2014/15 AHSMC Part 1 Problem 12

Problem 12 of 16

$P$ is a point inside an acute triangle $ABC$ . $D$ , $E$ and $F$ are the feet of the perpendiculars from $P$ on $BC$ , $CA$ and $AB$ respectively. If $BD = 1$ , $DC = 10$ , $CE = 6$ , $EA = 9$ and $AF = 13$ , the length of $FB$ is

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Solution

By Pythagoras' Theorem, $BD^2 - DC^2 = (BP^2 - PD^2) - (CP^2 - PD^2) = BP^2 - CP^2$ . Similarly, $CE^2 - EA^2 = CP^2 - AP^2$ and $AF^2 - BF^2 = AP^2 - BP^2$ . Summing these equations yields $(BD^2 + CE^2 + AF^2) - (DC^2 + EA^2 + FB^2) = 0.$ Hence $FB^2 = (1^2 + 6^2 + 13^2) - (10^2 + 9^2) = 5^2$ so that $FB = 5$ . The answer is (E).

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