Let $B$ be a point on a circle $k_1, A \ne B$ be a point on the tangent to the circle at $B$, and $C$ a point not lying on $k_1$ for which the segment $AC$ meets $k_1$ at two distinct points. Circle $k_2$ is tangent to line $AC$ at $C$ and to $k_1$ at point $D$, and does not lie in the same half-plane as $B$. Prove that the circumcenter of triangle $BCD$ lies on the circumcircle of $\vartriangle ABC$
Problem
Source: Croatia TST 2003 p2
Tags: Circumcenter, circumcircle, geometry
jrsbr
01.04.2023 00:52
Let $P$ and $Q$ be the intersections of $AC$ with $k_1$ and $M$ be the midpoint of the arc $PQ$ that doesn't contain $B$. By homothety, since the tangent to $k_1$ through $M$ is parallel to $PQ$ and $D$ is the center of the homothety that takes $k_1$ to $k_2$, we have $C$, $D$ and $M$ collinear. Now let $F$ be the intersection of $BM$ and $AC$. We invert by $B$ and, as usual, denote $X'$ as $X$s image under the inversion. See that $P'$, $F'$, $Q'$ and $C'$ are collinear. Furthermore, $BP'Q'F'$ is cyclic, and thus $F'$ is the midpoint of the arc $P'Q'$ that doesn't contain $B$. Thus, by shooting lemma, since $BC'M'D'$ is cyclic and $B$, $M'$ and $F'$ are collinear, we must have $F'$, $D'$ and $C'$ collinear. Now, back to the problem, we conclude that $BCDF$ is cyclic. If $O$ is the center of $(BCF)$, we have $$\angle BAC+\angle COB=\angle BAC+2(\angle BCQ+\angle QBF+\angle QBC)=\angle BAC+2\angle BQP+\angle QBP$$$$=\angle BAP+\angle PBA+\angle BQP+\angle QBP=\angle QBP+\angle BPQ+\angle PQB=180^{\circ},$$and it follows that $O$ lies on $(ABC)$. $\blacksquare$
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