The quadrilaterals $A B C D$ and $A C B E$ are parallelograms. The circle tangent to the side $A B$ of triangle $A B C$ passes through $E$, and the circle tangent to the side $A C$ passes through $D$. Calculate the angles of triangle $A B C$.

Let's proceed step by step to solve this problem.
In parallelograms $A D \# B C \# E A$ (including direction), so $D$ and $E$ are reflections of each other with $A$ as the center.

The two circles are inscribed in the angle domain of $\angle BAC$ (less than $180^\circ$) and in the two adjacent angle domains along the rays $AB$ and $AC$; these two domains are vertical angle domains. Thus, the two circles are images of each other under a similarity transformation centered at $A$ with some ratio, and the rays $AD$ and $AE$ are also images of each other.

![](https://cdn.mathpix.com/cropped/2024_05_02_a0e36259fb09b1a6ce2fg-1.jpg?height=539&width=1201&top_left_y=435&top_left_x=451)

The ray $AD$ intersects the circle tangent at $D$ a second time between $A$ and $D$, because otherwise the parallelogram $ABCD$ and the circle would have only one common point at $D$, and the circle would not touch the segment $AC$ inside the parallelogram. Similarly, $E$ is the farther intersection point of the circle tangent to $AB$ and the ray $AE$.

According to the above, points $D$ and $E$ are also corresponding points in the mentioned central similarity, so $DA = EA$, and the magnification ratio is 1, meaning the radii of the two circles are equal. Therefore, the line connecting the centers $O$ and $O'$ is parallel to $BC$, and it passes through $A$, since this line is the bisector of the external angles at $A$, and both $D$ and $E$ lie on it. Thus, the alternate interior angles $\angle ABC = \angle ACB$, making triangle $ABC$ isosceles: $AB = AC$. To find the angles, we need to determine the ratio of the base to the legs.

Let the midpoint of $BC$ be $F$, and the point of tangency of the circle tangent to $AC$ on the extension of $BC$ be $G$. Then, as is known, $BG$ is equal to half the perimeter of triangle $ABC$, and with the usual notation for sides and angles,

\[
\begin{gathered}
a = BC = AD = AO + OD = FG + OG = (BG - BF) + AF = \frac{a + 2b}{2} - \frac{a}{2} + \sqrt{b^2 - \frac{a^2}{4}} \\
(a - b)^2 = b^2 - \frac{a^2}{4}, \quad (a \neq 0) \\
\frac{a / 2}{b} = \cos \beta = \frac{4}{5}, \quad \beta = \gamma = 36^\circ 52', \quad \alpha = 106^\circ 16'
\end{gathered}
\]

This completes the solution.

Remark. It is customary to call the vertex $D$ of the parallelogram $ABCD$ the (one of the) external centroid of triangle $ABC$, as this is the center of mass of a system of masses where $+1$ mass is placed at $A$ and $C$, and $-1$ mass is placed at $B$. In this terminology, one part of the problem's statement can be phrased as: the circle tangent to the side opposite $B$ passes through the external centroid opposite $B$.

After this, we can point to the origin of the problem. The subject of our 1966 competition was the examination of triangles in various directions for which the incircle passes through the centroid. One such triangle was examined in problem 1805, along with an additional interesting property, and the question of generalization to external centroids and external tangent circles arose. This time, we did not take a numerical shortcut but considered an isosceles triangle (using two external centroids). - We recommend to the interested readers to determine the relationship between the sides with the requirement that one of the external tangent circles passes through the corresponding external centroid.[^0]

[^0]:    ${ }^{1}$ Solution: K. M. L. 45. (1972) 12.