5.3. Find the angle $C$ of triangle $ABC$, if the distance from vertex $C$ to the orthocenter of the triangle is equal to the radius of the circumscribed circle.
5.3. Answer. $60^{\circ}$.

1st solution. Let $O$ be the center of the circumcircle of triangle $ABC$, $M$ be the orthocenter, and $CC'$ be the diameter of the circumcircle (see figure a). Inscribed angles $ABC$ and $AC'C$ are equal, so the acute angles $MCB$ and $ACO$, which complement them to $90^{\circ}$, are also equal.

![](https://cdn.mathpix.com/cropped/2024_05_21_20eb9f6ef2c568605a63g-44.jpg?height=442&width=418&top_left_y=514&top_left_x=156)
a)

![](https://cdn.mathpix.com/cropped/2024_05_21_20eb9f6ef2c568605a63g-44.jpg?height=425&width=407&top_left_y=513&top_left_x=629)

![](https://cdn.mathpix.com/cropped/2024_05_21_20eb9f6ef2c568605a63g-44.jpg?height=37&width=42&top_left_y=954&top_left_x=827)

Fig. 5.3.

Let $K$ be the midpoint of $AC$, and $H$ be the foot of the altitude dropped from $A$ to $BC$. The median $KH$ of the right triangle $ACH$ is equal to $AC / 2 = KC$ (it is the radius of the circumcircle of $ACH$).

Both of these facts are true for any triangle $ABC$. Additionally, in this triangle, $KC = CH$. This follows from the equality of the right triangles $KCO$ and $CHM$ ($CM = CO$ by the condition and $\angle KCO = \angle HCM$ as proven). Therefore, triangle $KCH$ is equilateral, and the desired angle $C$ is $60^{\circ}$.

2nd solution. This solution is based on the following observation: in any triangle, the distance from the orthocenter to a vertex is twice the distance from the center of the circumcircle to the opposite side (in figure b, $CM = 2OC_1$). This fact is incidentally obtained during the proof of Euler's theorem (see the solution to problem 4.8). Now, from the condition of the problem, it immediately follows that in the right triangle $BOC_1$, the leg $OC_1$ is equal to $BO / 2$, and thus $\angle BOC_1 = 60^{\circ}$. But $\angle BCA = \angle BOC_1$, since both the inscribed angle $BCA$ and the central angle $BOC_1$ measure half the arc $BA$. Therefore, $\angle BCA = 60^{\circ}$.