Let's calculate the angles of the triangle if one of its sides is $\sqrt{3}$ times larger than the distance from the opposite vertex to the orthocenter, and another side is $1+\sqrt{2}+\sqrt{3}$ times larger than the radius of the inscribed circle.
Among the data, only ratios are mentioned, so we can choose, for example, the radius $\varrho$ of the inscribed circle of the sought triangle $A B C$. Let the first mentioned side be $B C=\sqrt{3} A M$, where $M$ is the orthocenter of the triangle, and the second side be $A B=1+\sqrt{2}+\sqrt{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_7c3e4b51ddf2c51d5df1g-1.jpg?height=444&width=764&top_left_y=304&top_left_x=667)

Let the point opposite to $B$ on the circumcircle of the triangle be $B^{\prime}$. Then $B^{\prime} A$ is perpendicular to $A B$, and thus parallel to $C M$, similarly $B^{\prime} C \| A M$, so $A, C, B^{\prime}$, and $M$ are the vertices of a parallelogram, and $A M=B^{\prime} C$. This is also valid if $B^{\prime}$ coincides with $A$ or $C$, because in this case there is a right angle at $C$ or $A$ in the triangle, so $M$ falls into the vertex of the right angle. In our problem, $B^{\prime}=C$ is not possible, because if $A M=0$, then $B C=0$.

Accordingly, in the right triangle $B B^{\prime} C$, the leg $B C$ is $\sqrt{3}$ times the leg $B^{\prime} C$, so the triangle is similar to the parts of an isosceles triangle cut by its height, hence the angle at $B^{\prime}$ is $60^{\circ}$. If $A$ is on the arc of $B C$ containing $B^{\prime}$, then $\alpha=B A C \varangle=60^{\circ}$, and if it is on the other arc, then $\alpha=120^{\circ}-\mathrm{az} 1$. as shown in the first figure at $A^{*}$ and $M^{*}$.

Knowing $\alpha$, we can construct the triangle, and in the calculation, we can follow the construction. We construct a circle with radius $\varrho=1$ that is tangent to the sides of the angle $\alpha$ at vertex $A$, and measure $A B$ on one side of the angle. The line $B C$ is given by the second tangent from $B$ to the circle, and a solution exists if this intersects the other side of $\alpha$. In the case of $\alpha=60^{\circ}$, we can easily find the angles: for the point of tangency $T$, $A T=\sqrt{3}$, which can be the first part of $A B$. Taking the second part $T U=1$, the remaining part $U B=\sqrt{2}$ is equal to $U O$, so the triangle $O U B$ is isosceles, hence the angle $O B U$ is half of the exterior angle $O U T$. The angle $C B A$ is twice the angle $O B U$ because $B C$ is the reflection of $B A$ over the axis $B O$, so $\beta=C B A \varangle=O U T \varangle=45^{\circ}$. Therefore, the angles of the triangle $A B C$ are $60^{\circ}, 45^{\circ}, 75^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_7c3e4b51ddf2c51d5df1g-1.jpg?height=461&width=702&top_left_y=1472&top_left_x=706)

In the case of $\alpha=120^{\circ}$, $A T=1 / \sqrt{3}$, and thus from the triangle $B O T$,

$$
\operatorname{ctg} \frac{\beta}{2}=T B / O T=1+\sqrt{2}+\sqrt{3}-1 / \sqrt{3} \approx 3.569
$$

and in the second solution, the angles are $120^{\circ}, 31.3^{\circ}, 28.7^{\circ}$

Kiss Katalin (Makó, József A. g. III. o. t.)