## Task 4 - 330824

For any triangle $A B C$, let $H$ be the foot of the perpendicular from $A$ to $B C$, and let $W$ be the intersection of $B C$ with the angle bisector through $A$.

a) What must the size of the angle $\angle W A H$ be in an isosceles triangle $A B C$ with $\overline{A C}=\overline{B C}$, where the interior angle $\angle A C B$ is $48^{\circ}$?

b), c) Are there isosceles triangles $A B C$ with $\overline{A C}=\overline{B C}$, in which the angle $\angle W A H$

b) is $12^{\circ}$,

c) is $60^{\circ}$?

Determine all possible values for the size of the base angle $\angle B A C$ in such a triangle!
a) According to the Base Angle Theorem and the Interior Angle Theorem (left figure),

$$
2 \cdot \angle B A C + \angle A C B = 180^{\circ} \quad \text { so } \quad \angle B A C = 90^{\circ} - \frac{1}{2} \angle A C B = 90^{\circ} - 24^{\circ} = 66^{\circ}
$$

Since $A W$ bisects the angle $\angle B A C$, we have $\angle B A W = 33^{\circ}$. (1)

According to the Interior Angle Theorem for triangle $A B H$,

$$
\angle B A H = 90^{\circ} - \angle A B H = 90^{\circ} - 66^{\circ} = 24^{\circ}
$$

The angle $\angle A B C$ is less than $90^{\circ}$, so $H$ lies on the same side of $A B$ as $W$. Therefore, from (1) and (2), $\angle W A H = 33^{\circ} - 24^{\circ} = 9^{\circ}$.

b) If $A B C$ is an isosceles triangle with $A C = B C$ and $\angle W A H = 12^{\circ}$, then:

Since the base angle $\angle A B C$ is less than $90^{\circ}$, $H$ lies on the same side of $A B$ as $W$, either on the segment $B W$ (left figure) or on its extension beyond $W$ (right figure). In these two cases, with the upper or lower sign,

$$
\angle B A H = \angle B A W + \angle W A R \quad \text { with } \alpha = \angle B A C \quad \text { so } \angle B A H = \frac{\alpha}{2} \mp 12^{\circ}
$$

Thus, the Interior Angle Theorem for triangle $A B H$ states

$$
\begin{gathered}
\left(\frac{\alpha}{2} \mp 12^{\circ}\right) + \alpha = 90^{\circ} \\
\frac{3}{2} \alpha = 90^{\circ} \pm 12^{\circ} \\
\alpha = 60^{\circ} \pm 8^{\circ}
\end{gathered}
$$

Indeed, there are isosceles triangles with $\alpha = 68^{\circ}$ and $\alpha = 52^{\circ}$ as base angles (since in both cases $\alpha < 90^{\circ}$); for these triangles, (4) holds, so (3) and thus $\angle W A H = 12^{\circ}$.

c) The same reasoning as in b) leads to $\alpha = 60^{\circ} \pm 40^{\circ}$.

Of these two values, $\alpha = 100^{\circ}$ is excluded because a base angle cannot be greater than $90^{\circ}$. However, there are isosceles triangles with $\alpha = 20^{\circ}$ as the base angle.

For these triangles, $\angle H A B = 70^{\circ}$ and $\angle H A W = 60^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0828.jpg?height=297&width=794&top_left_y=545&top_left_x=631)

Thus, it is shown: There are triangles as described in c); the base angle $\alpha = \angle B A C$ of such triangles is exactly $\alpha = 20^{\circ}$.

Solutions from the 2nd Round 1983 adopted from [5]

### 5.35.3 3rd Round 1993, Class 8