Trigonometry+ 08: UNIZOR.COM - Math+ & Problems - Trigonometry

Notes to a video lecture on http://www.unizor.com

Trigonometry+ 08


Problem A

Prove the following inequality
cos(36°) ≥ tan(36°)

Hint A
Find the point where left side equals to the right side and compare it with 36°.
Use the following values:
π/5≅0.628 and
arcsin(½(√5−1))≅0.666.


Problem B

Solve the equation
a·sin²(x) + b·sin(x)·cos(x) +
+ c·cos²(x) = d
where a ≠ d.

Hint B
For the right side of this equation use the identity
sin²(x) + cos²(x) = 1

Answer B
x = arctan{R/[2(a−d)]}+π·N
where
R=[−b±√b²−4·(a−d)·(c−d)]
and N is any integer number.


Problem C

Solve the following system of equations
tan(x)·tan(y) = 3
sin(x)·sin(y) = 3/4

Hint C
Convert this system into form
cos(x+y)=...
cos(x−y)=...

Solution C
Since tan()=sin()/cos(), substitute the second equation's value 3/4 into numerator of the first and invert the fraction
(3/4)/[cos(x)·cos(y)]=3
cos(x)·cos(y) = 1/4
As we know, cos(x+y) =
cos(x)·cos(y) − sin(x)·sin(y)
and cos(x−y) =
cos(x)·cos(y) + sin(x)·sin(y)
Since sin(x)·sin(y) = 3/4
and cos(x)·cos(y) = 1/4
we can find
cos(x+y) = 1/4 − 3/4 = −1/2
cos(x−y) = 1/4 + 3/4 = 1
Function cos() is periodical with a period of 2π.
The equation cos(x+y)=−1/2 has two solutions for (x+y) within an interval [0,2π]:
x + y = ±2π/3
The equation cos(x−y)=1 has one solution for (x−y) within an interval [0,2π]:
x − y = 0
Adding periodicity, we come up with two systems of equations, each depending on some integer parameters
x + y = 2π/3 + 2π·M
x − y = 2π·N
where M and N are any integers, and
x + y = −2π/3 + 2π·M
x − y = 2π·N
Each one of these systems can be easily solved by adding and subtracting the equations, which leads to the first series of solutions
x₁ = π/3 + π·(M+N)
y₁ = π/3 + π·(M−N)
and the second series of solutions
x₂ = −π/3 + π·(M+N)
y₂ = −π/3 + π·(M−N)
In both series M and N can independently take any integer value.

Note C
Since original system of equation contained tan(x) and tan(y), we have to make sure that by getting rid of cos() in the denominator we have not added extraneous solutions.
Function cos() is zero at π/2+π·K, where K can be any integer number. If any of our solutions falls in this set, it must be excluded. Fortunately, none of our solutions coincides with this set.

Trigonometry+ 07: UNIZOR.COM - Math+ & Problems - Trigonometry

Notes to a video lecture on http://www.unizor.com

Trigonometry+ 07


Problem A

Prove the following identity
2·arccos[√(1+x)/2] = arccos(x)

Proof A

By definition of function arccos(x), it's an angle in interval [0,π], whose cosine is x.
That is, cos(arccos(x))=x.
Therefore, we have to prove that cosine of the left side of an equality above equals to x.
Let's use a known identity
cos(2α)=2cos²(α)−1
Now
cos{2·arccos[√(1+x)/2]} =
= 2cos²{arccos[√(1+x)/2]}−1 =
= 2·[√(1+x)/2]² = x


Problem B
Simplify the expression
tan[½arctan(x)]

Hint B
Express tan(φ/2) in terms of tan(φ).

Solution B
Let angle φ = arctan(x).
We know that, by definition of function arctan(), its domain is all real values of x, its value, angle φ, is in the interval from −π/2 to π/2 and tan(φ)=x.
Hence, this problem can be formulated as
If a tangent of an angle φ is x, what is the tangent of the half of this angle?
Our first task is to express a tangent of the half of an angle in terms of a tangent of the whole angle.
As we know from UNIZOR.COM - Math 4 Teens - Trigonometry - Sum of Angles Problem 1 : tan(φ/2),
tan(φ)=2tan(φ/2)/[1−tan²(φ/2)]
Let's resolve this equation for tan(φ/2) in terms of tan(φ)
If A = 2B/(1−B²) then
A·B² + 2·B −A = 0
Solving this for B, we obtain
B_1,2 = (−2±√4+4A²)/2A =
= (−1±√1+A²)/A
Using this for A=tan(φ) and B=tan(φ/2), we get the expression for tan(φ/2) in terms of tan(φ).
An important detail is that on interval (−π/2,π/2) the sign of tan(φ) and tan(φ/2) are the same (positive for positive angle and negative for negative angle).
Therefore, sign ± in the formula above should be replaced with + and the final formula expressing tan(φ/2) in terms of function tan(φ) is
tan(φ/2) =
= [−1+√1+tan²(φ)] /tan(φ)
Since φ=arctan(x) and tan(φ)=x, we can state the following:
tan[½arctan(x)] =
= [−1+√1+x²] /x
The value x=0 should be excluded from this formula, we can, obviously, say that in this case
tan[½arctan(x)] = 0


Problem C
Prove geometrically that for an acute angle θ, measured in radians, the following inequalities are true:
sin(θ) ≤ θ ≤ tan(θ)

Hint
Use the unit circle and a geometric interpretation of components of the inequality to be proven.

