[ Angles between angle bisectors $]$

The sides of a triangle are 1 and 2, and the angle between them is $60^{\circ}$. A circle is drawn through the incenter of this triangle and the endpoints of the third side. Find its radius.
Apply the formula $a=2 R \sin \alpha$.

## Solution

Let $O$ be the center of the inscribed circle of triangle $ABC$ with sides $AC=1, AB=2$ and angle $CAB$ equal to $60^{\circ}$. By the cosine theorem, we find that $BC=\sqrt{3}$. Therefore, triangle $ABC$ is a right triangle, $\angle ACB=90^{\circ}, \angle ABC=30^{\circ}$. Since $O$ is the intersection point of the angle bisectors of triangle $ABC$, then

$$
\angle BOC=90^{\circ}+\frac{1}{2} \angle CAB=90^{\circ}+30^{\circ}=120^{\circ}.
$$

If $R$ is the desired radius, then

$$
R=\frac{BC}{2 \sin \angle BOC}=\frac{\sqrt{3}}{2 \sin 120^{\circ}}=1.
$$

## Answer

![](https://cdn.mathpix.com/cropped/2024_05_06_cb0f0d605674d837f9abg-36.jpg?height=97&width=1189&top_left_y=180&top_left_x=17)

The segments connecting the feet of the altitudes of an acute triangle are 8, 15, and 17. Find the area of the triangle.

## Hint

Let $AD, BE$, and $CF$ be the altitudes of an acute triangle $ABC$. Prove that triangle $AEF$ is similar to triangle $ABC$ with a similarity coefficient of $\cos \angle A$; express the angles of triangle $ABC$ in terms of the angles of the right triangle $DEF$.

## Solution

Let $AD, BE$, and $CF$ be the altitudes of an acute triangle $ABC; DF=8, EF=15, DE=17$. Since $8^{2}+15^{2}=17^{2}$, triangle $DEF$ is a right triangle, $\angle DFE=90^{\circ}$.

Denote the angles $A, B$, and $C$ of triangle $ABC$ by $\alpha, \beta$, and $\gamma$ respectively. Since side $AC$ is seen from points $F$ and $D$ at a right angle, these points lie on the circle with diameter $AC$, thus,

$$
\angle BDF=180^{\circ}-\angle CDF=\angle CAF=\alpha
$$

and triangle $BDF$ is similar to triangle $BAC$ (by two angles) with a similarity coefficient of $BD / AB=\cos \angle ABD=\cos \beta$.

Similarly, we can prove that

$$
\angle CDE=\alpha, \angle CED=\beta, \angle AEF=\beta, \angle AFE=\gamma, \angle BFD=\gamma
$$

triangle $CDE$ is similar to triangle $CAB$ with a similarity coefficient of $\cos \gamma$, and triangle $AEF$ is similar to triangle $ABC$ with a similarity coefficient of $\cos \alpha$.

Let the area of triangle $ABC$ be $S$. Then the areas of triangles $AEF, BDF$, and $CDE$ are $S \cdot \cos ^{2} \alpha, S \cdot \cos ^{2} \beta$, and $S \cdot \cos ^{2} \gamma$ respectively. The area of the right triangle $DEF$ is $8 \cdot 15 / 2=60$. We have the equation

$$
S=S \cdot \cos ^{2} \alpha+S \cdot \cos ^{2} \beta+S \cdot \cos ^{2} \gamma+60
$$

from which

$$
S=60 /\left(1-\cos ^{2} \alpha-\cos ^{2} \beta-\cos ^{2} \gamma\right)
$$

From the right triangle $DEF$, we find that

$$
\cos \angle EDF=DF / DE=8 / 17, \cos \angle DEF=EF / DE=15 / 17, \cos \angle DFE=0.
$$

Since

$$
\begin{aligned}
& \angle EDF=180^{\circ}-\angle BDF-\angle CDE=180^{\circ}-2 \alpha \\
& \angle DEF=180^{\circ}-\angle DEC-\angle AEF=180^{\circ}-2 \beta \\
& \angle DFE=180^{\circ}-\angle AFE-\angle BFD=180^{\circ}-2 \gamma
\end{aligned}
$$

then

$$
\begin{gathered}
8 / 17=\cos \angle EDF=\cos \left(180^{\circ}-2 \alpha\right)=-\cos 2 \alpha \\
15 / 17=\cos \angle DEF=\cos \left(180^{\circ}-2 \beta\right)=-\cos 2 \beta \\
0=\cos \angle DFE=\cos \left(180^{\circ}-2 \gamma\right)=-\cos 2 \gamma
\end{gathered}
$$

From this, we find that

$$
\cos ^{2} \alpha=(1+\cos 2 \alpha) / 2=(1-8 / 17) / 2=9 / 34
$$

$$
\begin{gathered}
\cos ^{2} \beta=(1+\cos 2 \beta) / 2=(1-15 / 17) / 2=2 / 34 \\
\cos ^{2} \gamma=(1+\cos 2 \gamma) / 2=(1-0) / 2=1 / 2
\end{gathered}
$$

Therefore,

$$
S=60 /(1-9 / 34-2 / 34-1 / 2)=340
$$

## Answer

340.