>2·cos(a)·sin(b) =<br/>
= sin(a+b)−sin(a−b)</i><br/>
<br/>
Let's use this formula to transform our series into only a couple of members.<br/>
First, we multiply all members of <i><b>S</b></i><sub>cos</sub> (that is, each member of this series) by <i><b>2·sin(y/2)</b></i>. and use the trigonometric identity above.<br/>
<i><b>2·S</b></i><sub>cos</sub><i><b>(x,y,n)·sin(y/2) =<br/>
= </b></i><font size=5>Σ</font><sub>k∈[0,n]</sub>[<i><b>sin(x+(2k+1)·y/2) −<br/>
− sin(x+(2k−1)·y/2)</b></i>]<i><b> =<br/>
= sin(x+(2n+1)·y/2) −<br/>
− sin(x−y/2)</b></i><br/>
<br/>
Therefore,<br/>
<i><b>S</b></i><sub>cos</sub><i><b>(x,y,n) =<br/>
= </b></i>[<i><b>sin(x+(2n+1)·y/2) −<br/>
− sin(x−y/2)</b></i>]<i><b> <font size=4>/</font><br/>
<font size=4>/</font></b></i> [<i><b>2·sin(y/2)</b></i>]<br/>
<br/>
Of cause, both formulas for <i><b>S</b></i><sub>sin</sub> and <i><b>S</b></i><sub>cos</sub> are correct only if <i><b>sin(y/2)≠0</b></i>, that is <i><b>y≠2πN</b></i>, where <i><b>N</b></i> - any integer number.<br/>
If, however, <i><b>y=2πN</b></i>, then <i><b>sin(x+k·y)=sin(x)</b></i> and <i><b>cos(x+k·y)=cos(x)</b></i>, so<br/>
<i><b>S</b></i><sub>sin</sub><i><b> = (n+1)·sin(x)</b></i><br/>
<i><b>S</b></i><sub>cos</sub><i><b> = (n+1)·cos(x)</b></i><br/>
<br/>
<br/>
<i>Solution A2</i><br/>
<br/>
A different and, arguably, more elegant approach to solve this problem would be use the Euler formula<br/>
<i>e<sup>i·φ</sup> = cos(φ)+i·sin(φ)</i><br/>
[See lecture on <i><a href="#" onClick="window.open('http://unizor.com', '_blank')">UNIZOR.COM</a> - Math 4 Teens - Trigonometry - Complex Numbers and Trigonometry - Euler's Formula</i>]<br/>
<br/>
Using it, we can express <i><b>cos(x+k·y)</b></i> and <i><b>sin(x+k·y)</b></i> as real and imaginary parts of<br/>
<i><b>e<sup>i·(x+k·y)</sup> = e<sup>i·x</sup>·(e<sup>i·y</sup>)<sup><sup>k</sup></sup></b></i><br/>
<br/>
The sum of these expression by <i>k</i> from <i>0</i> to <i>n</i> can be calculated as a geometric series with the first term <i>a=e<sup>i·x</sup></i> and factor <i>r=e<sup>i·y</sup></i>.<br/>
[See lecture on <i><a href="#" onClick="window.open('http://unizor.com', '_blank')">UNIZOR.COM</a> - Math 4 Teens - Algebra - Sequence and Series - Progression Series</i>]<br/>
This sum in terms of <i>a</i> and <i>r</i> is<br/>
<i>S = a·(r<sup>n+1</sup>−1)/(r−1)</i><br/>
<br/>
Therefore, our sum is equal to<br/>
<font size=5>Σ</font><sub>k∈[0,n]</sub><i><b>e<sup>i·x</sup>·(e<sup>i·y</sup>)<sup><sup>k</sup></sup> =<br/>
= e<sup>i·x</sup>·</b></i>[<i><b>(e<sup>i·y</sup>)<sup><sup>n+1</sup></sup>−1</b></i>]<i><b> <font size=5>/</font>(e<sup>i·y</sup>−1) =<br/>
= </b></i>[<i><b>e<sup>i·(x+y·(n+1))</sup>−e<sup>i·x</sup></b></i>]<i><b> <font size=5>/</font>(e<sup>i·y</sup>−1)</b></i><br/>
<br/>
What remains is to convert this expression into canonical representation of complex numbers as <i>A+B·i</i>. Then <i>A</i> will represent the sum of cosines, and <i>B</i> will represent a sum of sines.<br/>
<br/>
First, get rid of <i>i</i> in the denominator by multiplying both numerator and denominator by <i><b>(e<sup>−i·y</sup>−1)</b></i>.<br/>
In the denominator the result will be<br/>
<i><b>(e<sup>i·y</sup>−1)·(e<sup>−i·y</sup>−1) =<br/>
= e<sup>0</sup>−e<sup>−i·y</sup>−e<sup>i·y</sup>+1 =<br/>
= 2 − e<sup>i·y</sup> − e<sup>−i·y</sup> =<br/>
= 2 − cos(y) − i·sin(y) −<br/>
− cos(−y) − i·sin(−y) =<br/>
= 2 −2cos(y) = 4·sin²(y/2)</b></i><br/>
As we see, there is only real part in the denominator, an imaginary terms cancel each other.<br/>
<br/>
In the numerator we will have<br/>
</b></i>[<i><b>e<sup>i·(x+y·(n+1))</sup>−e<sup>i·x</sup></b></i>]<i><b>·(e<sup>−i·y</sup>−1) =<br/>
= e<sup>i·(x+y·n)</sup> − e<sup>i·(x+y·(n+1))</sup> −<br/>
− e<sup>i·(x−y)</sup> + e<sup>i·x</sup> =<br/>
= cos(x+y·n)−cos(x+y·(n+1))−<br/>
−cos(x−y) + cos(x) +<br/>
+ i·</b></i>[<i><b>sin(x+y·n)−sin(x+y(n+1))−<br/>
−sin(x−y)+sin(x)</b></i>]<i><b></b></i><br/>
<br/>
Let's deal with real part that gives the sum of cosines.<br/>
We will use another trigonometric identity which can be easily proved using<br/>
<i>α=½(α+β)+½(α−β)</i> and<br/>
<i>β=½(α+β)−½(α−β)</i>:<br/>
<i>cos(α) − cos(β) =<br/>
= −2sin(½(α+β))·sin(½(α−β))</b></i><br/>
Using this identity, the numerator for a sum of cosines is<br/>
<i><b>cos(x+y·n)−cos(x+y·(n+1))−<br/>
−cos(x−y) + cos(x) =<br/>
= 2·sin(x+½(2n+1)·y)·sin(½y) −<br/>
− 2·sin(x−½y)·sin(½y) =<br/>
= 2·sin(½y)·<br/>
·</b></i>[<i><b>sin(x+½(2n+1)·y) −<br/>
− sin(x−½y)</b></i>]<br/>
<br/>
Combining this numerator and calculated above denominator <i><b>4·sin²(y/2)</b></i>, we obtain<br/>
<i><b>S</b></i><sub>cos</sub><i><b>(x,y,n) =<br/>
= </b></i>[<i><b>sin(x+(2n+1)·y/2) −<br/>
− sin(x−y/2)</b></i>]<i><b>·<br/>
·2·sin(y/2) <font size=4>/</font></b></i> [<i><b>4·sin²(y/2)</b></i>]<i><b> =<br/>
= </b></i>[<i><b>sin(x+(2n+1)·y/2) −<br/>
− sin(x−y/2)</b></i>]<i><b> <font size=4>/</font><br/>
<font size=4>/</font></b></i> [<i><b>2·sin(y/2)</b></i>]<br/>
which corresponds completely to <i>Solution B1</i>.<br/>
<br/>
Similarly, the imaginary part of a numerator can be used to calculate the sum of sines.<br/>
We will use yet another trigonometric identity, which can be derived analogously to a difference of cosines above:<br/>
<i>sin(α) − sin(β) =<br/>
= 2sin(½(α−β))·cos(½(α+β))</b></i><br/>
Using this identity, the numerator for a sum of sines is<br/>
<i><b>sin(x+y·n)−sin(x+y(n+1))−<br/>
−sin(x−y)+sin(x) =<br/>
=−2·sin(½y)·cos(x+½(2n+1)·y)+<br/>
+ 2·sin(½y)·cos(x−½y) =<br/>
= 2·sin(½y)·<br/>
·</b></i>[<i><b>cos(x−½y) −<br/>
− cos(x+½(2n+1)·y)</b></i>]<br/>
<br/>
Combining this numerator and calculated above denominator <i><b>4·sin²(y/2)</b></i>, we obtain<br/>
<i><b>S</b></i><sub>sin</sub><i><b>(x,y,n) =<br/>
= </b></i>[<i><b>cos(x−y/2) −<br/>
− cos(x+(2n+1)·y/2)</b></i>]<i><b>·<br/>
·2·sin(y/2) <font size=4>/</font></b></i> [<i><b>4·sin²(y/2)</b></i>]<i><b> =<br/>
= </b></i>[<i><b>cos(x−y/2) −<br/>
− cos(x+(2n+1)·y/2)</b></i>]<i><b> <font size=4>/</font><br/>
<font size=4>/</font></b></i> [<i><b>2·sin(y/2)</b></i>]<br/>
which corresponds completely to <i>Solution B1</i>.<br/>
<br/>
Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-37970751248013949002024-02-21T07:44:00.000-08:002024-02-21T07:44:34.695-08:00Logic+ 04: UNIZOR.COM - Math+ & Problems - Logic<iframe width="480" height="270" src="https://youtube.com/embed/a0D8Bg1Fv68?si=AfpECV5CRCricoxt" frameborder="0"></iframe>
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<br/>
<u> Logic+ 04</u><br />
<br />
<i>Problem A</i><br/>
<br/>
A traveler has to stay in some city for <i>7</i> days.
A restaurant owner near a hotel where a traveler stays does not accept credit cards, and a traveler has no local currency to pay for food.<br/>
<br/>
However, a traveler has a silver chain of <i>7</i> links, and a restaurant owner agreed to take one chain link for each dinner.<br/>
<br/>
What is the minimum number of chain links that must be cut open to be able to pay for each dinner separately?<br/>
<br/>
<i>Answer A</i><br/>
Only one link needs to be cut open, link #3.<br/>
<br/>
<br/>
<i>Problem B</i><br/>
<br/>
Same as <i>Problem A</i>, but the number of days is <i><b>N</b></i>, and the number of links in a chain is the same <i><b>N</b></i>.<br/>
<br/>
<br/>
<i>Answer B</i><br/>
<br/>
Assume, a traveler cuts open only <i><b>K</b></i> links (<i><b>K</b></i> must be small relatively to <i><b>N</b></i>).<br/>
Then he can pay for <i>1</i>, <i>2</i>, <i>3</i>,..., <i>K</i> dinners using only these links.<br/>
<br/>
For the next dinner he needs a segment of chain that contains <i><b>K+1</b></i> links.<br/>
With this additional segment of chain he can pay for <i>K+1</i>, <i>K+2</i>,..., <i>2K+1</i> dinners.<br/>
<br/>
For the next dinner he needs a segment of <i>2K+2</i> links.<br/>
Using <i>K</i> individual links that were cut, a segment of a chain with <i>K+1</i> links and another segment with <i>2K+2</i> links a traveler pay for any number of dinners from 1 to <i>K+(K+1)+(2K+2)=4K+3</i>.<br/>
<br/>
For the next dinner he needs a segment of <i>4K+4</i> links.<br/>
Using <i>K</i> individual links that were cut, a segment of a chain with <i>K+1</i> links, a segment with <i>2K+2</i> links and a segment with <i>4K+4</i> links a traveler pay for any number of dinners from 1 to <i>K+(K+1)+(2K+2)+(4K+4)=8K+7</i>.<br/>
<br/>
For the next dinner he needs a segment of <i>8K+8</i> links.<br/>
Using <i>K</i> individual links that were cut, a segment of a chain with <i>K+1</i> links, a segment with <i>2K+2</i> links, a segment with <i>4K+4</i> links and a segment with <i>8K+8</i> links a traveler pay for any number of dinners from 1 to <i>K+(K+1)+(2K+2)+(4K+4)+(8K+8)=16K+15</i>.<br/>
<br/>
Let's generalize it.<br/>
When <i><b>K</b></i> links were cut, it separated a chain into <i><b>K+1</b></i> segments as follows.<br/>
<br/>
Segment #1 should have<br/>
<i><b>K+1</b></i> links in it.<br/>
Segment #2 should have<br/>
<i><b>2K+2=2·(K+1)</b></i> links in it.<br/>
Segment #3 should have<br/>
<i><b>4K+4=4·(K+1)</b></i> links in it.<br/>
Segment #4 should have<br/>
<i><b>8K+8=8·(K+1)</b></i> links in it.<br/>
Segment #5 should have<br/>
<i><b>16K+16=16·(K+1)</b></i> links in it.<br/>
etc.<br/>
The last segment #(<i>K+1</i>) should have <i><b>2<sup>K</sup>·(K+1)</b></i> links in it.<br/>
<br/>
All individual links that were cut splitting a chain into segments and all segments listed above allow to pay for the following number of dinners.<br/>
<i>K +<br/>
+ (K+1) +<br/>
+ 2·(K+1) +<br/>
+ 4·(K+1)+<br/>
+...+<br/>
+ 2<sup>K</sup>·(K+1) =<br/>
= K + (2<sup>K+1</sup> −1)·(K+1) =<br/>
= (K+1)·2<sup>K+1</sup> − 1</i><br/>
<br/>
If <i><b>N=(K+1)·2<sup>K+1</sup>−1</b></i> then the minimum number of links to cut is <i><b>K</b></i>.<br/>
If <i><b>N</b></i> is not such a number, we have to find <i><b>K</b></i> such that <i><b>(K+1)·2<sup>K+1</sup>−1</b></i> is greater than <i><b>N</b></i>. That <i><b>K</b></i> is the minimum number of links to cut.<br/>
<br/>
Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-61165764289024450852024-02-17T07:21:00.000-08:002024-02-17T07:21:12.828-08:00Arithmetic+ 04: UNIZOR.COM - Math+ &Problems - Arithmetic<iframe style="background-image:url(https://i.ytimg.com/vi/IubirxKtSs0/hqdefault.jpg)" width="480" height="270" src="https://youtube.com/embed/IubirxKtSs0?si=iqZchLcZp7PNiurO" frameborder="0"></iframe>
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<br/>
<u> Arithmetic+ 04</u><br />
<br />
<i>Problem A</i><br/>
Consider two numbers:<br/>
<i>X=<b>11111111</b></i><br/>
(digit 1 repeats 8 times) and<br/>
<i>Y=<b>11111...11111</b></i><br/>
(digit 1 repeats 100 times).<br/>
What is their greatest common factor?<br/>
<br/>
<i>Hint A</i><br/>
Both numbers can be represented as a sum<br/>
<i>10<sup>0</sup>+10<sup>1</sup>+...+10<sup>k</sup></i><br/>
where <i>k=7</i> for number <i>X</i> and <i>k=99</i> for number <i>Y</i>.<br/>
From this it's easy to see that<br/>
<i>Y = X·n + 1111·10<sup>96</sup></i><br/>
where <i>n</i> is some natural number.<br/>
<br/>
<i>Answer A</i><br/>
The greatest common factor of numbers <i>X</i> and <i>Y</i> is <i><b>1111</b></i>.<br/>
<br/>
<br/>
<i>Problem B</i><br/>
Prove that a remainder of the division of some natural number by <i>9</i> is the same as a remainder of the division by <i>9</i> of the sum of this number's digits.<br/>
<br/>
<i>Example B</i><br/>
Take, for instance, number <i>2024</i> and divide it by <i>9</i>. <br/>
<i><b>2024÷9 = 224 (8)</b></i><br/>
(remainder is <i><b>8</b></i>)<br/>
Now with the sum this number's digits:<br/>
<i><b>(2+0+2+4)÷9 = 8÷9 = 0(8)</b></i><br/>
(the same remainder <i><b>8</b></i>)<br/>
<br/>
<i>The rule of divisibility by 9</i><br/>
Consequently, if the sum of digits of a natural number is divisible by <i>9</i>, then the number itself is divisible by <i>9</i>.<br/>
<br/>
<i>Hint B</i><br/>
Represent a number as a sum of its digits, each multiplied by a factor <i>10<sup>n</sup></i>, where <i>n=0</i> for the right most digit, <i>n=1</i> for the next digit to the left etc.<br/>
<br/>
<br/>
<i>Problem C</i><br/>
Prove that if the sum of digits of some natural number <i><b>N</b></i> is the same as the sum of digits of the number <i><b>2·N</b></i> then number <i><b>N</b></i> is divisible by <i><b>9</b></i>.<br/>
<br/>
<i>Example C</i><br/>
For number <i><b>279</b></i> the sum of digits is <i><b>2+7+9=18</b></i>.<br/>
Double this number:<br/>
<i><b>2·279 = 558</b></i><br/>
Sum of digits of <i><b>558</b></i> is also <i><b>18</b></i>.<br/>
The number <i><b>279</b></i> is indeed divisible by <i><b>9</b></i>:<br/>
<i><b>279÷9 = 31 (0)</b></i><br/>
(remainder is <i><b>0</b></i>)<br/>
<br/>
<i>Hint C</i><br/>
Use the result of the <i>Problem B</i> above and consider the difference between <i><b>2·N</b></i> and <i><b>N</b></i>.<br/>
<br/>
<i>Proof C</i><br/>
<i><b>N÷9 = n (r)</b></i><br/>
(here <i><b>n</b></i> is a quotient and <i><b>r</b></i> is a remainder with the value from <i>0</i> to <i>8</i>)<br/>
Therefore,
<i><b>N = 9·n + r</b></i><br/>
Analogously,<br/>
<i><b>(2·N)÷9 = m (r)</b></i><br/>
(same remainder <i><b>r</b></i>)<br/>
<i><b>2·N = 9·m + r</b></i><br/>
Subtract formula for <i><b>N</b></i> from formula for <i><b>2·N</b></i><br/>
<i><b>2·N − N = N = <br/>
= (9·m + r) − (9·n + r) =<br/>
= 9·(m−n)</b></i><br/>
Hence<br/>
<i><b>N = 9·(m−n)</b></i><br/>
that is, <i><b >N</b></i> is divisible by <i><b>9</b></i>.<br/>
<br/>
<i>Note</i><br/>
The same result (divisibility by <i>9</i>) can be obtained under weaker conditions.<br/>
The theorem can be stated as follows.<br/>
Prove that if the sum of digits of some natural number <i><b>N</b></i> divided by <i>9</i> gives the same remainder as the sum of digits of the number <i><b>2·N</b></i> divided by <i>9</i>, then number <i><b>N</b></i> is divisible by <i><b>9</b></i>.<br/>
Obviously, that remainder in both cases will be zero. No other equal remainder can be encountered for sums of digits of <i><b>N</b></i> and <i><b>2·N</b></i> divided by <i>9</i>.<br/>
<br/>
<i>Example CC</i><br/>
For number <i><b>144</b></i> the sum of digits is <i><b>1+4+4=9</b></i>.<br/>
Double this number:<br/>
<i><b>2·144 = 288</b></i><br/>
Sum of digits of <i><b>288</b></i> is <i><b>18</b></i>.<br/>
Sums of digits are different (though, both are divisible by <i>9</i>), but the number <i><b>144</b></i> is indeed divisible by <i><b>9</b></i>:<br/>
<i><b>144÷9 = 16 (0)</b></i><br/>
(remainder is <i><b>0</b></i>)<br/>
You can easily prove it or watch the proof in the lecture.<br/>
<br/>
Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-72316912739436395002024-02-12T09:13:00.000-08:002024-02-12T09:13:12.322-08:00Logic+ 03: UNIZOR.COM - Math+ & Problems - Logic<iframe width="480" height="270" src="https://youtube.com/embed/cQckFbEtjb4?si=GZiPCpWHs7yGo6aW" frameborder="0"></iframe>
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<br/>
<u> Logic+ 03</u><br />
<br />
<i>Problem A</i><br/>
<br/>
There are two cities:<br/>
CityOfTruth, where all people always tell the truth and<br/>
CityOfLies, where all people always lie.<br/>
<br/>
A traveler intends to go to CityOfTruth. He comes to a fork, one road leading to CityOfTruth, another - to CityOfLies and he has to choose which way to go.<br/>
<br/>
Right there he meets a person who, apparently, lives in one of these cities. So a traveler can ask him the direction to CityOfTruth.<br/>
