Let $A$ and $C$ be two points on a circle $X$ so that $AC$ is not diameter and $P$ a segment point on $AC$ different from its middle. The circles $c_1$ and $c_2$, inner tangents in $A$, respectively $C$, to circle $X$, pass through the point $P$ ¸ and intersect a second time at point $Q$. The line $PQ$ intersects the circle $X$ in points $B$ and $D$. The circle $c_1$ intersects the segments $AB$ and $AD$ in $K$, respectively $N$, and circle $c_2$ intersects segments $CB$ and ¸ $CD $ in $L$, respectively $M$. Show that: a) the $KLMN$ quadrilateral is isosceles trapezoid; b) $Q$ is the middle of the segment $BD$. Proposed by Thanos Kalogerakis
Problem
Source: Romanian "Stars of Mathematics" Junior 2019 P2
Tags: geometry
Ricochet
04.12.2019 21:13
For A) just take homothety centered at $A$ that sends $c_1$ to $(X)$ to see that $KN \| BD$ and then homethety centered at $C$ that sends $c_2$ to $(X)$ to see that $LM \| BD$ $\implies KN \| LM$. Also notice that $PKNQ$ and $PLMQ$ are both isosceles trapezoids, so $PK= QN$ and $PL=QM$. Also $\angle MQN= 360^{\circ}- \angle MQP- \angle NQP= 360^{\circ}- \angle LPQ- \angle KPQ=\angle LPK$ so $\triangle MQN \cong \triangle LPK$ and $MN=LK$ and since $KN \| LM$ we are done.
Ricochet
04.12.2019 21:33
For B) first we prove that $QNDM$ is cyclic, for this just notice that since $B$ and $D$ are on the radical axis of $c_1$ and $c_2$ then $KLCA$ and $NMCA$ are cyclic. Angle chase to get that $KL$ and $MN$ are actually common tangents of $c_1$ and $c_2$ , now angle chase again to get the desired cyclic. Now notice that homothety at $A$ gives us $KP \|BC$ and $PN\| CD$ ,and homothety at $C$ gives us $PL \| AB$ and $PM \| AD$. So $BKPL$ and $PMDN$ are both parallelograms. So $MQ=LP=BK$ and $MD=PN=QK$. But we also have $\angle BKQ=\angle QNA=\angle DMQ$ (because of the cyclic $QNDM$). So $\triangle DMQ \cong \triangle BKQ$ and $DQ=BQ$.
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