Proof

Since the radius of a circle is 1, the length of an arc AB is θ - a measure in radians of an angle ∠AOB.
By definitions of sin() and cos(), abscissa of point A (segment OD) equals to cos(θ) and the ordinate of point A (segment AD) equals to sin(θ).
Comparing the length of segment AD (that is, sin(θ)) and arc AB (that is, θ), taking into consideration that AD is a perpendicular to radius OB, while arc AB is a curve from the same original point A to OB, we conclude that the length of AD is less or equal to the length of arc AB with equality held only if point A coincides with point B, that is when angle θ is zero.
Therefore, we have proven that sin(θ) is less or equal to θ.
By definition of function tan(), it's a ratio of sin() to cos().
In our case it's a ratio of AD to OD.
Draw a perpendicular to OB at point B, and let C be an intersection of this perpendicular with a continuation of OA.
Since ΔOAD is similar to ΔOCB, the ratio of AD to OD is the same as the ratio of CB to OB. Since OB=1, tan(θ)=CB.
Area of circular sector AOB is less than area of right triangle COB.
Area of circular sector AOB of radius 1 equals to the area of a circle (that is, π) times θ/(2π), that is ½θ.
Area of a right triangle COB equals to ½CB·OB=½tan(θ).
Comparing these two areas, we conclude that θ is less or equal to tan(θ) with equality held only if point C coincides with point B, that is when angle θ is zero.

Trigonometry+ 06: UNIZOR.COM - Math+ & Problems - Trigonometry

Notes to a video lecture on http://www.unizor.com

Trigonometry+ 06


Problem A

Given ∠α, ∠β and ∠γ are acute angles of a triangle.
Prove that
cos(α)+cos(β)+cos(γ) ≤ 3/2

Hint A
Using α + β + γ = π
reduce the left side of the inequality to a function of sin(½γ).

Solution A
Recall the transformation of a sum of two cosines into a product of other cosines
cos(α)= cos[½(α+β)+½(α−β)]=
= cos(½(α+β))·cos(½(α−β)) −
− sin(½(α+β))·sin(½(α−β))
cos(β)= cos[½(α+β)−½(α−β)]=
= cos(½(α+β))·cos(½(α−β)) +
+ sin(½(α+β))·sin(½(α−β))
Therefore,
cos(α) + cos(β) =
= 2·cos(½(α+β))·cos(½(α−β)) =
= 2·cos(½(π−γ))·cos(½(α−β)) =
= 2·cos(½π−½γ))·cos(½(α−β))
But cos(½π−½γ) = sin(½γ).
Also, cos(½(α−β)) ≤ 1.
Therefore,
cos(α) + cos(β) ≤
≤ 2·sin(½γ))
Hence,
cos(α)+cos(β)+cos(γ) ≤
≤ 2·sin(½γ)) + cos(γ) =
= 2·sin(½γ)) + cos(2·½γ) =
= 2·sin(½γ)) + 1 − 2·sin²(½γ) =
= −2X² + 2X + 1
where X=sin(½γ)
The quadratic function −2X²+2X+1 has a maximum of 3/2 at X=½.


Problem B

Given
0 ≤ α₁ ≤ α₂ ≤ ...≤ α_n ≤ ½π.

Let A =

Σi∈[1,n]sin(αi) Σi∈[1,n]cos(αi)

Prove: tan(α₁) ≤ A ≤ tan(α_n)


Problem C

Solve the equation
sin(A·x) + sin(B·x) = 0
where A and B are some real numbers.

Hint C
Convert the left side of an equation into a product.

Solution
Recall the transformation of a sum of two sines into a product of other trigonometric functions
sin(α)= sin[½(α+β)+½(α−β)]=
= sin(½(α+β))·cos(½(α−β)) +
+ cos(½(α+β))·sin(½(α−β))
sin(β)= sin[½(α+β)−½(α−β)]=
= sin(½(α+β))·cos(½(α−β)) −
− cos(½(α+β))·sin(½(α−β))
Therefore,
sin(α) + sin(β) =
= 2·sin(½(α+β))·cos(½(α−β))
Using this for α=A·x and β=B·x, the left side of our equation can be invariantly transformed into
2·sin(½(A+B)x)·cos(½(A−B)x)
It can be equal to zero if
sin(½(A+B)x) = 0
from which follows
½(A+B)x = π·N
x = 2πN/(A+B)
(for A≠−B and where N is any integer)
or if
cos(½(A−B)x) = 0
from which follows
½(A−B)x = ½π+π·N
x = π·(2πN+1)/(A−B)
(for A≠B and where N is any integer)

Algebra+ 06: UNIZOR.COM - Math+ &Problems - Algebra

Notes to a video lecture on http://www.unizor.com

Algebra+ 06

Problem A

Prove that sum of square roots of 2, 3 and 5 is an irrational number.

Hint A
Assume, this sum is rational, that is
√2 + √3 + √5 = p/q
where p and q are integer numbers without common divisors (if they do, we can reduce the fraction by dividing a numerator p and denominator q by a common divisor without changing the value of a fraction).
Then simplify the above expression by getting rid of square roots and prove that p must be an even number and, therefore, can be represented as p=2r.
Then prove that q must be even as well, and, therefore, p and q have a common divisor 2, which we assumed they don't.


Problem B

Given a polynomial represented as a product
P = (5−4x)¹⁰⁰⁰·(3x−4)¹⁰⁰¹
Assume, we open all the parenthesis and combine all similar terms to express this polynomial in a canonical form
P = Σ_n∈[0,2001]A_n·x²⁰⁰¹⁻ⁿ
What would be a sum of all the coefficients A_n?

Hint B:
Do not attempt to use Newton's binomial and find the answer by explicitly performing all the operations to convert the given expression into canonical polynomial form.
There is a better and very quick way.

Answer B:
Sum of all the coefficients will be equal to −1.


Problem C

Find all prime integer x and y, for which the following is true
13·(x + y) = 3·(x² − x·y + y²)

Hint C
Express the given equation as a quadratic equation for x with coefficients as functions of y.
Then, to have solutions for x, a discriminant must be non-negative, which reduces the possible values for y to be in an interval (0,10).

