Tuesday, June 30, 2015

Optimization Contest Problem: Find the maximum of f(x): Why The Second Solution Fails?

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

Previously, we discussed how to avoid using the surest and safest way of approaching this particular problem with the calculus method and opted for the inequality method.

To recap, here is what we have gotten:

Monday, June 29, 2015

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

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

In my previous post(optimization-contest-problem), I showed you the way of approaching this particular problem with the most comfortable and familiar method (calculus method) that it would spring first to students' mind, a method that students' from all four corners of the world are inclined to use.

Sunday, June 28, 2015

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

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

This is an excellent problem that could help in effective questioning so to get students fully understand some particular mathematics concepts and how to resolve optimization problem beautifully yet effectively.

But one may be tempted to solve it using the calculus method, as that is what teachers taught them what to do whenever we are to ask to find the maximum of a function, yes, calculus method works wonder and I will let you to decide, after showing you ways of approaching this problem so that you know how much this problem means as the best quality problem that enable us to activate and connect various areas in our brain and therefore think creatively.

Thursday, June 25, 2015

USA Mathematical Olympiad 1989 Problem (Third Method)

USA Mathematical Olympiad 1989 Problem (Third Method)

Which is larger, the real root of $x + x^2 + ... + x^8 = 8 - 10x^9$, or the real root of $x + x^2 + ... + x^{10} = 8 - 10x^{11}$?

My solution:

First, notice that

$1-x^9=(1-x)(x+x^2+\cdots+x^8)$

$1-x^9=(1-x)(8-10x^9)$

$1-x^9=8-10x^9-8x+10x^{10}$

$10x^{10}-9x^9-8x+7=0$

Whereas

$1-x^{11}=(1-x)(x+x^2+\cdots+x^{10})$

$1-x^{11}=(1-x)(8-10x^{11})$

$1-x^{11}=8-10x^{11}-8x+10x^{12}$

$10x^{12}-9x^{11}-8x+7=0$

So I let

$f(x)=10x^{10}-9x^9-8x+7$ and $g(x)=10x^{12}-9x^{11}-8x+7$.

Descartes's Rule says $f(x)$ has at most two positive real roots and obviously $x=1$ is one of the root of $f(x)$ and the Intermediate Value theorem tells us the other root lies between $(0, 0.95)$ since $f(0)\cdot f(0.95)=7(-0.28)\lt 0$.

Similarly, Descartes's Rule says $g(x)$ has two positive real roots and obviously $x=1$ is one of the root of $g(x)$ and the Intermediate Value theorem tells us the other root lies between $(0, 0.95)$ since $g(0)\cdot g(0.95)=7(-0.316)\lt 0$.

Observe also that

\begin{align*}g(x)&=10x^{12}-9x^{11}-8x+7\\&=x^2(10x^{10}-9x^9-8x+7)-8x+7+8x^3-7x^2\\&=x^2f(x)+(8x-7)(x^2-1)\end{align*}

If $a$ is a root of the function of $f$, that means $f(a)=0$, then we have:

\begin{align*}g(a)&=a^2f(a)+(8a-7)(a^2-1)\\&=0+(8a-7)(a^2-1)\\&=(8a-7)(a^2-1)\end{align*}

And here is a rough sketch of the graph $g(a)=(8a-7)(a^2-1)$

So, if $\dfrac{7}{8}\lt a\lt 1$, then $g(a)\lt 0$ and it follows that the real root of $x + x^2 + ... + x^8 = 8 - 10x^9$ is greater than the real root of $x + x^2 + ... + x^{10} = 8 - 10x^{11}$.

Whereas if $0\lt a\lt \dfrac{7}{8}$, then $g(a)\gt 0$, that will suggest the real root of $x + x^2 + ... + x^8 = 8 - 10x^9$ is less than the real root of $x + x^2 + ... + x^{10} = 8 - 10x^{11}$.

Now, our objective is to find out whether the root of the function of $f$ is greater than or less than $\dfrac{7}{8}$ so we can determine if the real root of $f(x)$ or $g(x)$ is greater.

