In the acute triangle $ABC$, let's take points $C_{1}$, $A_{1}$, and $B_{1}$ on sides $AB$, $BC$, and $CA$ respectively, such that the angles $CC_{1}B$, $AA_{1}C$, and $BB_{1}A$ are all equal to a given acute angle $\varphi$, which is greater than any angle of the triangle. Let the intersection of lines $AA_{1}$ and $BB_{1}$ be $C_{2}$, the intersection of lines $BB_{1}$ and $CC_{1}$ be $A_{2}$, and the intersection of lines $CC_{1}$ and $AA_{1}$ be $B_{2}$. Determine the area of triangle $A_{2}B_{2}C_{2}$ using suitable defining data of triangle $ABC$ and the angle $\varphi$.
The sides of the triangle are conventionally denoted. Since $\alpha$ and $\beta$ are acute angles, $A$ and $B$ are on opposite sides of the foot $C_{0}$ of the altitude from $C$, and since $C C_{1} B \varangle=\varphi<90^{\circ}$, $C_{1}$ is an interior point of the segment $A C_{0}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_ebfea863a9ca540cfd71g-1.jpg?height=650&width=724&top_left_y=263&top_left_x=690)

By the exterior angle theorem, $A C C_{1} \varangle=\varphi-\alpha$, and since the roles of the letters are cyclically symmetric in the construction, $B A A_{1} \varangle=\varphi-\beta$, $A_{1}$ is on the segment $B C$, and $B_{2}$ is inside the triangle. Furthermore, in the triangle $A C_{1} B_{2}$, $A B_{2} C_{1} \varangle=A_{1} B_{2} C \varangle=\beta$, and in the triangle $A B_{2} C$, $B_{2} A C \varangle=\pi-\varphi-\gamma$, and with these,

\[
\begin{aligned}
B_{2} C=\frac{A C}{\sin (\pi-\beta)} \sin (\varphi+\gamma) & =2 r \sin (\varphi+\gamma) \\
B_{2} A & =2 r \sin (\varphi-\gamma)
\end{aligned}
\]

where $2 r$ is the diameter of the circumcircle of triangle $A B C$. From the latter, by cyclic permutation of the letters,

\[
A_{2} C=2 r \sin (\varphi-\gamma)
\]

Let us form the following difference:

\[
B_{2} C-A_{2} C=2 r(\sin (\varphi+\gamma)-\sin (\varphi-\gamma))=4 r \cos \varphi \sin \gamma=2 A B \cos \varphi
\]

This is positive, so $A_{2}$ is on the segment $B_{2} C$. By cyclic permutation of the letters, $C_{2}$ is on the segment $A_{2} B$, and $B_{2}$ is on the segment $C_{2} A$, so the triangle $A_{2} B_{2} C_{2}=H_{2}$ has the angle at $B_{2}$ identical to the angle $A_{1} B_{2} C \varangle=\beta$. By cyclic permutation, $A_{2} C_{2} B_{2} \varangle=\gamma$, so the triangle $H_{2}$ is similar to the original triangle $A B C=H$ - with $A_{2}$ corresponding to $A$ - and it is inside $H$.

We also obtained above that $A_{2} B_{2}=A B \cdot 2 \cos \varphi$, so $H_{2}$ is the image of $H$ scaled down by a factor of $2 \cos \varphi$. Then the area $t_{2}$ of $H_{2}$ is obtained by multiplying the area $t$ of $H$ by $4 \cos ^{2} \varphi$. With this, we have essentially solved the problem.

A concern might arise that $t$ does not uniquely determine $H$. This can be resolved by substituting any set of data that determines the area, for example,

\[
t_{2}=2 a b \sin \gamma \cos ^{2} \varphi
\]

The largest angle of $H$ is at least $60^{\circ}$, and since $\varphi$ is larger than this, $0<\cos \varphi<1 / 2$, so the factor $4 \cos ^{2} \varphi$ is less than 1. (This also follows from the fact that, as we have stated, $H_{2}$ is inside $H$.)

As $\varphi$ approaches $90^{\circ}$, $H_{2}$ collapses to the foot of the altitude.

Kiss Emil (Budapest, Fazekas M. Gyak. Gimn., III. o. t.)