Answer C
Only a pair of numbers 2 and 7 satisfies the condition of this problem.
So, the solutions are (x=2,y=7) or (x=7,y=2).

Algebra+ 05: UNIZOR.COM - Math+ & Problems- Algebra

Notes to a video lecture on http://www.unizor.com

Algebra+ 05

Problem A

Given a system of two equations with three unknown variables x, y and z:
x + y + z = A
x⁻¹ + y⁻¹ + z⁻¹ = A⁻¹
Prove that one of the unknown variables equals to A.

Hint A

System of equations
x + y = p
x · y = q
fully defines a pair of numbers (generally speaking, complex numbers) as solutions to a quadratic equation
X² − p·X + q.
Indeed, if X₁ and X₂ are the solution of the equation, then, according to the Vieta's Theorem,
X₁ + X₂ = −(−p) = p and
X₁ · X₂ = q
(See a lecture Math 4 Teens - Algebra - Quadratic Equations - Lecture on UNIZOR.COM)

The only unresolved issue is:
which unknown variable takes which value from a pair
is it (x=X₁,y=X₂) or (x=X₂,y=X₁).

From this follows that, if the following system of equations is given
x + y = a + b
x · y = a · b
then either (x=a,y=b)
or (x=b,y=a).

Proof A

x + y = A − z
x⁻¹ + y⁻¹ = A⁻¹ − z⁻¹
None of the unknown variables can be equal to zero, since each is represented in the second equation in the denominator.
Therefore, we can multiply the second equation by x·y getting
y + x = x·y/A − x·y/z

Using the first equation, substitute x+y into the second getting a system of equations
x + y = A − z
A − z = x·y·(1/A − 1/z)
or
x + y = A − z
x·y = (A−z)/(1/A − 1/z)
or
x + y = A − z
x·y = −A·z
or
x + y = A + (−z)
x·y = A·(−z)
Therefore, either x=A, y=−z
or y=A, x=−z.


Problem B

Prove that
(x + y)⁴ ≤ 8·x⁴ + 8·y⁴

Proof B

Let's start with analysis of this problem.
Assume, this inequality (call it "statement A") is true and make invariant (reversible and equivalent) transformations to it, trying to get to an obviously true statement B.
Then, using the fact that our transformations were invariant, we can say that we can start with obviously true statement B and, using the reverse transformations, derive statement A, that is we will prove that A is true.

Notice that, if we divide both sides of this inequality by y⁴ and assign t=x/y, we will reduce the number of variables from two to one, which seems to simplify the task.
Dividing by positive y⁴ is an invariant transformation of an inequality, except a case of y=0. The case of y=0 can be considered separately, and in this case the inequality is obviously true since x⁴ ≤ 8·x⁴.

After dividing by y⁴ and substituting t=x/y the new inequality looks like
(t + 1)⁴ ≤ 8t⁴ + 8
which seems to be simpler to prove.

Let's open the parenthesis and bring all items to one side of an inequality - obviously invariant transformation
7t⁴ − 4t³ − 6t² − 4t + 7 ≥ 0
Notice that the sum of coefficients of a polynomial on the left is zero. That means that t=1 is a root of this polynomial, that is it's equal to zero for t=1.

Recall the Fundamental Theorem of Algebra (see Math 4 Teens course on UNIZOR.COM, menu items Algebra - Fundamental Theorem of Algebra and its Corollary 1) that states that if x=a is a root of a polynomial P⁽ⁿ⁾(x) of n^th degree, then this polynomial is divisible by x−a, that is
P⁽ⁿ⁾(x) = (x−a)·Q⁽ⁿ⁻¹⁾(x)
where Q⁽ⁿ⁻¹⁾(x) is a polynomial of a degree lower by 1 than P⁽ⁿ⁾(x).

Therefore, since t=1 is a root of the polynomial of the 4th degree above, we can represent that polynomial as (t−1) multiplied by another polynomial of the 3rd degree.
7t⁴ − 4t³ − 6t² − 4t + 7 =
= (t − 1)·(7t³ + 3t² − 3t − 7)

Consider the polynomial
7t³ + 3t² − 3t − 7
The sum of its coefficient is zero too. Therefore, we can represent it as a product of (t−1) and a polynomial of the second degree
7t³ + 3t² − 3t − 7 =
= (t − 1)·(7t² + 10t + 7)

So, the inequality we have to prove was transformed into this one:
(t − 1)²·(7t² + 10t + 7) ≥ 0
In this inequality the member (t−1)² is always greater or equal to zero.
Quadratic polynomial 7t²+10t+7 has discriminant Δ=10²−4·7·7=−96, which is negative and, consequently, it has no roots, it's always not equal to zero.
It can only be greater than zero since the coefficient at t² is positive.
Therefore, this polynomial is always greater than zero.
That concludes the analysis of our problem.

The proof proper is to start from an obviously truthful statement
(t − 1)²·(7t² + 10t + 7) ≥ 0
and transform it into
(t + 1)⁴ ≤ 8t⁴ + 8
Replacing t with x/y (recall, a trivial case y=0 was already checked, so now we assume that y≠0) and multiplying by y⁴ finishes the proof.


Problem C

Prove the following inequality
x¹² − x⁹ + x⁴ − x + 1 > 0

Proof C

Consider a polynomial
x¹² − x⁹ + x⁴ − x
It can be invariantly transformed into
x⁹·(x³ − 1) + x·(x³ − 1) or
(x³ − 1)·(x⁹ + x) or
(x³ − 1)·x·(x⁸ + 1)

This polynomial has only two roots: x=0 and x=1
As easily checked, values outside interval (0,1) are non-negative and inside this interval the values of a polynomial are negative.
Since we are interested in the values of this polynomial +1, the only interval where it's not obvious whether after adding 1 it is positive or not is inside the interval (0,1).

