In a right-angled triangle, the orthocenter divides one of the altitudes in the ratio $1: 2$, and another in the ratio $2: 3$, in both cases the larger segment falls on the side of the vertex. Calculate the angles of the triangle.

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Note: The original problem statement is about a right-angled triangle, but the context suggests it might be a non-right-angled triangle with a specific orthocenter property. The translation maintains the original problem's wording and structure.
In triangle $ABC$, let the starting vertices of the altitudes be $A$ and $B$, and their feet be $A_1$ and $B_1$, respectively, with the orthocenter being $M$.

![](https://cdn.mathpix.com/cropped/2024_05_02_38be380be40979d55ad5g-1.jpg?height=379&width=422&top_left_y=266&top_left_x=838)

Then, with the notation

$$
A_1 M = x \quad \text{and} \quad B_1 M = 2y,
$$

we have

$$
M A = 2x, \quad \text{and} \quad M B = 3y
$$

and based on the similarity of triangles $M B A_1$ and $M A B_1$,

$$
\frac{M B}{M A} = \frac{M A_1}{M B_1}, \quad 6y^2 = 2x^2, \quad x : y = \sqrt{3} : 1
$$

From this, using the usual notation for angles, we have

$$
\cos \gamma = \cos \angle A_1 M B = \frac{x}{3y} = \frac{1}{\sqrt{3}}, \quad \gamma = 54^\circ 44',
$$

and in triangle $M B A_1$,

$$
B A_1 = \sqrt{9y^2 - x^2} = \sqrt{3x^2 - x^2} = x \sqrt{2}
$$

thus, from triangle $A B A_1$,

$$
\tan \beta = \frac{A A_1}{B A_1} = \frac{3}{\sqrt{2}}, \quad \beta = 64^\circ 36'
$$

finally, $\alpha = 60^\circ 40'$.

Remarks. 1. Those who recognize that since $M$ trisects $A A_1$, just as the centroid $S$ does the medians, will immediately see that the Euler line of our triangle is parallel to side $BC$. And if they happen to know that this parallelism condition is $1 - \tan \beta \cdot \tan \gamma = 3$, they can find $\beta$ from $\tan \gamma = \sqrt{2}$.

2. We constructed the triangle from these data in Exercise 809, K. M. L. 27 (1963), p. 25.[^0]

[^0]:    ${ }^{1}$ See, for example: Interesting Mathematical Practice Problems, Part II (compiled by Jenő Tolnai), Tankönyvkiadó, Budapest, 1971, p. 43, Problem 361.