Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.7 National olympiad Prove it North Macedonia

The circles $k_1$ , $k_2$ intersect at points $A$ and $B$ . A line through $B$ intersects the circles $k_1$ and $k_2$ for the second time at points $C$ and $D$ , respectively, in such a way that $C$ lies outside of $k_2$ , and $D$ lies outside of $k_1$ . Let $M$ be the point of intersection of the tangents to $k_1$ and $k_2$ drawn through $C$ and $D$ , respectively, and $AM \cap CD = \{P\}$ . The tangent drawn through $B$ to $k_1$ intersects $AD$ in point $L$ , and the tangent drawn through $B$ to $k_2$ intersects $AC$ in point $K$ . Let $KP \cap MD = \{N\}$ and $LP \cap MC = \{Q\}$ . Show that the quadrilateral $MNPQ$ is a parallelogram.

Solution

**Due to symmetry reasons, it is enough to show that $KP \parallel MC$ holds.**

First we show that the quadrilateral $ACMD$ is inscribed. (1 point) Namely, let us notice that $B$ lies on the line segment $\overline{CD}$ , and $A$ and $M$ are on different sides of the line $CD$ . From $\angle BDM = \angle DAB$ and $\angle BCM = \angle BAC$ , it follows that $\angle DAC = \angle DAB + \angle BAC = \angle BDM + \angle BCM = 180^\circ - \angle DMC$ . (2 points)

Second, we will show that $B$ and $P$ lie on the same arc which passes through points $A$ and $K$ . (1 point) For that purpose, we consider two cases:

first case: the point $P$ lies on the segment $BC$ ; let us notice that points $A$ and $B$ are on the same side of the line $KP$ . Let $E$ denote the intersection of the lines $KB$ and $DM$ . We have the sequence of equalities $\angle KBP = \angle DBE = \angle BDE = \angle CDM = \angle CAM = \angle KAP$ . Then, from $\angle KBP = \angle KAP$ it follows that the quadrilateral $AKPB$ is inscribed. (1 point)

second case: the point $P$ lies on the segment $AC$ ; this time the points $A$ and $B$ are on different sides of the line $KP$ . Again, let $E$ be the intersection of $KB$ and $DM$ . We have the following sequence of equalities $180^\circ - \angle KBP = \angle DBE = \angle BDE = \angle CDM = \angle CAM = \angle KAP$ , from where it follows that the quadrilateral $AKBP$ is inscribed. (1 point)

Therefore we get $\angle APK = \angle ABK = \angle ADB = \angle ADC = \angle AMC$ , with which we confirm that $KP \parallel MC$ . (2 points)

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