## Task 1 - 180831

Inside an acute triangle $ABC$, whose interior angles have the sizes $\alpha, \beta, \gamma$, there is a point $P$ such that $\overline{P A}=\overline{P B}=\overline{P C}$ holds. The sizes of the angles $\angle P A B, \angle P A B$ respectively $\angle P A C$ are denoted by $\delta, \epsilon$ respectively $\eta$.

a) Calculate $\delta, \epsilon$ and $\eta$ for the case where $\alpha=70^{\circ}$ and $\beta=80^{\circ}$!

b) Determine a formula for $\delta$ in terms of $\alpha, \beta$ and $\gamma$, as well as a formula for $\epsilon$ and a formula for $\eta$!
a) According to the theorem of the sum of interior angles, applied to triangle $ABC$, we have $\gamma=180^{\circ}-\alpha-\beta=30^{\circ}$. Consequently,

$$
\begin{gathered}
\delta=\frac{1}{2}\left(70^{\circ}+80^{\circ}-30^{\circ}\right)=60^{\circ} \quad ; \quad \epsilon=\frac{1}{2}\left(80^{\circ}+30^{\circ}-70^{\circ}\right)=20^{\circ} \\
\eta=\frac{1}{2}\left(30^{\circ}+70^{\circ}-80^{\circ}\right)=10^{\circ}
\end{gathered}
$$

b) Since the triangles $ABP$, $BCP$, and $CAP$ are isosceles by assumption, the base angle theorem gives

$$
\begin{gathered}
\angle PBA=\angle PAB=\delta \quad, \quad \angle PCB=\angle PBC=\epsilon \\
\angle PAC=\angle PCA=\eta
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0703.jpg?height=623&width=380&top_left_y=745&top_left_x=1432)

Since $P$ lies inside the triangle, it follows that

$$
\alpha=\delta+\eta \quad, \quad \beta=\delta+\epsilon \quad, \quad \gamma=\epsilon+\eta
$$

thus $\alpha+\beta-\gamma=2 \delta$,

$$
\delta=\frac{1}{2}(\alpha+\beta-\gamma) \quad, \quad \epsilon=\frac{1}{2}(\beta+\gamma-\alpha) \quad, \quad \eta=\frac{1}{2}(\gamma+\alpha-\beta)
$$