Circles $\Omega_{1},$ centre $Q,$ and $\Omega_{2},$ centre $R,$ touch externally at $B .$ A third circle, $\Omega_{3},$ which contains $\Omega_{1}$ and $\Omega_{2},$ touches $\Omega_{1}$ and $\Omega_{2}$ at $A$ and $C,$ respectively. Point $C$ is joined to $B$ and the line $B C$ is extended to meet $\Omega_{3}$ at $D$. Prove that $Q R$ and $A D$ intersect on the circumference of $\Omega_{1}$.
Problem
Source: Thirty Third Irish Mathematical Olympiad 2020 P3/10
Tags: geometry, touching circles, Inversion
franchester
20.07.2020 21:12
Cute. Draw the common tangent between $\Omega_1$ and $\Omega_2$, and let the segments formed containing $\Omega_1$ and $\Omega_2$ be called $A$ and $B$, respectively. It's well known that $CB$ and $AB$ will pass through the midpoints of the arcs of segments $A$ and $B$, respectively. By definition, then, $D$ is the midpoint of the arc of segment $A$. Let $X$ be the other arc midpoint. Clearly $DX$ is a diameter of $\Omega_3$. Because of this, $\angle DAX=\angle DAB=90^\circ$. Note that $QR$ is simply the diameter of $\Omega_1$ through $B$. If we let $B'$ be its antipode, then we have $\angle BAB'=90^{\circ}=\angle BAD$ which implies $B'$ lies on $AD$, so $AD$ and $QR$ intersect on $\Omega_1$, as desired.
circlethm
20.07.2020 21:58
Solution (Not mine) If we invert all points with respect to the circle with center $B$, we get the following problem. Problem. Two parallel lines $\Omega_1'$ and $\Omega_2'$ have a point $b$ in between them. A circle $\Omega_3'$ is tangent to $\Omega_1'$ and $\Omega_2'$ at $A'$ and $C'$ respectively. Line $BC'$ meets $\Omega_3'$ again at $D'$. The line $Q' R'$ maps to the line perpendicular to both $\Omega_1'$ and $\Omega_2'$ that goes through $B$ after it has been inverted. Prove that this line and circle $(BA'D')$ meet again on $\Omega_1'$. Let $l$ be the line $QR$ inverted with respect to $B$. Then $l \perp \Omega_1'$. Since $A'C'$ is a diameter of $\Omega_3'$, we know $C'D' \perp D' A'$. Let $l$ meet $\Omega_1'$ at $U'$. Since $l \perp \Omega_1'$, we know $\angle BU'A' = 90^\circ$. As $C'G' \perp D'A'$, we have $BD' \perp D'A'$ so $\angle BD'A = 90^\circ$. Since $\angle BU'A' = 90^\circ = \angle BD'A$, we know $U'$, $A'$, $D'$ are concyclic. In the original problem, we would have defined $U$ to also be where $QR$ and $\Omega_1$ meet, so that gives $B$, $U'$, $A'$, $D'$ are concyclic, so $U$, $A$ and $D$ are collinear, so we are done.
Miku3D
13.02.2021 05:43
Very kawaii problem
Throughout this solution, we use $\measuredangle$ to denote directed angles modulo $180^{\circ}$.
Let $QR$ intersect $\Omega_{1}$ at point $X$. It suffices to show that points $D,X,A$ are collinear.
Clearly, $BX$ is a diameter of $\Omega_{1}$, so $\measuredangle BAX = 90^{\circ}$. Thus, it suffices to show that $\measuredangle BAD = 90^{\circ}$.
Let line $AB$ meet $\Omega_{3}$ a second time a point $E$ and $O$ be the center of $\Omega_{3}$. Clearly, points $Q,B,R$, points $O,Q,A$, and points $O,R,C$ are collinear.
Now it suffices to show that $DE$ is a diameter of $\Omega_{3}$, or equivalently, $D,O,E$ collinear. But note that
$$\measuredangle OEB+ \measuredangle EBD + \measuredangle BDO$$$$=\measuredangle OEA + \measuredangle ABC + \measuredangle CDO$$$$=\measuredangle EAO + \measuredangle ABC + \measuredangle OCD$$$$=\measuredangle BAQ + \measuredangle ABC + \measuredangle RCB$$$$=\measuredangle QBA + \measuredangle ABC + \measuredangle CBR$$$$=\measuredangle QBR = 0^{\circ}$$and
$$\measuredangle OEB+ \measuredangle EBD + \measuredangle BDO + \measuredangle DOE = 0^{\circ}$$
Thus, we get $\measuredangle DOE = 0^{\circ}$, which implies that points $D,O,E$ are collinear. It then follows that $DE$ is a diameter of $\Omega_{3}$. Thus, $\measuredangle BAD = \measuredangle EAB = 90^{\circ} = \measuredangle BAX$, so $D,X,A$ collinear. Therefore, lines $QR$ and $AD$ intersect at $X$, which lies on the circumference of $\Omega_ {1}$. $\blacksquare$
DrYouKnowWho
03.04.2021 08:13
Define the following homotheties. $\mathcal{H}_A$ centered at $A$ sending $\Omega_1\to \Omega_3$, $\mathcal{H}_C$ centered at $C$ sending $\Omega_3\to \Omega_2$, $\mathcal{H}:= \mathcal{H}_A\circ\mathcal{H}_C$ sending $\Omega_1\to\Omega_2$. Now notice that if we let $T$ denote the second intersection of $AD$ with $\Omega_1$, we have $\mathcal{H}(T) = \mathcal{H}_C(\mathcal{H}_A(T)) = \mathcal{H}_C(D) = B$. And also, this implies that $T$,$B$ and the two centers of $\Omega_1$ and $\Omega_2$ are collinear since $B\in QR$, as desired.