## Task 4 - 330724

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

a) What must the base angle $\angle B A C$ of an isosceles triangle $A B C$ with $A C = B C$ be for the points $H$ and $W$ to coincide?

b) What must the angle $\angle W A H$ in an isosceles triangle $A B C$ with $A C = B C$ be if the base angle $\angle B A C$ is $70^{\circ}$?

c) Determine all possible values for the base angle $\angle B A C$ in an isosceles triangle $A B C$ with $A C = B C$ for which the angle $\angle W A H$ is $15^{\circ}$.
a) From the assumption $H=W$, it follows that triangle $ABC$ is also isosceles with $AB=AC$, and due to the assumption $AC=BC$, it is even equilateral. Therefore, $\angle BAC=60^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0547.jpg?height=505&width=822&top_left_y=1769&top_left_x=617)

b) See Figure a): Since $AW$ bisects the angle $\angle BAC$, it follows that $\angle BAW=35^{\circ}$. (1)

By the base angle theorem, $\angle ABC=70^{\circ}$; by the interior angle theorem for triangle $ABH$, it further follows that $\angle BAH=90^{\circ}-\angle ABH=20^{\circ}$. (2)

The angle $\angle ABC$ is less than $90^{\circ}$, so $H$ lies on the same side of $AB$ as $W$. Therefore, from (1) and (2), $\angle WAH=35^{\circ}-20^{\circ}=15^{\circ}$.

c) If $ABC$ is an isosceles triangle with $AC=BC$ and $\angle W A H=15^{\circ}$, then it follows: Since the base angle $\angle ABC$ is less than $90^{\circ}$, $H$ lies on the same side of $AB$ as $W$, so either on the segment $BW$ or on its extension beyond $W$ (see Figures a and b).

In these two cases, with the upper or lower sign, $\angle BAH=\angle BAW \mp \angle W A H$, with $\alpha=\angle BAC$ thus

$$
\angle BAH=\frac{\alpha}{2} \mp 15^{\circ}
$$

According to the interior angle theorem for triangle $ABH$,

$$
\left(\frac{\alpha}{2} \mp 15^{\circ}\right)+\alpha=90^{\circ} \quad \Rightarrow \quad \alpha=60^{\circ} \pm 10^{\circ}
$$

Indeed, there are isosceles triangles with $\alpha$ as the base angle measure for both $\alpha=70^{\circ}$ and $\alpha=50^{\circ}$ (since in both cases $\alpha<90^{\circ}$); for these triangles, (4) holds, thus (3) and therefore $\angle W A H=15^{\circ}$. This shows that exactly the two values $70^{\circ}$ and $50^{\circ}$ are possible as the base angle measure in a triangle of the type mentioned in c).

Solutions from the 2nd Round 1993 adopted from [5]

### 4.35.3 3rd Round 1993, Class 7