Estonian Mathematical Olympiad

Fig. 18 Fig. 19 Fig. 20

Solution: Let the foot of the altitude drawn from vertex $B$ of triangle $ABC$ be $K$. By symmetry, $\angle B{A}^{\prime }C=\angle BAC$ and $\angle B{B}^{\prime }C=\angle K{B}^{\prime }C=\angle KBC$.

a) By assumptions, $\angle BAC=\angle KBC$ (Fig. 18). As the angle at vertex $K$ of triangle $KBC$ is right, $\angle KBC<90^{\circ }$. If the largest angle of triangle $ABC$ were at vertex $A$, the triangle would have to be acute. This would mean that $K$ would lie on line segment $AC$, whence $\angle ABC>\angle KBC=\angle BAC$. Consequently, the largest angle of triangle $ABC$ cannot be at vertex $A$.

b) If the angle at vertex $B$ of triangle $ABC$ is right then it is the largest angle of the triangle (Fig. 19). Triangles $BAC$ and $KBC$ are similar, whence $\angle BAC=\angle KBC$ and $\angle B{A}^{\prime }C=\angle B{B}^{\prime }C$.

c) Let $ABP$ be an equilateral triangle with midpoint $C$ (Fig. 20). Obviously $P$ is the reflection of $A$ through line $BC$, as well as the reflection of $B$ through line $AC$. Hence defining $A^{\prime }=B^{\prime }=P$ fulfills the conditions of the problem. The angle at vertex $C$ of triangle $ABC$ has size $120^{\circ }$, whence it must be the largest.

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Alternative solution.

a) By symmetry, $\angle B{A}^{\prime }C=\angle BAC$. Hence the assumptions imply $\angle BAC=\angle B{B}^{\prime }C$. Suppose that the largest angle of the triangle $ABC$ is at the vertex $A$. Then the angle at the vertex $C$ must be acute, implying that points $A$ and $B^{\prime }$ lie at the same side of the line $BC$. Now $\angle BAC=\angle B{B}^{\prime }C$ implies that $B^{\prime }$ lies on the circumcircle of the triangle $ABC$ (Fig. 21). Hence $AC$ is the perpendicular bisector of the chord $B{B}^{\prime }$ of the circumcircle of the triangle $ABC$, meaning that it passes through the circumcentre of the triangle $ABC$. Consequently, $AC$ is the diameter of the circumcircle of the triangle $ABC$. Hence the triangle $ABC$ has right angle at the vertex $B$ which contradicts the assumption that the largest angle is at the vertex $A$.

Fig. 21 Fig. 22 Fig. 23

b) If the angle at vertex $B$ of triangle $ABC$ is right then it is the largest angle of the triangle (Fig. 22). By symmetry, $\angle A{B}^{\prime }C=\angle ABC=90^{\circ }$. Hence points $A,B,C$, and $B^{\prime }$ are concyclic, whereby $A$ and $B^{\prime }$ lie on the same side from line $BC$. Consequently $\angle B{A}^{\prime }C=\angle BAC=\angle B{B}^{\prime }C$.

c) Choose points $B$ and $C$ on a circle $c$ with centre $P$ in such a way that $\angle BPC<90^{\circ }$. Let $B^{\prime }$ be the reflection of $B$ through line $CP$. Let $A^{\prime }$ be a point on circle $c$ such that $\angle A^{\prime }CB=180^{\circ }-\angle PCB$ (Fig. 23). This choice is possible as moving with point $A^{\prime }$ along the longer arc $BC$ of circle $c$ gives rise to all values of $\angle A^{\prime }CB$ below $180^{\circ }-\frac{1}{2}\angle BPC$, but $180^{\circ }-\angle PCB=90^{\circ }+\frac{1}{2}\angle BPC<180^{\circ }-\frac{1}{2}\angle BPC$. Let now $A$ be the reflection of $A^{\prime }$ through line $BC$. As $\angle ACB=\angle A^{\prime }CB=180^{\circ }-\angle PCB$, point $A$ lies on line $CP$, i.e., lines $CP$ and $AC$ coincide. The conditions of the problem are met, since both $B{A}^{\prime }C$ and $B{B}^{\prime }C$ are angles inscribed in circle $c$ subtended by equal chords.