<br/>
The problem is, the person can live in either of these cities and nobody knows whether he tells the truth or lies.<br/>
<br/>
Can a traveler ask this person a question in such a way that there will be no doubts about which way leads to CityOfTruth?<br/>
<br/>
<br/>
<i>Answer A</i><br/>
<br/>
The question might be: "Could you show me which way leads to a city where you live?"<br/>
<br/>
If this person lives in CityOfTruth and always tells the truth, he will point to CityOfTruth.<br/>
<br/>
If this person lives in CityOfLies and always lies, he will point to the same CityOfTruth, because pointing to CityOfLies would be the truth, which he never tells..<br/>
<br/>
In any case, he will correctly point to a direction the travel wants to know.<br/>
<br/>
<br/>
<i>Problem B</i><br/>
<br/>
A man (<i><b>M</b></i>), a wolf (<i><b>W</b></i>), a goat (<i><b>G</b></i>) and a cabbage (<i><b>C</b></i>) have to cross a river from side A to side B.<br/>
There is a boat that can hold a man and either one of the others.<br/>
<br/>
The problem is, in the absence of a man a wolf might eat a goat, a goat might eat a cabbage. But in the man's presence nobody eats anything. A wolf does not eat cabbage under any circumstances.<br/>
<br/>
How can they all safely cross the river from side A to side B, so everyone is alive and well at the end?<br/>
<br/>
<br/>
<i>Answer B</i><br/>
<br/>
1. A→(<i><b>M+G</b></i>)→B<br/>
2. B→(<i><b>M</b></i>)→A<br/>
3. A→(<i><b>M+W</b></i>)→B<br/>
4. B→(<i><b>M+G</b></i>)→A<br/>
5. A→(<i><b>M+C</b></i>)→B<br/>
6. B→(<i><b>M</b></i>)→A<br/>
7. A→(<i><b>M+G</b></i>)→B<br/>
<br/>
<br/>
<i>Problem C</i><br/>
<br/>
A person has a revolver designed for 6 bullets, but there is only one loaded and the chamber is spun.<br/>
<br/>
He has two attempts to hit a target.<br/>
He shoots once, but no bullet comes out.<br/>
<br/>
He wants to maximize his chances of making a real shot to hit a target.<br/>
What would be a better choice for him before the second shot, to spin or not to spin a chamber?<br/>
<br/>
<br/>
<i>Answer C</i></br/>
<br/>
Not to spin gives a one in five chances to shoot a bullet.<br/>
Spinning gives a one in six chances.<br/>
So, not to spin gives a better chance to make a real shot and, hopefully, hit a target.<br/>
<br/>
<br/>
<i>Problem D</i><br/>
<br/>
Two players, A and B, are playing against each other.<br/>
Each game results in one person winning and another losing.<br/>
The loser pays $1 to a winner.<br/>
Initially, both players came with $100 each.<br/>
<br/>
At end it appears that player A won 10 games and player B ended with $120.<br/>
How many games did they play?<br/>
<br/>
<i>Answer D</i><br/>
<br/>
They played 40 games.<br/>
<br/>
<br/>
<i>Problem E</i><br/>
<br/>
Three wise men (very smart indeed) after discussing some very important subject fell asleep.<br/>
<br/>
Some foolish child was passing along and decided, as a joke, to put some black shoe wax on their foreheads.<br/>
<br/>
All three wise men woke up at the same time and each of them, seeing black spots on two others' foreheads, started to laugh.<br/>
<br/>
But after a short time one of them (a bit wiser than others) stopped laughing, realizing that his own forehead also has a black spot.<br/>
<br/>
What was his logic?<br/>
<br/>
<br/>
<i>Answer E</i><br/>
<br/>
Let's call the wise men <i><b>WiseA</b></i> (that's the one who stopped laughing first), <i><b>WiseB</b></i> and <i><b>WiseC</b></i>.<br/>
<br/>
<i><b>WiseA</b></i> thinks as follows.<br/>
<br/>
<font color=brown>Assume, my forehead is clean.<br/>
Then <i><b>WiseB</b></i> and <i><b>WiseC</b></i> can see only spots on the foreheads of each other.<br/>
<br/>
Considering <i><b>WiseB</b></i> is very smart, seeing that <i><b>WiseC</b></i> is laughing and seeing no black spot on my forehead, he would immediately realize that <i><b>WiseC</b></i> is laughing at him, because he also has a black spot on his forehead. Then <i><b>WiseB</b></i> would stop laughing.<br/>
<br/>
Since <i><b>WiseB</b></i> still laughs, he must see a black spot on my forehead too. So, I better stop laughing and go to wash my face.</font><br/>
<br/>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-50563643406101574892024-02-11T07:41:00.000-08:002024-02-11T07:41:00.779-08:00Algebra+ 03: UNIZOR.COM - Math+ & Problems - Algebra<iframe style="background-image:url(https://i.ytimg.com/vi/85H_nKUTi5E/hqdefault.jpg)" width="480" height="270" src="https://youtube.com/embed/85H_nKUTi5E?si=EMgXk3Nd0CnXCDi_" frameborder="0"></iframe>
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<br/>
<u> Algebra+ 03</u><br />
<br />
<i>Problem A</i><br/>
<br/>
Let <i><b>n</b></i> be any natural number/<br/>
<br/>
Define two functions of <i><b>n</b></i>:<br/>
<i><b>F(n) = 1 − 1/2 + 1/3 -…+<br/>
+ 1/(2n-1) − 1/2n</b></i><br/>
and<br/>
<i><b>G(n) = 1/(n+1) + 1/(n+2) +…+<br/>
+ 1/(2n-1) + 1/2n</b></i><br/>
<br/>
Prove that for any natural <i><b>n</b></i> the following equality is true:<br/>
<i><b>F(n)=G(n)</b></i>.<br/>
<br/>
For example, you can manually check the following equalities:<br/>
for <i><b>n=1</b></i> (<i><b>2n=2</b></i>):<br/>
<i>1 − 1/2 = 1/2</i><br/>
for <i><b>n=2</b></i> (<i><b>2n=4</b></i>):<br/>
<i>1 − 1/2 + 1/3 − 1/4 =<br/>
= 1/3 + 1/4</i><br/>
for <i><b>n=3</b></i> (<i><b>2n=6</b></i>):<br/>
<i>1− 1/2 + 1/3 − 1/4 +<br/>
+ 1/5 − 1/6 =<br/>
= 1/4 + 1/5 + 1/6</i><br/>
<br/>
<br/>
<i>Answer A</i><br/>
<br/>
Here is a straight forward proof by induction.<br/>
<br/>
Check it for <i><b>n=1</b></i>:<br/>
<i><b>F(1) = 1−1/2 = 1/2</b></i><br/>
<i><b>G(1) = 1/2</b></i><br/>
As we see, <i><b>F(1) = G(1)</b></i><br/>
<br/>
Assume, <i><b>F(n)=G(n)</b></i>.<br/>
Let's switch to <i><b>n+1</b></i><br/>
<i><b>F(n+1) =<br/>
= F(n) + 1/(2n+1) − 1/(2n+2)</b></i><br/>
<i><b>G(n+1) = G(n) − 1/(n+1) +<br/>
+ 1/(2n+1)+1/(2n+2)</b></i><br/>
<br/>
When we switch from <i><b>n</b></i> to <i><b>n+1</b></i>, function <i><b>F(n)</b></i> has changed by<br/>
<i><b>f(n+1) = 1/(2n+1) − 1/(2n+2)</b></i><br/>
So, <i><b>F(n+1)=F(n)+f(n+1)</b></i>.<br/>
<br/>
Analogously, function <i><b>G(n)</b></i> has changed by<br/>
<i><b>g(n+1) = −1/(n+1) +<br/>
+ 1/(2n+1) + 1/(2n+2)</b></i><br/>
So, <i><b>G(n+1)=G(n)+g(n+1)</b></i>.<br/>
<br/>
If <i><b>f(n+1)</b></i>, an increment of <i><b>F(n)</b></i>, and <i><b>g(n+1)</b></i>, an increment of <i><b>G(n)</b></i>, are equal, assuming <i><b>F(n)=G(n)</b></i>, we can conclude that <i><b>F(n+1)=G(n+1)</b></i>, which proves that <i><b>F(n)=G(n)</b></i> for any natural number <i><b>n</b></i>.<br/>
<br/>
Let's check the equality<br/>
<i><b>f(n+1) = g(n+1)</b></i><br/>
<br/>
Indeed, transforming an expression to a common denominator, we get<br/>
<i><b>f(n+1) = 1<font size=5>/</font></b></i>[<i><b>(2n+1)·(2n+2)</b></i>]<br/>
<br/>
For <i><b>g(n+1)</b></i> the common denominator is<br/>
<i><b>(n+1)(2n+1)(2n+2)</b></i><br/>
The numerator is<br/>
<i><b>−(2n+1)(2n+2)+(n+1)(2n+2)+<br/>
+(n+1)(2n+1) =<br/>
= −4n² − 2n − 4n − 2 +<br/>
+ 2n² + 2n + 2n + 2 +<br/>
+ 2n² + n + 2n + 1 =<br/>
= n+1</b></i><br/>
<br/>
Dividing this numerator of <i><b>g(n+1)</b></i> by its denominator obtained above and cancelling <i><b>n+1</b></i> from both, we will get<br/>
<i><b>g(n+1) = 1<font size=5>/</font></b></i>[<i><b>(2n+1)(2n+2)</b></i>]<br/>
<br/>
This is exactly what we obtained for <i><b>f(n+1)</b></i>.<br/>
<br/>
END OF PROOF<br/>
<br/>
<br/>
<i>Problem B</i><br/>
<br/>
Two cars are approaching an intersection on two perpendicular roads.<br/>
In the beginning at time <i><b>t=0</b></i> car #1 is at distance <i><b>a</b></i> from the intersection.<br/>
The car #2 (on a perpendicular road) at <i><b>t=0</b></i> is at distance <i><b>b</b></i> from the intersection.<br/>
Both cars move with the same speed <i><b>V</b></i>.<br/>
For traffic safety, assume that <i><b>a≠b</b></i>, so there will be no collision.<br/>
<img src='http://www.unizor.com/Pictures/ProblemsAlgebra_03B.png' style='width:200px;height:200px;'><br/>
<br/>
At what time the distance between the cars will be minimal and what that minimal distance will be?<br/>
<br/>
<br/>
<i>Answer B</i><br/>
<br/>
Let <i><b>d(t)</b></i> be a distance between the cars at any moment <i><b>t</b></i>.<br/>
Then<br/>
<i><b>d²(t) = (a−v·t)² + (b−v·t)²</b></i><br/>
<br/>
Minimum of <i><b>d(t)</b></i> and <i><b>d²(t)</b></i> occur for the same moment in time <i><b>t</b></i>.<br/>
So, our task is to find minimum of <i><b>d²(t)</b></i>, which is a quadratic function of <i><b>t</b></i>.<br/>
<br/>
The time when the cars are closest to each other is<br/>
<i><b>t<sub>min</sub> = (a+b)<font size=4>/</font>(2V)</b></i><br/>
<br/>
The closest distance between the car at that time is<br/>
<i><b>d<sub>min</sub> = |a−b| <font size=4>/</font>√<span style='text-decoration:overline'>2</span></b></i><br/>
<br/>
<br/>
Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-85112939950620560352024-02-04T08:15:00.000-08:002024-02-04T08:15:12.266-08:00Geometry+ 05: UNIZOR.COM - Math+ &Problems - Geometry<iframe style="background-image:url(https://i.ytimg.com/vi/ynjmEKS5O0A/hqdefault.jpg)" width="480" height="270" src="https://youtube.com/embed/ynjmEKS5O0A?si=Cu1vG4_twAuM4FMk" frameborder="0"></iframe>
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<br/>
<u> Geometry+ 05</u><br />
<br />
<br />
<i>Problem A</i><br/>
<br/>
Construct a quadrilateral <i><b>ABCD</b></i> by its 4 sides <i><b>AB</b></i>, <i><b>BC</b></i>, <i><b>CD</b></i>, <i><b>DA</b></i> and an angle <i><b>φ</b></i> between opposite sides <i><b>AB</b></i> and <i><b>CD</b></i>.<br/>
<br/>
<i>Hint A</i><br/>
<img src='http://www.unizor.com/Pictures/ProblemsGeo_05A.png' style='width:200px;height:180px;'><br/>
Find point <i><b>P</b></i> such that <i><b>BP</b></i> is parallel and congruent to <i><b>CD</b></i>.<br/>
Consider Δ<i><b>ABP</b></i>.<br/>
<br/>
<br/>
<i>Problem B</i><br/>
<br/>
Given a circle of radius <i><b>R</b></i> and <i>n</i>-sided regular polygon inscribed into it.<br/>
Let <i><b>P</b></i> be any point on this circle.<br/>
Find a sum of squares of distances from this point <i><b>P</b></i> to all vertices of a polygon.<br/>
<br/>
<i>Hint B</i><br/>
(a) Geometrical solution for even number <i><b>n</b></i> of vertices of a regular polygon can be obtained by adding pairs of distances from <i><b>P</b></i> to <i><b>i</b></i><sup>th</sup> and (<i><b>i+</b>½<b>n</b></i>)<sup>th</sup> vertices.<br/>
(b) General solution can be obtained if using vectors from the center of a circle to all its vertices and to point <i><b>P</b></i>.<br/>
<br/>
<i>Answer</i><br/>
Sum of squares of distances from point <i><b>P</b></i> to all vertices of a polygon equals to <i><b>2nR²</b></i>.<br/>
<br/>
<br/>
<i>Problem C</i><br/>
<br/>
Given an equilateral triangle Δ<i><b>ABC</b></i>.<br/>
Extend side <i><b>AC</b></i> beyond vertex <i><b>C</b></i> to point <i><b>D</b></i> and build another equilateral triangle <i><b>CDE</b></i> with point <i><b>E</b></i> on the same side from <i><b>AD</b></i> as point <i><b>B</b></i>.<br/>
Connect points <i><b>A</b></i> and <i><b>E</b></i>. Let point <i><b>M</b></i> be a midpoint of segment <i><b>AE</b></i>.<br/>
Connect points <i><b>B</b></i> and <i><b>D</b></i>. Let point <i><b>N</b></i> be a midpoint of segment <i><b>BD</b></i>.<br/>
<img src='http://www.unizor.com/Pictures/ProblemsGeo_05C.png' style='width:200px;height:120px;'><br/>
Prove that triangle Δ<i><b>CMN</b></i> is equilateral.<br/>
<br/>
<i>Hint C</i><br/>
Triangles Δ<i><b>ACE</b></i> and Δ<i><b>BCD</b></i> are congruent.<br/>
<br/>
Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-63566201060034305702024-01-28T07:54:00.000-08:002024-01-28T07:56:18.648-08:00GeoTheorem+ 1: UNIZOR.COM - Math+ & Problems - Geometry<iframe style="background-image:url(https://i.ytimg.com/vi/64BgGbcRl0E/hqdefault.jpg)" width="480" height="270" src="https://youtube.com/embed/64BgGbcRl0E?si=FvclYAMEcQqAaAx2" frameborder="0"></iframe>
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<br/>
<u> Geometry+ GeoTheorem 1</u><br />
<br />
<i>Theorem</i><br/>
<br/>
Prove that <b>non-null</b> vector <i><b><span style='text-decoration:overline'>n</span></b></i>(<i><b>a,b,c</b></i>) in three-dimensional Cartesian coordinates <i>OXYZ</i> is <b>normal</b> (perpendicular) to a plane <i><b>α</b></i> described by an equation<br/>
<i><b>a·x + b·y + c·z + d = 0</b></i><br/>
where <i><b>a</b></i>, <i><b>b</b></i>, <i><b>c</b></i>, <i><b>d</b></i> are real numbers.<br/>
<br />
<i>Proof</i><br/>
<br/>
CASE 1 (easy) - Constant <i><b>d</b></i> in an equation that describes plane <i><b>α</b></i> equals to zero.<br/>
<br/>
The equation for plane <i><b>α</b></i> looks in this case as<br/>
<i><b>a·x + b·y + c·z = 0</b></i><br/>
<br/>
Then plane <i><b>α</b></i> must go through the origin of coordinates <i><b>O</b></i>(<i><b>0,0,0</b></i>) because this point satisfies the equation for <i><b>α</b></i>.<br/>
<br/>
Consider a vector from an origin of coordinate <i><b>O</b></i>(<i><b>0,0,0</b></i>) to <b>any other point</b> on a plane <i><b>Q</b></i>(<i><b>x,y,z</b></i>).<br/>
Obviously, non-null vector <i><b><span style='text-decoration:overline'>OQ</span></b></i>(<i><b>x,y,z</b></i>) is lying fully within plane <i><b>α</b></i> because both its ends - points <i><b>O</b></i> and <i><b>Q</b></i> lie within this plane.<br/>
<br/>
We can interpret the equation<br/>
<i><b>a·x + b·y + c·z = 0</b></i><br/>
that describes plane <i><b>α</b></i> in this case as a scalar product of non-null vector <i><b><span style='text-decoration:overline'>n</span></b></i>(<i><b>a,b,c</b></i>) and non-null vector <i><b><span style='text-decoration:overline'>OQ</span></b></i>(<i><b>x,y,z</b></i>).<br/>
<br/>
Since this scalar product between vector <i><b><span style='text-decoration:overline'>n</span></b></i>(<i><b>a,b,c</b></i>) and any vector <i><b><span style='text-decoration:overline'>OQ</span></b></i>(<i><b>x,y,z</b></i>) lying within plane <i><b>α</b></i> is zero, vector <i><b><span style='text-decoration:overline'>n</span></b></i>(<i><b>a,b,c</b></i>) is perpendicular to plane <i><b>α</b></i>.<br/>
<br/>
CASE 2 - Constant <i><b>d</b></i> is not equal to zero.<br/>
<br/>
Consider two planes defined by two equations<br/>
Plane <i><b>α</b></i>: <i><b>a·x + b·y + c·z + d = 0</b></i><br/>
Plane <i><b>β</b></i>: <i><b>a·x + b·y + c·z = 0</b></i><br/>
<br/>
Since <i><b>d ≠ 0</b></i>, any point that satisfies one of these equations will not satisfy another.<br/>
Therefore, these planes do not have any common points and, therefore, are parallel.<br/>
<br/>
We have already proven that <b>non-null</b> vector <i><b><span style='text-decoration:overline'>n</span></b></i>(<i><b>a,b,c</b></i>) is perpendicular to plane <i><b>β</b></i> (see CASE 1 above).<br/>
Consequently, this vector <i><b><span style='text-decoration:overline'>n</span></b></i> is perpendicular to <i><b>α</b></i> as well, that is <i><b><span style='text-decoration:overline'>n</span></b></i> is <b>normal</b> to <i><b>α</b></i>.<br/>
<br/>
End of the proof that <i><b><span style='text-decoration:overline'>n</span></b></i><b>⊥</b><i><b>α</b></i>.<br/>
