## Thursday, April 30, 2015

### Algebraic Method to Tackle the Mock APMO Problem

There exists another way to tackle the previously discussed AMPO mock problem (Asian Pacific Mathematics Olympiad Mock Problem ).

Find [MATH]\sum_{x=0}^{101}\dfrac{\dfrac{2x}{101}-1}{\dfrac{3x^2}{10201}-\dfrac{3x}{101}+1}[/MATH].

In case you are not well prepared to attack the problem analytically, you could still tackle it algebraically, that is purely allowable and no one will ever say algebraic method is not awesome!

For simplicity's sake, we let $x_i=\dfrac{i}{101}$ and $f(x)=\dfrac{\dfrac{2x}{101}-1}{\dfrac{3x^2}{10201}-\dfrac{3x}{101}+1}$, we then have:

### Asian Pacific Mathematics Olympiad Mock Problem

Find [MATH]\sum_{x=0}^{101}\frac{\frac{2x}{101}-1}{\frac{3x^2}{10201}-\frac{3x}{101}+1}[/MATH].

At first glance, this looks like a very demanding hard challenge, and you don't even know where to start working on the problem.

But this is necessarily a delicious problem in terms of how well we can milk it for all its worth to benefit students. In case  you might not already know, I am the advocate for teaching students and equip them with analytical thinking and critical problem solving skills, so, this problem can help me to achieve my aim.

## Wednesday, April 29, 2015

### Prove $\tan 55^{\circ}\tan 65^{\circ}\tan 75^{\circ}=\tan85^{\circ}$ (Method II)

$\tan 55^{\circ}\tan 65^{\circ}\tan 75^{\circ}$

$=\tan 55^{\circ}\tan 65^{\circ}\tan 75^{\circ}\cdot \dfrac{\tan85^{\circ}}{\tan85^{\circ}}$

$=\tan85^{\circ}\cdot \dfrac{\tan 55^{\circ}\tan 65^{\circ}\tan 75^{\circ}}{\tan85^{\circ}}$

$=\tan85^{\circ}\cdot \dfrac{\sin 55^{\circ}\sin 65^{\circ}\sin 75^{\circ}(\cos85^{\circ})}{\cos 55^{\circ}\cos 65^{\circ}\cos 75^{\circ}\sin85^{\circ}}$

$=\tan85^{\circ}\cdot \dfrac{(\sin 55^{\circ}\sin 65^{\circ})(\sin 75^{\circ}\cos85^{\circ})}{(\cos 55^{\circ}\cos 65^{\circ})(\cos 75^{\circ}\sin85^{\circ})}$

### Prove $\tan 55^{\circ}\tan 65^{\circ}\tan 75^{\circ}=\tan85^{\circ}$ (Method I)

Let's begin with the left side:

[MATH]\tan\left(55^{\circ} \right)\tan\left(65^{\circ} \right)\tan\left(75^{\circ} \right)[/MATH]

Using the product to sum identity for the tangent function:

[MATH]\tan(\alpha)\tan(\beta)=\frac{\cos(\alpha-\beta)-\cos(\alpha+\beta)}{\cos(\alpha-\beta)-\cos(\alpha+\beta)}[/MATH]

we may write:

[MATH]\tan\left(65^{\circ} \right)\tan\left(55^{\circ} \right)=\frac{\cos\left(10^{\circ} \right)-\cos\left(120^{\circ} \right)}{\cos\left(10^{\circ} \right)+\cos\left(120^{\circ} \right)}[/MATH]

## Tuesday, April 28, 2015

### Calculate $\displaystyle \lim_{x \to \infty} (\sin \sqrt{x+1}-\sin \sqrt{x})$

Calculate $\displaystyle \lim_{x \to \infty} (\sin \sqrt{x+1}-\sin \sqrt{x})$.

Let:

[MATH]L=\lim_{x\to\infty}\left(\sin\left(\sqrt{x+1} \right)-\sin\left(\sqrt{x} \right) \right)[/MATH]

Application of the sum-to-product identity:

## Monday, April 27, 2015

### Find all the possible values for: $1+a+a^2+\cdots+a^{2011}$

Let $x=a$ be a solution of the equation $x^{2012}-7x+6=0$. Find all the possible values for: $1+a+a^2+\cdots+a^{2011}$.

We're told $x=a$ is a solution of the equation $x^{2012}-7x+6=0$, therefore we have $a^{2012}-7a+6=0$.

