We have drawn the angle trisectors of the $A B C$ triangle. Let us call the first trisector of the angle at vertex $A$ the ray that forms half the angle with $A B$ as it does with $A C$. Similarly, the first trisector of angle $B$ forms half the angle with $B C$ as it does with $B A$, and the first trisector of angle $C$ forms half the angle with $C A$ as it does with $C B$. Calculate the angles of the triangle if the two angles of the triangle formed by the first trisectors are $45^{\circ}$ and $55^{\circ}$. What are the angles of the triangle formed by the second trisectors?
Let the first trisectors from vertices $A, B, C$ be $a^{\prime}, b^{\prime}, c^{\prime}$, and the second trisectors be $a^{\prime \prime}, b^{\prime \prime}, c^{\prime \prime}$. Furthermore, let the intersection of the rays $a^{\prime}, b^{\prime}$ be $C^{\prime}$, the intersection of $b^{\prime}, c^{\prime}$ be $A^{\prime}$, and the intersection of $c^{\prime}, a^{\prime}$ be $B^{\prime}$. We need to find the relationships between the angles of the triangles $A B C=H$ and $A^{\prime} B^{\prime} C^{\prime}=H^{\prime}$. For this, we need to clarify the relative positions of $H$ and $H^{\prime}$. We will show that $H^{\prime}$ is inside $H$, specifically on the same side of each first trisector as the second trisector (as a ray) from the vertex of the angle; for example, $H^{\prime}$ is on the same side of $a^{\prime}$ as $a^{\prime \prime}$. Let the intersection points of the internal angle bisectors from $A, B, C$ with the opposite sides be $A_{1}$, $B_{1}$, $C_{1}$, and the common point of the angle bisectors be $O$.

![](https://cdn.mathpix.com/cropped/2024_05_02_e734568d0aea22e7695cg-1.jpg?height=422&width=656&top_left_y=445&top_left_x=729)

Then $a^{\prime}$ is in the angle region $B A A_{1}$, $b^{\prime}$ is in the angle region $C B B_{1}$, so $C^{\prime}$ is in their common part, in the triangle $A_{1} O B$, which is part of the triangle $A B C$. Similarly, $A^{\prime}$ is in the triangle $B_{1} O C$, and $B^{\prime}$ is in the triangle $C_{1} O A$. From this, we also see that, for example, the points on $a^{\prime}$ and $b^{\prime}$ are in the order $A, B^{\prime}, C^{\prime}$, and $B, C^{\prime}, A^{\prime}$, so $C^{\prime}$ indeed separates $B$ and $A^{\prime}$.

Accordingly, the angle $\gamma^{\prime}$ at $C^{\prime}$ in $H^{\prime}$ is the exterior angle of the triangle $A B C^{\prime}$, so with the usual notation, $\gamma^{\prime}=(\alpha+2 \beta) / 3$. Similarly, the angles at $A^{\prime}$ and $B^{\prime}$ are $\alpha^{\prime}=(\beta+2 \gamma) / 3$ and $\beta^{\prime}=(\gamma+2 \alpha) / 3$. - Assuming the given angles $\alpha^{\prime}$ and $\beta^{\prime}$ of $H^{\prime}$, we need to solve the following system of equations for $\alpha, \beta, \gamma$:

$$
\begin{aligned}
& \beta+2 \gamma=3 \alpha^{\prime} \\
& \gamma+2 \alpha=3 \beta^{\prime} \\
& \alpha+\beta+\gamma=180^{\circ}
\end{aligned}
$$

From (2), $\gamma=3 \beta^{\prime}-2 \alpha$, and with this, from (1), $\beta=3 \alpha^{\prime}-6 \beta^{\prime}+4 \alpha$, and with both, from (3):

$$
\alpha=60^{\circ}-\alpha^{\prime}+\beta^{\prime} \text {, so } \beta=240^{\circ}-\alpha^{\prime}-2 \beta^{\prime}, \gamma=-120^{\circ}+2 \alpha^{\prime}+\beta^{\prime} \text {. }
$$

In each of these formulas, the coefficients of $\alpha^{\prime}$ and $\beta^{\prime}$ are different, so we need to consider both orders of the given numerical values for $\alpha^{\prime}$ and $\beta^{\prime}$.

$$
\begin{aligned}
& \alpha^{\prime}=45^{\circ} \text { and } \beta^{\prime}=55^{\circ} \text { give: } \alpha=70^{\circ}, \quad \beta=85^{\circ}, \quad \gamma=25^{\circ}, \\
& \alpha^{\prime}=55^{\circ} \text { and } \beta^{\prime}=45^{\circ} \text { give: } \alpha=50^{\circ}, \quad \beta=95^{\circ}, \quad \gamma=35^{\circ},
\end{aligned}
$$

Thus, we have two solutions for $H$, in which the traversal of $H^{\prime}$ is opposite.

For the angles of the triangle $H^{\prime \prime}$ formed by the trisectors $a^{\prime \prime}, b^{\prime \prime}, c^{\prime \prime}$, similarly to the above,

$$
\alpha^{\prime \prime}=(2 \beta+\gamma) / 3, \quad \beta^{\prime \prime}=(2 \gamma+\alpha) / 3, \quad \gamma^{\prime \prime}=(2 \alpha+\beta) / 3
$$

therefore, expressing each with $\alpha^{\prime}$ and $\beta^{\prime}$ according to (4),

$$
\begin{aligned}
& \alpha^{\prime \prime}=\left[\left(480^{\circ}-2 \alpha^{\prime}-4 \beta^{\prime}\right)-120^{\circ}+2 \alpha^{\prime}+\beta^{\prime}\right] / 3=120^{\circ}-\beta^{\prime}, \\
& \beta^{\prime \prime}=\alpha^{\prime}+\beta^{\prime}-60^{\circ}, \quad \gamma^{\prime \prime}=120^{\circ}-\alpha^{\prime} .
\end{aligned}
$$

According to these, swapping the order of $\alpha^{\prime}$ and $\beta^{\prime}$ swaps $\alpha^{\prime \prime}$ and $\gamma^{\prime \prime}$, while $\beta^{\prime \prime}$ remains unchanged. In other words, the angles of $H^{\prime \prime}$ are the same in the two solutions for $H$, but their traversal is opposite. The values of the angles, in order of magnitude: $40^{\circ}, 65^{\circ}, 75^{\circ}$.

István Deák (Budapest, Vörösmarty M. g. I. o. t.)