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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 x_{1} = π/3 + π·(M+N) y_{1} = π/3 + π·(M−N)
and the second series of solutions x_{2} = −π/3 + π·(M+N) y_{2} = −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.

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.

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 ≤ α_{1} ≤ α_{2} ≤ ...≤ α_{n} ≤ ½π.

Let A =

Σ_{i∈[1,n]}sin(α_{i})

Σ_{i∈[1,n]}cos(α_{i})

Prove: tan(α_{1}) ≤ 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)

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 qhave 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}·x^{2001−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).

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 X_{1} and X_{2} are the solution of the equation, then, according to the Vieta's Theorem, X_{1} + X_{2} = −(−p) = p and X_{1} · X_{2} = 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=X_{1},y=X_{2}) or (x=X_{2},y=X_{1}).

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·x^{4} + 8·y^{4}

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 y^{4} 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 y^{4} 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 x^{4} ≤ 8·x^{4}.

After dividing by y^{4} and substituting t=x/y the new inequality looks like (t + 1)^{4} ≤ 8t^{4} + 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 7t^{4} − 4t^{3} − 6t^{2} − 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. 7t^{4} − 4t^{3} − 6t^{2} − 4t + 7 =
= (t − 1)·(7t^{3} + 3t^{2} − 3t − 7)

Consider the polynomial 7t^{3} + 3t^{2} − 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 7t^{3} + 3t^{2} − 3t − 7 =
= (t − 1)·(7t^{2} + 10t + 7)

So, the inequality we have to prove was transformed into this one: (t − 1)^{2}·(7t^{2} + 10t + 7) ≥ 0
In this inequality the member (t−1)^{2} is always greater or equal to zero.
Quadratic polynomial 7t^{2}+10t+7 has discriminant Δ=10^{2}−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 t^{2} 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}·(7t^{2} + 10t + 7) ≥ 0
and transform it into (t + 1)^{4} ≤ 8t^{4} + 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 y^{4} finishes the proof.

Problem C

Prove the following inequality x^{12} − x^{9} + x^{4} − x + 1 > 0

Proof C

Consider a polynomial x^{12} − x^{9} + x^{4} − x
It can be invariantly transformed into x^{9}·(x^{3} − 1) + x·(x^{3} − 1) or (x^{3} − 1)·(x^{9} + x) or (x^{3} − 1)·x·(x^{8} + 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) x^{12} − x^{9} + x^{4} − x + 1 =
= x^{12} + x^{4}·(1−x^{5}) + (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.

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.