Inside interval (0,1)
x¹² − x⁹ + x⁴ − x + 1 =
= x¹² + x⁴·(1−x⁵) + (1−x)
with every item in parenthesis and every other participant in the above expression is positive, which results in a positive value of an entire expression.

End of proof.

Geometry+ 07: UNIZOR.COM - Math+ & Problems - Geometry

Notes to a video lecture on http://www.unizor.com

Geometry+ 07

Problem A

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

Logic+ 06: UNIZOR.COM - Math+ & Problems - Logic

Notes to a video lecture on http://www.unizor.com

Logic+ 06

Problem A

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 5 there 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 6 there 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+1 there 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.

Hint C: Follow the logic of a proof in Problem B.

Arithmetic+ 06: UNIZOR.COM - Math+ &Problems - Arithmetic

Notes to a video lecture on http://www.unizor.com

Arithmetic+ 06

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−1 is 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=2^k·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.

Logic+ 05: UNIZOR.COM - Math+ & Problems - Logic

Notes to a video lecture on http://www.unizor.com

Logic+ 05

Problem A

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 least 1 time;
there were N₂ people who won at least 2 times;
there were N₃ people who won at least 3 times;
etc.
there were N_n people who won at least n times;
and nobody won more than n times.

How many times casino has lost in roulette?

Hint A:
N₁ ≥ N₂ ≥ N₃ ≥...≥ N_n

Answer A
Casino has lost
N₁+N₂+N₃+...+N_n 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.

Geometry+ 06: UNIZOR.COM - Math+ & Problems - Geometry

Notes to a video lecture on http://www.unizor.com

Geometry+ 06

Problem A

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.

Trigonometry+ 04: UNIZOR.COM - "Math+ & Problems" - "Trigonometry"

Notes to a video lecture on http://www.unizor.com

Trigonometry+ 04

Problem A

Find the sums
Σ_{k∈[0,n−1]}sin²(x+k·π/n)

Solution A

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
Σ_{k∈[0,n−1]}
½(1−cos(2x+2k·π/n)) =
= n/2 −
− ½Σ_{k∈[0,n−1]}cos(2x+2k·π/n)

To calculate Σ_{k∈[0,n−1]} above, let's use the result of the previous lecture Trigonometry 03 that proved the following
Σ_k∈[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
Σ_{k∈[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
Σ_{k∈[0,n−1]}
½(1−cos(2x+2k·π/n)) = n/2

Answer A
Σ_{k∈[0,n−1]}sin²(x+k·π/n) = n/2


Problem B

Given a regular n-sided polygon with vertices A₁, A₂,...,A_n 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 A_k
S = Σ_k∈[1,n](PA_k)².

Hint:
Let ∠POA_k = φ_k.
Then φ_k = φ₁ + (k−1)·2π/n.
Use the results of Problem A.

Answer
S = 2nR²

Algebra+ 04: UNIZOR.COM - Math+ & Problems - Algebra

Notes to a video lecture on http://www.unizor.com

Algebra+ 04

Problem A

Prove that
if X+Y+Z=1
then X²+Y²+Z² ≥ 1/3.

Hint A1
X² + Y² ≥ 2·X·Y.

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.

Answer B
X_1,2,3,4 = ±1/2 ± i·√3/2
X_5,6,7,8 = ±√3/2 ± i/2


Problem C

Find a number from 10 to 99 knowing that its square equals to sum of its digits in cube.

Hint C
Actually, we have to solve the equation
(10·X+Y)² = (X+Y)³
for natural
1 ≤ X ≤ 9 and 0 ≤ Y≤ 9.

Answer C
The only solution is number 27.
Check:
27² = 729
(2+7)³ = 729

Arithmetic+ 05: UNIZOR.COM - Math+ & Problems - Arithmetic

Notes to a video lecture on http://www.unizor.com

Arithmetic+ 05

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 10^N+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 S_k=Σ_i∈[1,k]N_i
where k=1, 2,...,100.
If there is one particular number S_k 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 S_k is divisible by 100.
Let R_k is a remainder of division of S_k by 100.
Note that there are 99 possible remainders from 1 to 99.

Trigonometry+ 02: UNIZOR.COM - Math+ & Problems - Trigonometry

Notes to a video lecture on http://www.unizor.com

Trigonometry+ 02


Problem A

Prove the following trigonometric identities using the Euler formula
e^i·φ = cos(φ)+i·sin(φ)
[See lecture on UNIZOR.COM - Math 4 Teens - Trigonometry - Complex Numbers and Trigonometry - Euler's Formula]

sin(x+y) =
= sin(x)·cos(y) + cos(x)·sin(y)

cos(x+y) =
= cos(x)·cos(y) − sin(x)·sin(y)

Proof A
cos(x+y) + i·sin(x+y) =
= e^i·(x+y) = e^i·x·e^i·y =
= [cos(x) + i·sin(x)] ·
· [cos(y) + i·sin(y)]=
=cos(x)·cos(y)+i²sin(x)·sin(y)+
+i·sin(x)·cos(x)+i·cos(x)·sin(y)=
=cos(x)·cos(y)−sin(x)·sin(y)+
+i·[sin(x)·cos(y)+cos(x)·sin(y)]


Problem B

Prove the following trigonometric identities

tan(x+y) =

tan(x)+tan(y) 1−tan(x)·tan(y)

cot(x+y) =

cot(x)·cot(y)−1 cot(x)+cot(y)

Proof B

tan(x+y) =

sin(x+y) cos(x+y)

=	sin(x)·cos(y) + cos(x)·sin(y) cos(x)·cos(y) − sin(x)·sin(y)

Divide numerator and denominator by cos(x)·cos(y)

=	sin(x)/cos(x) + sin(y)/cos(y) 1−sin(x)·sin(y)/(cos(x)·cos(y))

Cotangent formula is proved analogously.


