## Saturday, May 16, 2015

### Mathematical Problem Solving Skill

In this blog post, I will show one really insightful method given by one very intelligent U.K. mathematician on how he used his own elegant way to prove that $\tan^2 20^{\circ}+\tan^2 40^{\circ}+\tan^2 80^{\circ}=33$.

He first noticed that

$\tan 3(20^\circ)=\tan 60^\circ=\sqrt{3}$

$\tan 3(40^\circ)=\tan 120^\circ=-\sqrt{3}$

$\tan 3(80^\circ)=\tan 240^\circ=\sqrt{3}$

This is saying if $x=20^\circ,\,40^\circ,\,80^\circ$, each of them satisfies $\tan 3x=\pm \sqrt{3}$, or more usefully, $\tan^2 3x=3$.

That is to say, we could use the triple angle formula for $\tan x$ in our proof.

He then take the square of both sides of the trigonometric triple angle formula for $\tan x$ before using it for the proof of this Olympiad Math Trigonometry problem:

$\tan 3x=\dfrac{3\tan x-\tan^3 x}{1-3\tan^2 x}$

$\sqrt{3}=\dfrac{3\tan x-\tan^3 x}{1-3\tan^2 x}$

$\sqrt{3}^2=\left(\dfrac{\tan x(3-\tan^2 x)}{1-3\tan^2 x}\right)^2$

$3(1-3\tan^2 x)^2=\tan^2 x(3-\tan^2 x)^2$

He realized if he then let $t=\tan^2 x$, the equation above becomes

$3(1-3t)^2=t(3-t)^2$

Expanding and grouping the like terms yields a cubic equation:

$3(1-6t+9t^2)=t(9-6t+t^2)$

$3-18t+27t^2=9t-6t^2+t^3$

$t^3-33t^2+27t-3=0$

This cubic equations has roots $\tan^2 20^\circ, \tan^2 40^\circ$ and $\tan^2 80^\circ$ and hence the sum of the roots is $33$.

That is, $\tan^2 20^{\circ}+\tan^2 40^{\circ}+\tan^2 80^{\circ}=33$.

I want so much to post a shout-out to my math friend, a retired math professor from the University of Leeds, London, to show my appreciation to him because he is the sweet friend of mine who helped me to figure out the reasoning behind the solution provided by the original solver.