Library / Saudi Arabian Mathematical Competitions /104 of 395

Saudi Arabian Mathematical Competitions, 2012

Geometry Difficulty 5.1 AIME, harder Prove it Saudi Arabia

Point $P$ lies inside quadrilateral $ABCD$ such that $\widehat{APD} = \widehat{BPC} = 90^\circ$ and $AP \cdot DP = BP \cdot CP$ . Let $O$ denote the circumcenter of triangle $CDP$ . Prove that line $OP$ bisects segment $AB$ .

Solution

Let $M$ be the midpoint of $AB$ , and let $E$ be the point on line $BP$ such that $AE \parallel MP$ . Then $P$ is the midpoint of $EB$ . Since $\frac{AP}{PB} = \frac{CP}{PD}$ , we have $\frac{AP}{PE} = \frac{CP}{PD}$ . Also, note that

$\widehat{EPC} = \widehat{APD} = 90^\circ$ , so $\widehat{EPA} = \widehat{DPC}$ . These last two facts imply that triangles $APE$ and $CPD$ are similar.

From this we conclude that $\widehat{PAE} = \widehat{PCD}$ . Since $AE \parallel MP$ , $\widehat{MPA} = \widehat{PAE} = \widehat{PCD}$ . Take a point $O'$ on ray $MP$ past $P$ . Then

$\widehat{O'PD} = 180^\circ - \widehat{APD} - \widehat{MPA} = 90^\circ - \widehat{PCD} = \widehat{OPD}.$

Therefore $M, P, O$ are collinear, which completes the proof.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.