
Problem 11.2. On each side of a triangle with angles $30^{\circ}, 60^{\circ}$ and $90^{\circ}$ and hypotenuse 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 triangle $A B C$ with angles $\alpha, \beta$ and $\gamma$. 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 with $\Varangle A_{1} C_{1} B_{1}=90^{\circ}$. Suppose that the circle $k$ of diameter $A_{1} B_{1}$ intersects $A B$ at point $X \neq C_{1}$. It is easy to see that if $\alpha \leq 90^{\circ}$ and $\beta \leq 90^{\circ}$ then $X$ is an interior point for line segment $A B$. Draw a tangent $l$ to $k$ which is parallel to $A B$ and the tangential point $Y$ belongs to the smaller of the arcs $\widehat{C_{1} X}$. Denote the intersecting points of $l$ and lines $A C$ and $B C$ by $B_{2}$ and $A_{2}$ respectively. Consider homothety of center $C$ for which the image of $B_{2}\left(A_{2}\right)$ is the point $A(B)$. It is obvious that the image of $\triangle B_{1} A_{1} Y$ is inscribed in $\triangle A B C$ and its hypotenuse is less than $B_{1} A_{1}$. Therefore wlog we may suppose that $A B$ is tangent to $k$. In this case if $\Varangle B_{1} A_{1} C_{1}=\delta$ then $\Varangle B_{1} C_{1} A=\delta$.

The Sine Low for $\triangle A C_{1} B_{1}$ and $\triangle B C_{1} A_{1}$ gives

$$
A C_{1}=\frac{B_{1} A_{1} \sin \delta \sin (\alpha+\delta)}{\sin \alpha} ; B C_{1}=\frac{B_{1} A_{1} \cos \delta \sin \left(90^{\circ}-\delta+\beta\right)}{\sin \beta}
$$

Since $A B=A C_{1}+C_{1} B$ we obtain

$$
A B=\frac{B_{1} A_{1} \sin \delta \sin (\alpha+\delta)}{\sin \alpha}+\frac{B_{1} A_{1} \cos \delta \sin \left(90^{\circ}-\delta+\beta\right)}{\sin \beta}
$$

This equality is equivalent to

$$
2 A B=A_{1} B_{1}(\operatorname{cotg} \alpha+\operatorname{cotg} \beta+2 \sin 2 \delta-(\operatorname{cotg} \alpha-\operatorname{cotg} \beta) \cos 2 \delta)
$$

From the other hand

$$
\begin{gathered}
2 \sin 2 \delta-(\operatorname{cotg} \alpha-\operatorname{cotg} \beta) \cos 2 \delta= \\
\sqrt{4+(\operatorname{cotg} \alpha-\operatorname{cotg} \beta)^{2}} \sin (2 \delta-\phi) \leq \sqrt{4+(\operatorname{cotg} \alpha-\operatorname{cotg} \beta)^{2}}
\end{gathered}
$$

where

$$
\cos \phi=\frac{2}{\sqrt{4+(\operatorname{cotg} \alpha-\operatorname{cotg} \beta)^{2}}}, \sin \phi=\frac{\operatorname{cotg} \alpha-\operatorname{cotg} \beta}{\sqrt{4+(\operatorname{cotg} \alpha-\operatorname{cotg} \beta)^{2}}}
$$

It is easily seen that such an angle $\delta$ always exists.

Finally, the least value is

$$
B_{1} A_{1}=\frac{2 A B}{\operatorname{cotg} \alpha+\operatorname{cotg} \beta+\sqrt{4+(\operatorname{cotg} \alpha-\operatorname{cotg} \beta)^{2}}}
$$

It is clear that if $B_{1} A_{1}=\frac{2 A B}{\operatorname{cotg} \alpha+\operatorname{cotg} \beta+\sqrt{4+(\operatorname{cotg} \alpha-\operatorname{cotg} \beta)^{2}}}$ and $2 \delta-\phi=90^{\circ}$ then (1) holds true and therefore there exists a triangle with $B_{1} A_{1}=\frac{2 A B}{\operatorname{cotg} \alpha+\operatorname{cotg} \beta+\sqrt{4+(\operatorname{cotg} \alpha-\operatorname{cotg} \beta)^{2}}}$.

It remains to compute the above expression for the three possible ways to inscribe a right triangle in triangle of angles $30^{\circ}, 60^{\circ}$ and $90^{\circ}$ and hypotenuse 1. We obtain $\frac{\sqrt{39}-\sqrt{3}}{12}, \frac{\sqrt{21}-3}{4}, \frac{\sqrt{3}}{4}$. The least value is $\frac{\sqrt{39}-\sqrt{3}}{12}$ which is the answer of the problem.
