[ Orthocenter and orthic triangle ]  [ [ median line of a triangle ]

Segments connecting the feet of the altitudes of an acute-angled triangle are 8, 15, and 17. Find the radius of the circumscribed circle around the triangle.

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Continue the heights $A D, B E$ and $C F$ of the given triangle $A B C$ until they intersect the circumscribed circle at points $A^{\prime}, B^{\prime}$ and $C^{\prime}$ respectively. Then the triangle $A^{\prime} B^{\prime} C^{\prime}$ is similar to triangle $D E F$ with a similarity coefficient of 2.

## Solution

Let $A D, B E$ and $C F$ be the heights of the acute-angled triangle $A B C; H$ be the point of their intersection (orthocenter); $D F=$ $8, E F=15, D E=17$. Since $8^{2}+15^{2}=17^{2}$, triangle $D E F$ is a right triangle, $\angle D F E=90^{\circ}$.

Extend the segments $A D, B E$ and $C F$ until they intersect the circumscribed circle of triangle $A B C$ at points $A^{\prime}, B^{\prime}$ and $C$ respectively. Since the point symmetric to the orthocenter of the triangle lies on the circumscribed circle, $F E, D E$ and $D F$ are the midlines of triangles $B^{\prime} H C^{\prime}, A^{\prime} H B^{\prime}$ and $A^{\prime} H C^{\prime}$ respectively. From this, it follows that triangle $A^{\prime} B^{\prime} C^{\prime}$ is similar to triangle $D E F$ with a similarity coefficient of 2. Therefore, triangle $A^{\prime} B^{\prime} C^{\prime}$ is also a right triangle, its hypotenuse $A^{\prime} B^{\prime}$ is 34, and the radius of the circumscribed circle is 17.

![](https://cdn.mathpix.com/cropped/2024_05_06_9466e5d28b3947d5c4a5g-15.jpg?height=512&width=694&top_left_y=224&top_left_x=682)

Answer

17.