<br/>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-8186560483613083112024-01-27T13:40:00.000-08:002024-01-27T13:40:38.101-08:00Geometry+ 04: UNIZOR.COM - Math+ & Problems - Geometry<iframe width="480" height="270" src="https://youtube.com/embed/USlxZ6Gx39w?si=yGy8FLhPNKglPumI" frameborder="0"></iframe>
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<br/>
<u> Geometry+ 04</u><br />
<br />
<br />
This lecture is dedicated to planes, straight lines and their perpendicularity in three-dimensional space with Cartesian coordinates.<br/>
<br/>
<i>Problem A</i><br/>
<br/>
Given a point <i><b>P</b></i>(<i><b>a,b,c</b></i>) in three-dimensional Cartesian space not coinciding with the origin of coordinates <i><b>O</b></i>(<i><b>0,0,0</b></i>).<br/>
<br/>
What will be an equation for coordinates <i><b>x</b></i>, <i><b>y</b></i> and <i><b>z</b></i> that describes a plane going through origin of coordinates <i><b>O</b></i>(<i><b>0,0,0</b></i>) and is perpendicular to line <i><b>OP</b></i>?<br/>
<br/>
<i>Solution A</i><br/>
<br/>
Let's use Vector Algebra techniques to solve this problem.<br/>
<br/>
Consider a non-null vector <i><b><span style='text-decoration:overline'>OP</span></b></i> originated at point <i><b>O</b></i>(<i><b>0,0,0</b></i>) with endpoint at <i><b>P</b></i>(<i><b>a,b,c</b></i>) and a plane that goes through point <i><b>O</b></i>(<i><b>0,0,0</b></i>) perpendicularly to vector <i><b><span style='text-decoration:overline'>OP</span></b></i> at this point.<br/>
So, vector <i><b><span style='text-decoration:overline'>OP</span></b></i> is <b>normal</b> to this plane.<br/>
<br/>
Consider <b>any</b> point <i><b>Q</b></i>(<i><b>x,y,z</b></i>) on this plane and vector <i><b><span style='text-decoration:overline'>OQ</span></b></i>.<br/>
This vector is lying completely within a plane because both its ends <i><b>O</b></i> and <i><b>Q</b></i> belong to this plane.<br/>
<br/>
Therefore, since vector <i><b><span style='text-decoration:overline'>OP</span></b></i> is, by assumption, perpendicular to the plane, it's perpendicular to vector <i><b><span style='text-decoration:overline'>OQ</span></b></i> lying within this plane.<br/>
<br/>
Two perpendicular vectors have their scalar product equal to zero.<br/>
Therefore, <i><b><span style='text-decoration:overline'>OP</span></b></i> <b>·</b> <i><b><span style='text-decoration:overline'>OQ</span></b></i> <i><b>= 0</b></i><br/>
<img src='http://www.unizor.com/Pictures/ProblemsGeo_04A.png' style='width:200px;height:210px;'><br/>
<br/>
Let's express this in coordinates.<br/>
Vector <i><b><span style='text-decoration:overline'>OP</span></b></i> in coordinate form is </b></i>(<i><b>a,b,c</b></i>).<br/>
Since coordinates of point <i><b>Q</b></i> are (<i><b>x,y,z</b></i>), vector <i><b><span style='text-decoration:overline'>OQ</span></b></i> in coordinate form is (<i><b>x,y,z</b></i>).<br/>
<br/>
Now the scalar product of vectors <i><b><span style='text-decoration:overline'>OP</span></b></i> and <i><b><span style='text-decoration:overline'>OQ</span></b></i> is<br/>
<i><b><span style='text-decoration:overline'>OP</span></b></i> <b>·</b> <i><b><span style='text-decoration:overline'>OQ</span> =<br/>
= a·x + b·y + c·z</b></i><br/>
which is supposed to be equal to <b>zero</b> since these vectors are perpendicular.<br/>
<br/>
Taking into account that there is <b>one and only one</b> plane perpendicular to a given vector <i><b><span style='text-decoration:overline'>OP</span></b></i> at its origin <i><b>O</b></i>(<i><b>0,0,0</b></i>) and we have chosen <b>any</b> point <i><b>Q</b></i>(<i><b>x,y,z</b></i>) on this plane, the equation of a plane should be<br/>
<i><b>a·x+b·y+c·z = 0</b></i><br/>
<br/>
CONCLUSION 1<br/>
If vector (<i><b>a,b,c</b></i>) is <b>normal</b> to a plane going through the origin of coordinates </b></i>(<i><b>0,0,0</b></i>) then the equation of this plane is<br/>
<i><b>a·x+b·y+c·z = 0</b></i><br/>
<br/>
CONCLUSION 2<br/>
The plane described by an equation<br/>
<i><b>a·x+b·y+c·z = 0</b></i><br/>
is perpendicular to vector (<i><b>a,b,c</b></i>).<br/>
<br/>
<br/>
<i>Problem B</i><br/>
<br/>
Given a point <i><b>P</b></i>(<i><b>a,b,c</b></i>) in three-dimensional Cartesian space not coinciding with the origin of coordinates <i><b>O</b></i>(<i><b>0,0,0</b></i>).<br/>
<br/>
What will be an equation for coordinates <i><b>x</b></i>, <i><b>y</b></i> and <i><b>z</b></i> that describes a plane going through point <i><b>P</b></i> and is perpendicular to a line connecting this point with the origin of coordinates <i><b>O</b></i>?<br/>
<br/>
<i>Solution B</i><br/>
<br/>
Let's use Vector Algebra techniques to solve this problem.<br/>
<br/>
Consider a non-null vector <i><b><span style='text-decoration:overline'>OP</span></b></i> originated at point <i><b>O</b></i>(<i><b>0,0,0</b></i>) with endpoint at <i><b>P</b></i>(<i><b>a,b,c</b></i>) and a plane that is assumed to be perpendicular to <i><b><span style='text-decoration:overline'>OP</span></b></i> at point <i><b>P</b></i>.<br/>
So, vector <i><b><span style='text-decoration:overline'>OP</span></b></i> is <b>normal</b> to this plane.<br/>
<br/>
Consider <b>any</b> point <i><b>Q</b></i>(<i><b>x,y,z</b></i>) on this plane and vector <i><b><span style='text-decoration:overline'>PQ</span></b></i>.<br/>
This vector is lying completely within a plane because both its ends <i><b>P</b></i> and <i><b>Q</b></i> belong to this plane.<br/>
<br/>
Therefore, since vector <i><b><span style='text-decoration:overline'>OP</span></b></i> is, by assumption, perpendicular to the plane, it's perpendicular to vector <i><b><span style='text-decoration:overline'>PQ</span></b></i> lying within this plane.<br/>
<br/>
Two perpendicular vectors have their scalar product equal to zero.<br/>
Therefore, <i><b><span style='text-decoration:overline'>OP</span></b></i> <b>·</b> <i><b><span style='text-decoration:overline'>PQ</span></b></i> <i><b>= 0</b></i><br/>
<img src='http://www.unizor.com/Pictures/ProblemsGeo_04B.png' style='width:200px;height:210px;'><br/>
<br/>
Let's express this in coordinates.<br/>
Vector <i><b><span style='text-decoration:overline'>OP</span></b></i> in coordinate form is </b></i>(<i><b>a,b,c</b></i>).<br/>
Since coordinates of point <i><b>Q</b></i> are (<i><b>x,y,z</b></i>), vector <i><b><span style='text-decoration:overline'>OQ</span></b></i> in coordinate form is (<i><b>x,y,z</b></i>).<br/>
Vector <i><b><span style='text-decoration:overline'>PQ</span></b></i> can be represented as a difference between <i><b><span style='text-decoration:overline'>OQ</span></b></i> and <i><b><span style='text-decoration:overline'>OP</span></b></i>.<br/>
Therefore,<br/>
<i><b><span style='text-decoration:overline'>PQ</span> = </b></i>(<i><b>x−a, y−b, z−c</b></i>).<br/>
<br/>
Now the scalar product of vectors <i><b><span style='text-decoration:overline'>OP</span></b></i> and <i><b><span style='text-decoration:overline'>PQ</span></b></i> is<br/>
<i><b><span style='text-decoration:overline'>OP</span></b></i> <b>·</b> <i><b><span style='text-decoration:overline'>PQ</span> =<br/>
= a·(x−a) + b·(y−b) + c·(z−c) =<br/>
= a·x+b·y+c·z − (a²+b²+c²)</b></i><br/>
which is supposed to be equal to <b>zero</b> since these vectors are perpendicular.<br/>
<br/>
Taking into account that there is <b>one and only one</b> plane perpendicular to a given vector <i><b><span style='text-decoration:overline'>OP</span></b></i> at its endpoint <i><b>P</b></i>(<i><b>a,b,c</b></i>) and we have chosen <b>any</b> point <i><b>Q</b></i>(<i><b>x,y,z</b></i>) on this plane, the equation of a plane should be<br/>
<i><b>a·x+b·y+c·z − (a²+b²+c²) = 0</b></i><br/>
or<br/>
<i><b>a·x+b·y+c·z = a²+b²+c²</b></i><br/>
<br/>
CONCLUSION 1<br/>
If vector (<i><b>a,b,c</b></i>) is <b>normal</b> to a plane going through its endpoint </b></i>(<i><b>a,b,c</b></i>) then the equation of this plane is<br/>
<i><b>a·x+b·y+c·z = a²+b²+c²</b></i><br/>
<br/>
CONCLUSION 2<br/>
The plane described by an equation<br/>
<i><b>a·x+b·y+c·z = a²+b²+c²</b></i><br/>
is perpendicular to vector (<i><b>a,b,c</b></i>) at its endpoint </b></i>(<i><b>a,b,c</b></i>).<br/>
<br/>
<br/>
<i>Problem C</i><br/>
<br/>
Given a plane <i><b>α</b></i> in three-dimensional Cartesian space going through the origin of coordinates <i><b>O</b></i>(<i><b>0,0,0</b></i>) and described by an equation<br/>
<i><b>a·x + b·y + c·z = 0</b></i><br/>
<br/>
What will be the coordinates <i><b>p</b></i>, <i><b>q</b></i> and <i><b>r</b></i> of an endpoint of a <b>normal</b> (perpendicular to plane <i><b>α</b></i>) vector originated at point <i><b>O</b></i> in terms of coefficients <i><b>a,b,c</b></i> defining the plane <i><b>α</b></i>?<br/>
<br/>
<i>Solution C</i><br/>
<br/>
Obviously, point <i><b>O</b></i>(<i><b>0,0,0</b></i>) belongs to plane <i><b>α</b></i> because substituting <i><b>x=0</b></i>, <i><b>y=0</b></i> and <i><b>z=0</b></i> transforms the equation into an identity <i><b>0=0</b></i>.<br/>
<br/>
Let's choose any other point <i><b>Q</b></i> with coordinates (<i><b>x,y,z</b></i>) satisfying the equation of a plane <i><b>α</b></i><br/>
<i><b>a·x + b·y + c·z = 0</b></i><br/>
<br/>
Consider vector <i><b><span style='text-decoration:overline'>OQ</span>=(x,y,z)</b></i>.<br/>
Its origin is point <i><b>O</b></i> that belongs to our plane <i><b>α</b></i>.<br/>
Its endpoint <i><b>Q</b></i>(<i><b>x,y,z</b></i>) also belongs to this plane since plane <i><b>α</b></i> is a locus of all points satisfying the above equation.<br/>
Therefore, vector <i><b><span style='text-decoration:overline'>OQ</span></b></i> lies within our plane <i><b>α</b></i>.<br/>
<br/>
Assume, vector <i><b><span style='text-decoration:overline'>OP</span>=(p,q,r)</b></i> is normal to plane <i><b>α</b></i> at point <i><b>O</b></i>.<br/>
Its origin is the same point <i><b>O</b></i> as the origin of vector <i><b><span style='text-decoration:overline'>OQ</span>=(x,y,z)</b></i> that belongs to our plane.<br/>
Its endpoint <i><b>P</b></i>(<i><b>p,q,r</b></i>) lies somewhere in space.<br/>
<br/>
Since <i><b><span style='text-decoration:overline'>OP</span> </i>⊥<i> α</b></i>, <i><b><span style='text-decoration:overline'>OP</span></b></i> is perpendicular to any vector lying within plane <i><b>α</b></i>, including vector <i><b><span style='text-decoration:overline'>OQ</span></b></i>.<br/>
<br/>
Recall that necessary and sufficient condition for two vectors to be perpendicular to each other is their scalar product is zero.<br/>
<i><b><span style='text-decoration:overline'>OP</span> · <span style='text-decoration:overline'>OQ</span> = 0</b></i><br/>
Hence,<br/>
<i><b>p·x + q·y + r·z = 0</b></i><br/>
<br/>
The above equation is very much alike the equation of the plane<br/>
<i><b>a·x + b·y + c·z = 0</b></i><br/>
Therefore, we should not look very far for values of <i><b>p</b></i>, <i><b>q</b></i> and <i><b>r</b></i>, but just choose <i><b>p=a</b></i>, <i><b>q=b</b></i> and <i><b>r=c</b></i>.<br/>
<br/>
Now the equation of a plane <i><b>α</b></i> can be interpreted as a scalar product of two vectors <i><b><span style='text-decoration:overline'>OP</span>=(a,b,c)</b></i> and <i><b><span style='text-decoration:overline'>OQ</span>=(x,y,z)</b></i>.<br/>
Since this scalar product is equal to zero, vectors <i><b><span style='text-decoration:overline'>OP</span></b></i> and <i><b><span style='text-decoration:overline'>OQ</span></b></i> are perpendicular to each other.<br/>
<br/>
Since point <i><b>Q</b></i>(<i><b>x,y,z</b></i>) can be <b>any</b> point on a plane vector <i><b><span style='text-decoration:overline'>OP</span>=(a,b,c)</b></i> is perpendicular to the whole plane <i><b>α</b></i> defined by equation<br/>
<i><b>a·x + b·y + c·z = 0</b></i><br/>
<br/>
<i>Answer C</i><br/>
The coordinates of an endpoint of a <b>normal</b> (perpendicular to plane <i><b>α</b></i>) vector originated at point <i><b>O</b></i> are <i><b>(a,b,c)</b></i>.<br/>
<br/>
<br/>
<i>Problem D</i><br/>
<br/>
Given a plane <i><b>α</b></i> in three-dimensional Cartesian space described by a general equation<br/>
<i><b>a·x + b·y + c·z + d = 0</b></i><br/>
<br/>
Point <i><b>O</b></i>(<i><b>0,0,0</b></i>) is an origin of coordinates.<br/>
Assume, point <i><b>P</b></i>(<i><b>p,q,r</b></i>) belongs to plane <i><b>α</b></i> and vector <nobr><i><b><span style='text-decoration:overline'>OP</span>=(p,q,r)</b></i></nobr> is <b>normal</b> (perpendicular) to this plane.<br/>
<br/>
What will be the coordinates <i><b>p</b></i>, <i><b>q</b></i> and <i><b>r</b></i> of an endpoint of vector <i><b><span style='text-decoration:overline'>OP</span></b></i> in terms of coefficients <i><b>a,b,c,d</b></i> defining the plane <i><b>α</b></i>?<br/>
<br/>
<i>Solution D</i><br/>
<br/>
We have two conditions for coordinates <i><b>p</b></i>, <i><b>q</b></i> and <i><b>r</b></i> to satisfy:<br/>
1. Point <i><b>P</b></i>(<i><b>p,q,r</b></i>) belongs to plane <i><b>α</b></i>, which means<br/>
<i><b>a·p + b·q + c·r + d = 0</b></i><br/>
2. Vector <i><b><span style='text-decoration:overline'>OP</span></b></i> is <b>normal</b> (perpendicular) to plane <i><b>α</b></i>.<br/>
<br/>
Consider <b>any</b> point <i><b>Q</b></i>(<i><b>x,y,z</b></i>) on plane <i><b>α</b></i> and vector <i><b><span style='text-decoration:overline'>PQ</span></b></i>.<br/>
This vector is lying completely within a plane because both its ends <i><b>P</b></i> and <i><b>Q</b></i> belong to this plane.<br/>
<br/>
Therefore, since vector <i><b><span style='text-decoration:overline'>OP</span></b></i> is, by assumption, perpendicular to the plane, it's perpendicular to vector <i><b><span style='text-decoration:overline'>PQ</span></b></i> lying within this plane.<br/>
<br/>
Two perpendicular vectors have their scalar product equal to zero.<br/>
Therefore, <i><b><span style='text-decoration:overline'>OP</span></b></i> <b>·</b> <i><b><span style='text-decoration:overline'>PQ</span></b></i> <i><b>= 0</b></i><br/>
<br/>
Let's transform this into coordinates.<br/>
Vector <i><b><span style='text-decoration:overline'>PQ</span></b></i> is a difference between <i><b><span style='text-decoration:overline'>OQ</span></b></i> and <i><b><span style='text-decoration:overline'>OP</span></b></i>.<br/>
Therefore, the coordinates of <i><b><span style='text-decoration:overline'>PQ</span></b></i> are<br/>
(<i><b>x−p, y−q, z−r</b></i>)<br/>
<br/>
Since <i><b><span style='text-decoration:overline'>OP</span></b></i> <b>⊥</b> <i><b><span style='text-decoration:overline'>PQ</span></b></i>, their scalar product is zero<br/>
<i><b>p·(x−p)+q·(y−q)+r·(z−r) = 0</b></i><br/>
or<br/>
<i><b>p·x+q·y+r·z−(p²+q²+r²) = 0</b></i><br/>
<br/>
From the above equation we have to find <i><b>p</b></i>, <i><b>q</b></i> and <i><b>r</b></i> knowing that <i><b>x</b></i>, <i><b>y</b></i> and <i><b>z</b></i> satisfy the equation of a plane <i><b>α</b></i><br/>
<i><b>a·x + b·y + c·z + d = 0</b></i><br/>
<br/>
Simple guess <i><b>p=a</b></i>, <i><b>q=b</b></i> and <i><b>r=c</b></i> is a good idea, but not a solution since<br/>
<i><b>d ≠ −(p²+q²+r²)</b></i><br/>
<br/>
To correct this situation, let's scale the vector (<i><b>p,q,r</b></i>)<b>=</b>(<i><b>a,b,c</b></i>) by some coefficient <i><b>k</b></i> to satisfy both conditions for a solution listed above.<br/>
<br/>
First, let's find a scale factor <i><b>k</b></i> to make sure that endpoint <i><b>P</b></i>(<i><b>k·a,k·b,k·c</b></i>) of vector <i><b><span style='text-decoration:overline'>OP</span></b></i> belongs to plane <i><b>α</b></i>.<br/>
<br/>
