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Given any circle with a center at point O, its diameter MN and any point P on this circle not coinciding with the ends M, N of a given diameter.
Let point Q be a projection of point P on a diameter MN.
This point Q divides diameter MN into two parts: MQ = a and QN= b
Prove that
(1) Radius of a OP circle is an arithmetic average of a and b.
(2) Projection segment PQ is a geometric average of a and b.
(3) Based on these proofs, conclude that geometric average of two non-negative real numbers is less or equal to their arithmetic average. Proof:
(1) Diameter MN of this circle
is 2r=a+b.
Therefore, r = ½(a+b) - arithmetic average of a and b.
(2) ΔMPN ∝ ΔPQN ∝ ΔMQP
as right triangles with congruent angles. ⇒ c/b=a/c ⇒ c² = a·b ⇒ c = √a·b - geometric average of a and b
(3) Cathetus c is smaller than hypotenuse r in ΔPQO.
Therefore, √a·b = c ≤ a = ½(a+b)
End of the proof.
Problem B
Prove the following theorem:
If a median and an angle bisector in a triangle coincide, then this triangle is isosceles. Important
Triangle ΔACM has two sides congruent to corresponding two sides of triangle ΔBCM because they share side CM and AM=BM since CM is a median.
They also have congruent angles ∠ACM=∠BCM since CM is an angle bisector. But we cannot use a theorem about triangles ΔACM and ΔBCM being congruent by two sides and an angle because the angle is not between two congruent sides.
See the lecture Geometry+ 01 of this course, Problem A as an illustration of a case when two sides of one triangle are equal to two sides of another one and all angles of the first triangle are equal to all angles of the second, yet these triangles are not congruent.
Hint
Extend segment CM beyond point M to point N such that CM=MN.
Prove that ΔAMN = ΔBCM.
Then prove that ΔACM is isosceles.
Problem C
Given a convex polygon with N vertices.
Draw all the possible diagonals in it.
(1) Assuming that no three diagonals intersect at the same point, how many points of intersection between diagonals will exist?
(2) What is the sum of all angles inside this polygon formed by all its intersecting diagonals and sides?
Hint C
For N=6 the number of intersections is 15.
Consider quadrilaterals formed by any four vertices.
Answer C
The number of intersections between the diagonals is n = (N!)/ [(4!)·(N−4)!]
The total sum of all angles inside a polygon is equal to (2n+N−2)·π
There are 5 towns.
Some of them are connected by direct roads, that is by roads not going through other towns.
It's known that among any group of 4 towns out of these 5there is always one town connected by direct roads with each of the other 3 towns of this group.
Prove that there is at least one town connected with all 4 others by direct roads.
Proof A
Choose any 4 towns from given 5 as the first group towns.
One of these towns is connected to 3 others, as the problem states. Let's call this town A and the others will be B, C and D.
Let's use the term connector to describe a town in a group that is connected by direct roads to all other towns in the same group.
So, A is a connector in the first group.
The fifth town that is not in this group will be called E.
So far, the following roads are established: AB, AC and AD.
Now consider the second group of 4 towns {B, C, D, E}.
As you see, we eliminated from the first group town A that had roads to other 3 towns of this group and added town E which was not in it.
One of the towns in this second group should be connected by direct roads with each other 3 towns, as the problem states.
Here we have different cases, which we consider separately.
Case A1
If B, C or D is the town connected to all 3 others in this second group, then this is the town connected to all 4 other towns since it is also connected to A, as we know from the analysis of the first group.
Here is the road map if B is the town in this group that is connected to all others.
Analogous situation would be if C or D are connected to 3 other towns in this group.
This is the logical end of a proof for this case.
Case A2
If E is the town connected to 3 others by direct road, the list of roads is expended by adding the following roads: EB, EC and ED.
Now the roads map looks like this
Consider the next group of 4 towns: A, B, C and E.
One of the towns in this third group should be connected by direct roads with each other 3 towns, as the problem states.
Case A2.1
If A or E is connected to all other towns in this third group, it implies the existence of a road AE and A is the town connected to all 4 other towns. This is the logical end of a proof in this case.
