# 2015 USAMO Problems/Problem 2

### Problem

Quadrilateral is inscribed in circle with and . Let be a variable point on segment . Line meets again at (other than ). Point lies on arc of such that is perpendicular to . Let denote the midpoint of chord . As varies on segment , show that moves along a circle.

### Solution 1

We will use coordinate geometry.

Without loss of generality, let the circle be the unit circle centered at the origin, , where .

Let angle , which is an acute angle, , then .

Angle , . Let , then .

The condition yields: (E1)

Use identities , , , we obtain . (E1')

The condition that is on the circle yields , namely . (E2)

is the mid-point on the hypotenuse of triangle , hence , yielding . (E3)

Expand (E3), using (E2) to replace with , and using (E1') to replace with , and we obtain , namely , which is a circle centered at with radius .

### Solution 2

Let the midpoint of be . We claim that moves along a circle with radius .

We will show that , which implies that , and as is fixed, this implies the claim.

by the median formula on .

by the median formula on .

.

As , from right triangle .

By , .

Since is the circumcenter of , and is the circumradius, the expression is the power of point with respect to . However, as is also the power of point with respect to , this implies that .

By ,

Finally, by AA similarity ( and ), so .

By , , so , as desired.

## Solution 3(synthetic)

To begin with, we connect and we construct the nine-point circle of centered at .

Lemma : . We proceed on a directed angle chase. We get , so and the desired result follows by side length ratios.

Lemma : The locus of as moves along is a circle centered about . We add the midpoint of , , and let the circumradius of be . Taking the power of with respect to , we get Hence, , which remains constant as moves.

Next, consider the homothety of scale factor about mapping to . This means that the locus of is a circle as well.

Finally, we take a homothety of scale factor about mapping to . Hence, the locus of is a circle, as desired. - Spacesam

### Fake Solution

Note that each point on corresponds to exactly one point on arc . Also notice that since is the diameter of , is always a right angle; therefore, point is always . WLOG, assume that is on the coordinate plane, and corresponds to the origin. The locus of , since the locus of is arc , is the arc that is produced when arc is dilated by with respect to the origin, which resides on the circle , which is produced when is dilated by with respect to the origin. By MSmathlete1018