Substituting <i><b>x=k·a</b></i>, <i><b>y=k·b</b></i> and <i><b>z=k·c</b></i> into equation that defines plane <i><b>α</b></i>, we obtain an equation for <i><b>k</b></i>:<br/>
<i><b>a·k·a + b·k·b + c·k·c + d = 0</b></i><br/>
<br/>
From this equation<br/>
<i><b>k = −d<font size=4>/</font>(a²+b²+c²)</b></i><br/>
This value of <i><b>k</b></i> is sufficient to put our point <i><b>P</b></i>(<i><b>p,q,r</b></i>) on the plane <i><b>α</b></i>.<br/>
<br/>
Let's check that defined this way vectors <i><b><span style='text-decoration:overline'>OP</span></b></i> and <i><b><span style='text-decoration:overline'>PQ</span></b></i> are perpendicular to each other for any point <i><b>Q</b></i>(<i><b>x,y,z</b></i>) that satisfies the equation defining plane <i><b>α</b></i><br/>
<i><b>a·x + b·y + c·z + d = 0</b></i><br/>
<br/>
Indeed,<br/>
<i><b><span style='text-decoration:overline'>OP</span></b></i> <b>·</b> <i><b><span style='text-decoration:overline'>PQ</span></b></i> <b>=<br/>
=</b> (<i><b>k·a, k·b, k·c</b></i>) ·<br/>
· (<i><b>x−k·a, y−k·b, z−k·c</b></i>) <b>=<br/>
=</b> <i><b>k·(x·a + y·b + z·c) −<br/>
− k²·(a² + b² + c²)</b></i><br/>
where<br/>
<i><b>a·x + b·y + c·z + d = 0</b></i> and<br/>
<i><b>k = −d<font size=4>/</font>(a²+b²+c²)</b></i><br/>
<br/>
Substitute<br/>
<i><b>a·x + b·y + c·z = −d</b></i> and<br/>
<i><b>(a²+b²+c²) = −d·k</b></i><br/>
getting<br/>
<i><b><span style='text-decoration:overline'>OP</span></b></i> <b>·</b> <i><b><span style='text-decoration:overline'>PQ</span> =<br/>
= k·(−d) −k²·(−d/k) = 0</b></i><br/>
<br/>
So, since point <i><b>Q</b></i>(<i><b>x,y,z</b></i>) was chosen arbitrarily on plane <i><b>α</b></i>, vector <i><b><span style='text-decoration:overline'>OP</span></b></i> is perpendicular to <b>any</b> vector on a plane and, therefore, to plane <i><b>α</b></i> itself.<br/>
<br/>
Taking into account that <i><b>P </i>∈<i> α</b></i>, it proves that the coordinates <i><b>p</b></i>, <i><b>q</b></i> and <i><b>r</b></i> of an endpoint of vector <i><b><span style='text-decoration:overline'>OP</span></b></i> are<br/>
<i><b>p = −d·a/(a²+b²+c²)</b></i><br/>
<i><b>q = −d·b/(a²+b²+c²)</b></i><br/>
<i><b>r = −d·c/(a²+b²+c²)</b></i><br/>
<br/>
Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-32740333758585890362024-01-23T08:01:00.000-08:002024-01-23T08:01:14.366-08:00Geometry+ 03: UNIZOR.COM - Math+ & Problems - Geometry<iframe width="480" height="270" src="https://youtube.com/embed/i-GNJ8gVoRM?si=meWkiSoWiuh8xdqT" frameborder="0"></iframe>
<i>Notes to a video lecture on http://www.unizor.com</i><br/>
<br/>
<u> Geometry+ 03</u><br />
<br />
<br />
This lecture is dedicated to problems of triangle construction by its certain elements.<br/>
<br/>
The tools of construction are a ruler to draw straight lines and a compass to draw circles.<br/>
<br/>
We will use the following naming rules.<br/>
Vertices of triangle will be call by upper case Latin letters <i><b>A</b></i>, <i><b>B</b></i> and <i><b>C</b></i>.<br/>
Sides will be called by lower case Latin letters corresponding to opposite vertices: side <i><b>a</b></i> is opposite to vertex <i><b>A</b></i> etc.<br/>
Angles will be called by lower case Greek letters corresponding to names of their vertices: angle <i><b>α</b></i> is at vertex <i><b>A</b></i> etc.<br/>
Medians are named <i><b>m</b></i> with a subscript of a side onto which they fall: median <i><b>m<sub>a</sub></b></i> is from vertex <i><b>A</b></i> to side <i><b>a</b></i> etc.<br/>
Altitudes are named <i><b>h</b></i> with similar subscripts, like <i><b>h<sub>a</sub></b></i> etc.<br/>
Angle bisectors are named <i><b>l<sub>a</sub></b></i> etc.<br/>
Radius of a circumscribed circle of a triangle is named <i><b>R</b></i>.<br/>
Radius of an inscribed circle is named <i><b>r</b></i>.<br/>
<br/>
<i>Problem A</i><br/>
Construct a triangle by its three altitudes <i><b>h<sub>a</sub></b></i>, <i><b>h<sub>b</sub></b></i> and <i><b>h<sub>c</sub></b></i>.<br/>
<br/>
<i>Analysis A</i><br/>
<br/>
As we know, a product of a side by an altitude falling on it is a double area of a triangle.<br/>
Therefore,<br/>
<i><b>a·h<sub>a</sub> = b·h<sub>b</sub> = c·h<sub>c</sub></b></i><br/>
Hence,<br/>
<i><b>b = a·h<sub>a</sub>/h<sub>b</sub></b></i><br/>
<i><b>c = a·h<sub>a</sub>/h<sub>c</sub></b></i><br/>
<br/>
Let's construct a triangle similar to Δ<i><b>ABC</b></i> defined by sides <i><b>a</b></i>, <i><b>b</b></i> and <i><b>c</b></i> by <b>choosing any segment</b> <i><b>x</b></i> and defining two other segments <i><b>y</b></i> and <i><b>z</b></i> using the equations similar to above.<br/>
<i><b>y ≝ x·h<sub>a</sub>/h<sub>b</sub></b></i><br/>
<i><b>z ≝ x·h<sub>a</sub>/h<sub>c</sub></b></i><br/>
Segments <i><b>y</b></i> and <i><b>z</b></i> can be easily constructed from these definitions, knowing <i><b>x</b></i> (arbitrarily chosen) and given altitudes.<br/>
<br/>
Let <i><b>x/a=k</b></i> be a scaling factor between arbitrarily chosen segment <i><b>x</b></i> and side <i><b>a</b></i> of triangle Δ<i><b>ABC</b></i>.<br/>
From this follow these relationships:<br/>
<i><b>x = a·k</b></i><br/>
<i><b>y = x·h<sub>a</sub>/h<sub>b</sub> = a·k·h<sub>a</sub>/h<sub>b</sub> = b·k</b></i><br/>
<i><b>z = x·h<sub>a</sub>/h<sub>c</sub> = a·k·h<sub>a</sub>/h<sub>c</sub> = c·k</b></i><br/>
The scaling factor <i><b>k</b></i> is the same for <i><b>x/a=k</b></i>, <i><b>y/b=k</b></i> and <i><b>z/c=k</b></i>.<br/>
<br/>
Therefore, triangle Δ<i><b>XYZ</b></i> constructed from three segments <i><b>x</b></i>, <i><b>y</b></i> and <i><b>z</b></i> is <b>similar</b> to triangle Δ<i><b>ABC</b></i> with segments <i><b>a</b></i>, <i><b>b</b></i> and <i><b>c</b></i> we have to construct.<br/>
<br/>
From similarity of triangles follows the congruence of corresponding angles<br/>
<i><b>
</b></i>∠<i><b>BAC = </b></i>∠<i><b>α = </b></i>∠<i><b>YXZ<br/>
</b></i>∠<i><b>CBA = </b></i>∠<i><b>β = </b></i>∠<i><b>ZYX<br/>
</b></i>∠<i><b>ACB = </b></i>∠<i><b>γ = </b></i>∠<i><b>XZY</b></i><br/>
<br/>
Therefore, our analysis shows that by constructing Δ<i><b>XYZ</b></i> we get all angles of Δ<i><b>ABC</b></i>.<br/>
<br/>
This is the end of analysis, as the construction of triangle Δ<i><b>ABC</b></i>, knowing its three angles and altitudes, is straight forward.<br/>
<br/>
<i>Solution A</i><br/>
<img src='http://www.unizor.com/Pictures/ProblemsGeo_03A.png' style='width:200px;height:210px;'><br/>
To enlarge this picture, right click on it and choose "Open image in new tab"<br/>
<br/>
Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-78088772192196250462024-01-22T07:44:00.000-08:002024-01-22T07:44:40.408-08:00Geometry+ 02: UNIZOR.COM - Math+ & Problems - Geometry<iframe width="480" height="270" src="https://youtube.com/embed/5ZSPOFIHKGs?si=K4306JdJeIyzP02F" frameborder="0"></iframe>
<i>Notes to a video lecture on http://www.unizor.com</i><br/>
<br/>
<u> Geometry+ 02</u><br />
<br />
<br />
This lecture is dedicated to problems of triangle construction by its certain elements.<br/>
<br/>
The tools of construction are a ruler to draw straight lines and a compass to draw circles.<br/>
<br/>
We will use the following naming rules.<br/>
Vertices of triangle will be call by upper case Latin letters <i><b>A</b></i>, <i><b>B</b></i> and <i><b>C</b></i>.<br/>
Sides will be called by lower case Latin letters corresponding to opposite vertices: side <i><b>a</b></i> is opposite to vertex <i><b>A</b></i> etc.<br/>
Angles will be called by lower case Greek letters corresponding to names of their vertices: angle <i><b>α</b></i> is at vertex <i><b>A</b></i> etc.<br/>
Medians are named <i><b>m</b></i> with a subscript of a side onto which they fall: median <i><b>m<sub>a</sub></b></i> is from vertex <i><b>A</b></i> to side <i><b>a</b></i> etc.<br/>
Altitudes are named <i><b>h</b></i> with similar subscripts, like <i><b>h<sub>a</sub></b></i> etc.<br/>
Angle bisectors are named <i><b>l<sub>a</sub></b></i> etc.<br/>
Radius of a circumscribed circle of a triangle is named <i><b>R</b></i>.<br/>
Radius of an inscribed circle is named <i><b>r</b></i>.<br/>
<br/>
<i>Problem A</i><br/>
Construct a triangle by its two sides <i><b>a</b></i>, <i><b>b</b></i> and a median onto its third side <i><b>m<sub>c</sub></b></i>.<br/>
<img src='http://www.unizor.com/Pictures/ProblemsGeo_02A.png' style='width:200px;height:100px;'><br/>
<br/>
<i>Problem B</i><br/>
Construct a triangle by its side <i><b>a</b></i> and two medians <i><b>m<sub>a</sub></b></i> and <i><b>m<sub>b</sub></b></i>.<br/>
<img src='http://www.unizor.com/Pictures/ProblemsGeo_02B.png' style='width:200px;height:100px;'><br/>
<i>Hint</i>: Construct Δ<i><b>BQR</b></i> where <i><b>BR=a/2</b></i>, <i><b>BQ=2m<sub>b</sub>/3</b></i> and <i><b>QR=m<sub>a</sub>/3</b></i>.<br/>
<br/>
<i>Problem C</i><br/>
Construct a triangle by its three medians <i><b>m<sub>a</sub></b></i>, <i><b>m<sub>b</sub></b></i> and <i><b>m<sub>c</sub></b></i>.<br/>
<img src='http://www.unizor.com/Pictures/ProblemsGeo_02C.png' style='width:200px;height:100px;'><br/>
<i>Hint</i>: Construct Δ<i><b>PQC</b></i> where <i><b>CQ=2m<sub>a</sub>/3</b></i>, <i><b>PQ=2m<sub>b</sub>/3</b></i> and <i><b>CP=2m<sub>c</sub>/3</b></i>.<br/>
<br/>
<i>Problem D</i><br/>
Construct a triangle by its side <i><b>a</b></i> and two altitudes onto other sides <i><b>h<sub>b</sub></b></i> and <i><b>h<sub>c</sub></b></i>.<br/>
<img src='http://www.unizor.com/Pictures/ProblemsGeo_02D.png' style='width:200px;height:160px;'><br/>
<br/>
Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-19171079864967127162024-01-20T06:59:00.000-08:002024-01-28T08:41:25.805-08:00Trigonometry+ 01: UNIZOR.COM - Math+ & Problems - Trigonometry<iframe style="background-image:url(https://i.ytimg.com/vi/bZWT2U6pcRE/hqdefault.jpg)" width="480" height="270" src="https://youtube.com/embed/bZWT2U6pcRE?si=ZQGxkD_9ieHTevUA" frameborder="0"></iframe>
<i>Notes to a video lecture on http://www.unizor.com</i><br/>
<br/>
<u> Trigonometry+ 01</u><br />
<br />
<br />
<i>Problem A</i><br/>
<br/>
Using the Euler Formula<br/>
<i>e<sup>i·x</sup> = cos(x)+i·sin(x)</i><br/>
prove the formulas for sine and cosine of a sum of two angles.<br/>
<br/>
<i>Solution A</i><br/>
<br/>
The Euler formula states that the real part of <i>e<sup>i·x</sup></i> is <i>cos(x)</i> and its imaginary part s <i>i·sin(x)</i><br/>
<br/>
From <i><b>i² = −1</b></i>,<br/>
<i><b>e<sup>i·(α+β)</sup> = cos(α+β)+i·sin(α+β)</b></i>,<br/>
and <i><b>a<sup>(p+q)</sup> = a<sup>p</sup>·a<sup>q</sup></b></i><br/>
follows:<br/>
<i><b>cos(α+β)+i·sin(α+β) =<br/>
= e<sup>i·(α+β)</sup> = e<sup>i·α</sup>·e<sup>i·β</sup> =<br/>
= </b></i>[<i><b>cos(α)+i·sin(α)</b></i>]<i><b> ·<br/>
· </b></i>[<i><b>cos(β)+i·sin(β)</b></i>]<i><b> =<br/>
= cos(α)·cos(β)−sin(α)·sin(β) +<br/>
+ i·cos(α)·sin(β)+i·sin(α)·cos(β)</b></i><br/>
<br/>
The real part of the last expression is<br/>
<i><b>cos(α)·cos(β) − sin(α)·sin(β)</b></i><br/>
Its imaginary part is<br/>
<i><b>i·</b></i>[<i><b>cos(α)·sin(β) + sin(α)·cos(β)</b></i>]<br/>
<br/>
Therefore,<br/>
<i><b>cos(α+β)</b></i><br/>
(the real part of <i><b>e<sup>i·(α+β)</sup></b></i>)<br/>
equals to<br/>
<i><b>cos(α)·cos(β) − sin(α)·sin(β)</b></i>,<br/>
hence<br/>
<i><b>cos(α+β) =<br/>
= cos(α)·cos(β) − sin(α)·sin(β)</b></i><br/>
<br/>
Analogously,<br/>
<i><b>i·sin(α+β)</b></i><br/>
(the imaginary part of <i><b>e<sup>i·(α+β)</sup></b></i>)<br/>
equals to<br/>
<i><b>i·</b></i>[<i><b>cos(α)·sin(β) + sin(α)·cos(β)</b></i>],<br/>
hence<br/>
<i><b>sin(α+β) =<br/>
= cos(α)·sin(β) + sin(α)·cos(β)</b></i><br/>
<br/>
<br/>
<i>Problem B</i><br/>
<br/>
Using the Euler Formula<br/>
<i>e<sup>i·x</sup> = cos(x)+i·sin(x)</i><br/>
prove the formulas for derivatives of sine and cosine functions.<br/>
<br/>
<i>Solution B</i><br/>
<br/>
Let's differentiate the Euler's formula<br/>
<i>e<sup>i·x</sup> = cos(x)+i·sin(x)</i><br/>
<br/>
On the left side the result is<br/>
<i>d<b>/</b>d<b>x</b></i>[<i><b>e<sup>i·x</sup></b></i>]<i><b> = i·e<sup>i·x</sup> =<br/>
= i·</b></i>[<i><b>cos(x) + i·sin(x)</b></i>]<i><b> =<br/>
= −sin(x) + i·cos(x)</b></i><br/>
<br/>
On the right side the result of differentiation is<br/>
<i>d<b>/</b>d<b>x</b></i>[<i><b>cos(x)</b></i>]<i><b> + i·</b>d<b>/</b>d<b>x</b></i>[<i><b>sin(x)</b></i>]</b></i><br/>
<br/>
The results of differentiation of left and right sides of the Euler's formula must be equal to each other:<br/>
<i>d<b>/</b>d<b>x</b></i>[<i><b>cos(x)</b></i>]<i><b> + i·</b>d<b>/</b>d<b>x</b></i>[<i><b>sin(x)</b></i>]<i><b> =<br/>
= −sin(x) + i·cos(x)</b></i><br/><br/>
When two complex numbers are equal to each other, their real and imaginary parts must be equal.<br/>
Therefore,<br/>
<i>d<b>/</b>d<b>x</b></i>[<i><b>cos(x)</b></i>]<i><b> = −sin(x)</b></i><br/>
<i>d<b>/</b>d<b>x</b></i>[<i><b>sin(x)</b></i>]<i><b> = cos(x)</b></i><br/>
<br/>
Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-67449672073809981552024-01-17T07:12:00.000-08:002024-01-19T07:25:18.992-08:00Arithmetic+ 03: UNIZOR.COM - Math+ & Problems - Arithmetic<iframe width="480" height="270" src="https://youtube.com/embed/AOWKGv-aTnM?si=jjTUiiXmC70i9CA2" frameborder="0"></iframe>
<i>Notes to a video lecture on http://www.unizor.com</i><br/>
<br/>
<u> Arithmetic+ 03</u><br />
<br />
<br />
<i>Problem A</i><br/>
<br/>
I have purchased a share of some company XYZ for price <i><b>P<sub>0</sub>=$100</b></i>.<br/>
After a month the price was lower by <i><b>ρ=10%</b></i>.<br/>
Another month later the price went up by the same percentage <i><b>ρ=10%</b></i> and I sold this share.<br/>
Have I lost or gain money, or got even?<br/>
<br/>
<i>Solution A</i><br/>
<br/>
After the first month the price dropped from <i><b>P<sub>0</sub>=100</b></i> by <i><b>ρ=10%</b></i>, that is by <i><b>P<sub>0</sub>·0.1</b></i> and was<br/>
<i><b>P<sub>1</sub> = P<sub>0</sub> − P<sub>0</sub>·ρ =<br/>
= 100 − 100·0.1 = 90</b></i><br/>
<br/>
After the second month the price went up from <i><b>P<sub>1</sub>=90</b></i> by <i><b>ρ=10%</b></i>, that is by <i><b>P<sub>1</sub>·0.1</b></i> and was<br/>
<i><b>P<sub>2</sub> = P<sub>1</sub> + P<sub>1</sub>·ρ =<br/>
= 90 + 90·0.1 = 99</b></i><br/>
<br/>
Therefore, since I bought a share for <i><b>P<sub>0</sub>=$100</b></i> and sold it for <i><b>P<sub>2</sub>=$99</b></i>, my net result is<br/>
Δ<i><b>P = P<sub>2</sub> − P<sub>0</sub> = −$1</b></i><br/>
which is a <i><b><font color=red>loss</font></b></i>.<br/>
<br/>
<b>Notice that percentage of my loss was equal to percentage of my gain, but in dollars it was larger because the same percentage was applied to a larger amount on the way down, than to a smaller amount on the way up</b>.<br/>
<br/>
<br/>
<i>Problem B</i><br/>
<br/>
I have purchased a share of some company XYZ for price <i><b>P<sub>0</sub>=$100</b></i>.<br/>
After a month the price was higher by <i><b>ρ=10%</b></i>.<br/>
Another month later the price went down by the same percentage <i><b>ρ=10%</b></i> and I sold this share.<br/>
Have I lost or gain money, or got even?<br/>
<br/>
<i>Solution B</i><br/>
<br/>
After the first month the price went up from <i><b>P<sub>0</sub>=100</b></i> by <i><b>ρ=10%</b></i>, that is by <i><b>P<sub>0</sub>·0.1</b></i> and was<br/>
<i><b>P<sub>1</sub> = P<sub>0</sub> + P<sub>0</sub>·ρ =<br/>
= 100 + 100·0.1 = 110</b></i><br/>
<br/>
After the second month the price went down from <i><b>P<sub>1</sub>=110</b></i> by <i><b>ρ=10%</b></i>, that is by <i><b>P<sub>1</sub>·0.1</b></i> and was<br/>
