# Solving $\arctan(a) + \arctan(b) + \arctan(c) = \pi$ for $0 < a < b < c < 10$

This is a trigonometry math contest problem.

Which ordered triple of numbers $(a,b,c)$ with $0 < a < b < c < 10$ satisfies the equation

$$\arctan(a) + \arctan(b) + \arctan(c) = \pi\quad ?$$

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Presumably contest questions are to be done by the contestants, and not by us here. – GEdgar Feb 19 '12 at 3:24
Would you care to specify what type of numbers you are looking for. Do they have to be real or integers or what? – E.O. Feb 19 '12 at 3:48
You should at least name the contest. – Blue Feb 19 '12 at 4:31

Start with the trigonometric identity $$\tan(\alpha+\beta+\gamma) = \frac{\tan\alpha+\tan\beta+\tan\gamma-\tan\alpha\tan\beta\tan\gamma}{1-\tan\alpha\tan\beta-\tan\alpha\tan\gamma-\tan\beta\tan\gamma}.$$ Let $\alpha$, $\beta$, $\gamma$ be respectively the arctangents of $a$, $b$, and $c$. Then $$\frac{a+b+c-abc}{1-ab-ac-bc} = \tan\pi=0.$$ A fraction is $0$ only if the numerator is $0$. So we need $$a+b+c=abc.$$ Since $1+2+3=1\cdot2\cdot3$, we have at least one solution. I'm not actually sure if there are others.

Haste makes waste: Of course there are other solutions! If we perturb $\alpha$, $\beta$, and $\gamma$ a little bit while keeping the sum of the angles equal to $\pi$, i.e. slightly alter the shape of the triangle, we get another solution. There are as many solutions as there are shapes of acute triangles in which all of the angles are smaller than $\arctan(10)$.

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Any other triple of integers will leave the original left-hand side larger than $\pi$. So this is the only solution that uses integers. – alex.jordan Feb 19 '12 at 3:54
A related fact (with direct connection to $a+b+c=abc$): If $\alpha$, $\beta$, and $\gamma$ are the three angles in a triangle, $\tan\alpha+\tan\beta+\tan\gamma=\tan\alpha\tan\beta\tan\gamma$. – Isaac Feb 19 '12 at 3:56

If we're working over the reals, let $a$, $b$, and $c$ be the tangents of the smallest, middle, and largest angles, respectively, in a wide variety of acute triangles (acute guarantees $0<a<b<c$, but not all acute triangles will have $c<10$).

Over the integers, $(a,b,c)=(1,2,3)$ is the only solution. Since $\arctan(2)>\frac{\pi}{3}$ and the arctangent function is strictly increasing, if $a\ge2$, the sum would have to be greater than $\pi$, so $a=1$ and $\arctan(1)=\frac{\pi}{4}$. Now, $\arctan(b)+\arctan(c)=\frac{3\pi}{4}$ and $\arctan(3)>\frac{3\pi}{8}$, so if $b\ge3$, $\arctan(b)+\arctan(c)>\frac{3\pi}{4}$, so $b=2$. This leaves $\arctan(2)+\arctan(c)=\frac{3\pi}{4}$. Taking the tangent of both sides and applying the tangent of a sum identity, $\frac{2+c}{1-2c}=-1\implies c=3$.

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