Case A2.2
If B or C is connected to all other 3 towns in this third group, it means including a road BC into our map
In this case consider the fourth group A, C, D and E.
If A or E are the towns with three connections, we have to add road AE, and each one of them would be connected to all other 4 towns.
If we chose to connect C and D for the same purpose, then C will be connected to all other 4 towns.
In any case, the condition that in each group of 4 out of 5 towns there is one town connected by a direct road to 3 others is sufficient for existing of one town connected by direct roads to all other 4 towns.
End of Proof.
Problem B
(continuation of Problem A)
There are 6 towns.
Some of them are connected by direct roads, that is by roads not going through other towns.
It's known that among any group of 4 towns out of these 6there is always one town connected by direct roads with each of the other 3 towns of this group.
Prove that there is at least one town connected with all 5 others by direct roads.
Proof B
Choose any 5 towns from 6 as the first group.
According to Problem A, there must be at least one town in this group connected to 4 others. Let's call this town A and the others will be B, C, D and E.
The sixth town that is not in this group will be called F.
So far, the following roads are established: AB, AC, AD and AE.
Our second group of 5 towns excludes A and includes F.
As before, according to Problem A, it must have a town connected to other 4 towns in this second group.
If this one town is one of those that participated in the first group, that is B, C, D or E, the problem is solved because this town also was connected to A and, therefore, we found a town connected to all 5 other towns.
Assume, it's town F that connected to other 4 towns of the second group.
This necessitates adding to our map the roads FB, FC, FD and FE.
Our third group includes towns B, C, D and E.
One of this group of 4 towns must be connected to 3 others. But any of these 4 towns is connected to A and F, which makes it connected to all 6 towns.
End of proof.
Problem C
(continuation of Problem B using a method of induction)
There are N+1 towns.
Some of them are connected by direct roads, that is by roads not going through other towns.
It's known that among any group of 4 towns out of these N+1there is always one town connected by direct roads with each of the other 3 towns of this group.
Assume that it's also known that among any group of K towns out of these N+1, where K is less or equal to N, there is always one town connected by direct roads with each of the other K−1 towns of this group.
Prove that there is at least one town connected with all N others by direct roads.
Problem A
A positive integer number in decimal notation contains only digits 1 and 0.
The digit 1 occurs 111 times, while the digit 0 occurs an unknown number of times.
Can this number be a square of another integer number?
Hint A: Use the rules of divisibility.
Answer A: No.
Problem B
This problem is based on Problems 04(B,C) of this course Math+ & Problems.
Prove that if the sum of digits of some natural number N is the same as the sum of digits of the number k·N, where k−1is not divisible by 3, then number N is divisible by 9.
Hint B: Problem 04(B) 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.
Therefore, both N and k·N have the same remainder if divided by 9.
Problem C
Given a number N with a decimal representation 999...9 that contains k digits 9.
Assume for definitiveness, k is a prime number.
Find a number whose decimal representation contains only digits 1 that is divisible by N.
Answer C: 111...1 should contain 9·k digits 1.
Example C: For N=99 (k=2) the number 111...1 that contains 9·2=18 digits 1 is divisible by N.
Problem D
Consider the number N=(k+1)·(k+2)·...·(2k−1)·(2k)
How many 2's, depending on k, are in the representation of this number as a product of prime numbers?
Hint D: Notice that N=(2k)!/(k!)
Then N=(2k)! can be represented as a product of only odd numbers by only even numbers.
Answer D: N=2k·M where M is an odd number,
so the number of 2's in the representation of number N as a product of prime numbers is k.
A casino manager, analyzing the results of operations of roulette tables during a period of one day, comes to the following statistics:
there were N1 people who won at least1 time;
there were N2 people who won at least2 times;
there were N3 people who won at least3 times;
etc.
there were Nn people who won at leastn times;
and nobody won more than n times.
How many times casino has lost in roulette?
Hint A: N1 ≥ N2 ≥ N3 ≥...≥ Nn
Answer A
Casino has lost N1+N2+N3+...+Nn games.
Problem B
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?
Answer B: 8 rooks.
