Estonian Math Competitions

Answer: $45^{\circ }$, $90^{\circ }$, $135^{\circ }$.

Let $\alpha =\angle BAC$. Both triangles $BEA$ and $BEC$ have right angle at vertex $E$. Hence these triangles can be isosceles only if their other angles have size $45^{\circ }$.

Suppose that $BEC$ is isosceles (see figure below). Then $\angle BCE=45^{\circ }$. Since $\angle BCE=\angle BCA=\frac{180^{\circ }-\alpha }{2}$, we have $\alpha =180^{\circ }-2\cdot 45^{\circ }=90^{\circ }$.

Suppose now that $BEA$ is isosceles. Then $\angle BAE=45^{\circ }$. If $E$ lies on the line segment $AC$ (see figure below) then $\angle BAE=\angle BAC=\alpha$, implying $\alpha =45^{\circ }$. If $E$ lies outside the line segment $AC$ (see figure below) then $\angle BAE=180^{\circ }-\angle BAC=180^{\circ }-\alpha$, implying $\alpha =180^{\circ }-45^{\circ }=135^{\circ }$.







---

Alternative solution.

Let $\alpha =\angle BAC$. Then $\angle ABC=\angle ACB=\frac{180^{\circ }-\alpha }{2}$ and $\angle EBC=90^{\circ }-\frac{180^{\circ }-\alpha }{2}=\frac{\alpha }{2}$. Both triangles $BEA$ and $BEC$ have right angle at vertex $E$. Hence these triangles can be isosceles only if their other two angles have equal size.

Suppose that $BEC$ is isosceles. Then $\frac{180^{\circ }-\alpha }{2}=\frac{\alpha }{2}$, implying $\alpha =90^{\circ }$.

Suppose now that $BEA$ is isosceles. If the triangle $ABC$ is acute, we must have $\angle BAE=\alpha$ and $\angle ABE=\frac{180^{\circ }-\alpha }{2}-\frac{\alpha }{2}=90^{\circ }-\alpha$. Thus $\alpha =90^{\circ }-\alpha$, implying $\alpha =45^{\circ }$. If the triangle $ABC$ is obtuse then $\angle BAE=180^{\circ }-\alpha$ and $\angle ABE=\frac{\alpha }{2}-\frac{180^{\circ }-\alpha }{2}=\alpha -90^{\circ }$. Hence $180^{\circ }-\alpha =\alpha -90^{\circ }$, implying $\alpha =135^{\circ }$. The triangle $ABC$ is not right as, otherwise, the triangle $BEA$ would have two right angles.