A point $C$ lies on a line segment $AB$ between $A$ and $B$ and circles are drawn having $AC$ and $CB$ as diameters. A common tangent to both circles touches the circle with $AC$ as diameter at $P \ne C$ and the circle with $CB$ as diameter at $Q \ne C$. Prove that $AP, BQ$ and the common tangent to both circles at $C$ all meet at a single point which lies on the circumference of the circle with $AB$ as diameter.
Problem
Source: 2021 Irish Mathematical Olympiad P8
Tags: concurrency, concurrent, geometry
cadaeibf
30.05.2021 23:07
Let us invert at $C$ with any radius ($X\mapsto X^*$). Then $(AC)$ and $(CB)$ go to two parallel lines, perpendicular to line $A^*B^*C$, and line $PQ$ goes to one of the two circles tangent to the two lines and passing through $C$. Since the two tangents to $(CP^*Q^*)$ are parallel, they are perpendicular to $P^*Q^*$. Therefore, by taking the intersection $X^*$ of $P^*Q^*$ and the line through $C$ parallel to $A^*P^*$ and $B^*Q^*$, we see that $\angle P^*X^*C=\angle CX^*Q^*=\frac{\pi}{2}$, which means that $X^*\in (CP^*)=(CP^*A^*)$ and $X^*\in (CQ^*)=(CQ^*B^*)$. Furthermore, since $CP^*Q^*$ is the reflection of $X^*A^*B^*$ with respect to the axis of $CX^*$, $X^*$ belongs to the circle of diameter $A^*B^*$. Therefore, reinverting, $X$ lies on the common internal tangent of $(AC)$ and $(CB)$, on lines $AP$ and $BQ$ and on circle of diameter $AB$
circlethm
27.07.2021 23:44
Solution. Let $X$ be the intersection of $PQ$ and the tangent at $C$. Since the tangent at $C$ is the radical axis of the circles $(APC)$ and $(CQB)$, we have $XP = XQ$, and since it is also a tangent, we have $XP = XQ = XC$. Thus $X$ is the centre of $(PCQ)$. Let $Y$ be the intersection of $AP$ and $BQ$. Since $\angle CPY = \angle YQC = 90^\circ$, $(PCQY)$ is cyclic, with diameter $CY$. But we know that $X$ is the centre of $(PCQ)$, and thus $X$ lies on $CY$, so $CY$ is the tangent at $C$, and hence $AP, BQ$ and the tangent at $C$ all meet at $Y$. Finally, to see that $Y$ lies on the circle with diameter $AB$, it suffices to show that $\angle AYB = 90^\circ$. Indeed, we note that since $CX \perp AB$, $AB$ is a tangent to $(PCQ)$ at $C$. Thus $\angle PCA = \angle XQC = \angle QBC$, and thus $\triangle APC \sim \triangle CQB \sim AYB$, implying that $\angle AYB = 90^\circ$.
Jia-Jia-Jia-Jia
21.12.2022 19:48
Let $S$ be the exsimilicenter of the two circles. $D = AP \cap BQ$. A homothety at $S$ sending $(APC)$ to $(CQB)$ implies $AP \parallel CQ$ and $BQ \parallel CP$. Thus $BPDQ$ is a parallelogram. Since $\angle DPC = 180^\circ - \angle APC = 90^\circ$, $BPDQ$ is a rectangle (so $ADB = 90^\circ$). Now since $BPDQ$ cyclic and $PQ$ tangent to $(BC)$, \[\angle DCA = \angle DCP + \angle PCA = \angle PCA + \angle DQP = \angle PCA + 180^\circ - \angle BQP = \angle PCA + 90^\circ - \angle PQC = \angle PCA + 90^\circ - \angle DBA = 90^\circ\]Hence $CD \perp AB$, $CD$ is tangent to $(AC)$ and $(BC)$.