Example 22. The centers of the inscribed and circumscribed circles are symmetric with respect to one of the sides of the triangle. Find the angles of the triangle.

Given (Fig. 25): $\triangle A B C, \omega_{1}\left(O_{1},\left|O_{1} C\right|\right)$ - circumscribed circle, $\omega_{2}\left(O_{2},\left|O_{2} M\right|\right)$ - inscribed circle, $\left(O_{1} O_{2}\right) \perp[A C]$, $\left|O_{1} M\right|=\left|O_{2} M\right|, M \in[A C]$.

To find: $\widehat{B A C}, \widehat{A B C}, \widehat{B C A}$.
Solution. First, we will prove that triangle $ABC$ is isosceles. Indeed, since point $O_1$ is the center of circle $\omega_1$, we have $|O_1A| = |O_1C|$. Since $(O_1O_2) \perp (AC)$ and $|O_1M| = |O_2M|$, we have $\widehat{O_2CM} = \widehat{O_1CM}$. Similarly, $\widehat{O_2AM} = \widehat{O_1AM}$. But point $O_2$ is the center of circle $\omega_2$, i.e., $\widehat{BCO_2} = \widehat{O_2CM}$, and therefore $\widehat{BCM} = 2 \widehat{O_1CM}$. Similarly, $\widehat{BAM} = 2 \widehat{O_1AM}$, and thus $\widehat{BCM} = \widehat{BAM}$.

Thus, each of the points $O_1$, $O_2$, and $B$ is equidistant from points $A$ and $C$. Therefore, if we extend the segment $O_1B$ to intersect the circle $\omega_1$ at some point $D$, we get the segment $BD$ as the diameter of this circle, and then $\widehat{BCD} = 90^\circ$. Moreover, as can be easily noticed, $\widehat{BDC} = \widehat{BAC} = \widehat{BCA}$, $\widehat{O_1CD} = \widehat{BDC}$, and, as determined above, $\widehat{O_1CM} = \frac{1}{2} \widehat{BCM}$.

Thus,

$$
\widehat{BCM} + \widehat{O_1CM} + \widehat{O_1CD} = 90^\circ
$$

or $\frac{5}{2} \widehat{BCM} = 90^\circ$, from which $\widehat{BCM} = 36^\circ$.

Thus, we have obtained: $\widehat{BCA} = \widehat{BAC} = 36^\circ$ and, consequently, $\widehat{ABC} = 108^\circ$. (Note that the diagram for this problem can only be drawn approximately before the angles of triangle $ABC$ are determined.)

We solved this example purely geometrically. For comparison, let's briefly outline an algebraic solution.

Let $\widehat{O_1CM} = x$. Then, as shown, $\widehat{BCO_2} = \widehat{O_2CM} = x$.

Introduce an auxiliary parameter, setting $|O_1M| = a$, and note that $|O_1C| = |O_1B|$. From the right triangles $O_1CM$ and $B_2P$, we have respectively:

$$
\begin{aligned}
& |O_1C| = \frac{a}{\sin x}, \quad \text{and } |O_1B| = |O_1M| + |MB|, \text{ where } |O_1M| = a, \\
& |MB| = |MO_2| + |O_2B| = a + |O_2B|; \\
& |BO_2| = \frac{a}{\cos 2a}.
\end{aligned}
$$

Thus,

$$
|O_1B| = 2a + \frac{a}{\cos 2x}
$$

and then the equality $|O_1C| = |O_1B|$ takes the form:

$$
\frac{a}{\sin x} = 2a + \frac{a}{\cos 2x}
$$

or

$$
\cos 2x = 2 \sin x \cdot \cos 2x + \sin x
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_318d1c34bf6979fc9790g-041.jpg?height=460&width=466&top_left_y=250&top_left_x=344)

Fig. 26

Solving this equation (it reduces to a cubic equation:

$4 \sin^3 x - 2 \sin^2 x - 3 \sin x + 1 = 0$), we find:

$$
\sin x = \frac{-1 + \sqrt{5}}{4}
$$

from which $x = \arcsin \frac{-1 + \sqrt{5}}{4}$. There are two other values of $\sin x$ that are extraneous in the context of the problem. Referring to tables, we get: $x = 18^\circ$. Thus, $\widehat{BCA} = \widehat{BAC} = 36^\circ$ and $\widehat{ABC} = 108^\circ$.

We provided purely geometric and algebraic methods of solving in examples 16 (p. 30) and 22. The choice of one or the other method of solution is, to some extent, a matter of taste for the person solving the problem.

Without favoring either of these methods of solution, let's consider more examples.