72nd Czech and Slovak Mathematical Olympiad

Since the line segments $A{B}^{\prime }$, $A{C}^{\prime }$ have the same length $|A{B}^{\prime }|+|A{C}^{\prime }|$, $A{B}^{\prime }{C}^{\prime }$ is an isosceles triangle with base $B^{\prime }{C}^{\prime }$. It means that the perpendicular bisector of $B^{\prime }{C}^{\prime }$ coincides with the bisector of the angle $BAC$. Let $S\ne A$ be the intersection of this bisector with the circumcircle of $ABC$. If we prove that $S$ is a circumcenter of $A{B}^{\prime }{C}^{\prime }$ we will be finished. Since $S$ lies on the perpendicular bisector of $B^{\prime }{C}^{\prime }$, we have $|S{B}^{\prime }|=|S{C}^{\prime }|$. So, it remains to prove that $|SA|=|S{C}^{\prime }|$.

From congruence of inscribed angles $SAB$ and $SAC$ it follows that $S$ is the midpoint of the arc $BC$, and therefore $|BS|=|CS|$. From the cyclic quadrilateral $ABSC$ we have $|\angle ACS|=180^{\circ }-|\angle SBA|=|\angle C^{\prime }BS|$. Together with equality $|CA|=|B{C}^{\prime }|$ we get that triangles $SAC$ and $S{C}^{\prime }B$ are congruent by condition $SAS$ and therefore $|SA|=|S{C}^{\prime }|$.

Figure 1

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Alternative solution.

Let us define the point $S$ as in the first solution. This time we verify the desired equality $|SA|=|S{C}^{\prime }|$ by showing that $S$ lies on the perpendicular bisector of $A{C}^{\prime }$.

In the special case where $|AB|=|AC|$ holds, the midpoint of $A{C}^{\prime }$ is $B$ (according to the construction of $C^{\prime }$); therefore it suffices to verify that the angle $ABS$ is right. However, this follows from the fact that the cyclic quadrilateral $ABSC$ is then composed of two identical triangles $ABS$ and $ACS$, so the angles at their opposite vertices $B$ and $C$ are identical and therefore right.

When $|AB|\ne |AC|$, we can without loss of generality assume that $|AB|>|AC|$ as in figure 2. Here $P$ and $Q$ denote the perpendicular projections of $S$ onto lines $AB$ and $AC$, respectively. Thanks to our assumption $|AB|>|AC|$ the point $P$ lies inside the segment $AB$, while the point $Q$ lies on the opposite ray to the ray $CA$. We prove that $P$ is the midpoint of $A{C}^{\prime }$.

BPS and CQS, also the angles PSB and QSC have the same size. In addition, we have $|PS|=|QS|$, since $S$ lies on the angle bisector of $C^{\prime }A{B}^{\prime }$. We thus obtain that triangles $PBS$ and $QCS$ are congruent according to condition ASA. Hence, equality $|BP|=|CQ|$ follows. Moreover, from the identical rectangular triangles $ASP$ and $ASQ$ we also have $|AP|=|AQ|$, so together it yields $$|AP|=|AQ|=|AC|+|CQ|=|C^{\prime }B|+|BP|=|C^{\prime }P|.$$ Thus, $P$ is indeed the midpoint of $A{C}^{\prime }$, and the proof is complete.

Figure 2

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Alternative solution.

This time we denote by $S$ the circumcenter of $A{B}^{\prime }{C}^{\prime }$ and we prove that points $A$, $B$, $S$, and $C$ lie on one circle. According to the introduction of the first solution, we know that $A{B}^{\prime }{C}^{\prime }$ is an isosceles triangle with base $B^{\prime }{C}^{\prime }$, hence its circumcenter $S$ lies on the angle bisector of $C^{\prime }A{B}^{\prime }$, that is $BAC$. The points $B$ and $C$ therefore lie in the opposite half-planes with the boundary line $AS$, therefore it is sufficient to verify $|\angle ABS|=180^{\circ }-|\angle ACS|$.

By equalities $|A{C}^{\prime }|=|A{B}^{\prime }|$ and $|AS|=|C^{\prime }S|=|B^{\prime }S|$, the triangles $C^{\prime }AS$ and $A{B}^{\prime }S$ are isosceles and congruent. Thus, if we rotate triangle $C^{\prime }AS$ around $S$ about the oriented angle $C^{\prime }SA$ we obtain triangle $A{B}^{\prime }S$. Since $B$ lies on $C^{\prime }A$, $C$ lies on $A{B}^{\prime }$ and at the same time $|C^{\prime }B|=|AC|$, this rotation maps $B$ onto $C$ and thus the angle $C^{\prime }BS$ to the angle $ACS$. Therefore, $|\angle C^{\prime }BS|=|\angle ACS|$, which already follows $|\angle ABS|=180^{\circ }-|\angle C^{\prime }BS|=180^{\circ }-|\angle ACS|$ as we promised to show.

Figure 3