Problem 1053

The internal angles of triangle $ABC$ are denoted in the usual way, i.e., $\alpha, \beta$, $\gamma$. The angle bisectors of the internal angles of triangle $ABC$ are the axes of symmetry of these angles, so $\Varangle B A D=\Varangle D A C=\frac{\alpha}{2}$, etc. Triangles $DEF, DGH, DIJ$ are equilateral, and points $A$, $B, C$ are the midpoints of the sides opposite the common vertex $D$. Therefore, lines $DA, DB, DC$ are the axes of symmetry of these triangles, and $\Varangle E D A=\Varangle A D F=30^{\circ}$, etc. (see the figure on the next page).

The measure of angle $ADC$ is $30^{\circ}+51^{\circ}+30^{\circ}=111^{\circ}$, and for the sum of the internal angles in triangle $ADC$ we have

$$\frac{\alpha}{2}+111^{\circ}+\frac{\gamma}{2}=180^{\circ}$$

which simplifies to $\alpha+\gamma=138^{\circ}$. Simultaneously, for the internal angles of triangle $ABC$ we have

$$\alpha+\beta+\gamma=180^{\circ}$$

or $\alpha+\gamma=180^{\circ}-\beta$. From the two expressions for $\alpha+\gamma$, we get $138^{\circ}=180^{\circ}-\beta$, thus

$$\beta=42^{\circ} \text {. }$$

Similarly, we express the measure of angle $BDC$ as $30^{\circ}+66^{\circ}+30^{\circ}=126^{\circ}$, and from the sum of the internal angles in triangle $BDC$,

$$\frac{\beta}{2}+126^{\circ}+\frac{\gamma}{2}=180^{\circ}$$

we determine $\beta+\gamma=108^{\circ}$. Simultaneously, from (2) we have $\beta+\gamma=180^{\circ}-\alpha$. Altogether, we get $108^{\circ}=180^{\circ}-\alpha$, thus

$$\alpha=72^{\circ}$$

By substituting $\alpha$ or $\beta$ into one of the previous equations, we can express the remaining angle $\gamma$. For example, substituting into (2) gives $72^{\circ}+42^{\circ}+\gamma=180^{\circ}$, and thus

$$\gamma=66^{\circ}$$

The measures of the internal angles of triangle $ABC$ are $72^{\circ}, 42^{\circ}$, and $66^{\circ}$.

!

Notes. In addition to conditions (1) and (3), we can also use a similar condition for triangle $ADB$. For this purpose, it is necessary to express the angle $FDG$ as the supplementary angle to the full angle with vertex $D$, whose measure is $360^{\circ}-3 \cdot 60^{\circ}-51^{\circ}-66^{\circ}=63^{\circ}$. Then the measure of angle $ADB$ is $30^{\circ}+63^{\circ}+30^{\circ}=123^{\circ}$, and for the sum of the internal angles in triangle $ADB$ we have

$$\frac{\alpha}{2}+123^{\circ}+\frac{\beta}{2}=180^{\circ}$$

Any three of the four equations (1) to (4) uniquely determine the solution, and the problem can also be solved this way. For example, the system of equations (1), (3), (4) is equivalent to the system

$$\alpha+\gamma=138^{\circ}, \quad \beta+\gamma=108^{\circ}, \quad \alpha+\beta=114^{\circ} .$$

Subtracting the third equation from the first gives $\gamma-\beta=24^{\circ}$. Subtracting this equation from the second gives $2 \beta=84^{\circ}$, thus $\beta=42^{\circ}$. Substituting this result into the second or third equation determines $\gamma=66^{\circ}$, and $\alpha=72^{\circ}$, respectively.

Evaluation. 2 points for initial observations related to the angle bisectors; 2 points for determining auxiliary angles and setting up equations; 2 points for solving and quality of the commentary.