Library / Ukrainian National Mathematical Olympiad, Third Round, First Tour /12 of 19

Ukrainian National Mathematical Olympiad, Third Round, First Tour

Geometry Difficulty 5.8 AIME, harder Prove it Ukraine

Let $P$ be a point of the intersection of diagonals of the cyclic quadrilateral $ABCD$ . The circumscribed circles of $\triangle APD$ and $\triangle BPC$ intersect the line $AB$ at points $E$ and $F$ correspondingly. $Q$ and $R$ are projections of the point $P$ onto the lines $FC$ and $DE$ . Prove that $AB \parallel QR$ .

Solution

Denote by $d(Z, XY)$ the distance from the point $Z$ to the line $XY$ . From the equality of inscribed angles it follows that (fig. 6)
$\angle PEB = \angle ADB = \angle ACB = \angle PFA \Rightarrow PE = PF.$
Furthermore,
$\angle EDP = \angle BAP = \angle BDC, \quad \angle FCP = \angle PBA = \angle DCP,$
so $P$ is the intersection of bisectors of angles $\angle EDC$ and $\angle DCF$ , so $d(P, DE) = d(P, DC) = d(P, CF)$ , so $PQ = PR$ . Then $\triangle FPQ \cong \triangle EPR$ as right triangles with equal cathetus and hypotenuse. From the equality $PQ = PR$ it follows that $\angle PQR = \angle PRQ$ , so $\angle FQR = \angle QRE$ . Similarly, $\angle QFE = \angle REF$ . This means that $EQRF$ is an isosceles trapezoid, so $QR \parallel EF$ , as desired.

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.