Problem C

Prove the following trigonometric identities

sin(x) =

2·tan(½x) 1+tan²(½x)

cos(x) =

1−tan²(½x) 1+tan²(½x)

Proof C
sin(x) = sin(½x + ½x)
= 2sin(½x)·cos(½x)
= 2[sin(½x)/cos(½x)]·cos²(½x)
= 2·tan(½x) · cos²(½x)
At the same time
1/[1+tan²(½x)]
= 1/[1+sin²(½x)/cos²(½x)]
= cos²(½x)/[cos²(½x)+sin²(½x)]
= cos²(½x)

Trigonometry+ 03: UNIZOR.COM - Math+ & Problems - Trigonometry

Notes to a video lecture on http://www.unizor.com

Trigonometry+ 03


Problem A

Find the sums
S_sin(x,y,n)= Σ_k∈[0,n]sin(x+k·y)
S_cos(x,y,n)= Σ_k∈[0,n]cos(x+k·y)
as algebraic expression with arguments x, y and n.


Solution A1

Let's start with a sum of sines.
Recall the formula for a cosine of a sum of two angles
cos(a+b) =
= cos(a)·cos(b) − sin(a)·sin(b)
From this formula it's easy to derive another trigonometric identity formula
2·sin(a)·sin(b) =
= cos(a−b)−cos(a+b)

Let's use this formula to transform our series into only a couple of members.
Multiply S_sin (that is, each member of this series) by 2·sin(y/2) and use the trigonometric identity above.

Then after this multiplication by 2·sin(y/2) the k^th member of S_sin would look like
2·sin(x+k·y)·sin(y/2) =
= cos(x+k·y−y/2) −
− cos(x+k·y+y/2) =
= cos(x+(2k−1)·y/2) −
− cos(x+(2k+1)·y/2)

Thus, each member of our series is converted into a pair of cosines with a minus in-between:
2·S_sin(x,y,n)·sin(y/2) =
= Σ_k∈[0,n][cos(x+(2k−1)·y/2) −
− cos(x+(2k+1)·y/2)] =
= [cos(x−y/2)−cos(x+y/2)] +
+ [cos(x+y/2)−cos(x+3y/2)] +
+ [cos(x+3y/2)−cos(x+5y/2)] +
etc. and the last two members of a series are
+ [cos(x+(2n−3)/2) −
− cos(x+(2n−1)y/2)] +
+ [cos(x+(2n−1)/2) −
− cos(x+(2n+1)y/2)]

It's easy to see that in this series every members, except the first and the last, cancel each other and the result is
2·S_sin(x,y,n)·sin(y/2) =
= cos(x−y/2) −
− cos(x+(2n+1)·y/2)

Therefore,
S_sin(x,y,n) =
= [cos(x−y/2) −
− cos(x+(2n+1)·y/2)] /
/ [2·sin(y/2)]

The sum of cosines can be treated analogously.
Recall the formula for a sine of a sum of two angles
sin(a+b) =
= sin(a)·cos(b) + cos(a)·sin(b)
From this formula it's easy to derive another trigonometric identity formula
2·cos(a)·sin(b) =
= sin(a+b)−sin(a−b)

Let's use this formula to transform our series into only a couple of members.
First, we multiply all members of S_cos (that is, each member of this series) by 2·sin(y/2). and use the trigonometric identity above.
2·S_cos(x,y,n)·sin(y/2) =
= Σ_k∈[0,n][sin(x+(2k+1)·y/2) −
− sin(x+(2k−1)·y/2)] =
= sin(x+(2n+1)·y/2) −
− sin(x−y/2)

Therefore,
S_cos(x,y,n) =
= [sin(x+(2n+1)·y/2) −
− sin(x−y/2)] /
/ [2·sin(y/2)]

Of cause, both formulas for S_sin and S_cos are correct only if sin(y/2)≠0, that is y≠2πN, where N - any integer number.
If, however, y=2πN, then sin(x+k·y)=sin(x) and cos(x+k·y)=cos(x), so
S_sin = (n+1)·sin(x)
S_cos = (n+1)·cos(x)


Solution A2

A different and, arguably, more elegant approach to solve this problem would be use the Euler formula
e^i·φ = cos(φ)+i·sin(φ)
[See lecture on UNIZOR.COM - Math 4 Teens - Trigonometry - Complex Numbers and Trigonometry - Euler's Formula]

Using it, we can express cos(x+k·y) and sin(x+k·y) as real and imaginary parts of
e^i·(x+k·y) = e^i·x·(e^i·y)^k

The sum of these expression by k from 0 to n can be calculated as a geometric series with the first term a=e^i·x and factor r=e^i·y.
[See lecture on UNIZOR.COM - Math 4 Teens - Algebra - Sequence and Series - Progression Series]
This sum in terms of a and r is
S = a·(rⁿ⁺¹−1)/(r−1)

Therefore, our sum is equal to
Σ_k∈[0,n]e^i·x·(e^i·y)^k =
= e^i·x·[(e^i·y)ⁿ⁺¹−1] /(e^i·y−1) =
= [e^{i·(x+y·(n+1))}−e^i·x] /(e^i·y−1)

What remains is to convert this expression into canonical representation of complex numbers as A+B·i. Then A will represent the sum of cosines, and B will represent a sum of sines.

First, get rid of i in the denominator by multiplying both numerator and denominator by (e^−i·y−1).
In the denominator the result will be
(e^i·y−1)·(e^−i·y−1) =
= e⁰−e^−i·y−e^i·y+1 =
= 2 − e^i·y − e^−i·y =
= 2 − cos(y) − i·sin(y) −
− cos(−y) − i·sin(−y) =
= 2 −2cos(y) = 4·sin²(y/2)
As we see, there is only real part in the denominator, an imaginary terms cancel each other.