It can be rewritten as

$a^{2012}-1-7a+6+1=0$

$a^{2012}-1-7a+7=0$

$(a^{2012}-1)-7(a-1)=0$

## Sunday, April 26, 2015

### Determine p, q, r, s if p+q+r+s=12, pqrs=27+pq+pr+ps+qr+qs+rs.

This is one very delicious problem that shows how inequality could come into the picture and help in solving system of equations.

Given that $p, q, r, s$ are all positive real numbers and they satisfy the system

$p+q+r+s=12$

$pqrs=27+pq+pr+ps+qr+qs+rs$

Determine $p, q, r$ and $s$.

Okay, our intuition tells us the logic doesn't follow for this problem because we all know very well solving a system of two nonlinear equations in four variables sounds pretty much like a mission impossible.

## Saturday, April 25, 2015

### Hard and Intriguing Indefinite Integral Problem (Second Method)

On this blog post, I will show another intelligent way to tackle the problem(Hard and Intriguing Indefinite Integral Problem ) that to compute the following indefinite integral:

[MATH]\int \dfrac{(3x^{10}+2x^8-2)\sqrt[4]{x^{10}+x^8+1}}{x^6} \,dx[/MATH]

If we rewrite the integrand such that we have:

### Putnam Optimization Hard Problem

Find the minimum of $|\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|$.

This is a horrible problem, you could have wasted an inordinate amount of your precious time finding how to, for example, express this given expression in terms of only one variable instead of the given six and then work from there because the methods that could be employed are not necessarily immediately obvious.  But that's life. If you want to acquire the skills of thinking flexibly and creatively, you need to solve complex and unfamiliar problems.

If you are a mathematics educator, then this kind of problem is what we should shower our students because we must acknowledge that the stronger performance of our students, the more it reflects our curricular emphasis and our schools’ efforts in developing problem solving skills in our students.

### Putnam Definite Integral Hard Problem: Evaluate $\displaystyle\int_0^{\frac{\pi}{2}}\frac{dx}{1+(\tan x)^{\sqrt{2}}}$.

Another hard definite integral but it is not as bad as it looks. Let's first try the substitution method:

Attempt I:

We are given to evaluate:

[MATH]I=\int_0^{\pi/2}\frac{1}{1+\left(\tan(x)\right)^{\sqrt{2}}}\,dx[/MATH]

If we use the substitution:

[MATH]u=\frac{\pi}{2}-x[/MATH]

### Hard Inequality Problem: Prove that $6<3^{\sqrt{3}}<7$

Prove that $6<3^{\sqrt{3}}<7$ from the simple and straightforward inequality where $1<\sqrt{3}<2$.

This is a particularly daunting mathematics challenge and you could spend days or even a week trying to prove the inequality, with no fruitful result.

Notice that if we exponentiate the given inequality $1<\sqrt{3}<2$ with base 3, we get:

$3^1<3^{\sqrt{3}}<3^2$

$3<3^{\sqrt{3}}<9$ (Compare it with the targeted inequality $6<3^{\sqrt{3}}<7$)

The lower and upper bound that we could get from the given inequality is far too low and high for the targeted inequality.

## Friday, April 24, 2015

### Hard and Intriguing Indefinite Integral Problem

Compute the following indefinite integral:

[MATH]\int \dfrac{(3x^{10}+2x^8-2)\sqrt[4]{x^{10}+x^8+1}}{x^6} \,dx[/MATH]

Oh boy! Good mathematical Olympiad or any contest problems are hard to come by and this one particularly helps to cultivate analytical thinking and this is also one completely unpredictable and not straightforward problem.

This problem is by all means designed specifically for students to apply the substitution method.

The hardest part is to decide which part of the integrand should be replaced by the substitution. But we have to try, diligently and persistently as we know perseverance will paid off in the end.

### Evaluate √((1000000)(1000001)(1000002)(1000003)+1)

Evaluate $\sqrt{(1000000)(1000001)(1000002)(1000003)+1}$ without the help of calculator.

At first glance, you might notice $1000000,\,1000001,\,1000002,\,1000003$ are the terms from arithmetic sequence and now, we have to compute the product of these four figures, but not to find the sum of the first four terms from that sequence.

But, let us look at it a lot more closely, we can actually now discern a pattern if we employ the heuristic strategy by using the substitution method which helps in solving this intriguing problem.

