MHB Find the greatest possible value of tan B.

[anemone]

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 $\tan B$.

 

[Opalg]

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 $\tan B$.

[sp]Let $n = 2015.$ Then $\tan B(1 + n\sin^2 A) = n\sin A\cos A$. Therefore $\tan B = \dfrac{n\sin A\cos A}{1 + n\sin^2 A}$ and so $$\cot B = \frac{1 + n\sin^2 A}{n\sin A\cos A} = \frac2{n\sin (2A)} + \tan A.$$ To maximise $\tan B$ we need to minimise $\cot B$ (the reason for doing this is that $\cot B$ is easier to differentiate than $\tan B$), so let's differentiate it: $$\frac d{dA}\cot B = -\frac{4\cos(2A)}{n\sin^2(2A)} + \frac1{\cos^2 A} = \frac{-\cos(2A) + n\sin^2A}{n\sin^2A\cos^2A}.$$ That is zero when $0 = -\cos(2A) + n\sin^2A = -1 + (n+2)\sin^2A$, so that $\sin A = \dfrac1{\sqrt{n+2}}$ and $\cos A = \dfrac{\sqrt{n+1}}{\sqrt{n+2}}.$

The corresponding (maximal) value of $\tan B$ is $$\frac{\frac{n\sqrt{n+1}}{n+2}}{1 + \frac{n}{n+2}} = \frac{n\sqrt{n+1}}{2(n+1)} = \frac n{2\sqrt{n+1}}.$$ With $n=2015$ that gives the maximum value of $\tan B$ as $\dfrac{2015}{2\sqrt{2016}} = \dfrac{2015}{24\sqrt{14}} \approx 22.4388.$[/sp]

 

[anemone]

[sp]Let $n = 2015.$ Then $\tan B(1 + n\sin^2 A) = n\sin A\cos A$. Therefore $\tan B = \dfrac{n\sin A\cos A}{1 + n\sin^2 A}$ and so $$\cot B = \frac{1 + n\sin^2 A}{n\sin A\cos A} = \frac2{n\sin (2A)} + \tan A.$$ To maximise $\tan B$ we need to minimise $\cot B$ (the reason for doing this is that $\cot B$ is easier to differentiate than $\tan B$), so let's differentiate it: $$\frac d{dA}\cot B = -\frac{4\cos(2A)}{n\sin^2(2A)} + \frac1{\cos^2 A} = \frac{-\cos(2A) + n\sin^2A}{n\sin^2A\cos^2A}.$$ That is zero when $0 = -\cos(2A) + n\sin^2A = -1 + (n+2)\sin^2A$, so that $\sin A = \dfrac1{\sqrt{n+2}}$ and $\cos A = \dfrac{\sqrt{n+1}}{\sqrt{n+2}}.$

The corresponding (maximal) value of $\tan B$ is $$\frac{\frac{n\sqrt{n+1}}{n+2}}{1 + \frac{n}{n+2}} = \frac{n\sqrt{n+1}}{2(n+1)} = \frac n{2\sqrt{n+1}}.$$ With $n=2015$ that gives the maximum value of $\tan B$ as $\dfrac{2015}{2\sqrt{2016}} = \dfrac{2015}{24\sqrt{14}} \approx 22.4388.$[/sp]

Spoiler

Aww Opalg! Thanks for participating in this other Olympiad Math problem from Hong Kong and your answer is of course correct! I will wait for a couple of days before posting the proposed solution.:)

 

[anemone]

Solution of other:

Spoiler

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

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

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

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

And we get

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

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

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