Problem C
Squares on a chessboard are enumerated as follows:
1st row: 1, 2,...,8
2nd row: 9, 10,...,16
3rd row: 17, 18,...,24
etc.
8th row: 57, 58,...,64
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.
What is the sum of numbers on the squares occupied by these rooks?
Hint:
Each number N on a square can be represented as N=(R−1)·8+C, where R is the row number (from 1 to 8) and C is the column number (also from 1 to 8).
Answer C: 260.
Problem D
There is a cup of coffee and a cup of milk. Assume, both cups contain the same amount of liquid.
A spoon of milk is taken from a milk cup and added to coffee.
Then, after stirring the coffee, a spoon of coffee with milk is taken and added to a milk cup.
Which is greater,
a concentration (in %) of milk in a coffee cup or
a concentration of coffee in a milk cup?
Answer D: They are the same.
Problem E
Given a table with each its cell containing some number.
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.
If a number is in the middle of a table, it has 4 neighbors (up, down, left, right).
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).
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).
Prove that all numbers in the table are equal to each other.
Hint E: Consider the neighbors of the smallest or the biggest number.
Given an isosceles triangle ΔABC with AB=BC and ∠ABC=20°.
Point D on side BC is chosen such that ∠CAD=60°.
Point E on side AB is chosen such that ∠ACE=50°.
Find angle ∠ADE.
Solution A
Problem B
Given two triangles ΔA1B1C1 and ΔA2B2C2 with the following properties:
(a) side A1B1 of the first triangle equals to side A2B2 of the second;
(b) angles opposite to these sides, ∠A1C1B1 and ∠A2C2B2, are equal to each other;
(c) bisectors of these angles, C1X1 and C2X2, are also equal to each other.
Prove that these triangles are congruent.
Proof
Put side A1B1 on top of A2B2 and draw a circle around ΔA1B1C1.
Obviously, point C2 would lie on this circle because of equality of angles opposite to equal sides.
Assume that points C1 and C2 do not coincide.
Extend bisectors C1X1 and C2X2 to intersection with a circle at point P. It must be the same point for both bisectors because each divides the arc ⌒A1B1 in half.
First, let's prove that triangle ΔPC1C2 is similar to triangle ΔPX1X2
Our proof will be based on congruency of two angles of one triangle to two corresponding angles of another.
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).
We will also use known theorems of Geometry:
(Theorem 1) 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.
(Theorem 2) An angle formed by two chords intersecting inside a circle is measured by half of a sum of two opposite arcs these chords cut.
These theorems were presented in UNIZOR.COM - Math 4 Teens - Geometry - Circles - Mini Theorems 1 and ...Mini Theorems 2.
1. ∠PC1C2 is measured by half of sum of arcs ⌒C2B2 and ⌒B2P;
∠PX2A2 is measured by half of sum of arcs ⌒C2B2 and ⌒PA2;
but arcs ⌒B2P and ⌒PA2 are equal, from which follows that
∠PC1C2 equal to ∠PX2A2.
2. ∠PC2C1 is measured by half of sum of arcs ⌒C1A2 and ⌒A1P;
∠PX1B1 is measured by half of sum of arcs ⌒C2B2 and ⌒PA2;
but arcs ⌒B2P and ⌒PA2 are equal, from which follows that
∠PC1C2 equal to ∠PX2A2.
Since two angles of ΔPC1C2 are congruent to two angles of ΔPX1X2, these triangles are similar.
Consequently, their sides are proportional PC1/PX2 = PC2/PX1
If angle bisectors C1X1 and C2X2 are equal, segments C1P and C2P must be equal to preserve the proportionality of corresponding sides of triangles ΔPC1C2 and ΔPX1X2.
Indeed, if C1X1=C2X2=d, PX1=x1 and PX2=x2, then (d+x1)/x2 = (d+x2)/x1 ⇒d·x1+x1² = d·x2+x2² ⇒d·(x1−x2) = (x2−x1)·(x1+x2) ⇒x1 = x2, because otherwise d would be negative −(x1+x2).
