A triangle's inscribed circle passes through the center of the circumscribed circle. One of the angles of the triangle is $60^{\circ}$. Calculate the largest angle of the triangle!
In triangle $ABC$, let the center of the inscribed circle be $O$, the center of the circumscribed circle be $K$, the corresponding radii be $r$ and $R$, and the distance between the two points be $d$.

According to Euler's Theorem 1, in any triangle,

$$
d^{2}=R^{2}-2 r R
$$

In our case, the inscribed circle passes through point $K$, so $d=r$, thus $r^{2}=R^{2}-2 r R$. From this, $R=(\sqrt{2}+1) r$, the other value of the ratio $R / r$ is negative.

# $1985-10-299-1$. eps 

Let $\angle ABC = \beta = 60^{\circ}$, and $\angle BAC = \alpha$ be the largest angle of the triangle to be found. Let $T$ be the foot of the perpendicular from point $O$ to side $BC$. In the right triangle $OBT$, $OB$ is the angle bisector of $\beta$, so $\angle OBT = 30^{\circ}$, and $BO = 2r$. Thus, in triangle $KOB$, we know all three sides in terms of $r$.

Triangle $ABC$ is acute-angled, since $K$ is an interior point, so $60^{\circ} \leq \alpha < 90^{\circ}$. $\angle BKC = 2\alpha$, since this is the central angle corresponding to the inscribed angle $\alpha$. In the isosceles triangle $BKC$, $\angle KBC = \angle KCB = \frac{180^{\circ} - 2\alpha}{2} = 90^{\circ} - \alpha \leq 30^{\circ}$.

Apply the Law of Cosines to triangle $OBK$:

$$
\begin{gathered}
\angle OBK = \angle OBT - \angle KBC = 30^{\circ} - (90^{\circ} - \alpha) = \alpha - 60^{\circ}. \\
\cos(\alpha - 60^{\circ}) = \frac{BK^2 + BO^2 - OK^2}{2 \cdot BK \cdot BO} = \frac{2 \sqrt{2} - 1}{2},
\end{gathered}
$$

from which $\alpha = 83.91^{\circ}$ (we did not convert the fractional degrees).

This value is appropriate, since from (2),

$$
OK^2 = d^2 = R^2 - 4r^2 - 2(2\sqrt{2} - 1)Rr = R^2 - 2rR
$$

by (1), from which $R = (\sqrt{2} + 1)r$, and substituting this into (1): $d^2 = (\sqrt{2} + 1)^2 r^2 - 2(\sqrt{2} + 1)r^2 = r^2$, so in the triangle whose angles are $83.91^{\circ}$, $60^{\circ}$, and $36.09^{\circ}$, the inscribed circle indeed passes through the center of the circumscribed circle.

Remark. The proof that the found triangle meets the requirement was missing from almost every solution.

[^0]:    ${ }^{1}$ The proof of the theorem can be found, for example, in Molnár Emil: Collection of Mathematical Competition Problems 1947-1970 (Tankönyvkiadó, 1974), p. 285.