# 2013 IMO Problems/Problem 4

## Contents

## Problem

Let be an acute triangle with orthocenter , and let be a point on the side , lying strictly between and . The points and are the feet of the altitudes from and , respectively. Denote by is [sic] the circumcircle of , and let be the point on such that is a diameter of . Analogously, denote by the circumcircle of triangle , and let be the point such that is a diameter of . Prove that and are collinear.

## Hint

Draw a good diagram, or use the one below. What do you notice? (In particular, what do you want to be true? How do you prove it true?)

## Solution 1

Let be the intersection of and other than .

*Lemma 1:* is on .

Proof: We have because they intercept semicircles. Hence, , so is a straight line.

*Lemma 2:* is on .

Proof: Let the circumcircles of and be and , respectively, and, as is cyclic (from congruent ), let its circumcircle be . Then each pair of circles' radical axises, and , must concur at the intersection of and , which is .

*Lemma 3:* is perpendicular to .

Proof: This is immediate from .

Let meet at , which is also the foot of the altitude to that side. Hence,

*Lemma 4:* Quadrilateral is cyclic.

Proof: We know that is cyclic because and , opposite and right angles, sum to . Furthermore, we are given that is cyclic. Hence, by Power of a Point,

The converse of Power of a Point then proves cyclic.

Hence, , and so is perpendicular to as well. Combining this with Lemma 3's statement, we deduce that are collinear. But, as is on (from Lemma 1), are collinear. This completes the proof.

--Suli 13:51, 25 August 2014 (EDT)

## Solution 2

Probably a simpler solution than above.

As above, let By Miquel is cyclic. Then since we know, because and that thus are collinear. There are a few ways to finish.

(a) so are collinear, as desired

(b) Since is cyclic we know which means so is on the radical axis, hence so lies on this line as well and we may conclude

--mathguy623 03:10, 12 August 2016 (EDT)