<i><b>P<sub>2</sub> = P<sub>1</sub> − P<sub>1</sub>·ρ =<br/>
= 110 − 110·0.1 = 99</b></i><br/>
<br/>
Therefore, since I bought a share for <i><b>P<sub>0</sub>=$100</b></i> and sold it for <i><b>P<sub>2</sub>=$99</b></i>, my net result is<br/>
Δ<i><b>P = P</sub>2</sub> − P<sub>0</sub> = −$1</b></i><br/>
which is a <i><b><font color=red>loss</font></b></i>.<br/>
<br/>
<b>Notice that percentage of my loss was equal to percentage of my gain, but in dollars it was larger because the same percentage was applied to a larger amount on the way down, than to a smaller amount on the way up</b>.<br/>
<br/>
<i>IMPORTANT CONCLUSION</i>:<br/>
If market price of a stock regularly goes up and down by the same percentage, the price per share goes down, investors lose money.<br/>
To be profitable, the price should go up more frequently or/and by a greater percentage than down.<br/>
<br/>
<br/>
<i>Problem C</i><br/>
<br/>
(a) Within the framework of Problems A, what the percentage of gain would neutralize the loss after price moving down by <i><b>ρ=10%</b></i>?<br/>
<br/>
(b) Within the framework of Problems B, what the percentage of loss would neutralize the gain after price moving up by <i><b>ρ=10%</b></i>?<br/>
<br/>
<i>Solution C</i><br/>
<br/>
(a) If we lost <i><b>ρ=10%</b></i>, the price became<br/>
<i><b>P<sub>1</sub> = P<sub>0</sub> − P<sub>0</sub>·ρ =<br/>
= 100 − 100·0.1 = 90</b></i><br/>
To return back to <i><b>$100</b></i> from <i><b>$90</b></i>, the growth in price <i><b>σ</b></i> should satisfy the equation<br/>
<i><b>100 = 90 + 90·σ</b></i><br/>
from which<br/>
<i><b>σ = 1/9 ≅ 0.1111</b></i><br/>
Therefore, if we add approximately <i><b>11.11%</b></i> to <i><b>P<sub>1</sub>=90</b></i>, the final price will be<br/>
<i><b>P<sub>2</sub> = 90 + 90·0.1111 =<br/>
= 99.999 ≅ 100</b></i><br/>
<br/>
(b) If we gain <i><b>ρ=10%</b></i>, the price became<br/>
<i><b>P<sub>1</sub> = P<sub>0</sub> + P<sub>0</sub>·ρ =<br/>
= 100 + 100·0.1 = 110</b></i><br/>
To return back to <i><b>$100</b></i> from <i><b>$110</b></i>, the diminishing in price <i><b>σ</b></i> should satisfy the equation<br/>
<i><b>100 = 110 − 110·σ</b></i><br/>
from which<br/>
<i><b>σ = 10/11 ≅ 0.0909</b></i><br/>
Therefore, if we subtract approximately <i><b>9.09%</b></i> from <i><b>P<sub>1</sub>=110</b></i>, the final price will be<br/>
<i><b>P<sub>2</sub> = 110 − 110·0.0909 =<br/>
= 99.999 ≅ 100</b></i><br/>
<br/>
<i>Answer C</i>:<br/>
<br/>
(a) To neutralize a loss of <i><b>10%</b></i> we need a larger (in %) gain of approximately <i><b>11.11%</b></i>.<br/>
(b) To neutralize a gain of <i><b>10%</b></i> we need a smaller (in %) loss of approximately <i><b>9.09%</b></i>.<br/>
<br/>
<br/>
<i>Problem D</i> (suggested by Max)<br/>
<br/>
Concentration of salt dissolved in water is measured in percentage of the mass of salt to the total mass of solution (salt + water).<br/>
<br/>
Initially, a glass contained <nobr><i><b>100</b> gram</i></nobr> of solution of salt in water with the concentration of salt <i><b>1%</b></i>.<br/>
<i>Question 1</i>: how much salt and how much water was in a glass?<br/>
<br/>
This same glass was standing on a table for a week and part of water has evaporated, so the concentration of salt after a week became <i><b>2%</b></i>.<br/>
<i>Question 2</i>: how much salt and how much water was in a glass after this process of evaporation?<br/>
<i>Question 3</i>: how much water has evaporated in a week?<br/>
<br/>
<i>Solution D</i><br/>
<br/>
If <i><b>100</b> gram</i> of solution contained <i><b>1%</b></i> of salt, the mass of salt was<br/>
<i><b>m<sub>salt</sub> = 100·0.01 = 1</b> (gram)</i><br/>
Hence, the mass of water was<br/>
<i><b>m<sub>water</sub> = 100 − 1 = 99</b> (gram)</i><br/>
<br/>
<i>Answer 1</i> (initially):<br/>
<i><b>m<sub>salt</sub> = 1</b> (gram)</i><br/>
<i><b>m<sub>water</sub> = 99</b> (gram)</i><br/>
<br/>
After some water has evaporated, the glass contained the same amount salt, that is <nobr><i><b>1</b> gram</i>.</nobr><br/>
Since we know that it constitutes <i><b>2%</b></i> of a solution, the total mass of solution must be <i><b>50</b> gram</i> because<br/>
<i><b>1/0.02 = 50</b> (gram)</i><br/>
<br/>
To make <i><b>50</b> gram</i> of solution with <i><b>1</b> gram</i> of salt, we need <nobr><i><b>49</b> gram</i></nobr> of pure water.<br/>
<br/>
<i>Answer 2</i> (after a week):<br/>
<i><b>m<sub>salt</sub> = 1</b> (gram)</i><br/>
<i><b>m<sub>water</sub> = 49</b> (gram)</i><br/>
<br/>
<i>Answer 3</i>:<br/>
That means, <nobr><i><b>99−49=50</b> gram</i></nobr> of water has evaporated in a week.<br/>
<br/>
Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-70809094827158698622024-01-11T07:41:00.000-08:002024-01-11T07:41:26.472-08:00Algebra+ 02: UNIZOR.COM - Math+ & Problems - Algebra<iframe width="480" height="270" src="https://youtube.com/embed/OGrXOStf2DQ?si=JJokrafbkaApwybF" frameborder="0"></iframe>
<i>Notes to a video lecture on http://www.unizor.com</i><br/>
<br/>
<u> Algebra+ 02</u><br />
<br />
<br />
<i>Problem</i><br/>
<br/>
Let {<i><b>x<sub>i</sub></b></i>} (where <i><b>i</b></i>∈[<i><b>1,n</b></i>] and <i><b>n</b></i> is any natural number) be a set of <i><b>n</b></i> real non-negative numbers.<br/>
<br/>
Their <b>arithmetic average</b><br/>(or <b>mean</b>) is defined as<br/>
<i><b>A<sub>n</sub> = </b></i>(<i><b>1/n</b></i>)<i><b>·</b></i><font size=4>Σ</font><sub><i><b>i</b></i>∈[<i><b>1,n</b></i>] </sub><i><b>x<sub>i</sub></b></i><br/>
(symbol <font size=4>Σ</font><sub><i><b>i</b></i>∈[<i><b>1,n</b></i>] </sub> is summation for all indices <i><b>i</b></i>∈[<i><b>1,n</b></i>])<br/>
<br/>
Their <b>geometric average</b> is defined as<br/>
<i><b>G<sub>n</sub> = </b></i>(<font size=4>Π</font><sub><i><b>i</b></i>∈[<i><b>1,n</b></i>] </sub><i><b>x<sub>i</sub></b></i>)<i><b><sup>1/n</sup></b></i><br/>
(symbol <font size=4>Π</font><sub><i><b>i</b></i>∈[<i><b>1,n</b></i>] </sub> is multiplication for all indices <i><b>i</b></i>∈[<i><b>1,n</b></i>])<br/>
<br/>
Prove that <i><b>A<sub>n</sub> ≥ G<sub>n</sub></b></i>.<br/>
<br/>
<i>Solution A</i><br/>
(by Cauchy)<br/>
<br/>
Our plan is multi-step.<br/>
First (step 1), we will check our inequality for <i><b>n=1</b></i> and prove it for <i><b>n=2</b></i>.<br/>
Then (step 2) we will prove that, if it's held for some <i><b>n</b></i>, it will be held for <i><b>2n</b></i>, which, if start with <i><b>n=2</b></i> would prove the inequality for <i><b>n=4</b></i>, <i><b>n=8</b></i> etc. - for all <i><b>n=2<sup>k</sup></b></i>, where <i><b>k</b></i> is any natural number.<br/>
Finally (step 3), we will prove for all other <i><b>n</b></i> by complementing any given <i><b>n</b></i> non-negative numbers {<i><b>x<sub>i</sub></b></i>} with some new numbers to make the total quantity of numbers equal to <i><b>2<sup>k</sup></b></i>.<br/>
<br/>
<i>Step 1</i><br/>
It's obvious that for <i><b>n=1</b></i> the inequality is held because<br/>
<i><b>A<sub>1</sub> = x<sub>1</sub>/1 = x<sub>1</sub></b></i><br/>
and<br/>
<i><b>G<sub>1</sub> = (x<sub>1</sub>)<sup>1</sup> = x<sub>1</sub></b></i><br/>
<br/>
For <i><b>n=2</b></i> let's prove it as follows.<br/>
Start from the original inequality that needs to be proven, do some <i>reversible invariant transformations</i> to get to an unconditionally true statement (this is the <b>analysis</b> stage) and then use the reversibility of invariant transformations to prove the original inequality by going backward through these transformations, beginning with final unconditionally true statement (the <b>proof</b> proper).<br/>
<br/>
Note that all our participating numbers are real non-negative, it's important when we talk about reversible invariant transformations that involve raising both sides of inequality to the power of 2.<br/>
<br/>
Original inequality to be proven:<br/>
(I) <i><b>(x<sub>1</sub>+x<sub>2</sub>)/2 ≥ (x<sub>1</sub>·x<sub>2</sub>)<sup>½</sup></b></i><br/>
Now do the following series of invariant transformations.<br/>
(II) Square both sides (it is an invariant reversible transformation for non-negative numbers).<br/>
<i><b>(x<sub>1</sub>+x<sub>2</sub>)²/4 ≥ x<sub>1</sub>·x<sub>2</sub></b></i><br/>
(III) Open parenthesis.<br/>
<i><b>x<sub>1</sub>²+2·x<sub>1</sub>·x<sub>2</sub>+x<sub>2</sub>² ≥ 4·x<sub>1</sub>·x<sub>2</sub></b></i><br/>
(IV) Bring all members to the left side.<br/>
<i><b>x<sub>1</sub>² − 2·x<sub>1</sub>·x<sub>2</sub> + x<sub>2</sub>² ≥ 0</b></i><br/>
(V) Notice, we have a full square on the left, let's transform it into an explicit square:<br/>
<i><b>(x<sub>1</sub> − x<sub>2</sub>)² ≥ 0</b></i><br/>
<b>The above is an unconditionally true statement</b>.<br/>
<br/>
<font color=blue><b>Therefore, the above analysis allows us to formulate the proof by starting from this unconditionally true statement (V) and invariantly transform it into (IV), (III), (II) and (I), which proves the original inequality</b>.</font><br/>
Step 1 of the proof is done.<br/>
<br/>
<i>Step 2</i><br/>
Let's prove that the quantity of non-negative numbers participating in arithmetic and geometric averaging can be doubled.<br/>
More precisely, we will prove that if the inequality is true for <i><b>n</b></i> elements, it's true for <i><b>2n</b></i> elements.<br/>
<br/>
Assume, <i><b>A<sub>n</sub> ≥ G<sub>n</sub></b></i> for some <i><b>n</b></i>.<br/>
Let's prove that <i><b>A<sub>2n</sub> ≥ G<sub>2n</sub></b></i>.<br/>
Divide our set of <i><b>2n</b></i> elements into two subsets:<br/>
subset #1 {<i><b>x<sub>i</sub></b></i>}<sub>i∈[1,n]</sub></b></i> and<br/>
subset #2 {<i><b>x<sub>i</sub></b></i>}<sub>i∈[n+1,2n]</sub></b></i><br/>
<br/>
Let <i><b>A<sub>n</sub><sup>(1)</sup></b></i> and <i><b>G<sub>n</sub><sup>(1)</sup></b></i> be arithmetic and geometric averages of the elements of subset #1 and<br/>
<i><b>A<sub>n</sub><sup>(2)</sup></b></i> and <i><b>G<sub>n</sub><sup>(2)</sup></b></i> be arithmetic and geometric averages of the elements of subset #2.<br/>
Let <i><b>A<sub>2n</sub></b></i> and <i><b>G<sub>2n</sub></b></i> be arithmetic and geometric averages of the elements of an entire set of <i><b>2n</b></i> elements.<br/>
<br/>
Each subset contains <i><b>n</b></i> elements (non-negative numbers), and we assumed that our original inequality is held for this quantity of elements.<br/>
Therefore,<br/>
<i><b>A<sub>n</sub><sup>(1)</sup> ≥ G<sub>n</sub><sup>(1)</sup></b></i> and<br/>
<i><b>A<sub>n</sub><sup>(2)</sup> ≥ G<sub>n</sub><sup>(2)</sup></b></i>.<br/>
<br/>
Obviously,<br/>
<i><b>A<sub>2n</sub> = </b></i>[<i><b>A<sub>n</sub><sup>(1)</sup>+A<sub>n</sub><sup>(2)</sup></b></i>] <i><b><font size=4>/</font>2</b></i><br/>
<i><b>G<sub>2n</sub> = </b></i>[<i><b>G<sub>n</sub><sup>(1)</sup>·G<sub>n</sub><sup>(2)</sup></b></i>]<i><b><sup>½</sup></b></i><br/>
<br/>
Using the above assumption about inequality between arithmetic and geometric averages of sets of <i><b>n</b></i> elements,<br/>
<i><b>A<sub>2n</sub> = </b></i>[<i><b>A<sub>n</sub><sup>(1)</sup>+A<sub>n</sub><sup>(2)</sup></b></i>] <i><b><font size=4>/</font>2 ≥ <br/>
≥ </b></i>[<i><b>G<sub>n</sub><sup>(1)</sup>+G<sub>n</sub><sup>(2)</sup></b></i>] <i><b><font size=4>/</font>2</b></i><br/>
Using the proven above inequality for only two elements <i><b>G<sub>n</sub><sup>(1)</sup></b></i> and <i><b>G<sub>n</sub><sup>(2)</sup></b></i>, we can state<br/>
[<i><b>G<sub>n</sub><sup>(1)</sup>+G<sub>n</sub><sup>(2)</sup></b></i>] <i><b><font size=4>/</font>2 ≥<br/>
≥ </b></i>[<i><b>G<sub>n</sub><sup>(1)</sup>·G<sub>n</sub><sup>(2)</sup></b></i>]<i><b><sup>½</sup> = G<sub>2n</sub></b></i><br/>
<br/>
Combination of the two inequalities obtained above proves that<br/>
<i><b>A<sub>2n</sub> ≥ G<sub>2n</sub></b></i><br/>
Step 2 of the proof is done, and we can state that, since the inequality was proven for <i><b>n=2</b></i>, it is true for <i><b>n=4,8,16...</b></i>, that is for any <i><b>n=2<sup>k</sup></b></i>.<br/>
<br/>
<i>Step 3</i><br/>
Let's prove the inequality for values of <i><b>n</b></i> between powers of 2.<br/>
Let the value of <i><b>n</b></i> be greater than <i><b>2<sup>k−1</sup></b></i> but less than <i><b>2<sup>k</sup></b></i>.<br/>
<i><b>A<sub>n</sub></b></i> and <i><b>G<sub>n</sub></b></i> are, as before, arithmetic and geometric averages of our <i><b>n</b></i> numbers.<br/>
<br/>
Let's complement our set of <i><b>n</b></i> numbers {<i><b>x<sub>i</sub></b></i>}<sub>i∈[1,n]</sub> with <i><b>m=2<sup>k</sup>−n</b></i> new numbers, each of which is equal to <i><b>G<sub>n</sub></b></i>.<br/>
Now we have a set of <i><b>2<sup>k</sup>=m+n</b></i> numbers, and we can apply the inequality between arithmetic and geometric averages proven to be true for such sets.<br/>
<br/>
Consider <i><b>A<sub>m+n</sub></b></i> for our new set that contains <i><b>n</b></i> original elements {<i><b>x<sub>i</sub></b></i>} and <i><b>m</b></i> new elements, each of which equals to <i><b>G<sub>n</sub></b></i>.<br/>
Obviously,<br/>
<i><b>A<sub>m+n</sub> = </b></i>[<i><b>n·A<sub>n</sub> + m·G<sub>n</sub></b></i>] <i><b><font size=4>/</font>(m+n)</b></i><br/>
<br/>
Now calculate <i><b>G<sub>m+n</sub></b></i> for our new set that contains <i><b>n</b></i> original elements {<i><b>x<sub>i</sub></b></i>} and <i><b>m</b></i> new elements, each of which equals to <i><b>G<sub>n</sub></b></i>.<br/>
Obviously,<br/>
<i><b>G<sub>m+n</sub> = </b></i>[<i><b>G<sub>n</sub><sup>n</sup> · G<sub>n</sub><sup>m</sup></b></i>]<i><b><sup>1/(m+n)</sup> =<br/>
= </b></i>[<i><b>G<sub>n</sub><sup>m+n</sup></b></i>]<i><b><sup>1/(m+n)</sup> = G<sub>n</sub></b></i><br/>
<br/>
Since <i><b>m+n=2<sup>k</sup></b></i> and our original inequality was proven for this quantity of participating non-negative numbers,<br/>
<i><b>A<sub>m+n</sub> ≥ G<sub>m+n</sub></b></i><br/>
Therefore,<br/>
[<i><b>n·A<sub>n</sub> + m·G<sub>n</sub></b></i>] <i><b><font size=4>/</font>(m+n) ≥ G<sub>n</sub></b></i><br/>
<br/>
Simple transformations of the last inequality result in the following.<br/>
<i><b>n·A<sub>n</sub> + m·G<sub>n</sub> ≥ (m+n)·G<sub>n</sub></b></i><br/>
<i><b>n·A<sub>n</sub> ≥ n·G<sub>n</sub></b></i><br/>
<i><b>A<sub>n</sub> ≥ G<sub>n</sub></b></i><br/>
Step 3 of the proof is done, and we can state that it was proven for any set of non-negative numbers that their arithmetic average is greater or equal to their geometric average.<br/>
<br/>
<i>Alternate step 3</i><br/>
Instead of complementing our original set of <i><b>n</b></i> numbers with <i><b>G<sub>n</sub></b></i>, we can complement it with <i><b>A<sub>n</sub></b></i> with similar results.<br/>
Then,<br/>
<i><b>A<sub>m+n</sub> = </b></i>[<i><b>n·A<sub>n</sub>+m·A<sub>n</sub></b></i>] <i><b><font size=4>/</font>(m+n) =<br/>
= A<sub>n</sub></b></i><br/>
<br/>
Now calculate <i><b>G<sub>m+n</sub></b></i> for our new set that contains <i><b>n</b></i> original elements {<i><b>x<sub>i</sub></b></i>} and <i><b>m</b></i> new elements, each of which equals to <i><b>A<sub>n</sub></b></i>.<br/>
<i><b>G<sub>m+n</sub> = </b></i>[<i><b>G<sub>n</sub><sup>n</sup> · A<sub>n</sub><sup>m</sup></b></i>]<i><b><sup>1/(m+n)</sup></b></i><br/>
<br/>
Since <i><b>m+n=2<sup>k</sup></b></i> and our original inequality was proven for this quantity of participating non-negative numbers,<br/>
<i><b>A<sub>m+n</sub> ≥ G<sub>m+n</sub></b></i><br/>
Therefore,<br/>