## Thursday, April 23, 2015

### Evaluate without calculator or logarithm help

On my previous blog post(Evaluate $12^{\frac{1-(x+y)}{2(1-y)}}$), we're asked, without using the calculator and the help from logarithm, evaluate $\large12^\frac{1-(x+y)}{2(1-y)}$ provided $3=60^x$ and $5=60^y$.

Let do this problem as we're told, where we could not borrow help from calculator nor logarithms.

If we multiply the two given exponential equations, we get:

$3\cdot 5=60^x\cdot 60^y$

$3\cdot 5=60^{x+y}$

$3\cdot 5=5^{x+y}\cdot 12^{x+y}$

$3=5^{x+y-1}\cdot 12^{x+y}(*)$

### Floor Function Problem...

Solve the following equation:

$\displaystyle \left\lfloor x+\frac{7}{3} \right\rfloor^2-\left\lfloor x-\frac{9}{4} \right\rfloor=16$

Note: $\displaystyle \lfloor x \rfloor$ denotes the largest integer not greater than $x$. This function, referred to as the floor function, is also called the greatest integer function, and its value at $x$ is called the integral part or integer part of $x$.

### Find The Area Of The Equilateral Triangle

Show that the curve $x^3+3xy+y^3=1$ has only one set of three distinct points, $P$, $Q$, and $R$ which are the vertices of an equilateral triangle, and find its area.

### Find The Sum Involving The Inverse Tangent Function

We are given to evaluate:

$S_n=\sum_{k=0}^n\left[\tan^{-1}\left(\frac{1}{k^2+k+1} \right) \right]$

### Find the value $12^\frac{1-(x+y)}{2(1-y)}$ given $3=60^x$ and $5=60^y$.

Without using the calculator and the help from logarithm, evaluate $\large12^\frac{1-(x+y)}{2(1-y)}$ if $3=60^x$ and $5=60^y$.

Aww, this problem looks easy peasy if we could solve it using the logarithm method as we can convert the exponential forms to logarithmic forms and get:

$\log_{60}3=x$ and $\log_{60}5=y$

## Wednesday, April 22, 2015

### Analytic Geometry: Orthogonal Trajectories

The problem is as follows:

a) Find the family of circles centered on the $y$-axis, that pass through the points $(\pm a,0)$, where $0<a\le r\in\mathbb{R}$.

b) Find the family of curves orthogonal to the family of circles found in part a).

Hint 1: Express the family of circles from part a) in the form:

$F(x,y)=C$

Hint 2: It can be shown that the orthogonal trajectories must satisfy:

$\frac{\partial F}{\partial y}dx-\frac{\partial F}{\partial x}dy=0$

### Probability: Coin Tossing

A while back I helped a student with a probability problem, and I took the problem, generalized it a bit, and wish to post it here. Here is the problem:

A coin has the probability $p$ of turning up heads when tossed. Suppose we toss the coin $2n$ times, where $n$ is a natural number. Compute the probability that the total number of heads is even.

### Trigonometric Inequality

Show that :

$\left( {\sin x + a\cos x} \right)\left( {\sin x + b\cos x} \right) \leq 1 + \left( \frac{a + b}{2} \right)^2$

### Simplify $\dfrac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125}}}$

This is also one good practice example that enlighten us we should keep an open-mindedness to solve for any given intriguing problem.

It'd be at first hard to foretell what would be the first step to start with this kind of problem. But, we know it won't be too much away to multiply the top and the bottom of the given fraction by some expression with the hope to simplify the nasty denominator as we have been given.

### Prove that $10152^8-10887^8+27195^8$ is divisible by $26460$.

The very first thing that we need to do is to factorize $26460$ as [MATH]\color{yellow}\bbox[5px,purple]{26460=2^2\cdot 3^3\cdot 5\cdot 7^2}[/MATH]. The next thing we have to accomplish is to show that $10152^8-10887^8+27195^8$ is divisible by $2^2\cdot 3^3\cdot 5\cdot 7^2$, either

1. all at once or
2. separately.

But, I am sure you will also work out the prime factors for the other three numbers

$10152=2^3\cdot 3^3\cdot 47$; $10887=3\cdot 19\cdot 191$; $27195=3\cdot 5\cdot 7^2\cdot 37$

### Prove $x^2+y^2< 1$ (Continued)

On this blog post, I will still discuss the problem how to prove $x^2+y^2< 1$ given $x,\,y>0$ and $x^3+y^3<x-y$.

