## Friday, July 31, 2015

### 21th USA Mathematical Olympiad 1992 Problem

In the (previous blog post), we were asked to prove a 21th USA Mathematical Olympiad 1992 Problem:

Let $k=1^{\circ}$, show that [MATH]\sum_{n=0}^{88}\dfrac{1}{\cos (nk) \cos(n+1)k}=\dfrac{\cos k}{\sin^2 k}[/MATH]

But we have already worked out an trigonometric identity where:

$\dfrac{\sin 1^{\circ}}{\sin x^{\circ} \sin (x^{\circ}+1^{\circ})}=\cot x^{\circ}-\cot (x^{\circ}+1^{\circ})$

## Thursday, July 30, 2015

### Analysis for Quiz 12: Brain Power Enrichment Quiz (I)

Analysis for Quiz 12: Brain Power Enrichment Quiz (I)

Question 1: If $a\gt b$ and $c\gt d$ are true, then we can say $a+c\gt b+d$ is always true.

A. Correct.
B. Incorrect.

Yes, that is always correct, regardless if $a,\,b,\,c,\,d$ are negative or positive real number.

Take for example, we have:

## Tuesday, July 28, 2015

### How to make use of 1 in helping us to solve International Math Olympiad problem?

Here is another blog post that describes how we could use the figure $1$ in solving intriguing International Olympiad Mathematical Problems.

According to wikipedia (number one (1)):

## Monday, July 27, 2015

### Algebraic Manipulation Skill II

Another algebraic manipulation skill that I want to share with students today:

Rewrite $a^2c^2+b^2c^2+a^2d^2+b^2d^2$ as the sum of two squares.

The vast majority students would say, hey! Isn't it obvious that we can factor out $a^2$ in $a^2c^2+a^2d^2$ and $b^2$ in $b^2c^2+b^2d^2$ such that we get:

## Saturday, July 25, 2015

### Learning Algebraic Manipulation Skills

Some algebraic manipulation skills that you could learn by watching me doing it and it is going to be very rewarding if you know when to utilize these skills and use them to your advantage.

I'm sure you're pretty familiar with factoring sum and difference of squares or even cubes and recognizing perfect squares at first glance, take for example, when you see

## Friday, July 24, 2015

### Analysis for Quiz 11: Creative Problem Solving

Analysis for Quiz 11: Creative Problem Solving

Question 1:  Do you think we can factor part of the expression in the given equation $4x^2+5xy-6y^2-4x+3y-1=0$?

A. Yes.
B. No.
C. Perhaps.

## Thursday, July 23, 2015

### Solve in positive integers the equation $ab(a+b-10)+21a-3a^2+16b-2b^2=60$: First Solution (Continued)

I want to apology for posting the unfinished blog post on last Saturday (solve-in-positive-integers-equation.html), and I am truly sorry for that.

Here is the other part of the solution that I unwittingly left out.

## Tuesday, July 21, 2015

### Potential Flaw in The Second Solution for the problem $ab(a+b-10)+21a-3a^2+16b-2b^2=60$.

Solve in positive integers the equation $ab(a+b-10)+21a-3a^2+16b-2b^2=60$.

In this blog post, I will discuss why the last step in my attempt to look for the positive integers $a$ and $b$ doesn't sound so right.

## Monday, July 20, 2015

### Solve in positive integers the equation $ab(a+b-10)+21a-3a^2+16b-2b^2=60$ Second Solution

Solve in positive integers the equation $ab(a+b-10)+21a-3a^2+16b-2b^2=60$.

Second solution:

This second solution is much neater than the first, but in solving mathematical problems, solving it is what matters, isn't it? :D

## Saturday, July 18, 2015

### Solve in positive integers the equation $ab(a+b-10)+21a-3a^2+16b-2b^2=60$: First Solution

Solve in positive integers the equation $ab(a+b-10)+21a-3a^2+16b-2b^2=60$.

## Friday, July 17, 2015

### Determine if $n^2-21n+111$ is or is not a perfect square (Why the second solution fails?)

Determine if $n^2-21n+111$ is or is not a perfect square.

Previously I wanted you to find the answer and explain back to me why the following solution cannot be deemed to be a sound solution.

## Thursday, July 16, 2015

### Determine if $n^2-21n+111$ is or is not a perfect square (Second Solution ?)

Determine if $n^2-21n+111$ is or is not a perfect square.

I want you to consider the following method of attacking the problem, and digest it, then think about it long enough so you have found the answer and explain back to me why it cannot deem to be a solution.

I first treat $n^2-21n+111$ as a square, says $m^2$ and I then rewrite $n^2-21n+111$ in the following fashion:

## Wednesday, July 15, 2015

### Determine if n²-21n+111 is or is not a perfect square.

Determine if $n^2-21n+111$ is or is not a perfect square.

