2019 JBMO TST- North Macedonia, problem 2

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2019 JBMO TST- North Macedonia, problem 2

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Source: 2019 JBMO TST- North Macedonia

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#1 h May 26, 2019, 1:32 PM • 1 Y

Y by Adventure10

Circles $\omega_{1}$ and $\omega_{2}$ intersect at points $A$ and $B$ . Let $t_{1}$ and $t_{2}$ be the tangents to $\omega_{1}$ and $\omega_{2}$ , respectively, at point $A$ . Let the second intersection of $\omega_{1}$ and $t_{2}$ be $C$ , and let the second intersection of $\omega_{2}$ and $t_{1}$ be $D$ . Points $P$ and $E$ lie on the ray $AB$ , such that $B$ lies between $A$ and $P$ , $P$ lies between $A$ and $E$ , and $AE = 2 \cdot AP$ . The circumcircle to $\bigtriangleup BCE$ intersects $t_{2}$ again at point $Q$ , whereas the circumcircle to $\bigtriangleup BDE$ intersects $t_{1}$ again at point $R$ . Prove that points $P$ , $Q$ , and $R$ are collinear.

This post has been edited 1 time. Last edited by Lukaluce, May 26, 2019, 2:15 PM

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#2 h May 26, 2019, 1:36 PM • 2 Y

Y by Adventure10, Mango247

Since $t_{2}$ is a tangent to $\omega_{2}$ at $A$ and since $BDRE$ is a cyclic quadrilateral, we get $\measuredangle REB = \measuredangle ADB =\measuredangle QAB$ , so $QA\parallel ER$ . Similarly, $AR\parallel EQ$ . Since $P$ is the midpoint of $AE$ , we get that $P$ is the intersection of the diagonals of $AQER$ (as $AQER$ is a parallelogram), and therefore, $P$ lies on $RQ$ .

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#3 h Jun 10, 2022, 2:40 AM

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Plssend the diagram once.

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