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Bulgarian Mathematical Olympiad

Geometry Difficulty 5.8 AIME, harder Prove it Bulgaria

Problem:

Given a right triangle $ABC$ ( $\angle ACB = 90^\circ$ ), let $CH$ , $H \in AB$ , be the altitude to $AB$ and $P$ and $Q$ be the tangent points of the incircle of $\triangle ABC$ to $AC$ and $BC$ , respectively. If $AQ \perp HP$ find the ratio $\frac{AH}{BH}$ .

Solution

Solution:

It follows from $AQ \perp HP$ that $\angle QAB = \angle PHC$ . On the other hand $\angle ABC = \angle ACH$ and therefore $\triangle ABQ \sim \triangle HCP$ . Thus, $\frac{AB}{BQ} = \frac{HC}{CP}$ . Using the standard notation for the elements of a triangle we obtain the following equalities:

$\begin{aligned} \frac{c}{p-b} = \frac{h}{r} &\Leftrightarrow \frac{c}{p-b} = \frac{2S/c}{S/p} \Leftrightarrow \frac{c}{p-b} = \frac{2p}{c} \\ &\Leftrightarrow \frac{2c}{a+c-b} = \frac{a+b+c}{c} \Leftrightarrow 2c^2 = (a+c)^2 - b^2 \\ &\Leftrightarrow c^2 = a^2 + 2ac - b^2 \Leftrightarrow b^2 = ac \end{aligned}$

since $c^2 = a^2 + b^2$ . Hence

$b^4 = a^2(a^2 + b^2) \Longleftrightarrow b^4 - a^2 b^2 - a^4 = 0$

Set $k = \frac{AH}{BH} = \frac{b^2 / c}{a^2 / c} = \frac{b^2}{a^2}$ . Then $k^2 - k - 1 = 0$ and we get $k = \frac{1 + \sqrt{5}}{2}$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.