In a right-angled triangle, the orthocenter is $M$, the center of the circumscribed circle is $O$, and the sides are $a<b<c$. The line of side $c$, the altitude to side $b$, and the line $M O$ form a triangle similar to the original triangle but with the opposite orientation. What are the angles of the triangle?
Solution. Let the angles of the triangle be $\alpha, \beta, \gamma$ as usual. Let the feet of the altitudes from sides $a, b$, and $c$ be $A_{T}, B_{T}$, and $C_{T}$, respectively. Let the reflection of vertex $M$ over side $c$ be $M^{\prime}$, and the intersection points of the ray $MO$ with side $c$ and segment $AM^{\prime}$ be $P$ and $Q$, respectively.

![](https://cdn.mathpix.com/cropped/2024_05_02_4f08d11149155a22b258g-1.jpg?height=542&width=545&top_left_y=304&top_left_x=779)

The "small" triangle in the problem is the triangle $BMP$.

Since $\angle ABM < \beta$, this angle can only be $\alpha$. Therefore, $\angle MPB = \beta$ and $\angle PMB = \gamma$.

Due to the reflection,

$$
MA = M^{\prime}A
$$

It is known that reflecting the orthocenter over a side of the triangle results in a point on the circumcircle: $\angle AM^{\prime}C = \angle ABC = \beta$ and $\angle BM^{\prime}C = \angle BAC = \alpha$.

Since

$$
\begin{aligned}
\angle QMM^{\prime} & = \angle PMC_{T} = 180^{\circ} - (\angle MPC_{T} + \angle MC_{T}P) = \\
& = 180^{\circ} - (\beta + 90^{\circ}) = 90^{\circ} - \beta
\end{aligned}
$$

and thus

$$
\angle M^{\prime}QM = 180^{\circ} - (\angle QMM^{\prime} + \angle QM^{\prime}M) = 180^{\circ} - ((90^{\circ} - \beta) + \beta) = 90^{\circ}
$$

it follows that $OQ \perp AM^{\prime}$. This implies that $MQ$ is the perpendicular bisector of segment $AM^{\prime}$, and thus

$$
AM = MM^{\prime}
$$

(1) and (2) together mean that $MA = M^{\prime}A = MM^{\prime}$, so triangle $AMM^{\prime}$ is equilateral, and thus $\angle AM^{\prime}M = \beta = 60^{\circ}$.

We also know that $\alpha = \angle BM^{\prime}C = \angle BMM^{\prime}$, so the sum of the angles in triangle $BMC_{T}$ is $2\alpha + 90^{\circ}$, from which $\alpha = 45^{\circ}$. This implies $\gamma = 75^{\circ}$.

The angles of the triangle are: $\alpha = 45^{\circ}, \beta = 60^{\circ}$, and $\gamma = 75^{\circ}$.