## 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:

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

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

$\tan B=\dfrac{2015\sin A \cos A}{1+2015\sin^2 A}$(*)

Using the double angle formulas for both sine and cosine functions that says:

$\sin 2A=2\sin A \cos A$; $\cos 2A=1-2\sin^2 A$

We turn the equation in (*) into:

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

Simplifying gives

$\tan B=\dfrac{2015\sin 2A}{2017-2015\cos 2A}$

We're asked to look for the maximal of $\tan B$, that is to say, the maximum value of the RHS of the equation above.

Let $y=\tan B$ such that we have:

$y=\dfrac{2015\sin 2A}{2017-2015\cos 2A}$

We adopt the calculus (differentiation method) to look for the maximum point for the function of $y$.

Let us find the first derivative of $y$, we have:

\begin{align*}y'&=\dfrac{(2017-2015\cos 2A)(2(2015)\cos 2A)-(2015\sin 2A)(2(2015)\sin 2A)}{(2017-2015\cos 2A)^2}\\&=\dfrac{2(2015)(2017\cos 2A-2015(\sin^2 2A+\cos^2 2A))}{(2017-2015\cos 2A)^2}\\&=\dfrac{2(2015)(2017\cos 2A-2015)}{(2017-2015\cos 2A)^2}\end{align*}

The stationary point(s) occur at $y'=0$, i.e. when:

$2017\cos 2A-2015=0$, which implies [MATH]\color{yellow}\bbox[5px,purple]{\cos A=\dfrac{2015}{2017}}[/MATH] (Or $A\approx 0.044536\,\,\text{rad}$) and [MATH]\color{yellow}\bbox[5px,red]{\sin A=\dfrac{\sqrt{2017^2-2015^2}}{2017}}[/MATH].

We need to determine if this value will lead to a maximum, minimum or inflexion point.

At maximum points, the gradient is positive just before the maximum, it is zero at the maximum and it is negative just after the maximum.

$A= 0.04452\,\,\text{rad}$ yields $y'\gt0$;

$A= 0.04454\,\,\text{rad}$ yields $y'\lt0$;

Therefore, we get a maximum point at [MATH]\color{yellow}\bbox[5px,purple]{\cos A=\dfrac{2015}{2017}}[/MATH] and [MATH]\color{yellow}\bbox[5px,red]{\sin A=\dfrac{\sqrt{2017^2-2015^2}}{2017}}[/MATH]. and hence this gives the maximal $\tan B$ as:

\begin{align*}(\tan B)_{\text{max}}&=\dfrac{2015\sin 2A}{2017-2015\cos 2A}\\&=\dfrac{2015\left(\dfrac{\sqrt{2017^2-2015^2}}{2017}\right)}{2017-2015\left(\dfrac{2015}{2017}\right)}\\&=\dfrac{2015\sqrt{2(4032)}}{2(4032)}\\&=\dfrac{2015}{\sqrt{2(4032)}}\\&=\dfrac{2015}{24\sqrt{14}}\end{align*}

and we're hence done.

The other approach, the more ingenious and clever strategy (provided by the smart solver) has successfully avoided the calculus route to determine the maximal of $\tan B$.