<i><b>A<sub>n</sub> ≥ </b></i>[<i><b>G<sub>n</sub><sup>n</sup> · A<sub>n</sub><sup>m</sup></b></i>]<i><b><sup>1/(m+n)</sup></b></i><br/>
<br/>
Simple transformations of the last inequality result in the following.<br/>
<i><b>A<sub>n</sub><sup>m+n</sup> ≥ G<sub>n</sub><sup>n</sup>·A<sub>n</sub><sup>m</sup></b></i><br/>
<i><b>A<sub>n</sub><sup>m</sup>·A<sub>n</sub><sup>n</sup> ≥ G<sub>n</sub><sup>n</sup>·A<sub>n</sub><sup>m</sup></b></i><br/>
<i><b>A<sub>n</sub><sup>n</sup> ≥ G<sub>n</sub><sup>n</sup></b></i><br/>
<i><b>A<sub>n</sub> ≥ G<sub>n</sub></b></i><br/>
<br/>
The equality between arithmetic and geometric averages is obvious if all numbers are equal to each other, that is <i><b>x<sub>i</sub>=x<sub>j</sub></b></i> for any indices <i><b>i</b></i> and <i><b>j</b></i>.<br/>
Symbolically, it looks like<br/>
<i><b>x<sub>i</sub>=x<sub>j</sub></b></i> ∀ <i><b>i,j</b></i>∈[<i><b>1,n</b></i>] ⇒ <i><b>A<sub>n</sub> = G<sub>n</sub></b></i><br/>
<br/>
<br/>
<i>Solution B</i><br/>
(by Polya)<br/>
<br/>
From the standard course of Math is known that the function <i><b>y=e<sup>x</sup></b></i> at point <i><b>x=0</b></i> has a value <i><b>y(0)=1</b></i> and has a tangential at <i><b>45°</b></i>, which looks like this.<br/>
<img src='http://www.unizor.com/Pictures/PolyaProof.png' style='width:200px;height:200px;'><br/>
Shifting the graphs by <i><b>1</b></i> to the right by replacing <i></b>x</b></i> with <i><b>x−1</b></i>, we will have the following picture<br/>
<img src='http://www.unizor.com/Pictures/PolyaProof_1.png' style='width:200px;height:200px;'><br/>
The last picture shows that the following inequality is held for all real values of <i><b>x</b></i><br/>
<i><b>e<sup>x−1</sup> ≥ x</b></i><br/>
<br/>
Consider now a set of <i><b>n</b></i> non-negative numbers {<i><b>x<sub>i</sub></b></i>} (where <i><b>i</b></i>∈[<i><b>1,n</b></i>] and <i><b>n</b></i> is any natural number) with <i><b>A</b></i> being their arithmetic average and <i><b>G</b></i> being their geometric average (we are omitting a subscript <i><b>n</b></i> in averages for brevity).<br/>
<br/>
Using an inequality above for <i><b>x=x<sub>i </sub>/A</b></i>, we obtain<br/>
<i><b>e<sup>(x<sub>i </sub>/A)−1</sup> ≥ x<sub>i </sub>/A</b></i> for any <i><b>i</b></i> or<br/>
<i><b>A·e<sup>(x<sub>i </sub>/A)−1</sup> ≥ x<sub>i </sub></b></i><br/>
<br/>
Multiplying the above inequality for all values of index <i><b>i</b></i> from <i><b>1</b></i> to <i><b>n</b></i>, we obtain<br/>
<i><b>A<sup>n</sup>·e<sup> Σ(x<sub>i </sub>/A)−n</sup> ≥ </i></b><font size=4>Π</font><b><i>(x<sub>i </sub>)</b></i><br/>
<br/>
Interestingly, the exponent in the expression above is zero.<br/>
Here is why<br/>
<font size=4>Σ</font><i><b>(x<sub>i </sub>/A) = (1/A)·</b></i><font size=4>Σ</font><i><b>(x<sub>i</sub>) =<br/>
= (1/A)·A·n = n</b></i>,<br/>
which makes the exponents in the inequality above equal to zero.<br/>
<br/>
Therefore, considering <i><b>e<sup>0</sup>=1</b></i>, the inequality above looks like<br/>
<i><b>A<sup>n</sup> ≥ </i></b><font size=4>Π</font><b><i>(x<sub>i </sub>)</b></i><br/>
<i><b>A ≥ </i></b><font size=4>[Π</font><b><i>(x<sub>i </sub>)</b></i><font size=4>]</font><i><b><sup>1/n</sup> = G</b></i><br/>
End of proof.
<br/>
Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-68738103935155119252024-01-09T07:41:00.000-08:002024-01-28T08:41:54.973-08:00Algebra+ 01: UNIZOR.COM - Math+ & Problems - Algebra<iframe width="480" height="270" src="https://youtube.com/embed/LLbF5vmp1Zk?si=YSTjHoAMHZrk3ySW" frameborder="0"></iframe>
<i>Notes to a video lecture on http://www.unizor.com</i><br/>
<br/>
<u> Algebra+ 01</u><br />
<br />
<br />
<i>Problem A</i><br/>
<br/>
Given an equation of parabola in Cartesian coordinates on a plane<br/>
<i><b>y = x² − A·x + A</b></i><br/>
where parameter <i><b>A</b></i> can be any real number.<br/>
Each value of parameter <i><b>A</b></i> defines some parabola.<br/>
<br/>
(A1) Find the point on a plane, through which go all the parabolas defined by all real values of parameter <i><b>A</b></i>.<br/>
<br/>
(A2) On what curve lie the vertices of all the parabolas defined by all real values of parameter <i><b>A</b></i>?
<br/>
<br/>
<i>Solution A</i><br/>
<br/>
A1<br/>
Assuming the statement of this problem implies that such a point, where all parabolas are crossing, exists, let's find it by equating any two values of <i><b>y</b></i> for different values of parameter <i><b>A</b></i>.<br/>
<br/>
Take, for example two parabolas for <i><b>A=0</b></i> and <i><b>A=1</b></i> and find their crosing point, that is such <i><b>x</b></i> for which corresponding quadratic forms equal to each other.<br/>
<i><b>A=0</i> ⇒ <i>y = x²</b></i><br/>
<i><b>A=1</i> ⇒ <i>y = x² − x + 1</b></i><br/>
<br/>
Equating both expressions gives<br/>
<i><b>x² = x² − x + 1</b></i><br/>
⇒ <i><b>x = 1</b></i><br/>
<br/>
Consider now this value <i><b>x=1</b></i>.<br/>
With this <i><b>x</b></i> the equation becomes independent of parameter <i><b>A</b></i>:<br/>
<i><b>y = 1² − A·1 + A = 1</b></i><br/>
for any value of parameter <i><b>A</b></i>.<br/>
<br/>
Therefore, any parabola defined by any value of <i><b>A</b></i> goes through point with coordinates <i><b>x=1</b></i>, <i><b>y=1</b></i> on a plane.<br/>
<img src='http://www.unizor.com/Pictures/CommonPointParabolas.png' style='width:200px;height:300px;'><br/>
<br/>
A2<br/>
We need a formula for a vertex of a parabola represented by a general quadratic form<br/>
<i><b>y = a·x² + b·x + c</b></i><br/>
If we do not remember the formula for X-coordinate of its vertex, recall that it is in the middle between the two roots of a quadratic equations<br/>
<i><b>a·x² + b·x + c = 0</b></i><br/>
The formula for its roots (and that is something to remember) is<br/>
<i><b>x<sub>1,2</sub> = </b></i><font size=4>[</font><i><b>−b ± √<span style='text-decoration:overline'>b²−4ac</span></b></i><font size=4>]</font><i><b> <font size=4>/</font>2a</b></i><br/>
Therefore, the X-coordinate of a vertex is in the middle between its roots<br/>
<i><b>x<sub>0</sub> = ½(x<sub>1</sub> + x<sub>2</sub>) = −b/2a</b></i><br/>
<br/>
Alternatively, using Calculus, we can find <i><b>x<sub>0</sub></b></i> by equating a derivative of a given quadratic form to zero:<br/>
<i><b>y' = 2ax + b = 0</b></i><br/>
Therefore,<br/>
<i><b>x<sub>0</sub> = −b/2a</b></i><br/>
<br/>
The Y-coordinate of a vertex is, therefore,<br/>
<i><b>y<sub>0</sub> = a·x<sub>0</sub>² + b·x<sub>0</sub> + c =<br/>
= a(b/2a)² + b(−b/2a) + c =<br/>
= b²/4a − b²/2a +c =<br/>
= −b²/4a + c</b></i><br/>
<br/>
In our case of a quadratic form<br/>
<i><b>y = x² − A·x + A</b></i><br/>
the values of coefficients are<br/>
<i><b>a = 1, b = −A, c = A</b></i><br/>
Therefore, the coordinates of a parabola's vertex for a specific value of parameter <i><b>A</b></i> is<br/>
<i><b>x<sub>0</sub> = A/2</b></i><br/>
<i><b>y<sub>0</sub> = −A²/4 + A</b></i><br/>
As parameter <i><b>A</b></i> changes, so do coordinates of a vertex forming some parametrically defined curve.<br/>
<br/>
To convert this parametric definition of a curve into the usual format <i><b>y<sub>0</sub>=f(x<sub>0</sub>)</b></i> we resolve parameter <i><b>A</b></i> in terms of <i><b>x<sub>0</sub></b></i> and substitute it into an expression for <i><b>y<sub>0</sub></b></i>.<br/>
Since<br/>
<i><b>x<sub>0</sub> = A/2</b></i><br/>
the value of <i><b>A</b></i> in terms of <i><b>x<sub>0</sub></b></i> is<br/>
<i><b>A = 2x<sub>0</sub></b></i><br/>
<br/>
Substitute it into an expression for <i><b>y<sub>0</sub></b></i>:<br/>
<i><b>y<sub>0</sub> = −4x<sub>0</sub>²/4 + 2x<sub>0</sub> =<br/>
= −x<sub>0</sub>² + 2x<sub>0</sub></b></i><br/>
which is a parabola with coefficients<br/>
<i><b>a = −1</b></i><br/>
<i><b>b = 2</b></i><br/>
<i><b>c = 0</b></i><br/>
<br/>
The standard characteristics of this parabola are as follows.<br/>
The "horns" are directed downwards.<br/>
Vertex is at<br/>
<i><b>x<sub>0</sub> = −b/2a = −2/−2 = 1</b></i><br/>
<i><b>y<sub>0</sub> = −b²/4a + c = −4/−4 + 0 = 1</b></i><br/>
Roots (X-intercepts) are<br/>
<i><b>x<sub>1,2</sub> = (−2±√<span style='text-decoration:overline'>4</span>)/−2</b></i><br/>
⇒ <i><b>x<sub>1</sub> = 0</b></i>, <i><b>x<sub>2</sub> = 2</b></i><br/>
<br/>
Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-15844449362917018922024-01-06T10:54:00.000-08:002024-01-28T08:42:17.069-08:00Geometry+ 01: UNIZOR.COM - Math+ & Problems - Geometry<iframe style="background-image:url(https://i.ytimg.com/vi/B_ud7NleeHE/hqdefault.jpg)" width="480" height="270" src="https://youtube.com/embed/B_ud7NleeHE?si=v24f3uXCAJ1WMF1-" frameborder="0"></iframe>
<i>Notes to a video lecture on http://www.unizor.com</i><br/>
<br/>
<u> Geometry+ 01</u><br />
<br />
<i>Problem A</i><br/>
<br/>
Can two sides and all angles of one triangle be equal to two sides and angles of another triangle, but triangles be different (that is, not congruent because the third sides of these triangles have different lengths)?<br/>
<br/>
If yes, are there any requirements on the sides of a triangle to assure the existence of a non-congruent to it another triangle with the same three angles and two sides?<br/>
<br/>
<i>Solution</i><br/>
<br/>
Here is an approximate picture that seems to illustrate that this is possible.<br/>
<br/>
Both triangles (red and blue) have the same angles <i><b>α</b></i>, <i><b>β</b></i> and <i><b>γ</b></i> and two equal sides <i><b><font color=red>a</font>=<font color=blue>a</font></b></i> and <i><b><font color=red>b</font>=<font color=blue>b</font></b></i>.<br/>
<img src='http://www.unizor.com/Pictures/UnequalTriangles.png' style='width:200px;height:240px;'><br/>
Point <i><b><font color=red>D</font></b></i> lies on the segment <i><b><font color=blue>AC</font></b></i>.<br/>
Point <i><b><font color=red>E</font></b></i> lies on the segment <i><b><font color=blue>BC</font></b></i>.<br/>
The length of segment <i><b><font color=blue>BC=a</font></b></i> is greater than the length of segment <i><b><font color=red>CE=b</font></b></i>.<br/>
<br/>
Sides are parallel: <i><b><font color=blue>AB</font> || <font color=red>DE</font><br/>
</i>⇒<i> <font color=blue>∠CAB</font>=<font color=red>∠CDE</font> = α<br/>
</i>⇒<i> <font color=blue>∠ABC</font>=<font color=red>∠DEC</font> = β</b></i><br/>
Therefore, triangles are similar:<br/>
<font color=blue>Δ<i><b>ABC</b></i></font> <b>∼</b> <font color=red>Δ<i><b>DEC</b></i></font><br/>
<br/>
Congruent sides:<br/>
<i><b><font color=blue>BC</font> ≅ <font color=red>DE</font> = a</b></i><br/>
<i><b><font color=blue>AC</font> ≅ <font color=red>CE</font> = b</b></i><br/>
<br/>
Now the question is, can we start with <b>any</b> triangle <font color=blue>Δ<i><b>ABC</b></i></font> and construct <font color=red>Δ<i><b>DEC</b></i></font> that is similar but not congruent to <font color=blue>Δ<i><b>ABC</b></i></font>.<br/>
<br/>
The answer is negative, <b>we cannot start with any triangle</b> <font color=blue>Δ<i><b>ABC</b></i></font>, but under certain conditions it is possible, and we will examine specific requirements to make it.<br/>
<br/>
Let's assume that <font color=blue>Δ<i><b>ABC</b></i></font> is defined by its three sides <font color=blue><i><b>BC=a</b></i></font>, <font color=blue><i><b>AC=b</b></i></font> and <font color=blue><i><b>AB=c</b></i></font>.<br/>
Triangle <font color=red>Δ<i><b>DEC</b></i></font> is similar to <font color=blue>Δ<i><b>ABC</b></i></font> and we know its two sides - <font color=red><i><b>DE=a</b></i></font> and <font color=red><i><b>CE=b</b></i></font>.<br/>
Assume, the third side of <font color=red>Δ<i><b>DEC</b></i></font> is <font color=red><i><b>CD=x</b></i></font>.<br/>
<br/>
From similarity of these triangles follows that the ratio of sides lying opposite to equal angles is the same:<br/>
<font color=blue><i><b>CB</b></i></font>/<font color=red><i><b>CE</b></i></font> = <font color=blue><i><b>CA</b></i></font>/<font color=red><i><b>CD</b></i></font> = <font color=blue><i><b>AB</b></i></font>/<font color=red><i><b>DE</b></i></font><br/>
Therefore,<br/>
<font color=blue><i><b>a</b></i></font>/<font color=red><i><b>b</b></i></font> = <font color=blue><i><b>b</b></i></font>/<font color=red><i><b>x</b></i></font> = <font color=blue><i><b>c</b></i></font>/<font color=red><i><b>a</b></i></font><br/>
<br/>
From the first equation follows that<br/>
<i><b>x = b²/a</b></i><br/>
From the second equation follows that<br/>
<i><b>x = a·b/c</b></i><br/>
Therefore,<br/>
<i><b>b²/a = x = a·b/c</b></i><br/>
⇒ <i><b>a²b = b²c</b></i><br/>
⇒ <i><b>a² = b·c</b></i><br/>
The last equation <i><b>a²=b·c</b></i> is the <b>necessary condition</b> required to construct triangle <font color=red>Δ<i><b>DEC</b></i></font>.<br/>
As we stated above, length <i><b>a</b></i> is greater than <i><b>b</b></i>. Therefore, to satisfy the last equation, length <i><b>c</b></i> must be greater than <i><b>a</b></i>, that is, <i><b>a</b></i> is somewhere in-between <i><b>b</b></i> and <i><b>c</b></i>.<br/>
<br/>
The condition <i><b>a² = b·c</b></i> with length <i><b>a</b></i> being somewhere between unequal <i><b>b</b></i> and <i><b>c</b></i> is also a <b>sufficient</b> one for the task.<br/>
<br/>
To prove it, assume that we have triangle <font color=blue>Δ<i><b>ABC</b></i></font> with unequal sides <font color=blue><i><b>a</b></i></font>, <font color=blue><i><b>b</b></i></font> and <font color=blue><i><b>c</b></i></font>, as on the picture above, that satisfy the condition <font color=blue><i><b>a² = b·c</b></i></font>.<br/>
<br/>
Using length <font color=blue><i><b>b</b></i></font> of side <i><b><font color=blue>AC</font></b></i>, we mark point <i><b><font color=red>E</font></b></i> on the side <i><b><font color=blue>BC</font></b></i> at distance <i><b><font color=blue>b</font>=<font color=red>b</font></b></i> from point <i><b>C</b></i>.<br/>
Next we draw a line parallel to <i><b><font color=blue>AB</font></b></i> through point <i><b><font color=red>E</font></b></i> that crosses <i><b><font color=blue>AC</font></b></i> at point <i><b><font color=red>D</font></b></i>.<br/>
<br/>
Consider triangle <font color=red>Δ<i><b>DEC</b></i></font>.<br/>
All its angles are, correspondingly, congruent to angles of triangle <font color=blue>Δ<i><b>ABC</b></i></font>, so these triangles are similar.<br/>
<br/>
Let's prove that the length of side <font color=red><i><b>DE</b></i></font> equals to the length of side <font color=blue><i><b>BC=a</b></i></font>. That will conclude the construction (and, therefore, existence) of a triangle with two sides and all angles equal to two sides and all angles of triangle <font color=blue>Δ<i><b>ABC</b></i></font>, but not congruent to it.<br/>
<br/>
Let the length of side <font color=red><i><b>DE</b></i></font> be an unknown <font color=red><i><b>y</b></i></font>.<br/>
From similarity of two triangle follows:<br/>
<i><b><font color=blue>a</font>/<font color=red>b</font> = <font color=blue>c</font>/<font color=red>y</font></b></i><br/>
⇒ <i><b><font color=red>y</font> = <font color=red>b</font>·<font color=blue>c/a</font></b></i><br/>
As we know, <i><b><font color=blue>b</font>=<font color=red>b</font></b></i> and, by assumption, <i><b><font color=blue>a²=b·c</font></b></i>.<br/>
Therefore,<br/>
<i><b><font color=red>y</font> = <font color=blue>a²/a = a</font></b></i><br/>
which is exactly what we had to prove.<br/>
<br/>
<i>Conclusion</i><br/>