## Tuesday, July 14, 2015

### Making decision by weighing the data

Solve for all 6 complex roots of the equation $x^6+10x^5+70x^4+288x^3+880x^2+1600x+1792=0$.

Previously we talked about how to assign $2^8(7)$ to be the constants for the three factors such that we have

$\small x^6+10x^5+70x^4+288x^3+880x^2+1600x+2^8(7)=(x^2+ax+?)(x^2+bx+?)(x^2+cx+?)$

We realized that we would get to correctly assign the number as the constants for the three factors if we have got lucky, what if we don't have the luck and have to exhaust all possibility?

As the math educator, I would ask you to stop solving for this problem for a short while and turn your focus to asking students the number of cases that are possible to rewrite $2^8(7)$ as the product of three positive integers.

Let your students to shift their mind from algebra to probability, this often generates a good course for students as in real life, we have to deal with the situation where we have to consider many aspects so to suggest good solutions than coming up with approach that would waste our precious time and effort hugely.

We could, list out all the possible cases and do it like follows:

For [MATH]\color{yellow}\bbox[5px,purple]{2^8(7)=1(7)(2^8)}[/MATH], we could split it down to such cases:

[MATH]\left.\begin{array}{I}\begin{align*}2^8(7)&=(2^1)(7)(2^7)\\&=(2^2)(7)(2^6)\\ &=(2^3)(7)(2^5)\\ &=\cancel{(2^5)(7)(2^3)}\end{align*}\end{array}\right\}\text{3 cases}[/MATH]

For [MATH]\color{yellow}\bbox[5px,green]{2^8(7)=1(7\cdot 2)(2^7)}[/MATH], we could split it down to such cases:

[MATH]\left.\begin{array}{I}\begin{align*}2^8(7)&=1(7\cdot 2)(2^7)\\&=(2^1)(7\cdot 2)(2^6)\\ &=(2^2)(7\cdot 2)(2^5)\\ &=(2^3)(7\cdot 2)(2^4)\\&=\cancel{(2^4)(7\cdot 2)(2^3)}\end{align*}\end{array}\right\}\text{4 cases}[/MATH]

For [MATH]\color{black}\bbox[5px,orange]{2^8(7)=1(7\cdot 2^2)(2^6)}[/MATH], we could split it down to such cases:

[MATH]\left.\begin{array}{I}\begin{align*}2^8(7)&=1(7\cdot 2^2)(2^6)\\&=(2^1)(7\cdot 2^2)(2^5)\\ &=(2^2)(7\cdot 2^2)(2^4)\\ &=(2^3)(7\cdot 2^2)(2^3)\\&=\cancel{(2^4)(7\cdot 2^2)(2^2)}\end{align*}\end{array}\right\}\text{4 cases}[/MATH]

For [MATH]\color{yellow}\bbox[5px,blue]{2^8(7)=1(7\cdot 2^3)(2^6)}[/MATH], we could split it down to such cases:

[MATH]\left.\begin{array}{I}\begin{align*}2^8(7)&=1(7\cdot 2^3)(2^5)\\&=(2^1)(7\cdot 2^3)(2^4)\\ &=(2^2)(7\cdot 2^3)(2^3)\\ &=(2^3)(7\cdot 2^3)(2^2)\\&=\cancel{(2^4)(7\cdot 2^3)(2^1)}\end{align*}\end{array}\right\}\text{4 cases}[/MATH]

For [MATH]\color{yellow}\bbox[5px,red]{2^8(7)=1(7\cdot 2^4)(2^4)}[/MATH], we could split it down to such cases:

[MATH]\left.\begin{array}{I}\begin{align*}2^8(7)&=1(7\cdot 2^4)(2^4)\\&=(2^1)(7\cdot 2^4)(2^3)\\ &=(2^2)(7\cdot 2^4)(2^2)\\ &=\cancel{(2^3)(7\cdot 2^4)(2^1)}\end{align*}\end{array}\right\}\text{3 cases}[/MATH]

For [MATH]\color{yellow}\bbox[5px,indigo]{2^8(7)=1(7\cdot 2^5)(2^3)}[/MATH], we could split it down to such cases:

[MATH]\left.\begin{array}{I}\begin{align*}2^8(7)&=1(7\cdot 2^5)(2^3)\\&=(2^1)(7\cdot 2^5)(2^2)\\ &=\cancel{(2^2)(7\cdot 2^5)(2^1)}\end{align*}\end{array}\right\}\text{2 cases}[/MATH]

For [MATH]\color{black}\bbox[5px,yellow]{2^8(7)=1(7\cdot 2^6)(2^2)}[/MATH], we could split it down to such cases:

[MATH]\left.\begin{array}{I}\begin{align*}2^8(7)&=1(7\cdot 2^6)(2^2)\\&=(2^1)(7\cdot 2^6)(2^1)\\ &=\cancel{(2^2)(7\cdot 2^5)(2^1)}\end{align*}\end{array}\right\}\text{2 cases}[/MATH]

For [MATH]\color{black}\bbox[5px,orange]{2^8(7)=1(7\cdot 2^7)(2)}[/MATH], we only have one case for that.

