Problem 9.5. Given a right triangle $A B C\left(\Varangle A C B=90^{\circ}\right)$, let $C H, H \in A B$, be the altitude to $A B$ and $P$ and $Q$ be the tangent points of the incircle of $\triangle A B C$ to $A C$ and $B C$, respectively. If $A Q \perp H P$ find the ratio $\frac{A H}{B H}$.

Stoyan Atanasov
9.5. It follows from $A Q \perp H P$ that $\Varangle Q A B=\Varangle P H C$. On the other hand $\Varangle A B C=\Varangle A C H$ and therefore $\triangle A B Q \sim$ $\triangle H C P$. Thus, $\frac{A B}{B Q}=\frac{H C}{C P}$. Using the standard notation for the elements of a triangle we obtain the following equalities:

![](https://cdn.mathpix.com/cropped/2024_06_03_f8ccd71b4db367e7c2a9g-190.jpg?height=353&width=510&top_left_y=128&top_left_x=1135)

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

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

$$
b^{4}=a^{2}\left(a^{2}+b^{2}\right) \Longleftrightarrow b^{4}-a^{2} b^{2}-a^{4}=0
$$

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