In the numerator we will have
[e^{i·(x+y·(n+1))}−e^i·x]·(e^−i·y−1) =
= e^i·(x+y·n) − e^{i·(x+y·(n+1))} −
− e^i·(x−y) + e^i·x =
= cos(x+y·n)−cos(x+y·(n+1))−
−cos(x−y) + cos(x) +
+ i·[sin(x+y·n)−sin(x+y(n+1))−
−sin(x−y)+sin(x)]

Let's deal with real part that gives the sum of cosines.
We will use another trigonometric identity which can be easily proved using
α=½(α+β)+½(α−β) and
β=½(α+β)−½(α−β):
cos(α) − cos(β) =
= −2sin(½(α+β))·sin(½(α−β))
Using this identity, the numerator for a sum of cosines is
cos(x+y·n)−cos(x+y·(n+1))−
−cos(x−y) + cos(x) =
= 2·sin(x+½(2n+1)·y)·sin(½y) −
− 2·sin(x−½y)·sin(½y) =
= 2·sin(½y)·
·[sin(x+½(2n+1)·y) −
− sin(x−½y)]

Combining this numerator and calculated above denominator 4·sin²(y/2), we obtain
S_cos(x,y,n) =
= [sin(x+(2n+1)·y/2) −
− sin(x−y/2)]·
·2·sin(y/2) / [4·sin²(y/2)] =
= [sin(x+(2n+1)·y/2) −
− sin(x−y/2)] /
/ [2·sin(y/2)]
which corresponds completely to Solution B1.

Similarly, the imaginary part of a numerator can be used to calculate the sum of sines.
We will use yet another trigonometric identity, which can be derived analogously to a difference of cosines above:
sin(α) − sin(β) =
= 2sin(½(α−β))·cos(½(α+β))
Using this identity, the numerator for a sum of sines is
sin(x+y·n)−sin(x+y(n+1))−
−sin(x−y)+sin(x) =
=−2·sin(½y)·cos(x+½(2n+1)·y)+
+ 2·sin(½y)·cos(x−½y) =
= 2·sin(½y)·
·[cos(x−½y) −
− cos(x+½(2n+1)·y)]

Combining this numerator and calculated above denominator 4·sin²(y/2), we obtain
S_sin(x,y,n) =
= [cos(x−y/2) −
− cos(x+(2n+1)·y/2)]·
·2·sin(y/2) / [4·sin²(y/2)] =
= [cos(x−y/2) −
− cos(x+(2n+1)·y/2)] /
/ [2·sin(y/2)]
which corresponds completely to Solution B1.

Logic+ 04: UNIZOR.COM - Math+ & Problems - Logic

Notes to a video lecture on http://www.unizor.com

Logic+ 04

Problem A

A traveler has to stay in some city for 7 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.

However, a traveler has a silver chain of 7 links, and a restaurant owner agreed to take one chain link for each dinner.

What is the minimum number of chain links that must be cut open to be able to pay for each dinner separately?

Answer A
Only one link needs to be cut open, link #3.


Problem B

Same as Problem A, but the number of days is N, and the number of links in a chain is the same N.


Answer B

Assume, a traveler cuts open only K links (K must be small relatively to N).
Then he can pay for 1, 2, 3,..., K dinners using only these links.

For the next dinner he needs a segment of chain that contains K+1 links.
With this additional segment of chain he can pay for K+1, K+2,..., 2K+1 dinners.

For the next dinner he needs a segment of 2K+2 links.
Using K individual links that were cut, a segment of a chain with K+1 links and another segment with 2K+2 links a traveler pay for any number of dinners from 1 to K+(K+1)+(2K+2)=4K+3.

For the next dinner he needs a segment of 4K+4 links.
Using K individual links that were cut, a segment of a chain with K+1 links, a segment with 2K+2 links and a segment with 4K+4 links a traveler pay for any number of dinners from 1 to K+(K+1)+(2K+2)+(4K+4)=8K+7.

For the next dinner he needs a segment of 8K+8 links.
Using K individual links that were cut, a segment of a chain with K+1 links, a segment with 2K+2 links, a segment with 4K+4 links and a segment with 8K+8 links a traveler pay for any number of dinners from 1 to K+(K+1)+(2K+2)+(4K+4)+(8K+8)=16K+15.

Let's generalize it.
When K links were cut, it separated a chain into K+1 segments as follows.

Segment #1 should have
K+1 links in it.
Segment #2 should have
2K+2=2·(K+1) links in it.
Segment #3 should have
4K+4=4·(K+1) links in it.
Segment #4 should have
8K+8=8·(K+1) links in it.
Segment #5 should have
16K+16=16·(K+1) links in it.
etc.
The last segment #(K+1) should have 2^K·(K+1) links in it.

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.
K +
+ (K+1) +
+ 2·(K+1) +
+ 4·(K+1)+
+...+
+ 2^K·(K+1) =
= K + (2^K+1 −1)·(K+1) =
= (K+1)·2^K+1 − 1

If N=(K+1)·2^K+1−1 then the minimum number of links to cut is K.
If N is not such a number, we have to find K such that (K+1)·2^K+1−1 is greater than N. That K is the minimum number of links to cut.

Arithmetic+ 04: UNIZOR.COM - Math+ &Problems - Arithmetic

Notes to a video lecture on http://www.unizor.com

Arithmetic+ 04

Problem A
Consider two numbers:
X=11111111
(digit 1 repeats 8 times) and
Y=11111...11111
(digit 1 repeats 100 times).
What is their greatest common factor?

Hint A
Both numbers can be represented as a sum
10⁰+10¹+...+10^k
where k=7 for number X and k=99 for number Y.
From this it's easy to see that
Y = X·n + 1111·10⁹⁶
where n is some natural number.

