# 2013 IMO Problems/Problem 4

## 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 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)

## Solution 3 (Complex Bash)

For any point let denote the complex number for point First off, let be the origin. Now, since is the diameter of the circumcircle of triangle , we must have Since angles inscribed in the same arc are congruent, This means that Combining this with the fact that is right, we find that Solving, we find that We wish to simplify first. Note that This means that This means that Now, since is a diameter of the circumcircle of triangle we must have Since angles inscribed in the same arc are congruent, This means that Combining this with the fact that is right, we find that Solving, we find that We wish to simplify first. Note that This means that This means that This means that the line through complex numbers and satisfy the equation If there is a fixed point to the line, then the real and imaginary values of the point must not contin . If the fixed point is then we have after comparing the coefficient of This means that From the Law of Sines, we find that Substituting this, we find that This means that Since the 's cancel out, we can just discard everything with in it. Thus, Since we have Thus, In conclusion, which is the fixed point always passes through. However, by inspection, Therefore, we conclude that are collinear for all acute triangles

~pinkpig