Previously, I have mentioned about how important it is for us to look at what are given, assimilate it next  and jot down if we are given something that is/are implicitly correct.

## Monday, April 20, 2015

### Prove that $x^2+y^2<1$.

Let $x,\,y>0$ be such that $x^3+y^3<x-y$. Prove that $x^2+y^2≤1$.

We have to rate this as a 5 star intriguing Mathematical Olympiad Inequality problem because it is a special problem that is designed for the application of heuristic skills.

I am going to show some sublimely insightful approaches of intelligent people here and as always, I hope you will to learn something awesome today!

### Heuristic method for solving $\sin^6 x+\cos^6 x=\dfrac{1}{4}$

Heuristic method for solving $\sin^6 x+\cos^6 x=\dfrac{1}{4}$:

Method III:

Have you ever amazed and marveled by the powerful heuristic skill (the use of substitution, like we used in method I in previous post(Vietnamese Mathematical Olympiad (Trigonometric) Problem of 1962 )) that aimed to help in trick our mind (in a good way) to see the whole problem more clearly and so we could simplify things further to ease our work?

You would be even more staggered by what I am going to reveal to you on this blog post, I will show you how we could develop an already helpful and pretty heuristic thinking!

## Sunday, April 19, 2015

### Vietnamese Mathematical Olympiad (Trigonometric) Problem of 1962

Solve the equation $\sin^6 x+\cos^6 x=\dfrac{1}{4}$.

This is one of the brilliant Mathematics Olympiad Contest Problems because we can show to the students how there are plenty of ways to attacking a good problem and how one approach is different from the other and how heuristic skill enable us to find solution quickly that save us time for more challenging problems!

## Friday, April 17, 2015

### Singapore Mathematics Olympiad: Techniques of Counting Problem

In a bag there are 120 green marbles, 80 blue marbles, 80 yellow marbles and 20 purple marbles. All the marbles are identical except for the colors. If you close your eyes, and pick marbles from the bag, what is the minimum number of marbles that you must pick in order to be sure of having at least two marbles that are of the same color?

### Maximize The Area Of The Lune

In the $xy$ plane, there are two circles, a larger one of radius 1 unit and a smaller one of radius $r$. The two circle intersect such that the two points of intersection are on a diameter of the smaller circle. Find the value of $r$ which maximizes the area that is inside the smaller circle, but is outside the larger circle.

### Maximizing The Trajectory Of A Projectile

Suppose we are asked to compute the launch angle which will maximize the arc-length of the trajectory for a projectile, assuming gravity is constant and is the only force on the projectile after the launch.

## Thursday, April 16, 2015

Previously we asked you, the educators or yourself, if you are a student to think of way(s) to find the ratio of $\dfrac{PR}{QR}$ in this thread (Olympiad Trigonometric Problem), we will now lead you to a credible way to solve for that intriguing hard trigonometric problem.

We in fact, should have the eagle eye sight and should be able to tell offhand that we suspect $\angle P=\angle Q$.

## Wednesday, April 15, 2015

### When is Cheryl's birthday?

Question ((Singapore and Asian Schools Math Olympiad test 2015))

Albert and Bernard just become friends with Cheryl, and they want to know when her birthday is. Cheryl gives them a list of 10 possible dates.
May 15             May 16         May 19
June 17             June 18
July 14              July 16
August 14       August 15      August 17

Cheryl then tells Albert and Bernard separately the month and the day of her birthday respectively.

Albert:  I don't know when Cheryl's birthday is, but I know that Bernard does not know too.
Bernard:  At first I don't know when Cheryl's birthday is, but I know now.
Albert:  Then I also know when Cheryl's birthday is.

So, when is Cheryl's birthday?

## Tuesday, April 14, 2015

### Olympiad Trigonometric Problem $\cos^{48}\dfrac{P}{2}\sin^{23}\dfrac{Q}{2}=\cos^{48}\dfrac{Q}{2}\sin^{23}\dfrac{P}{2}$

Great and professional teaching methodology could definitely help to mold one's thinking and stimulate the inquisitive mind of students. In order to achieve this noble aim, we need good resources, such as the state of the art technologies, professional educators, competitive peers, supportive family members and above all, the challenging and intriguing math problems!

I will present one really intriguing problem here, and let's see how we are going to milk it for all its worth...

In a triangle $PQR$ the following equality holds:

$\cos^{48}\dfrac{P}{2}\sin^{23}\dfrac{Q}{2}=\cos^{48}\dfrac{Q}{2}\sin^{23}\dfrac{P}{2}$

Find the value of $\dfrac{PR}{QR}$.