Similarly for [MATH]\color{white}\bbox[5px,purple]{2^8(7)=1(7\cdot 2^8)(1)}[/MATH], we only have one case for that.

So altogether, we would end up with a total of 26 possible cases to work.

Now, ask yourself by answer to yourself honestly, do you really want to work out all those 26 cases to look for the right answer to factor

$\small x^6+10x^5+70x^4+288x^3+880x^2+1600x+2^8(7)=(x^2+ax+?)(x^2+bx+?)(x^2+cx+?)$?

I'm fairly certain that you would beg for another alternative and put this one under the carpet by now. :D

## Monday, July 13, 2015

### Solve for all 6 complex roots of the equation $x^6+10x^5+70x^4+288x^3+880x^2+1600x+1792=0$ (Heuristic Solution)

Solve for all 6 complex roots of the equation $x^6+10x^5+70x^4+288x^3+880x^2+1600x+1792=0$.

My solution:

## Friday, July 10, 2015

### Solve for all 6 complex roots of the equation $x^6+10x^5+70x^4+288x^3+880x^2+1600x+1792=0$ (First attempt at solution)

Solve for all 6 complex roots of the equation $x^6+10x^5+70x^4+288x^3+880x^2+1600x+1792=0$.

The most common method that students would think of how to tackle this problem is by setting up three factors, toying around the possible candidates for the constants for each factor as the prime factorization of $1792=2^8(7)$.

## Thursday, July 9, 2015

### Prove that $ab\leq \dfrac {1}{8}$ (Third Solution)

Given $b^2-4ac$ is a real root of equation $ax^2+bx+c=0,\,\, (a\neq 0)$. Prove that $ab\leq \dfrac {1}{8}$.

Third solution:

## Wednesday, July 8, 2015

### Prove that $ab\leq \dfrac {1}{8}$ (Second Solution)

Given $b^2-4ac$ is a real root of equation :$ax^2+bx+c=0,\,\, (a\neq 0)$. Prove that $ab\leq \dfrac {1}{8}$.

Second solution is my solution: (I want to mention it here that the first solution is provided by a Taiwan friend of mine)

Multiply the equation of $ax^2+bx+c=0$ by $4a$, we get:

## Tuesday, July 7, 2015

### Prove that $ab\leq \dfrac {1}{8}$ (First Solution)

Given $b^2-4ac$ is a real root of equation :$ax^2+bx+c=0,\,\, (a\neq 0)$. Prove that $ab\leq \dfrac {1}{8}$.

Some students would find that this problem very confusing, as they have been told countless time that $b^2-4ac$  is actually a discriminant for the quadratic equation $ax^2+bx+c=0$.

In this instance, $b^2-4ac$ is both the discriminant and the root, they would think instantly something that looks like the following:

## Monday, July 6, 2015

### Fill up grids such that $\dfrac{\square}{\square \square}+\dfrac{\square}{\square \square} +\dfrac{\square}{\square\square}=1$

In the following grid, fill up the numbers from 1 to 9 (without repetition) such that their sum is $1$.

$\dfrac{\square}{\square \square}+\dfrac{\square}{\square \square} +\dfrac{\square}{\square\square}=1$

Solution provided by Mark, another contributor of this blog:

## Friday, July 3, 2015

### Analysis for Quiz 10: Training For Problem Solving Skills

The questions  I want to give something to ponder, something that is not immediately straightforward, and some cool and rational thought will be required to answer the question.

Admittedly, I've touched a modicum bit in this regard at this slideshow (http://masteringolympiadmathematics.blogspot.com/2015/06/slideshow-9-creative-teaching.html), you could, if you want to refer it and answer the following quiz questions.

Question 1: What is the next square number after $(x-6)^4$?

$((x-6)^2)^2+1$
$((x-6+1)^2)^2$
$((x-6)^2+1)^2$

## Wednesday, July 1, 2015

### Optimization Contest Problem: Find the maximum of f(x) (Heuristic Solution)

Find the maximum of $f(x) = \dfrac{x^4-x^2}{x^6+2x^3-1}$ where $x>1$.

If we want to maximize $f(x)=\dfrac{x^4-x^2}{x^6+2x^3-1}=\dfrac{1}{\left(\dfrac{x^6+2x^3-1}{x^4-x^2}\right)}$, this could be done if we are to find the minimum value for the expression $\dfrac{x^6+2x^3-1}{x^4-x^2}$.

Note that