$f(\dfrac{7}{8})=10(\dfrac{7}{8})^{10}-9(\dfrac{7}{8})^9-8(\dfrac{7}{8})+7=-0.075$

$f(\dfrac{6}{8})=10(\dfrac{6}{8})^{10}-9(\dfrac{6}{8})^9-8(\dfrac{6}{8})+7=0.8873$

Hence, we can say, based on the Intermediate Value theorem that the other positive real root of $f$, i.e. $a$, lies between $\dfrac{6}{8}$ and $\dfrac{7}{8}$, i.e. $a$ is less than $\dfrac{7}{8}$.

Therefore, we have proved that the real root of $x + x^2 + ... + x^8 = 8 - 10x^9$ is less than the real root of $x + x^2 + ... + x^{10} = 8 - 10x^{11}$.

Wednesday, June 24, 2015

USA Mathematical Olympiad 1989 Problem (Second Method)

Which is larger, the real root of $x + x^2 + ... + x^8 = 8 - 10x^9$, or the real root of $x + x^2 + ... + x^{10} = 8 - 10x^{11}$?

Previously we checked for the maximum and minimum points to ascertain if the first derivative of $f(x)=x+x^2+\cdots +x^8 +10x^9 - 8$ and $g(x)=x+x^2+\cdots +x^{10}+10x^{11}-8$ are always greater than or equal to zero.

Tuesday, June 23, 2015

Which is larger, the real root of $x + x^2 + ... + x^8 = 8 - 10x^9$, or the real root of $x + x^2 + ... + x^{10} = 8 - 10x^{11}$?

The upper part of the solution below is mine and the lower part belongs to the math professor from Arizona, USA.

Saturday, June 20, 2015

IMO contest problem (Second solution) Prove that $a^2+b^4=1994$

Let $a,\,b$ be positive integers with $b\gt3$ and $a^2+b^4=2((a-6)^2+(b+1)^2)$.

Prove that $a^2+b^4=1994$.

The solution below is provided by the retired math professor from the U.K. In my opinion, this solution is more easy to follow than the previously discussed, but credit must be given to both as they are equally insightful and neat solutions.

Friday, June 19, 2015

Let $a,\,b$ be positive integers with $b>3$ and $a^2+b^4=2((a-6)^2+(b+1)^2)$. Prove that $a^2+b^4=1994$.

Let $a,\,b$ be positive integers with $b>3$ and $a^2+b^4=2((a-6)^2+(b+1)^2)$.

Prove that $a^2+b^4=1994$.

This is a genuinely hard problem. If you are smart, you will wonder and ask your teacher if this question is saying there is only one solution for $(a,\,b)$ such that there is only a value for $a^2+b^4$, and that must be $1994$.

Yes, your intuition is right, the question is, how to find that only solution for both $a$ and $b$ from the only given single equation?

Thursday, June 18, 2015

Food for thought for both teachers and students

Food for thought:

It's rare that teachers will give something that are truly questioning and cultivating students' interest in math and science subjects. If math educators shift their teaching methodology towards more inquiry based teaching, and let students to take part by start thinking, improve their ability to think abstractly and multi-dimensionally, this will build strong foundation of math for students. This strong foundation will enable students to acquire higher-order skills needed for the 21st century and to becoming highly competent and extraordinary intelligent students that could always plan ahead, see the future consequences of an action, provide alternative explanation of events and reason more effectively.

Wednesday, June 17, 2015

Smartest Way of Solving Hardest Trigonometric Equation

Solve the trigonometric equation.

$\sqrt{2} \cos \left(\dfrac{x}{5}-\dfrac{\pi }{12}\right)-\sqrt{6}\sin \left(\dfrac{x}{5}-\dfrac{\pi}{12}\right)=2\left(\sin \left((\dfrac{x}{5}-\dfrac{2\pi}{3}\right)-\sin \left((\dfrac{3x}{5}+\dfrac{\pi}{6}\right)\right)$

An excellent observation and hence sublimely insightful solution provided by one mathematicans from the England:

Tuesday, June 16, 2015

Hardest Trigonometric Equation (First Solution)

Solve the trigonometric equation.

