Bay Area Mathematical Olympiad

Solution:

Translate $\triangle BDA$ horizontally until its side $BD$ coincides with side $EC$, and label the image of point $A$ by $A'$. We now have two triangles $\triangle ECA$ and $\triangle ECA'$ which share the same base $EC$, have the same height (equal to the height $AH$ of the original $\triangle ABC$), and equal angles $\angle EAC$ and $\angle EA'C$. The last implies that there is a circle $k$ passing through points $E, C, A'$ and $A$. The equal heights condition implies that $BC$ is parallel to $AA'$; yet the only trapezoids inscribed in a circle are isosceles. Therefore, $ECA'A$ is isosceles with $AE = A'C$. Since the diagonals of an isosceles trapezoid are equal, $EA' = CA$. Translating back to the original $\triangle ABC$, this means that $BA = CA$ and $\triangle ABC$ is isosceles.

Solution:

Without loss of generality, assume that the points $B, D, E$ and $C$ are arranged in this order on the line $\overleftrightarrow{BC}$; otherwise, switch points $D$ and $E$ in the remainder of the solution.

Suppose that $\triangle ABC$ is not isosceles. Without loss of generality, let $BA < CA$. Let $AL$ denote the angle bisector of $\angle BAC$, where $L$ lies on side $BC$. Reflect $\triangle ABL$ across $AL$ and denote by $B'$ and $D'$ the images of $B$ and $D$, respectively. Since $B'A = BA < CA$, $B'$ is inside side $CA$, and this forces the whole segment $LB'$ to be inside $\triangle LCA$; in particular, $D'$ is an interior point of segment $AE$.

Because of the reflection, $\triangle AD'B'$ has equal area as $\triangle ADB$. In its turn, $\triangle ADB$ has the equal area as $\triangle AEC$ (they have equal bases and heights). This implies that $\triangle AD'B'$ and $\triangle AEC$ have equal area, which contradicts the fact that one triangle is properly included in the other.

We conclude that our supposition is false. Therefore, $AB = AC$ and our triangle is isosceles as desired.

Solution:

Let $\angle BAD = \angle CAE = \theta$. In the end triangles, $\triangle ABD$ and $\triangle ACE$, and $\triangle ABC$, by the Law of Sines
$$\frac{a}{\sin \theta} = \frac{m}{\sin B} = \frac{n}{\sin C} \quad \text{and} \quad \frac{q}{\sin B} = \frac{p}{\sin C}$$
Then $mp = nq$. Suppose $p \neq q$. Then, without loss of generality, $p < q$, which implies $np < nq = mp$ and $n < m$. But if $p < q$ then $\angle C < \angle B$ which implies $\angle AED = \angle C + \theta < \angle B + \theta = \angle ADE$ and $m < n$.

This is a contradiction, so $p = q$ and $\triangle ABC$ is isosceles.

Solution:

Let $\angle BAD = \angle CAE = \theta$. $\triangle ABD$ and $\triangle ACE$ have equal areas since they have equal bases $(BD = EC)$ and the same altitude from $A$. Then
$$\frac{1}{2} p m \sin \theta = \frac{1}{2} q n \sin \theta$$
which implies $mp = nq$. Now use the same contradiction as in the above solution.