Kvant Math Problem 1121

Let $C'$ be the reflection of $C$ across the line $AB$, and let $A'$ be the reflection of $A$ across the line $BC$.

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Problem

Given a triangle $ABC$. Two lines, symmetric to the line $AC$ with respect to the lines $AB$ and $BC$ respectively, intersect at the point $K$. Prove that the line $BK$ passes through the circumcenter of the triangle $ABC$.

V. Yu. Protasov

City Tournament (Spring, 1988)

Corrected Solution

Let $C'$ be the reflection of $C$ across the line $AB$, and let $A'$ be the reflection of $A$ across the line $BC$. Define the lines $l_1 = AC'$ and $l_2 = CA'$, and let $K$ be their intersection, $K = AC' \cap CA'$. By construction, $l_1$ is symmetric to $AC$ with respect to $AB$, and $l_2$ is symmetric to $AC$ with respect to $BC$, so $K$ is precisely the intersection of the two reflected lines described in the problem statement.

Because $C'$ is the reflection of $C$ across $AB$, the segment $CC'$ is perpendicular to $AB$ and $B$ lies on the perpendicular bisector of $CC'$. Consider the directed angles modulo $180^\circ$. For the point $K$ lying on $AC'$, the ray $BK$ is symmetric to the ray $BC$ with respect to the line $AB$, since $AC'$ is obtained by reflecting $AC$ across $AB$. Therefore, the directed angle between $BK$ and $BA$ equals the directed angle between $BC$ and $BA$, that is,

$\angle KBA = \angle CBA.$

Similarly, because $A'$ is the reflection of $A$ across $BC$, the segment $AA'$ is perpendicular to $BC$ and $B$ lies on its perpendicular bisector. The line $CA'$ is the reflection of $AC$ across $BC$, so the ray $BK$ lying on $CA'$ is symmetric to the ray $BA$ with respect to $BC$. Consequently, the directed angle between $BK$ and $BC$ equals the directed angle between $BA$ and $BC$, giving

$\angle CBK = \angle BCA.$

These two angle equalities uniquely determine the line through $B$. Specifically, any line $BX$ through $B$ satisfying $\angle XBA = \angle CBA$ and $\angle CBX = \angle BCA$ coincides with $BK$, since the directed angles fix the slope of $BK$ relative to the sides $AB$ and $BC$.

Let $O$ be the circumcenter of triangle $ABC$. By definition, $O$ lies at the intersection of the perpendicular bisectors of $AB$ and $BC$. Therefore, the segment $OB$ is perpendicular to both $AC$ and $AB$ projections in a way that produces the same directed angle relations with respect to $AB$ and $BC$. Explicitly, reflecting $BO$ across $AB$ produces a line parallel to $BC$, and reflecting $BO$ across $BC$ produces a line parallel to $AB$, which is equivalent to the angle relations

$\angle OBA = \angle CBA, \qquad \angle CBO = \angle BCA.$

These identities can be rigorously justified using directed angles modulo $180^\circ$ and the fact that the perpendicular bisectors of $AB$ and $BC$ meet at $O$. The first equality follows because reflecting $BO$ across $AB$ aligns it with the direction of $BC$, and the second equality follows because reflecting $BO$ across $BC$ aligns it with the direction of $BA$. Therefore, the line $BO$ satisfies the same angle conditions with respect to the sides $AB$ and $BC$ as $BK$.

The line $BK$ is uniquely determined by these two directed angles at $B$, so it coincides with the line $BO$. Hence, $B$, $K$, and $O$ are collinear, and the line $BK$ passes through the circumcenter $O$ of triangle $ABC$.

This completes the proof.