# 1986 AIME Problems/Problem 2

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## Problem

Evaluate the product $(\sqrt 5+\sqrt6+\sqrt7)(-\sqrt 5+\sqrt6+\sqrt7)(\sqrt 5-\sqrt6+\sqrt7)(\sqrt 5+\sqrt6-\sqrt7)$.

## Solution

Simplify by repeated application of the difference of squares.

$\left((\sqrt{6} + \sqrt{7})^2 - \sqrt{5}^2\right)\left(\sqrt{5}^2 - (\sqrt{6} - \sqrt{7})^2\right)$
$= (13 + 2\sqrt{42} - 5)(5 - (13 - 2\sqrt{42}))$
$= (2\sqrt{42} + 8)(2\sqrt{42} - 8)$
$= (2\sqrt{42})^2 - 8^2 =$ $\boxed{104}$

## Solution 2

Notice that in a triangle with side lengths $2\sqrt5,2\sqrt6,$ and $2\sqrt7$, by Heron's formula, the area is the square root of what we are looking for. Let angle $\theta$ be opposite the $2\sqrt7$ side. By the Law of Cosines, $$\cos\theta=\frac{(2\sqrt5)^2+(2\sqrt{6})^2-(2\sqrt7)^2}{2\cdot 2\sqrt5\cdot2\sqrt6}=\frac{16}{8\sqrt{30}}=\sqrt{\frac{2}{15}}$$

So $\sin\theta=\sqrt{1-\cos^2\theta}=\sqrt{\frac{13}{15}}$. The area of the triangle is then $$\frac{\sin\theta}{2}\cdot 2\sqrt5\cdot 2\sqrt6=\sqrt{\frac{26}{30}}\cdot 2\sqrt{30}=\sqrt{104}$$

So our answer is $\left(\sqrt{104}\right)^2=\boxed{104}$