Two circles that are not equal are tangent externally at point $R$. Suppose point $P$ is the intersection of the external common tangents of the two circles. Let $A$ and $B$ are two points on different circles so that $RA$ is perpendicular to $RB$. Show that the line $AB$ passes through $P$.
Problem
Source: INAMO Shortlist 2015 G2
Tags: geometry, tangent circles, circles, perpendicular, collinear
amar_04
28.11.2019 14:54
INAMO Shortlist 2015 G2 wrote: Two circles that are not equal are tangent externally at point $R$. Suppose point $P$ is the intersection of the external common tangents of the two circles. Let $A$ and $B$ are two points on different circles so that $RA$ is perpendicular to $RB$. Show that the line $AB$ passes through $P$. Solution:- Let $\omega_1$ be the bigger circle and $\omega_2$ be the smaller circle and $O_1,O_2$ be the centers of $\omega_1,\omega_2$ respectively. Let $A\in\omega_1$ and $B\in\omega_2$ and let $PB\cap \omega_1=A'$. Let $PR\cap$ {$\omega_1,\omega_2$} be {$(X,Y)\neq R$}. So it suffices to prove that $A'\equiv A$. Note that $O_1,O_2\in PR$. Now it's obvious from the diagram that there exists a Homothety ($\Phi$) mapping $\omega_1$ to $\omega_2$ centered at $P$. So, $\begin{cases} \Phi:B\mapsto A' \\ \Phi:R\mapsto Y\end{cases}\implies A'Y\|BR\implies \angle A'RB=\angle YA'R=90^\circ=\angle ARB\implies \boxed{A'\equiv A}\blacksquare$
mmathss
28.11.2019 21:38
Easier than this.Suppose $X$ and $Y$ are the points of intersection of the external tangent with the circles.Then $\triangle XAR$ is similar to $\triangle YBR$.So homothety.