# 2004 IMO Shortlist Problems/G2

## Contents

## Problem

(*Kazakhstan*)
The circle and the line do not intersect. Let be the diameter of perpendicular to , with closer to than . An arbitrary point is chosen on . The line intersects at . The line is tangent to at , with and on the same side of . Let intersect at , and let intersect at . Prove that the reflection of in lies on the line .

*This was also a Problem 2 on the 2005 Greece TST, Problem 1 of Day 1 of the 2005 Moldova TST, and Problem 2 of the final exam of the 3rd 2005 Taiwan TST.*

## Solution

### Solution 1

We use directed angles mod .

Let meet at . The problem is equivalent to showing that lines and are parallel, which happens if and only if . But by cyclic quadrilaterals and vertical angles, . To prove , it suffices to show that triangles are similar. Since these triangles share a common angle, it then suffices to show , or .

By considering the power of the point with respect to , we see . Hence it suffices to show that . Let be the center of . Since , it follows that there is a spiral similarity mapping to , i.e., these triangles are similar. Since , it follows that . Q.E.D.

### Solution 2

We use directed angles mod .

**Lemma.** Let be two circles with centers , and common points . Let be a point on , and let be the second intersection of line and . Then .

*Proof.* Since are collinear, . But since lie on a circle with center , , as desired.

Let the center of be . Let be the reflection of across . It is sufficient to show that , since are collinear.

Since is symmetric about its diameter , lies on , and , so

.

If we consider the line which passes through and is parallel to , we see , since are semicircles. Thus

,

or . Since we also have , it follows that triangles are similar, with opposite orientation. In particular, , or , so are concyclic.

At this point, we note that , (since is inscribed in a semicircle), and . It follows that there is a spiral similarity centered at with rotation of mapping triangle to triangle and to a circle centered at with radius . Let be the second intersection of and . We note that must intersect at some point on the same side of as , since must be on the same side of as . By the lemma, , and since is on , . It follows that . Since is the intersection point of and , , and lies on . In particular, , and

.

Now, since are concyclic, as we noted above,

.

To summarize,

,

as desired. Q.E.D.

### Solution 3

We use projective geometry. Let be the reflection of over . Let be the intersection of two distinct parallels of ; let be the intersection of and ; let be the intersection of and ; and let be the intersection of and . It is sufficient to show that lies on .

We apply Pascal's Theorem for cyclic hexagons several times. By applying it to the degenerate hexagon , we see that are collinear, i.e., lies on . By applying the theorem to the hexagon , we see that are collinear, i.e., also lies on . Finally, by appling the theorem to , we see that are collinear, so lies on , as desired.

### Solution 4

Again, we use projective geometry. We send to infinity, so that becomes an ellpise with axis . A distortion then makes a circle again. We shall now refer to objects as their images under these transformations.

We know that is parallel to the tangent to at , and is parallel to . Now, since is a diameter, . Also, . Since equal angles inscribe equal arcs, we have . Since is the reflection of over , we have . This implies that and are parallel, i.e., they pass through the same point on the line at infinity as , i.e., . Q.E.D.

*Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.*