$A B C$ triangle has perpendiculars raised at vertex $A$ to $A B$ and $A C$, which intersect side $B C$ at points $B^{\prime}$ and $C^{\prime}$. Express the distance $B^{\prime} C^{\prime}$ in terms of the elements of the original triangle and calculate the radius of the inscribed circle of triangle $A B^{\prime} C^{\prime}$.
I. Solution. Let the sides of the triangle $ABC$ be: $a, b, c$; its angles: $\alpha, \beta, \gamma$. From the sine theorem applied to triangles $ABC'$ and $ACB'$:

$$
BC'=\frac{c \cos \alpha}{\cos \gamma} \text{ and } CB'=\frac{b \cos \alpha}{\cos \beta}
$$

thus

$$
\begin{gathered}
B' C'=C' B+a+CB' \\
=a+\frac{b \cos \alpha}{\cos \beta}+\frac{c \cos \alpha}{\cos \gamma} \\
=\frac{a \cos \beta \cos \gamma+b \cos \alpha \cos \gamma+c \cos \alpha \cos \beta}{\cos \beta \cos \gamma} \\
=\frac{a \cos \beta \cos \gamma+\cos \alpha(b \cos \gamma+c \cos \beta)}{\cos \beta \cos \gamma} \\
=\frac{a \cos \beta \cos \gamma+a \cos \alpha}{\cos \beta \cos \gamma}=\frac{a(\cos \alpha+\cos \beta \cos \gamma)}{\cos \beta \cos \gamma} \\
=\frac{a(\sin \beta \sin \gamma-\cos \beta \cos \gamma+\cos \beta \cos \gamma)}{\cos \beta \cos \gamma} \\
=a \tan \beta \tan \gamma
\end{gathered}
$$

Furthermore, let the center of the inscribed circle of triangle $AB'C'$ be $O$, and its radius $r$. Expressing the area of triangle $OC'B'$ in two ways, we get:

$$
\begin{gathered}
\frac{1}{2} C' B' \cdot r=\frac{1}{2} C' D \cdot B' D \cdot \sin \frac{B'+C'}{2} \\
C' B' \cdot r=\frac{B' C' \sin \frac{B'}{2}}{\sin \frac{B'+C'}{2}} \cdot \frac{B' C' \sin \frac{C'}{2}}{\sin \frac{B'+C'}{2}} \sin \frac{B'+C'}{2}
\end{gathered}
$$

from which

$$
r=\frac{B' C' \sin \frac{B'}{2} \sin \frac{C'}{2}}{\sin \frac{B'+C'}{2}}
$$

but

$$
B'=90^{\circ}-\beta, C'=90^{\circ}-\gamma
$$

thus

$$
r=\frac{a \tan \beta \tan \gamma \sin \left(45^{\circ}-\frac{\beta}{2}\right) \sin \left(45^{\circ}-\frac{\gamma}{2}\right)}{\cos \frac{\beta+\gamma}{2}}
$$

(Friedmann Bernát.)

II. Solution. Draw the altitudes $AD, BE$, and $CF$ of the triangle, which intersect at point $H$.

$$
AC' \| HB \text{ and } AB' \| HC
$$

therefore

$$
\triangle AC'B' \sim \triangle HBC
$$

thus

$$
C'B': a = AD: HD
$$

from which

$$
C'B'=a \frac{AD}{HD}
$$

but from the right-angled triangles $ADC$ and $HDC$:

$$
AD=CD \cdot \tan \gamma, HD=CD \cdot \tan HCD=CD \cdot \cot \beta
$$

and thus

$$
\frac{AD}{HD}=\tan \beta \tan \gamma
$$

therefore

$$
C'B'=a \tan \beta \tan \gamma
$$

(Weisz Lipót.)