$\small \sqrt{2} \cos \left(\dfrac{x}{5}-\dfrac{\pi }{12}\right)-\sqrt{6}\sin \left(\dfrac{x}{5}-\dfrac{\pi}{12}\right)=2\left(\sin \left(\dfrac{x}{5}-\dfrac{2\pi}{3}\right)-\sin \left(\dfrac{3x}{5}+\dfrac{\pi}{6}\right)\right)$

My solution:

By letting [MATH]A=\frac{x}{5}-\frac{\pi}{12}[/MATH] use the sum-to-product formula to simplify the LHS of the equation, I get:

Monday, June 15, 2015

Hardest Trigonometric Equation (2)

In today post, I will take up where we left off (effective-teaching-of-math), to continue solving the trigonometric equation

$\sqrt{2} \cos \left(\dfrac{x}{5}-\dfrac{\pi }{12}\right)-\sqrt{6}\sin \left(\dfrac{x}{5}-\dfrac{\pi}{12}\right)=2\left(\sin \left((\dfrac{x}{5}-\dfrac{2\pi}{3}\right)-\sin \left(\dfrac{3x}{5}+\dfrac{\pi}{6}\right)\right)$

The LHS of the equation has been simplified down to

Sunday, June 14, 2015

Effective Teaching of Math Using Hard Trigonometry Contest Problem

Many of the so-called practices/drills that are designed to achieve the learning for mastery are just near aimless activities, and genuine illumination of important mathematical ideas is rare, not to mention providing the opportunities of making students to think deeply, creatively and critically in generating solutions. There is a near obsession with calculators/ICT which preventing students from gaining a true mastery of basic skills. Overall, these practice problems are watered-down and less interesting math activities.

Saturday, June 13, 2015

Given $x,\,y,\,z$ are real such that $2x+y+z+14=2\sqrt{2x}+4\sqrt{y+1}+6\sqrt{z-1}$. Evaluate $\dfrac{x-y}{z}$.

Given $x,\,y,\,z$ are real such that $2x+y+z+14=2\sqrt{2x}+4\sqrt{y+1}+6\sqrt{z-1}$.

Evaluate $\dfrac{x-y}{z}$.

My solution:

This IMO Problem should look easy to you if you're being careful and look at the problem with your heart and mind, not merely with the eyes.

For me, I will try to group

Friday, June 12, 2015

Work backwards Putnam IMO Contest Problem

Putnam Contest Problem:

Evaluate [MATH]\int_{0}^{\dfrac{\pi}{2}} \dfrac{\cos^4 x+\sin x\cos^3x+\sin^2 x\cos^2 x+\sin^3x\cos x}{\sin^4 x+\cos^4 x+2\sin x\cos^3 x+2\sin^2 x\cos^2 x+2\sin^3 x\cos x}\,dx[/MATH].

Solution of mine, which is different from the previously posted method provided by Mark(putnam-contest-problem):

Wednesday, June 10, 2015

Creative Solution for IMO Trigonometry Problem

Let $A,\,B$ be acute angles such that $\tan B=2015\sin A \cos A-2015\sin^2 A \tan B$.

Find the greatest possible value of $\tan B$.

This blog post is to highlight the fact that if we're creative enough, we can avoid the tedious calculus method to look for the maximal of $\tan B$.

You have to be aware of a few things as well:

1.

When $B$ is an acute angle and if $\sin B\le \dfrac{m}{n}$, then  $\tan B\le \dfrac{m}{\sqrt{n^2-m^2}}$ must be true.

In other words, we obtain the maximal of $\tan B$ if we have obtained the maximal of $\sin B$.

Tuesday, June 9, 2015

IMO (Hong Kong) Trigonometric Problem (Modified)

Let $A,\,B$ be acute angles such that $\tan B=2015\sin A \cos A-2015\sin^2 A \tan B$.

Find the greatest possible value of $\tan B$.

This is a fun IMO problem, since it has many ways (all are nothing less than remarkable) to approach it and without any further ado, I will post with the first approach here:

$\tan B=2015\sin A \cos A-2015\sin^2 A \tan B$

$\tan B(1+2015\sin^2 A )=2015\sin A \cos A$

Monday, June 8, 2015

Another method to solve the summation series from Quiz 9

a
The summation of series from quiz 9 (quiz-9-mock-imo-algebra-contest) can be tackled using another method.

First, note that $n^2-2n+1,\,n^2-1,\,n^2+2n+1$ all can be factored as:

$n^2-2n+1=(n-1)^2$

$n^2-1=(n-1)(n+1)$

$n^2+2n+1=(n+1)^2$

4th June 2015's Edexcel exam paper: The Probability of Hannah eating two orange sweets

The probability problem in 4th June 2015's Edexcel exam paper by exam board Edexcel has left students feeling glum and helpless.

