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:


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 $\sin B$.


If $A,\,B$ are acute angles and if we have the equation

$\sin B=\dfrac{p}{q}\sin (2A+B)$, then

$\sin B \le \dfrac{p}{q}$ must hold.


The compound angle and product-to-sum formulas:

$\cos x\cos y-\sin x \sin y=\cos (x+y)$

$\cos x \sin y=\dfrac{\sin (x+y)-\sin(x-y)}{2}$

Okay, the above three are the important facts that would give us all that we need to solve this problem elegantly and easily:

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

$\dfrac{\sin B}{\cos B}=2015\sin A \cos A-2015\sin^2 A \left(\dfrac{\sin B}{\cos B}\right)$

$\begin{align*}\sin B&=2015\sin A \cos A\cos B-2015\sin^2 A\sin B\\&=2015\sin A( \cos A\cos B-\sin A\sin B)\\&=2015\sin A \cos (A+B)\\&=2015\left(\dfrac{\sin (2A+B)-\sin B}{2}\right)\end{align*}$

$\therefore \left(1+\dfrac{2015}{2}\right)\sin B=\dfrac{2015}{2}\sin (2A+B)$

And we get

$\sin B=\dfrac{2015}{2017}\sin (2A+B)$

Since $\sin B\le \dfrac{2015}{2017}$, that implies $\tan B\le \dfrac{2015}{\sqrt{2017^2-2015^2}}=\dfrac{2015}{24\sqrt{14}}$.

That is to say, the maximal $\tan B=\dfrac{2015}{24\sqrt{14}}$.

Equality is attained when $\sin (2A+B)=1$, that is when $2A+B$ is a right angle.

I am fairly certain that you will agree with me that this is an elegant and pretty approach!


  1. Yes, Isabelle has a knack for explaining things. :)

  2. Hi N Shan, thank you for your comment!

    Thanks MarkFL for your compliment! :D