2008 Mock ARML 1 Problems/Problem 8

Problem

For positive real numbers $a,b,c,d$ ,

$\begin{align*}2a^2 + \sqrt {(a^2 + b^2)(a^2 + c^2)} &= 2bc\\ 2a^2 + \sqrt {(a^2 + c^2)(a^2 + d^2)} &= 2cd\\ 2a^2 + \sqrt {(a^2 + d^2)(a^2 + b^2)} &= 2db\end{align*}$ $\sqrt {(a^2 + b^2)(a^2 + c^2)} + \sqrt {(a^2 + c^2)(a^2 + d^2)} + \sqrt {(a^2 + d^2)(a^2 + b^2)} = 2$

Compute $ab + ac + ad$ .

Solution

We consider a geometric interpretation, specifically with an equilateral triangle. Let the distances from the vertices to the incenter be $x$ , $y$ , and $z$ , and the tangents to the incircle be $b$ , $c$ , and $d$ . Then use Law of Cosines to express the sides in terms of $x$ , $y$ , and $z$ , and Pythagorean Theorem to express $x$ , $y$ , and $z$ in terms of $b$ , $c$ , $d$ , and the inradius $a$ . This yields the first three equations. The fourth is the result of the sine area formula for the three small triangles, and gives the area as $\frac {\sqrt {3}}{2}$ . The desired expression is $rs$ , which is also the area, so the answer is $\boxed{\frac {\sqrt {3}}{2}}$ .

Note that since the equations are symmetric in $b,c,d$ , we may consider $b=c=d$ ; the system reduces quickly, and we find that the desired sum is $\frac{\sqrt{3}}{2}$ .

2008 Mock ARML 1 (Problems, Source)
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2008 Mock ARML 1 Problems/Problem 8

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