2008 Mock ARML 1 Problems/Problem 8

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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 1

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 $\boxed{\frac {\sqrt {3}}{2}}$.

Solution 2

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

See also

2008 Mock ARML 1 (Problems, Source)
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