$3-$ [ Orthocenter and Orthic Triangle ]  Segments connecting the feet of the altitudes of an acute triangle form a right triangle with a hypotenuse of 10. Find the radius of the circumcircle of the original triangle.

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Let $H$ be the point of intersection of the altitudes $A A_{1}, B B_{1}, C C_{1}$ of triangle $A B C$, $\angle A_{1} C_{1} B_{1}=90^{\circ}$, $A_{1} B_{1}=10$; $A_{2}, B_{2}, C_{2}$ are the points of intersection of the extensions of the altitudes $A A_{1}, B B_{1}, C C_{1}$ with the circumcircle of triangle $A B C$. Then $A_{1}, B_{1}, C_{1}$ are the midpoints of segments $H A_{2}, H B_{2}, H C_{2}$. Therefore, $A_{1} B_{1}, B_{1} C_{1}, A_{1} C_{1}$ are the midlines of triangles $A_{2} \mathrm{HB}_{2}, \mathrm{~B}_{2} \mathrm{HC}_{2}, A_{2} \mathrm{HC}_{2}$, so the sides of triangle $A_{2} B_{2} C_{2}$ are parallel to the sides of triangle $A_{1} B_{1} C_{1}$, and $A_{2} B_{2}=2 A_{1} B_{1}, A_{2} C_{2}=2 A_{1} C_{1}, B_{2} C_{2}=2 B_{1} C_{1}$. Therefore, triangle $A_{2} B_{2} C_{2}$ is also a right triangle and its hypotenuse $A_{2} B_{2}$ is twice $A_{1} B_{1}$, i.e., 20. Therefore, the radius of the circumcircle of triangle $A_{2} B_{2} C_{2}$ (and thus of triangle $A B C$) is 10.

## Answer

10.

On the sides $B C, A C$, and $A B$ of triangle $A B C$, points $A_{1}, B_{1}$, and $C_{1}$ are located, respectively, such that $B A_{1}: A_{1} C = C B_{1}: B_{1} A = A C_{1}: C_{1} B = 2: 3$. Find the area of the triangle formed by the intersections of the lines $A A_{1}, B B_{1}$, and $C C_{1}$, given that the area of triangle $A B C$ is 1.

## Hint

Find the ratio in which the point of intersection divides the segments $B B_{1}$ and $C C_{1}$.

## Solution

Let $K$ be the point of intersection of segments $B B_{1}$ and $C C_{1}$. Draw a line through point $B$ parallel to $A C$ and extend $C C_{1}$ to intersect this line at point $T$. Let $C B_{1}=2 a, A B_{1}=3 a$. Triangles $B C_{1} T$ and $A C_{1} C$ are similar with a ratio of $\frac{3}{2}$. Therefore, $B T=\frac{3}{2} A C=\frac{3}{2} 5 a=\frac{15}{2} a$, and from the similarity of triangles $B K T$ and $B_{1} K C$, we find that

$$
\frac{B K}{K B_{1}}=\frac{B T}{C B_{1}}=\frac{\frac{15}{2} a}{2 a}=\frac{15}{4}
$$

Therefore,

$$
S_{\triangle C R_{1} X}=\frac{4}{19} S_{\triangle C B B_{1}}=\frac{4}{19} \cdot \frac{2}{5} S_{\triangle A B C}=\frac{8}{85}
$$

Similarly, we find that

$$
S_{\triangle A R M C_{1}}=S_{\triangle A M A_{1}}=\frac{8}{85}
$$

where $M$ is the point of intersection of $A A_{1}$ and $C C_{1}$, and $N$ is the point of intersection of $B B_{1}$ and $A A_{1}$. Therefore,

$$
\begin{gathered}
S_{\triangle M A N K}=S_{\triangle A B C}-S_{\triangle A E B_{1}}-S_{\triangle B C E_{1}}-S_{\triangle C A C_{1}}+S_{\triangle C R_{1} X}+S_{\triangle A C_{1} M}+S_{\triangle E A_{1} N}= \\
=1-\frac{2}{5}-\frac{2}{5}-\frac{2}{5}+\frac{8}{85}+\frac{8}{85}+\frac{8}{85}=1-\frac{6}{5}+\frac{24}{85}=\frac{24}{85}-\frac{1}{5}=\frac{1}{19}
\end{gathered}
$$

## Answer

$\frac{1}{19}$.