Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.8 National olympiad Prove it

Let $ABCD$ be a parallelogram such that $AC = BC$ . Let $P$ be a point located on the extension of the segment $[AB]$ beyond $B$ . Let $Q$ be the intersection point, other than $D$ , between the segment $[PD]$ and the circumcircle of $ACD$ . Let then $R$ be the intersection point, other than $P$ , between the segment $[PC]$ and the circumcircle of $APQ$ .
Prove that the lines $(AQ), (BR)$ , and $(CD)$ are concurrent.

Solution

First, we observe that

$(R A, R C)=(R A, R P)=(Q A, Q P)=(Q A, Q D)=(C A, C D)=(A C, A B)=(B A, B C)$

which means that the points $A, B, C$ and $R$ are concyclic.
Similarly, if we denote $X$ as the intersection point of the lines $(A Q)$ and $(C D)$ , we observe that

$(Q R, Q X)=(Q R, Q A)=(P R, P A)=(C R, C X)$

which means that the points $C, Q, R$ and $X$ are concyclic.
We conclude that

$(R C, R X)=(Q C, Q X)=(Q C, Q A)=(D C, D A)=(B A, B C)=(A C, A B)=(R C, R B)$

which means that, as announced, the points $B, R$ and $X$ are collinear.
!

Graders' Comments: If no student solved the problem in its entirety, a few brave students developed very interesting ideas: approaching the problem from the perspective of radical axes, showing that certain points were concyclic, etc. All students who submitted an attempt showed initiative, drew a figure on which they could reasonably make conjectures (and on which graders could follow their reasoning) and started a reasoning process aimed at reaching the solution. Even if it did not always earn points, this kind of effort will allow these students to progress over time.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.