## Monday, April 13, 2015

### $3^a+3^b+3^c = 7299$

Find the total number of positive integers ordered pairs of the equation $3^a+3^b+3^c = 7299$.

WLOG, let $c>b>a$.

Rewrite the RHS of the given equation as the product of two factors, we have:

$7299 = 3^a + 3^b + 3^c$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,=3^a\left(1 + \dfrac{3^b}{3^a} +\dfrac{3^c}{3^a}\right)$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,=3^a(1 + 3^{b-a} + 3^{c-a})$

## Saturday, April 11, 2015

### 16x²y²-48x²y+24xy²+100x²+16y²-72xy+150x-48y+100=28

Determine the pair(s) of real numbers $(a,\,b)$ that satisfy the equation $16x^2y^2-48x^2y+24xy^2+100x^2+16y^2-72xy+150x-48y+100=28$.

At first glance,  this seems like we need to use the modular arithmetic method to solve for $(x,\,y)$ but wait a minute! We need to find not the integers but real for $(x,\,y)$, so nope, modular arithmetic is out of the question...

## Friday, April 10, 2015

### $|x+|x|+m|+|x-|x|-m|=2$

Well, on my last blog post, I have shown with a strong Olympiad Math competition problem about how useful we have to always bear in mind that when $x$ is an solution to the function, says, $P(x)$, then we need to consider if $-x$ could also be a solution the that function of $P(x)$.

Fortunately that is not something that is hard to check. All that we need to do is to determine if $P(x)=P(-x)$.

I will show with one more challenging, hard and intriguing math competition problem why remembering to check with this fact could save our day from struggling with the problem.

Find all numbers of $m$ such that the equation $|x+|x|+m|+|x-|x|-m|=2$ has exactly three solutions.

### Evaluate $abcd$

Let $a, b, c, d$ be real numbers such that [MATH]a=\sqrt{4-\sqrt{5-a}}[/MATH], [MATH]b=\sqrt{4+\sqrt{5-b}}[/MATH], [MATH]c=\sqrt{4-\sqrt{5+c}}[/MATH] and [MATH]d=\sqrt{4+\sqrt{5+d}}[/MATH]. Calculate $abcd$.

This is one very intriguing problem because I am sure your instinct will tell you, No, please don't multiply all the given four expression out! That is a surest way to get into a big mess if you do that, and there got to be an easy way out for this problem.

## Thursday, April 9, 2015

### $\scriptsize \log_{24}(48)+\log_{12} (54)>4$ (Continued)

Have you tried out any other approaches that you could think of to solve the previous "$\log_{24}(48)+\log_{12} (54)>4$"?

At any rate, that is one very tough problem and it could cause you lose sleep over it if you're persistent enough and have the absolute determination to prove the statement. But, we know as it have been our experience that nothing would be more frustrating and aggravating that when we put up the titanic fight to solve for the problem, the attempts at the solution were never correct. After the complete failures, you began to wonder if the extra effort would hardly be worth it and if you should just ditch the problem in the first place once you realized there is nothing that you could have thought to tackle the problem successfully.

Yes, this seems like it would be a task of immense intelligence effort to prove for this logarithmic inequality where $\left(\log_{24}(48) \right)^2+ \left(\log_{12}(54) \right)^2 >4$.

We will lead you to the solution at this post and read on, as the solution will open your eyes:

## Wednesday, April 8, 2015

### Minimize The Crease

Consider a rectangular piece of paper of width $W$ laid on a flat surface. The lower left corner of the paper is bought over to the right edge of the paper, and the paper is smoothed flat creating a crease of length $L$, as in the diagram:

What is the minimal value of $L$ in terms of $W$?

### $\scriptsize \log_{24}(48)+\log_{12} (54)>4$

Prove that $\left(\log_{24}(48) \right)^2+ \left(\log_{12}(54) \right)^2 >4$.