Answer A
The greatest common factor of numbers X and Y is 1111.


Problem B
Prove 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.

Example B
Take, for instance, number 2024 and divide it by 9.
2024÷9 = 224 (8)
(remainder is 8)
Now with the sum this number's digits:
(2+0+2+4)÷9 = 8÷9 = 0(8)
(the same remainder 8)

The rule of divisibility by 9
Consequently, if the sum of digits of a natural number is divisible by 9, then the number itself is divisible by 9.

Hint B
Represent a number as a sum of its digits, each multiplied by a factor 10ⁿ, where n=0 for the right most digit, n=1 for the next digit to the left etc.


Problem C
Prove that if the sum of digits of some natural number N is the same as the sum of digits of the number 2·N then number N is divisible by 9.

Example C
For number 279 the sum of digits is 2+7+9=18.
Double this number:
2·279 = 558
Sum of digits of 558 is also 18.
The number 279 is indeed divisible by 9:
279÷9 = 31 (0)
(remainder is 0)

Hint C
Use the result of the Problem B above and consider the difference between 2·N and N.

Proof C
N÷9 = n (r)
(here n is a quotient and r is a remainder with the value from 0 to 8)
Therefore, N = 9·n + r
Analogously,
(2·N)÷9 = m (r)
(same remainder r)
2·N = 9·m + r
Subtract formula for N from formula for 2·N
2·N − N = N =
= (9·m + r) − (9·n + r) =
= 9·(m−n)
Hence
N = 9·(m−n)
that is, N is divisible by 9.

Note
The same result (divisibility by 9) can be obtained under weaker conditions.
The theorem can be stated as follows.
Prove that if the sum of digits of some natural number N divided by 9 gives the same remainder as the sum of digits of the number 2·N divided by 9, then number N is divisible by 9.
Obviously, that remainder in both cases will be zero. No other equal remainder can be encountered for sums of digits of N and 2·N divided by 9.

Example CC
For number 144 the sum of digits is 1+4+4=9.
Double this number:
2·144 = 288
Sum of digits of 288 is 18.
Sums of digits are different (though, both are divisible by 9), but the number 144 is indeed divisible by 9:
144÷9 = 16 (0)
(remainder is 0)
You can easily prove it or watch the proof in the lecture.

Logic+ 03: UNIZOR.COM - Math+ & Problems - Logic

Notes to a video lecture on http://www.unizor.com

Logic+ 03

Problem A

There are two cities:
CityOfTruth, where all people always tell the truth and
CityOfLies, where all people always lie.

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.

Right there he meets a person who, apparently, lives in one of these cities. So a traveler can ask him the direction to CityOfTruth.

The problem is, the person can live in either of these cities and nobody knows whether he tells the truth or lies.

Can a traveler ask this person a question in such a way that there will be no doubts about which way leads to CityOfTruth?


Answer A

The question might be: "Could you show me which way leads to a city where you live?"

If this person lives in CityOfTruth and always tells the truth, he will point to CityOfTruth.

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..

In any case, he will correctly point to a direction the travel wants to know.


Problem B

A man (M), a wolf (W), a goat (G) and a cabbage (C) have to cross a river from side A to side B.
There is a boat that can hold a man and either one of the others.

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.

How can they all safely cross the river from side A to side B, so everyone is alive and well at the end?


Answer B

1. A→(M+G)→B
2. B→(M)→A
3. A→(M+W)→B
4. B→(M+G)→A
5. A→(M+C)→B
6. B→(M)→A
7. A→(M+G)→B


Problem C

A person has a revolver designed for 6 bullets, but there is only one loaded and the chamber is spun.

He has two attempts to hit a target.
He shoots once, but no bullet comes out.

He wants to maximize his chances of making a real shot to hit a target.
What would be a better choice for him before the second shot, to spin or not to spin a chamber?


Answer C

Not to spin gives a one in five chances to shoot a bullet.
Spinning gives a one in six chances.
So, not to spin gives a better chance to make a real shot and, hopefully, hit a target.


Problem D

Two players, A and B, are playing against each other.
Each game results in one person winning and another losing.
The loser pays $1 to a winner.
Initially, both players came with $100 each.

At end it appears that player A won 10 games and player B ended with $120.
How many games did they play?

Answer D

They played 40 games.


Problem E

Three wise men (very smart indeed) after discussing some very important subject fell asleep.

Some foolish child was passing along and decided, as a joke, to put some black shoe wax on their foreheads.

All three wise men woke up at the same time and each of them, seeing black spots on two others' foreheads, started to laugh.

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.

What was his logic?


Answer E

Let's call the wise men WiseA (that's the one who stopped laughing first), WiseB and WiseC.

WiseA thinks as follows.

Assume, my forehead is clean.
Then WiseB and WiseC can see only spots on the foreheads of each other.

Considering WiseB is very smart, seeing that WiseC is laughing and seeing no black spot on my forehead, he would immediately realize that WiseC is laughing at him, because he also has a black spot on his forehead. Then WiseB would stop laughing.

Since WiseB still laughs, he must see a black spot on my forehead too. So, I better stop laughing and go to wash my face.

Algebra+ 03: UNIZOR.COM - Math+ & Problems - Algebra

Notes to a video lecture on http://www.unizor.com

Algebra+ 03

Problem A

Let n be any natural number/

Define two functions of n:
F(n) = 1 − 1/2 + 1/3 -…+
+ 1/(2n-1) − 1/2n
and
G(n) = 1/(n+1) + 1/(n+2) +…+
+ 1/(2n-1) + 1/2n

Prove that for any natural n the following equality is true:
F(n)=G(n).