There might exist a triangle with two sides and all angles equal to two sides and angles of a given triangle and not congruent it, but there is certain necessary and sufficient condition for its existence in terms of an equation between the sides <i><b>a,b,c</b></i> of a given triangle.<br/>
Namely, all sides <i><b>a,b,c</b></i> must be different and they must satisfy an equation <i><b>a²=b·c</b></i>.<br/>
<br/>
<i>Example</i><br/>
Consider a triangle with sides <font color=blue><i><b>a=6</b></i></font>, <font color=blue><i><b>b=4</b></i></font> and <font color=blue><i><b>c=9</b></i></font>.<br/>
It satisfies the conditions above (all sides are different and <font color=blue><i><b>6²=4·9</b></i></font>).<br/>
Now we know two sides of a triangle we have to construct: <font color=red><i><b>a=6</b></i></font> and <font color=red><i><b>b=4</b></i></font>.<br/>
The third side <font color=red><i><b>x</b></i></font> can be found from the similarity condition<br/>
<i><b><font color=red>x</font>/<font color=blue>b</font> = <font color=red>b</font>/<font color=blue>a</font></b></i><br/>
⇒ <i><b><font color=red>x</font> = <font color=red>b</font>·<font color=blue>b/a</font> = 16/6 = <font color=red>8/3</font></b></i><br/>
To prove similarity, let's check the proportionality of all sides.<br/>
<i><b><font color=red>(8/3)</font>/<font color=blue>4</font> = 2/3</b></i><br/>
<i><b><font color=red>4</font>/<font color=blue>6</font> = 2/3</b></i><br/>
<i><b><font color=red>6</font>/<font color=blue>9</font> = 2/3</b></i><br/>
<br/>
<i>Angle Restriction</i><br/>
An interesting restriction can be derived on the value of angle <font color=blue>∠<i><b>α</b></i></font> lying between the longest and the shortest sides of triangle <font color=blue>Δ<i><b>ABC</b></i></font> across side <font color=blue><i><b>a</b></i></font>.<br/>
As we agreed, to make possible the existence of triangle <font color=red>Δ<i><b>DEC</b></i></font> with two (but not all three) sides and all three angles equal to those of <font color=blue>Δ<i><b>ABC</b></i></font>, the necessary and sufficient condition is<br/>
<i><b>a² = b·c</b></i><br/>
At the same time, by the law of cosines,<br/>
<i><b>a² = b² + c² −2b·c·cos(α)</b></i><br/>
Therefore, we have an equality<br/>
<i><b>b·c = b² + c² −2b·c·cos(α)</b></i><br/>
or<br/>
<i><b>2b·c·cos(α) = b² + c² − b·c =<br/>
= (b − c)² + b·c</b></i><br/>
Since <i><b>(b−c)²</b></i> is always positive, we conclude that<br/>
<i><b>2b·c·cos(α) > b·c</b></i><br/>
or<br/>
<i><b>cos(α) > ½</b></i><br/>
<i><b>α < π/3 = 60°</b></i><br/>
Angle between the longest and the shortest sides of <font color=blue>Δ<i><b>ABC</b></i></font> should be less than <i><b>π/3=60°</b></i>.<br/>
Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-69865247649463557582023-12-23T08:00:00.000-08:002024-01-28T08:42:47.692-08:00Logic+ 02: UNIZOR.COM - Math+ & Problems - Logic<iframe style="background-image:url(https://i.ytimg.com/vi/KvoErf9OUuY/hqdefault.jpg)" width="480" height="270" src="https://youtube.com/embed/KvoErf9OUuY?si=x2IcUZJSZclnGhw4" frameborder="0"></iframe>
<i>Notes to a video lecture on http://www.unizor.com</i><br/>
<br/>
<u> Logic 02</u><br />
<br />
<i>Problem A</i><br/>
<br/>
The game starts with two sets of stones and two players.<br/>
At each turn a player can take any number of stones but only from one of the sets.<br/>
The winner is the player who takes the last stones.<br/>
<br/>
The player who should start the game can yield this right of the first move to his opponent. After that to take some stones on each move is the rule.<br/>
<br/>
What is the winning strategy for a person who starts the game?<br/>
<br/>
<i>Solution A</i><br/>
<br/>
The strategy to win for the first player is on every move to equalize the number of stones in two sets.<br/>
<br/>
If the initial number of stones is different in these two sets, the first move of the player who starts the game is to take a few stones from a bigger set to equalize the number of stones.<br/>
<br/>
His opponent, always facing two equal numbers of stones in two sets, cannot take the last stone and inevitably will make the number of stones in two sets different.<br/>
<br/>
If in the beginning the numbers of stones in these two sets are equal, the player who starts the game should yield the right of the first move to his opponent, achieving the same conditions as above.<br/>
<br/>
<br />
<i>Problem B</i><br/>
<br/>
There is a set of six points, no three points of this set lie on a straight line.<br/>
For clarity we can assume that they lie on a circle, though it's not essential.<br/>
<img src='http://www.unizor.com/Pictures/ConnectSix.png' style='width:200px;height:200px;'><br/>
<br/>
Consider all subsets of three points of this set of six (we will call these subsets <i>triplets</i>).<br/>
Let's call a <i>triplet</i>, that has one or more segments connecting its points, a <i>good triplet</i>.<br/>
If all three points of a <i>good triplet</i> are connected by three segments, we will call this triplet a <i>triangle</i>.<br/>
<img src='http://www.unizor.com/Pictures/Triplets.png' style='width:200px;height:500px;'><br/>
<br/>
<br/>
For example, on the picture below all <i>triplets</i> have at least one pair connected by a segment (triplet <i><b>ABC</b></i> has two segments <i><b>AB</b></i> and <i><b>AC</b></i>, triplet <i><b>CDF</b></i> has two segments <i><b>CF</b></i> and <i><b>DF</b></i>, triplet <i><b>ABF</b></i> has one segment <i><b>AB</b></i> etc.)<br/>
<img src='http://www.unizor.com/Pictures/ConnectSix0.png' style='width:200px;height:200px;'><br/>
Let's try to connect some points with segments in such a way that <u>in any <i>triplet</i> there would be a pair of points connected by a segment</u>.<br/>
That is, let's try to make all triplets <i>good</i>.<br/>
<br/>
Prove that, no matter how we try to accomplish this task of making all triplets <i>good</i> by connecting our six points, there would inevitably be a <i>triplet</i> with three segments connecting all its three points into a triangle (that is, there will be at least one <i>triangle</i>), like Δ<i><b>ACE</b></i> or Δ<i><b>CEF</b></i> in the example above.<br/>
<br/>
<i>Solution B</i><br/>
<br/>
First, let's prove the following auxiliary statement (<i>lemma</i>) using the following illustration.<br/>
<img src='http://www.unizor.com/Pictures/ConnectSixThreeSeg.png' style='width:200px;height:200px;'><br/>
<br/>
Assume that all <i>triplets</i> are <i>good</i> (as assumed in the <i>Problem B</i>) and some point (for example, <i><b>A</b></i> on the picture above) has three segments joining at it and connecting it to three other points (like on the picture above, point <i><b>A</b></i> is connected to <i><b>B</b></i>, <i><b>C</b></i> and <i><b>E</b></i>).<br/>
Then there exist a <i>triangle</i>.<br/>
<br/>
<i>Proof</i>:<br/>
Since a triplet of three connected points (<i><b>BCE</b></i> on the picture above) is, as all other triplets, assumed to be <i>good</i>, there exists an additional connecting segment between some of its points (between <i><b>C</b></i> and <i><b>E</b></i> on the picture above).<br/>
With three given segments going from a point with three connections (<i><b>AB</b></i>, <i><b>AC</b></i> and <i><b>AE</b></i> on the picture above) and one additional segment between some connected points (segment <i><b>CE</b></i> between <i><b>C</b></i> and <i><b>E</b></i> on the picture above), there is a triplet (in our example it is <i><b>ACD</b></i>) that forms a triangle.<br/>
<br/>
Consequently, to prove that in the process of connecting pairs of points to make all triplets <i>good</i> there would eventually have to be formed a <i>triangle</i>, it's sufficient to prove that there would eventually have to be a point with three segments connecting it to other three points.<br/>
<br/>
The plan is to<br/>
(1) count all the triplets that can be formed from the given six points and<br/>
(2) show that without connecting three points into a <i>triangle</i> or connecting some point to three others (which, as lemma above proved, leads to forming a <i>triangle</i>) it's impossible to make all counted in (1) triplets <i>good</i>.<br/>
<br/>
The total number of triplets is, obviously, the number of combinations of our six points in groups of three, which is<br/>
<i><b>C<sub>6</sub><sup>3</sup> = 6!<font size=4>/</font>(3!·3!) = 20</b></i><br/>
<br/>
Now let's proceed to make all triplets <i>good</i>, but obeying the rule of having <u>no more than two segments joining at any single point</u> (since, as we have proved, when three segments join in one point, triangle is unavoidable).<br/>
<br/>
We have to make <i><b>20</b></i> <i>good triplets</i>, but we will show that we cannot make more than <i><b>18</b></i> without breaking that rule, which would necessitate the existence of a point with three segments connecting it to three other points, which is sufficient condition for existence of a <i>triangle</i>.<br/>
<br/>
Let's start the process of connecting pairs of points to make <i>good triplets</i>.<br/>
<br/>
In the beginning there are no connecting segments, so we choose any two points (call them <i><b>A</b></i> and <i><b>B</b></i>) and connect them with a segment.<br/>
This is sufficient to make <i><b>4</b></i> new <i>good triplets</i>: <i><b>ABC</b></i>, <i><b>ABD</b></i>, <i><b>ABE</b></i> and <i><b>ABF</b></i>.<br/>
<img src='http://www.unizor.com/Pictures/Triplet1.png' style='width:200px;height:200px;'><br/>
<br/>
Now we have a choice of which points to connect next.<br/>
We can either<br/>
(a) choose a pair of new points to connect by a segment (for example, <i><b>D</b></i> and <i><b>E</b></i>) or<br/>
(b) choose one new (unconnected) point and one that already has a connecting segment (for example, <i><b>B</b></i> and <i><b>C</b></i>), or<br/>
(c) choose two points that already had connecting segments, which could happen on later steps.<br/>
<br/>
In case (a) we have <i><b>4</b></i> more <i>good triplets</i> (<i><b>DEA</b></i>, <i><b>DEB</b></i>, <i><b>DEC</b></i> and <i><b>DEF</b></i> in our example).<br/>
<img src='http://www.unizor.com/Pictures/Triplet2.png' style='width:200px;height:200px;'><br/>
In case (b) only <i><b>3</b></i> <i>good triplets</i> will be formed (<i><b>BCD</b></i>, <i><b>BCE</b></i> and <i><b>BCF</b></i> in our example).<br/>
<img src='http://www.unizor.com/Pictures/Triplet3.png' style='width:200px;height:200px;'><br/>
In case (c) only <i><b>2</b></i> <i>good triplets</i> will be formed (if segments <i><b>AB</b></i> and <i><b>CD</b></i> exist, and we connect points <i><b>B</b></i> and <i><b>C</b></i> with a segment, new <i>good triplets</i> <i><b>BCE</b></i> and <i><b>BCF</b></i> are formed).<br/>
<img src='http://www.unizor.com/Pictures/Triplet4.png' style='width:200px;height:200px;'><br/>
General consideration is that the more segments connect our six points - the more <i>good triplets</i> will be formed.<br/>
<br/>
Let's start with the strategy that assures the maximum number of segments - the one when there are two segments joining at each point, so the rule of not having three segments at any point is preserved.<br/>
<br/>
This can be accomplished if all points are connected into a closed chain:<br/>
<img src='http://www.unizor.com/Pictures/SixChained.png' style='width:200px;height:200px;'><br/>
Obviously, not all triplets are <i>good</i> in this case. For example, <i><b>ACE</b></i> or <i><b>BDF</b></i>.<br/>
<br/>
Another chained connection
<img src='http://www.unizor.com/Pictures/SixChained2.png' style='width:200px;height:200px;'><br/>
Here triplets <i><b>ABD</b></i> or <i><b>CEF</b></i> are not <i>good</i>.<br/>
<br/>
Let's count <i>good triplets</i> using the first example of a chained connection above, starting from segment <i><b>AB</b></i> and moving clockwise (the order is unimportant, we can connect <i><b>A-C-B-F-D-E-A</b></i>, as exemplified on the second picture, and the result will be the same).<br/>
Segment <i><b>AB</b></i>, being the first, produced <i><b>4</b></i> <i>good triplets</i>, as was stated above.<br/>
Then four consecutive segments <i><b>BC</b></i>, <i><b>CD</b></i>, <i><b>DE</b></i> and <i><b>EF</b></i> formed <i><b>3</b></i> <i>good triplets</i> each.<br/>
Final segment <i><b>FA</b></i> added <i><b>2</b></i> more <i>good triplets</i>.<br/>
The total number of <i>good triplets</i> is<br/>
<i><b>4+3+3+3+3+2=18</b></i><br/>
which is short of required <i><b>20</b></i>, which means, to make all triplets <i>good</i>, we need more segments, which will result in some points to be connected by three segments to three other points and forming of a <i>triangle</i> will be guaranteed.<br/>
<br/>
We can try to organize connections differently. Instead of putting all <i><b>6</b></i> points into a chain, we can do it with only <i><b>5</b></i> with points <i><b>A-B-C-D-E</b></i>, leaving point <i><b>F</b></i> outside.<br/>
<img src='http://www.unizor.com/Pictures/FiveChained.png' style='width:200px;height:200px;'><br/>
Similar counting of <i>good triplets</i> will result in the following:<br/>
Segments <i><b>AB</b></i>, as before, created <i><b>4</b></i> <i>good triplets</i>.<br/>
Segments <i><b>BC</b></i>, <i><b>CD</b></i>, <i><b>DE</b></i>, as before, created <i><b>3</b></i> <i>good triplets</i> each.<br/>
Finally, segment <i><b>EA</b></i> created two new <i>good triplets</i>: <i><b>EAC</b></i> and <i><b>EAF</b></i>.<br/>
The total number of <i>good triplets</i> is:<br/>
<i><b>4+3+3+3+2 = 15</b></i><br/>
<br/>
The last case is to leave two points out of a closed chain as follows.<br/>
<img src='http://www.unizor.com/Pictures/FourChained.png' style='width:200px;height:200px;'><br/>
In this case the number of <i>good triplets</i> is:<br/>
from <i><b>AB</b></i> - <i><b>4</b></i>,<br/>
from <i><b>BC</b></i> - <i><b>3</b></i>,<br/>
from <i><b>CD</b></i> - <i><b>3</b></i>,<br/>
from <i><b>DA</b></i> - <i><b>2</b></i> (<i><b>DAE</b></i>, <i><b>DAF</b></i>),<br/>
from <i><b>EF</b></i> - <i><b>4</b></i> (<i><b>EFA</b></i>, <i><b>EFB</b></i>, <i><b>EFC</b></i>, <i><b>EFD</b></i>)<br/>
The total number of <i>good triplets</i> is:<br/>
<i><b>4+3+3+2+4 = 16</b></i><br/>
Still less than <i><b>20</b></i>.<br/>
<br/>
There are no logically different cases of connecting the points deserving a consideration, so we came to<br/>
CONCLUSION:<br/>
to have all <i><b>20</b></i> triplets as <i>good</i>, we need more that two segments joining at each point; it is impossible to avoid <i>triangles</i>.<br/>
<br/>
Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-15823848586016706512023-12-09T07:27:00.000-08:002024-01-28T08:43:14.310-08:00Arithmetic+ 02: UNIZOR.COM - Math+ & Problems - Arithmetic<iframe style="background-image:url(https://i.ytimg.com/vi/tS2PDQvez_s/hqdefault.jpg)" width="480" height="270" src="https://youtube.com/embed/tS2PDQvez_s?si=WOfvC1uFrTl4UDDX" frameborder="0"></iframe>
<i>Notes to a video lecture on http://www.unizor.com</i><br/>
<br/>
<u> Arithmetic 02</u><br />
<br />
<br />
<i>Problem</i><br/>
<br/>
Prove that<br/>
<i><b><font size=4>S</font><font size=3><sub>N</sub></font> = </i><font size=5>Σ</font><i></b><sub>n∈[2,N]</sub><b>1/n</font></b></i><br/>
is not an integer number for any integer <i><b>N</b></i>.<br/>
<br/>
<i>Solution 1</i><br/>
<br/>