For any logarithms problem (be it solving a logarithmic equation or proving a logarithmic inequality), one would usually use the change of base formula to rewrite the two logarithms terms so that they have the same base. Also, one would be sorely tempted to rewrite the LHS of the inequality by breaking down the figures 12, 24, 48 and 52 into the product of prime numbers:

[MATH]\color{yellow}\bbox[5px,green]{12=2^2\cdot 3}[/MATH], [MATH]\color{yellow}\bbox[5px,purple]{24=2^3\cdot 3}[/MATH], [MATH]\color{yellow}\bbox[5px,blue]{48=2^4\cdot 3}[/MATH] and [MATH]\color{yellow}\bbox[5px,orange]{54=2\cdot 3 ^3}[/MATH]

## Tuesday, April 7, 2015

### Fun With "Normal Lines"

The curve in the figures above is the parabola $y=x^2$. Let us define a normal line as a line whose first quadrant intersection with the parabola is perpendicular to the parabola. Five normal lines are shown in the figures above.

For a while, the $x$-coordinate of the second quadrant intersection of a normal line with the parabola gets smaller as the $x$-coordinate of the first quadrant intersection gets smaller. But eventually a normal line's second quadrant intersection gets as small as it can get.

The extreme normal line is shown as a thick red line in the figures above. Once the normal lines pass the extreme normal line, the $x$-coordinate of the second quadrant intersections with the parabola start to increase.

The figures above show two pairs of normal lines. The two normal lines of a pair have the same second quadrant intersection with the parabola, but one is above the extreme normal line (in the first quadrant) and the other is below it.

(a) Find the normal line pair associated with a particular second quadrant intersection with the parabola.

(b) Hence, or otherwise, find the equation of the extreme normal line.

(c) Find the normal line that traps the smallest area between it and the parabola.

### How to obtain the cubic polynomial from our previous problem?

In the previous problem's post ((2) Evaluate The Sum Of 1/xy+z-1+1/yz+x-1+1/xz+y-1 ), we said if we let $k=\dfrac{1}{\dfrac{4}{x}+x-1}$, we will then have $k^3+\dfrac{2}{9}k^2-\dfrac{2}{81}k-\dfrac{4}{81}= 0$.

In this post, we will reveal the logic and reason behind it and guide you step-by-step in achieving the other cubic polynomial where $k^3+\dfrac{2}{9}k^2-\dfrac{2}{81}k-\dfrac{4}{81}= 0$.

### Find The Unknown Function...

Let $f(x)$ be an unknown function defined on $[0,\infty)$ with $f(0)=0$ and $f(x)\le x^2$ for all $x$. For each $0\le t$, let $A_t$ be the area of the region bounded by $y=x^2$, $y=ax^2$ (where $1<a$) and $y=t^2$. Let $B_t$ be the area of the region bounded by $y=x^2$, $y=f(x)$ and $x=t$. See the image below:

a) Show that if $A_t=B_t$ for some time $t$, then:

[MATH]\int_0^{t^2}\sqrt{y}-\sqrt{\frac{y}{a}}\,dy=\int_0^t x^2-f(x)\,dx[/MATH]

b) Suppose $A_t=B_t$ for all $0\le t$. Find $f(x)$.

c) What is the largest value that $a$ can have so that $0\le f(x)$ for all $x$?

## Monday, April 6, 2015

### (2) Evaluate The Sum Of 1/xy+z-1+1/yz+x-1+1/xz+y-1

Regarding the problem to evaluate:

[MATH]\frac{1}{xy+z-1}+\frac{1}{yz+x-1}+\frac{1}{xz+y-1}[/MATH]

given:

[MATH]\begin{cases}x+y+z=2 \\[3pt] x^2+y^2+z^2=3 \\[3pt] xyz=4 \\ \end{cases}[/MATH]

you might be wondering if the proposed solution is the only way to stab at the problem, or more specifically, if you solved it differently, but in a more tedious or long way, can you be proud of it?

### A Trigonometric Sum

It can be shown that the following sum:

[MATH]S=\sum_{k=1}^{89}\left(\sin^6\left(k^{\circ}\right)\right)[/MATH]

is rational. Find the value of $S$.

## Sunday, April 5, 2015

### Evaluate The Sum Of 1/xy+z-1+1/yz+x-1+1/xz+y-1

The first problem I am going to post here is perfect for training/study and to stretch and inspire the most able and willing secondary school pupils and help to develop one's ability and harness enthusiasm for mathematics.

Given [MATH]x,\;y,\;z\in\mathbb{C}[/MATH]

where:

[MATH]\begin{cases}x+y+z=2 \\[3pt] x^2+y^2+z^2=3 \\[3pt] xyz=4 \\ \end{cases}[/MATH]

Calculate [MATH]\frac{1}{xy+z-1}+\frac{1}{yz+x-1}+\frac{1}{xz+y-1}[/MATH].