For example, you can manually check the following equalities:
for n=1 (2n=2):
1 − 1/2 = 1/2
for n=2 (2n=4):
1 − 1/2 + 1/3 − 1/4 =
= 1/3 + 1/4
for n=3 (2n=6):
1− 1/2 + 1/3 − 1/4 +
+ 1/5 − 1/6 =
= 1/4 + 1/5 + 1/6


Answer A

Here is a straight forward proof by induction.

Check it for n=1:
F(1) = 1−1/2 = 1/2
G(1) = 1/2
As we see, F(1) = G(1)

Assume, F(n)=G(n).
Let's switch to n+1
F(n+1) =
= F(n) + 1/(2n+1) − 1/(2n+2)
G(n+1) = G(n) − 1/(n+1) +
+ 1/(2n+1)+1/(2n+2)

When we switch from n to n+1, function F(n) has changed by
f(n+1) = 1/(2n+1) − 1/(2n+2)
So, F(n+1)=F(n)+f(n+1).

Analogously, function G(n) has changed by
g(n+1) = −1/(n+1) +
+ 1/(2n+1) + 1/(2n+2)
So, G(n+1)=G(n)+g(n+1).

If f(n+1), an increment of F(n), and g(n+1), an increment of G(n), are equal, assuming F(n)=G(n), we can conclude that F(n+1)=G(n+1), which proves that F(n)=G(n) for any natural number n.

Let's check the equality
f(n+1) = g(n+1)

Indeed, transforming an expression to a common denominator, we get
f(n+1) = 1/[(2n+1)·(2n+2)]

For g(n+1) the common denominator is
(n+1)(2n+1)(2n+2)
The numerator is
−(2n+1)(2n+2)+(n+1)(2n+2)+
+(n+1)(2n+1) =
= −4n² − 2n − 4n − 2 +
+ 2n² + 2n + 2n + 2 +
+ 2n² + n + 2n + 1 =
= n+1

Dividing this numerator of g(n+1) by its denominator obtained above and cancelling n+1 from both, we will get
g(n+1) = 1/[(2n+1)(2n+2)]

This is exactly what we obtained for f(n+1).

END OF PROOF


Problem B

Two cars are approaching an intersection on two perpendicular roads.
In the beginning at time t=0 car #1 is at distance a from the intersection.
The car #2 (on a perpendicular road) at t=0 is at distance b from the intersection.
Both cars move with the same speed V.
For traffic safety, assume that a≠b, so there will be no collision.


At what time the distance between the cars will be minimal and what that minimal distance will be?


Answer B

Let d(t) be a distance between the cars at any moment t.
Then
d²(t) = (a−v·t)² + (b−v·t)²

Minimum of d(t) and d²(t) occur for the same moment in time t.
So, our task is to find minimum of d²(t), which is a quadratic function of t.

The time when the cars are closest to each other is
t_min = (a+b)/(2V)

The closest distance between the car at that time is
d_min = |a−b| /√2

Geometry+ 05: UNIZOR.COM - Math+ &Problems - Geometry

Notes to a video lecture on http://www.unizor.com

Geometry+ 05


Problem A

Construct a quadrilateral ABCD by its 4 sides AB, BC, CD, DA and an angle φ between opposite sides AB and CD.

Hint A

Find point P such that BP is parallel and congruent to CD.
Consider ΔABP.


Problem B

Given a circle of radius R and n-sided regular polygon inscribed into it.
Let P be any point on this circle.
Find a sum of squares of distances from this point P to all vertices of a polygon.

Hint B
(a) Geometrical solution for even number n of vertices of a regular polygon can be obtained by adding pairs of distances from P to i^th and (i+½n)^th vertices.
(b) General solution can be obtained if using vectors from the center of a circle to all its vertices and to point P.

Answer
Sum of squares of distances from point P to all vertices of a polygon equals to 2nR².


Problem C

Given an equilateral triangle ΔABC.
Extend side AC beyond vertex C to point D and build another equilateral triangle CDE with point E on the same side from AD as point B.
Connect points A and E. Let point M be a midpoint of segment AE.
Connect points B and D. Let point N be a midpoint of segment BD.

Prove that triangle ΔCMN is equilateral.

Hint C
Triangles ΔACE and ΔBCD are congruent.

GeoTheorem+ 1: UNIZOR.COM - Math+ & Problems - Geometry

Notes to a video lecture on http://www.unizor.com

Geometry+ GeoTheorem 1

Theorem

Prove that non-null vector n(a,b,c) in three-dimensional Cartesian coordinates OXYZ is normal (perpendicular) to a plane α described by an equation
a·x + b·y + c·z + d = 0
where a, b, c, d are real numbers.

Proof

CASE 1 (easy) - Constant d in an equation that describes plane α equals to zero.

The equation for plane α looks in this case as
a·x + b·y + c·z = 0

Then plane α must go through the origin of coordinates O(0,0,0) because this point satisfies the equation for α.

Consider a vector from an origin of coordinate O(0,0,0) to any other point on a plane Q(x,y,z).
Obviously, non-null vector OQ(x,y,z) is lying fully within plane α because both its ends - points O and Q lie within this plane.

We can interpret the equation
a·x + b·y + c·z = 0
that describes plane α in this case as a scalar product of non-null vector n(a,b,c) and non-null vector OQ(x,y,z).

Since this scalar product between vector n(a,b,c) and any vector OQ(x,y,z) lying within plane α is zero, vector n(a,b,c) is perpendicular to plane α.

CASE 2 - Constant d is not equal to zero.

Consider two planes defined by two equations
Plane α: a·x + b·y + c·z + d = 0
Plane β: a·x + b·y + c·z = 0

Since d ≠ 0, any point that satisfies one of these equations will not satisfy another.
Therefore, these planes do not have any common points and, therefore, are parallel.

We have already proven that non-null vector n(a,b,c) is perpendicular to plane β (see CASE 1 above).
Consequently, this vector n is perpendicular to α as well, that is n is normal to α.

End of the proof that n⊥α.