Among fractional numbers in the sum <i><b>S<sub>N</sub></b></i> there are those of type <i><b>1/(2<sup>k</sup>)</b></i>.<br/>
Let's choose among them the one with the maximum power <i><b>k</b></i>. This number is less than any other number of this type in a series and the closest to the end of a series among them. Let's call it the <i>least binary fraction</i>.<br/>
<br/>
For example, in the sum<br/>
<i><b>S<sub>10</sub> = </i><font size=5>Σ</font><i></b><sub>n∈[2,10]</sub><b>1/n</font></b></i><br/>
we choose <i><b>1/8=1/(2³)</b></i>.<br/>
<br/>
Now let's bring all the numbers in a series to the least common denominator (LCD).<br/>
For an example of <i><b>S<sub>10</sub></b></i> the <i>LCD=2520</i> and values of our fractions with this LCD are<br/>
<i>1/2 = 1260/2520</i><br/>
<i>1/3 = 840/2520</i><br/>
<i>1/4 = 630/2520</i><br/>
<i>1/5 = 504/2520</i><br/>
<i>1/6 = 420/2520</i><br/>
<i>1/7 = 360/2520</i><br/>
<i>1/8 = 315/2520</i><br/>
<i>1/9 = 280/2520</i><br/>
<i>1/10 = 252/2520</i><br/>
<br/>
Notice that the numerator of <i>1/8=315/2520</i> (a fraction of a type <i><b>1/(2<sup>k</sup>)</b></i> with maximum power <i><b>k</b></i>) equals <i>315</i>, which is odd, while all other numerators are even.<br/>
That means, the sum of numerators with this common denominator is odd, while the least common denominator <i>2520</i> is even, which means that the sum <i><b>S<sub>10</sub></b></i> cannot be an integer number.<br/>
<br/>
Actually, this situation will be similar with any number of members in a series, that is for any <i><b>N</b></i>.<br/>
Here is why (and this is the required proof).<br/>
<br/>
Let's analyze the process of determining the least common denominator (LCD).<br/>
<br/>
Any number in the denominators of our series can be uniquely represented as a product of prime numbers.<br/>
Then LCD is formed from all the prime numbers in representation of all denominators with the quantity of each prime number being equal to the largest number of these primes among all denominators.<br/>
<br/>
In our example of <i><b>S<sub>10</sub></b></i> we have the following representation of denominators:<br/>
<i>2 = 2</i><br/>
<i>3 = 3</i><br/>
<i>4 = 2·2</i><br/>
<i>5 = 5</i><br/>
<i>6 = 2·3</i><br/>
<i>7 = 7</i><br/>
<i>8 = 2·2·2</i><br/>
<i>9 = 3·3</i><br/>
<i>10 = 2·5</i><br/>
<br/>
As we see, the prime numbers we have to use are <i>2,3,5,7</i> and the number of these primes in the LCD should be:<br/>
prime <i>2</i> should be taken <i><b>3</b></i> times because of denominator <i>8</i>.<br/>
prime <i>3</i> should be taken <i><b>2</b></i> times because of denominator <i>9</i>.<br/>
prime <i>5</i> should be taken once.<br/>
prime <i>7</i> should be taken once.<br/>
Therefore, the LCD is<br/>
<i>2³·3²·5·7=2520</i><br/>
<br/>
In general case, in the representation of LCD the number of <i>2</i>'s is the largest in a member <i><b>1/(2<sup>k</sup>)</b></i> with the largest <i><b>k</b></i> that we called above the <i>least binary fraction</i>.<br/>
Therefore, in the representation of the LCD as a product of prime numbers there are exactly <i><b>k</b></i> <i>2</i>'s and, obviously. other prime numbers, which are all odd, of course.<br/>
<br/>
When we transform all the fractions in a series <i><b>S<sub>N</sub></b></i> to common denominator, we should multiply numerators (which are all <i>1</i>'s) by all primes in the LCD that are not in a representation of its denominator:<br/>
<br/>
In our example of <i><b>S<sub>10</sub></b></i><br/>
the numerator of a fraction <i>1/2</i> should be multiplied by <i>2³·3²·5·7/2 =<br/>
= 2²·3²·5·7 = 1260</i><br/>
the numerator of a fraction <i>1/3</i> should be multiplied by <i>2³·3²·5·7/3 =<br/>
= 2³·3·5·7 = 840</i><br/>
the numerator of a fraction <i>1/4</i> should be multiplied by <i>2³·3²·5·7/2² =<br/>
= 2·3²·5·7 = 630</i><br/>
the numerator of a fraction <i>1/5</i> should be multiplied by <i>2³·3²·5·7/5 =<br/>
= 2³·3²·7 = 504</i><br/>
the numerator of a fraction <i>1/6</i> should be multiplied by <i>2³·3²·5·7/(2·3) =<br/>
= 2²·3·5·7 = 420</i><br/>
the numerator of a fraction <i>1/7</i> should be multiplied by <i>2³·3²·5·7/7 =<br/>
= 2³·3²·5 = 360</i><br/>
the numerator of a fraction <i>1/8</i> should be multiplied by <i>2³·3²·5·7/2³ =<br/>
= 3²·5·7 = 315</i><br/>
the numerator of a fraction <i>1/9</i> should be multiplied by <i>2³·3²·5·7/3² =<br/>
= 2³·5·7 = 280</i><br/>
the numerator of a fraction <i>1/10</i> should be multiplied by <i>2³·3²·5·7/(2·5) =<br/>
= 2²·3²·7 = 252</i><br/>
<br/>
As a result, the only numerator without any <i>2</i>'s in its representation as a product of primes will be the one in the one we called the <i>least binary fraction</i>.<br/>
That's why this numerator will be odd, while all others (with one or more <i>2</i>'s in their prime representation) will be even.<br/>
Therefore, the sum of all these numerators with the least common denominator will be odd and not divisible by the LCD, which is even.<br/>
<br/>
<br/>
<i>Solution 2</i><br/>
<br/>
Among the fractions of our series there are those with prime denominators, which we will call <i>prime fractions</i>.<br/>
Let's choose among all <i>prime fractions</i> the one with the <u>largest prime denominator</u> <i><b>P<sub>max</sub></b></i> (hence, the smallest in value among all <i>prime fractions</i>) and call it the <i>least prime fraction</i>.<br/>
So, <i><b>P<sub>max</sub></b></i> is the largest prime number not exceeding <i><b>N</b></i>.<br/>
<br/>
At this point it's very important to state that all subsequent members of the series after <i><b>1/P<sub>max</sub></b></i> up to the last one <i><b>1/N</b></i> have denominators <u>not multiple of</u> <i><b>P<sub>max</sub></b></i>.<br/>
The reason why it is the case is based on so called <i>Bertrand's postulate</i> that states (and can be proven, but not at this moment) that for any integer <i><b>M</b></i> there exists a prime number <i><b>p</b></i> greater than <i><b>M</b></i> and less than <i><b>2·M−2</b></i>.<br/>
<br/>
If there is another fraction with denominator being a multiple of <i><b>P<sub>max</sub></b></i> than its denominator is, obviously, not less than <i><b>2·P<sub>max</sub></b></i> and, according to the <i>Bertrand's postulate</i>, there must be another prime number between <i><b>P<sub>max</sub></b></i> and <i><b>2·P<sub>max</sub></b></i>, so <i><b>P<sub>max</sub></b></i> is not the largest prime number not exceeding <i><b>N</b></i>.<br/>
<br/>
In our example with <i><b>S<sub>10</sub></b></i> <i><b>P<sub>max</sub>=7</b></i> and the <i>least prime fraction</i> is <i>1/7</i>.<br/>
<br/>
Now let's bring all the numbers in a series to the least common denominator (LCD) in order to sum up all numerators.<br/>
Since this <i>least prime fraction</i> has the largest prime denominator, this prime denominator will be represented only once in the LCD.<br/>
<br/>
In our example with <i><b>S<sub>10</sub></b></i> LCD is <i>2³·3²·5·7=2520</i> and, as we see, number <i>7</i> is presented only once.<br/>
<br/>
The next step to calculate <i><b>S<sub>N</sub></b></i> is to bring all fractions to the least common denominator by multiplying each numerator (all numerators are <i>1</i>'s) by all prime numbers in the LCD that are not in a prime representation of a corresponding denominator.<br/>
<br/>
Because the largest prime denominator in a series occurs only once in the LCD and does not occur in prime representation of any denominator other than that of the <i>least prime fraction</i>, this largest prime denominator will be represented in all new numerators of all fractions with common denominator except in the numerator of the <i>least prime fraction</i>.<br/>
<br/>
Consequently, all new numerators, except the one of the <i>least prime fraction</i> will be divisible by <i><b>P<sub>max</sub></b></i>.<br/>
<br/>
In our example with <i><b>S<sub>10</sub></b></i> and <i><b>P<sub>max</sub>=7</b></i> the prime number <i>7</i> occurs exectly once in every new numerator except for <i>1/7</i>.
<br/>
Therefore, the sum of these new numerators <u>will not be divisible</u> by <i><b>P<sub>max</sub></b></i> and the result of the summation cannot be an integer number.<br/>
<br/>
Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-44647760250809143312023-12-06T07:21:00.000-08:002024-01-28T08:43:40.804-08:00Arithmetic+ 01: UNIZOR.COM - Math+ & Problems - Arithmetic<iframe width="480" height="270" src="https://youtube.com/embed/9VWKeLhyzjM?si=uP1biB3sAdDzu2tj" frameborder="0"></iframe>
<i>Notes to a video lecture on http://www.unizor.com</i><br/>
<br/>
<u> Arithmetic 01</u><br />
<br />
<br />
<i>Problem A</i><br/>
<br/>
Given <i><b>n</b></i> pieces of rope.<br/>
Some of them are cut into <i><b>n</b></i> smaller pieces.<br/>
Then, again, some of the obtained pieces are cut into <i><b>n</b></i> smaller one.<br/>
This process is repeated a few times and, as a result, the final number of pieces is <i><b>N</b></i>.<br/>
Prove that <i><b>N−n</b></i> is divisible by <i><b>n−1</b></i>.<br/>
<br/>
<i>Solution A</i><br/>
<br/>
Assume, out of initial <i><b>n</b></i> pieces of rope we have chosen <i><b>k<sub>1</sub></b></i> to cut each into <i><b>n</b></i> pieces.<br/>
Then, after the first cut we will have the total number of pieces equal to<br/>
<i><b>N<sub>1</sub> = n − k<sub>1</sub> + n·k<sub>1</sub> =<br/>
= n + (n−1)·k<sub>1</sub></b></i><br/>
As we see, <i><b>N<sub>1</sub>−n</b></i> is divisible by <i><b>n−1</b></i>.<br/>
<br/>
Assume, out of obtained after the first step <i><b>N<sub>1</sub></b></i> pieces of rope we have chosen <i><b>k<sub>2</sub></b></i> to cut each into <i><b>n</b></i> pieces.<br/>
Then, after the second cut we will have the total number of pieces equal to<br/>
<i><b>N<sub>2</sub> = N<sub>1</sub> − k<sub>2</sub> + n·k<sub>2</sub> =<br/>
= N<sub>1</sub> + (n−1)·k<sub>2</sub> =<br/>
= n + (n−1)·k<sub>1</sub> + (n−1)·k<sub>2</sub> =<br/>
= n + (n−1)·(k<sub>1</sub>+k<sub>2</sub>)</b></i><br/>
As we see, <i><b>N<sub>2</sub>−n</b></i> is also divisible by <i><b>n−1</b></i>.<br/>
<br/>
Obviously, it's easy to prove by induction that after <i><b>s</b><sup>th</sup></i> step the number of obtained pieces of rope will be<br/>
<i><b>N<sub>s</sub> = n + (n−1)·(k<sub>1</sub>+k<sub>2</sub>+...+k<sub>s</sub> )</b></i><br/>
and <i><b>N<sub>s</sub>−n</b></i> will be divisible by <i><b>n−1</b></i>, which is what's required to prove.<br/>
<br/>
<br />
<i>Problem B</i><br/>
<br/>
One person thinks about some number from 0 to 1000.<br/>
Another person can ask questions about this number to determine it.<br/>
What is the minimum number of questions needed to find this number, if the answers can be either YES or NO?<br/>
<br/>
<i>Solution B</i><br/>
<br/>
Considering the answers are YES or NO, one of the ways is to find the binary expression for the number.<br/>
Since the number is less than <i><b>1024=2<sup>10</sup></b></i>, it's sufficient to ask <i><b>10</b></i> questions about each binary digit.<br/>
<br/>
So, the first question is: "Is the first digit in a binary representation of your number as 10 binary digits (including leading zeros) equal to <i><b>1</b></i>? Based on the answer, we determine this first digit to be <i><b>1</b></i> or <i><b>0</b></i>.<br/>
Then we repeat the same for the second binary digit etc.<br/>
<br/>
Alternatively, the number can be guessed by multiple steps of dividing the group of all possible numbers into two approximately equal parts.<br/>
Then the first question should be: "Is your number greater than <i><b>500</b></i>?".<br/>
If the answer is YES, the next question is "Is it greater than <i><b>750</b></i>?", but if the answer is NO, the next question is "Is it greater than <i><b>250</b></i>?".<br/>
And so on.<br/>
<br/>
<br />
<i>Problem C</i><br/>
<br/>
There is a set of <i><b>100</b></i> numbers, each from <i><b>1</b></i> to <i><b>9</b></i>.<br/>
The sum of these numbers is <i><b>789</b></i>.<br/>
Is it possible to choose <i><b>70</b></i> numbers out of this set with a sum less than <i><b>500</b></i>?<br/>
<br/>
<i>Solution C</i><br/>
<br/>
If we select <i><b>70</b></i> numbers with a total sum of less than <i><b>500</b></i>, the remaining <i><b>30</b></i> numbers must have sum not less than <i><b>789−500=289</b></i>.<br/>
But the maximum sum of <i><b>30</b></i> numbers, each of which is from <i><b>1</b></i> to <i><b>9</b></i>, is <i><b>9·30=270</b></i>, which is insufficient to cover <i><b>289</b></i> needed to complete the total sum to <i><b>789</b></i>.<br/>
<br/>
<i>Answer C</b></i><br/>
It is impossible to choose <i><b>70</b></i> numbers out of this set with a sum less than <i><b>500</b></i>.<br/>
<br/>
<br/>
<i>Problem D</i><br/>
<br/>
A regular clock with large round face and <i><b>12</b></i> numbers for <i><b>12</b></i> hours is used for this problem.<br/>
A coin is placed onto each number of this clock - a total of <i><b>12</b></i> coins.<br/>
In one move two coins should be moved from their numbers into a neighboring ones, but one of these coins should move clockwise, while another - counterclockwise.<br/>
<br/>
Is it possible to collect all coins on one number on the clock by performing th described moves?<br/>
<br/>
<i>Solution D</i><br/>
<br/>
Assign a value to each position of coins as follows.<br/>
If there are <i><b>n<sub>i</sub></b></i> coins on number <i><b>i</b></i>, the value of this coin position is<br/>
<i><b>V = </b></i><font size=5>Σ</font><i><b><sub>i∈[1,12]</sub> i·n<sub>i</sub></b></i><br/>
<br/>
Initially, when there is one coin per number on a clock, the total value is<br/>
<i><b>V=1+2+3+4+5+6+<br/>
+7+8+9+10+11+12=78</b></i>.<br/>
<br/>
In one step one coin that is on number <i><b>n</b></i> and steps clockwise will increase the value <i><b>V</b></i> of a position by <i><b>1</b></i> except when it moves from number <i><b>n=12</b></i> to <i><b>n=1</b></i>, in which case the value of a position will decrease by <i><b>11</b></i>.<br/>
<br/>
In one step one coin that is on number <i><b>n</b></i> and steps counterclockwise will decrease the value <i><b>V</b></i> of a position by <i><b>1</b></i> except when it moves from number <i><b>n=1</b></i> to <i><b>n=12</b></i>, in which case the value of a position will increase by <i><b>11</b></i>.<br/><br/>
The combination of these two steps, which constitutes a single move, will either retain the same value of a position (for example, <i><b>3→4, 10→9</b></i>) or increase the value by <i><b>12</b></i> (for example, <i><b>3→4, 1→12</b></i>) or decrease the value by <i><b>12</b></i> (for example, <i><b>4→3, 12→1</b></i>).<br/>
<br/>
In any case, no matter how many moves we make, the value of a position <i><b>V</b></i> is changing by a number that is a multiple of <i><b>12</b></i>.<br/>
<br/>
But, if we consider the value of a position when all <i><b>12</b></i> coins are on the same clock number <i><b>n</b></i>, the value of a position is <i><b>V=12·n</b></i> - a multiple of <i><b>12</b></i>.<br/>
<br/>
Starting from an initial position value of <i><b>78</b></i> (not a multiple of <i><b>12</b></i>) and changing this value by a number that is a multiple of <i><b>12</b></i> on each move, we will never come to a position whose value is a multiple of <i><b>12</b></i>.<br/>
<br/>
This proves the impossibility to gather all coins on the same clock number.<br/>
<br/>
Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0