From C1P = C2P follows ⇒ ⌒C1A1P = ⌒C2B2P ⇒ ⌒C1A1 = ⌒C2B2 ⇒C1A1 = C2B2
Analogously, ⇒C1B1 = C2A2
which proves that triangles ΔA1B1C1 and ΔA2B2C2 are congruent by three sides.
First, convert sin²(...) into cos(...) by using the identity cos(2φ) = cos²(φ) − sin²(φ) =
= 1 − 2sin²(φ)
from which follows sin²(φ) = ½(1 − cos(2φ))
Now our sum looks like this
Σk∈[0,n−1] ½(1−cos(2x+2k·π/n)) =
= n/2 −
− ½Σk∈[0,n−1]cos(2x+2k·π/n)
To calculate Σk∈[0,n−1] above, let's use the result of the previous lecture Trigonometry 03 that proved the following
Σk∈[0,n]cos(x+k·y) =
= [sin(x+(2n+1)·y/2) −
− sin(x−y/2)] /
/ [2·sin(y/2)]
Since in our sum we have n terms instead of n+1, as in the formula above, to use this formula in our case we have to use n−1 instead of n, which will result in
Σk∈[0,n−1]cos(x+k·y) =
= [sin(x+(2n−1)·y/2) −
− sin(x−y/2)] /
/ [2·sin(y/2)]
To use this formula in our case, we have to substitute 2x instead of x and 2π/n instead of y.
In this case the numerator in the formula above would be sin(x+(2n−1)·π/n) − sin(x−π/n)
The first term in the above expression for the numerator equals to sin(x+(2n−1)·π/n) =
= sin(x+2n·π/n−π/n) =
= sin(x−π/n+2π) = sin(x−π/n)
(since the value 2π is a period of function sin)
which makes the first term exactly the same as the second term and they will cancel each other resulting in zero in the numerator.
That leaves the sum in our case to be equal to
Σk∈[0,n−1] ½(1−cos(2x+2k·π/n)) = n/2
Answer A
Σk∈[0,n−1]sin²(x+k·π/n) = n/2
Problem B
Given a regular n-sided polygon with vertices A1, A2,...,An inscribed into a circle of radius R with a center at point O and any point P on this circle.
Calculate the sum of squares of distances from point P to all vertices Ak S = Σk∈[1,n](PAk)².
Hint:
Let ∠POAk = φk.
Then φk = φ1 + (k−1)·2π/n.
Use the results of Problem A.
Hint A2 X²+Y²+Z² =
= (X−a)²+(Y−a)²+(Z−a)²+
+2·a·(X+Y+Z)−3a² ≥
≥ 2a −3a²
Quadratic polynomial 2a−3a² has roots a=0 and a=2/3 and maximum at a=1/3 with value 2·(1/3)−3·(1/3)²=1/3.
Problem B
Solve an equation X8 + X4 + 1 = 0
Hint B1
Substitute y=X4.
Hint B2
Represent the left part as a product of 4 polynomials of a second degree by adding and subtracting X4.
Problem A
Prove that a remainder of division of any prime number P by 24 is a prime number, assuming P is greater than 24.
Hint A
If P is prime and R is a remainder of division of P by 24 then P=24·n+R where R≤23.
If R is not prime, it should be divisible by a product of, at least, two prime numbers.
Also, 24=2·2·2·3.
Problem B
Prove that for any natural number N the number 10N+18·N−1 is divisible by 27.
Hint B
First, prove a theorem that any natural number n divided by 3 (or 9) has the same remainder as a sum of its digits divided by 3 (or 9).
Obviously, the given number is divisible by 9.
Then prove that the number 111...111 consisting of N unit digits with their sum equal to N plus 2·N is divisible by 3 using a theorem proved above.
Problem C
Prove that from any set of 100 natural numbers N1, N2,... ,N100 it is possible to choose such a subset with a sum of its numbers divisible by 100.
Hint C
Let Sk=Σi∈[1,k]Ni
where k=1, 2,...,100.
If there is one particular number Sk divisible by 100, the components of this sum can be chosen as a subgroup and the problem is solved.
Assume, none of our 100 sums Sk is divisible by 100.
Let Rk is a remainder of division of Sk by 100.
Note that there are 99 possible remainders from 1 to 99.
