Let $C_1$ and $C_2$ be two concentric circles with $C_1$ inside $C_2$. Let $P_1$ and $P_2$ be two points on $C_1$ that are not diametrically opposite. Extend the segment $P_1P_2$ past $P_2$ until it meets the circle $C_2$ in $Q_2$. The tangent to $C_2$ at $Q_2$ and the tangent to $C_1$ at $P_1$ meet in a point $X$. Draw from X the second tangent to $C_2$ which meets $C_2$ at the point $Q_1$. Show that $P_1X$ bisects angle $Q_1P_1Q_2$.
Problem
Source: Canadian Junior Mathematical Olympiad - CJMO 2021 p1
Tags: geometry, concentric, angle bisector
MrOreoJuice
29.05.2021 18:41
Canada has junior Olympiads too? Anyways, let $O$ be the center of the concentric circles. $$\angle OQ_1X = \angle OQ_2X = 90^\circ$$$$\angle OP_1X = \angle OQ_2X = 90^\circ$$Thus all the $5$ points $O, P_1 , Q_1 , X , Q_2$ lie on a circle with diameter $OX$. $$\angle X P_1 Q_1 = \angle XQ_2 Q_1 = \angle X Q_1 Q_2 = \angle X P_1 Q_2.$$
parmenides51
29.05.2021 18:48
It started last year, this one is the second one, and it has common problems with Canadian MO. See here for the rest problems.
Bluesoul
02.03.2022 03:52
Notice $X,Q_1,P_1,O,Q_2$ are concyclic then angle chase with equal arcs($\widehat{XQ_2}=\widehat{XQ_1}$)
Mogmog8
05.03.2022 09:54
[asy][asy] size(7cm); import geometry; point O=(-3,1); point P_1=(3,1); point Q_2=(-8,7); circle C_1=circle(O,length(segment(O,P_1))); circle C_2=circle(O,length(segment(O,Q_2))); point P_2=intersectionpoints(C_1,P_1--Q_2)[1]; circle omega=circle(O,Q_2,P_1); point Q_1=intersectionpoints(C_2,omega)[0]; point X=intersectionpoints(bisector(Q_1,Q_2),omega)[1]; draw(C_1); draw(C_2); draw(omega,dotted); draw(P_1--Q_2); draw(Q_1--X--Q_2); draw(X--O); draw(O--P_1--Q_1); draw(O--Q_2); dot(O^^P_1^^Q_2^^P_2^^Q_1^^X); label("$O$",O,SW); label("$P_1$",P_1,SW); label("$Q_2$",Q_2,NW); label("$P_2$",P_2,N); label("$Q_1$",Q_1,SE); label("$X$",X,NE); label("$\mathcal{C}_1$",(-7,-3),N); label("$\mathcal{C}_2$",(-6.5,-6),N); [/asy][/asy] Let $O$ be the center of circles $\mathcal{C}_1$ and $\mathcal{C}_2.$ Since $\angle OQ_{\{1,2\}}X=\angle OP_1X=90,$ we know $OP_1Q_1XQ_2$ is cyclic. Hence, $$\angle Q_1P_1X=\tfrac{1}{2}\widehat{XQ_1}=\tfrac{1}{2}\widehat{XQ_2}=\angle XP_1Q_2.$$$\square$
parmenides51
06.05.2024 14:24
Notation: $(KL)$ stands for small arc $KL$
because $\angle OP_1x=\angle OQ_2X=90^o=\angle OQ_2X$,
we get that $OP_1Q_2XP_2$ is cyclic (with diameter $OX$)
$\angle P_2P_1X=(P_2X)/2=\angle P_2Q_2X =(*) \angle Q_2P_2X=(Q_2X)/2=\angle XP_1Q_2$
* because $Q_2X=P_2X$ as tangents of circle $C_2$ from point $P$
Fermat_Fanatic108
20.10.2024 20:20
[ See attachment ] It is clear that, $ \angle O Q_2 X = \angle O P_1 X = \angle O Q_1 X = 90^\circ $ Thus , we find that $ X Q_1 P_1 O Q_2$ lie on a circle Since, $ Q_1 P_1 Q_2 X $ is a cyclic quad $ \implies \angle X Q_1 Q_2 = \angle X P_1 Q_2 $ $\hspace{48mm} $ $ \implies \angle Q_1 Q_2 X = \angle Q_1 P_1 X$ Since, $ X Q_1 = X Q_2 \implies \angle X Q_1 Q_2 = \angle X Q_2 Q_1 $ This tells us, $ \angle X P_1 Q_2 = \angle X P_1 Q_1 $ $ \blacksquare $
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