This is the problem in full:

There are $n$ sweets in a bag. $6$ of the sweets are orange. The rest of the sweets are yellow. Hannah takes at random a sweet from the bag. She eats the sweet. Hannah then takes at random another sweet from the bag. She eats the sweet. The probability that Hannah eats two orange sweets is $\dfrac{1}{3}$.

a. Show that $n^2-n-90=0$.

b. Solve $n^2-n-90=0$ to find the value of $n$.

Here is how to tackle the problem using the tree diagram.

Sunday, June 7, 2015

Analysis: Quiz 9 Mock IMO Algebra Contest

Sum the series below:

[MATH]\sum_{n=4}^{999}\dfrac{1}{\sqrt[3]{n^2-2n+1}+\sqrt[3]{n^2+2n+1}+\sqrt[3]{n^2-1}}.[/MATH]

Question 1: Do you think the expression on the denominator can be factored?
Yes.

No.

This is a really rich way to asking problem so we make the students to think deep before simply giving out the answer yes or no so easily.

Friday, June 5, 2015

Third Method of Solving IMO Optimization Contest Problem: Find the minimum value of $xy$,

Third Method of Solving IMO Optimization Contest Problem:

Find the minimum value of $xy$, given that $x^2+y^2+z^2=7$, $xy+xz+yz=4$, and $x, y$ and $z$ are real numbers.

This third method is provided by Mark, another blog contributor and he approached the problem using the well-known Lagrange multipliers method:

Thursday, June 4, 2015

Second Method of Solving IMO Optimization Contest Problem: Find minimum $xy$

Second Method of Solving IMO Optimization Contest Problem:

Find the minimum value of $xy$, given that $x^2+y^2+z^2=7$, $xy+xz+yz=4$, and $x, y$ and $z$ are real numbers.

The second method is the solution provided by a well-known mathematics retired professor from the University in the U.K.

Wednesday, June 3, 2015

IMO Optimization Contest Problem: Find the minimum value of $xy$, given that $x^2+y^2+z^2=7$, $xy+xz+yz=4$, and $x, y$ and $z$ are real numbers.

IMO Optimization Contest Problem:

Find the minimum value of $xy$, given that $x^2+y^2+z^2=7$, $xy+xz+yz=4$, and $x, y$ and $z$ are real numbers.

My solution:

From the well-known identity

$(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)$

and the given values for $x^2+y^2+z^2=7$ and $xy+xz+yz=4$, we get:

$(x+y+z)^2=7+2(4)$

Tuesday, June 2, 2015

Putnam Contest Problem: Evaluate [MATH]\tiny \int_{0}^{\dfrac{\pi}{2}} \dfrac{\cos^4 x+\sin x\cos^3x+\sin^2 x\cos^2 x+\sin^3x\cos x}{\sin^4 x+\cos^4 x+2\sin x\cos^3 x+2\sin^2 x\cos^2 x+2\sin^3 x\cos x}\,dx[/MATH].

Putnam Contest Problem:

Evaluate [MATH] \int_{0}^{\dfrac{\pi}{2}} \dfrac{\cos^4 x+\sin x\cos^3x+\sin^2 x\cos^2 x+\sin^3x\cos x}{\sin^4 x+\cos^4 x+2\sin x\cos^3 x+2\sin^2 x\cos^2 x+2\sin^3 x\cos x}\,dx[/MATH].

Solution provided by Mark, another contributor of this blog.

Let:

[MATH]\small I= \int_{0}^{\frac{\pi}{2}} \frac{\cos^4(x)+\sin(x)\cos^3(x)+\sin^2(x)\cos^2(x)+\sin^3(x)\cos(x)}{\sin^4(x)+\cos^4(x)+2\sin(x)\cos^3 (x)+2\sin^2(x)\cos^2(x)+2\sin^3(x)\cos(x)}\,dx-(1)[/MATH]

Using the identity:

Monday, June 1, 2015

China IMO problem: Simplify $(ay+bx)^3-(ax+by)^3+(a^3-b^3)(x^3-y^3)$

Simplify $(ay+bx)^3-(ax+by)^3+(a^3-b^3)(x^3-y^3)$.

We know this is a trick problem because it couldn't be required us to expand all the three terms and then collect like terms to simplify it.

So, if we don't expand the expression, how are we going to simplify it?