Track / Stage 5 / 392 of 400 #992 of 1964

Problem 992

AIME late

Geometry Difficulty 6.0 Prove it

Points $P$ and $Q$ are on the diagonals $[A C]$ and $[B D]$ , respectively, of a quadrilateral $A B C D$ such that $\frac{A P}{A C}+\frac{B Q}{B D}=1$ . The line $(P Q)$ intersects the sides $[A D]$ and $[B C]$ at points $M$ and $N$ . Show that the circles $A M P, B N Q, D M Q$ , and $C N P$ are concurrent.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Let $E$ be the intersection of $(A C)$ and $(B D)$ . According to Miquel, the circles $A M P$ , $D M Q$ , $E P Q$ , $E A D$ are concurrent, as well as $B N Q$ , $C N P$ , $E P Q$ , $E B C$ . It remains to show that $E P Q$ , $E A D$ , $E B C$ are concurrent. Let $F$ be the second point of intersection of the circles $E B C$ , $E A D$ . Then $F$ is the center of the direct similarity $s$ which maps $[B D]$ to $[C A]$ . Since $\frac{A P}{A C}=\frac{D Q}{B D}$ , $s$ maps $Q$ to $P$ and $D$ to $A$ , so $F$ lies on the circle $E P Q$ .

## 4 Group D: Arithmetic

## 1 Tuesday 18 morning: Igor Kortchemski

NB. Additional exercises compared to what was covered during the session have been added in section 2, as well as the (very useful) method of Dan Schwarz.

## First Part

The first part of the session consisted of reviewing some basic tools and reflexes through exercises.

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