
Problem 10b.2. On each side of a right isosceles triangle with legs of length 1 is chosen a point such that the triangle formed from these three points is a right triangle. What is the least value of the hypotenuse of this triangle?


Solution: Consider a right isosceles triangle $A B C$ with right angle at $C$. Let $A_{1} \in B C, B_{1} \in C A$ and $C_{1} \in A B$ be such that $\triangle A_{1} B_{1} C_{1}$ is right triangle.

1. Suppose $\Varangle B_{1} A_{1} C_{1}=90^{\circ}$. Assume that the circle $k$ with diameter $B_{1} C_{1}$ intersects $B C$ at point $X \neq A_{1}$. It is easy to be seen that $X$ is an interior point for the line segment $B C$. Draw a tangent $l$ to $k$ which is parallel to $B C$ and the tangential point $Y$ belongs to the smaller of the arcs $\widehat{A_{1} X}$. Denote the intersecting points of $l$ with $A B$ and $A C$ by $C_{2}$ and $B_{2}$ respectively. Consider a homothety of center $A$ such that the image of $B_{2}\left(C_{2}\right)$ is $C(B)$. It is clear that the image of $\triangle B_{1} C_{1} Y$ is inscribed in $\triangle A B C$ and its hypotenuse is less than $B_{1} C_{1}$. Therefore wlog we may assume that $B C$ is tangent to $k$. Thus, if $\Varangle C_{1} B_{1} A_{1}=\alpha$, then $\Varangle C_{1} A_{1} B=\alpha$. From the Sine Low for $\triangle B C_{1} A_{1}$ we obtain $B A_{1}=\frac{B_{1} C_{1} \sin \alpha \sin \left(\alpha+45^{\circ}\right)}{\sin 45^{\circ}}=$ $B_{1} C_{1} \sin \alpha(\sin \alpha+\cos \alpha)$. Hence

(1) $1=B A_{1}+A_{1} C=B_{1} C_{1} \cos \alpha \sin \alpha+B_{1} C_{1} \sin \alpha(\sin \alpha+\cos \alpha)$

Therefore $1=B_{1} C_{1}\left(\sin ^{2} \alpha+\sin 2 \alpha\right) \Leftrightarrow B_{1} C_{1}(1+2 \sin 2 \alpha-$ $\cos 2 \alpha)=2$. From the other hand $2 \sin 2 \alpha-\cos 2 \alpha=$

$\sqrt{5}\left(\frac{2}{\sqrt{5}} \sin 2 \alpha-\frac{1}{\sqrt{5}} \cos 2 \alpha\right)=\sqrt{5} \sin (2 \alpha-\delta) \leq \sqrt{5}$, where $\delta$ is such that $\sin \delta=\frac{1}{\sqrt{5}}$ and $\cos \delta=\frac{2}{\sqrt{5}}$. Note that the equality holds when $2 \alpha-\delta=90^{\circ}$ and since $\delta\frac{\sqrt{5}-1}{2}$ we obtain that the least possible value equals $\frac{\sqrt{5}-1}{2}$.
