# Difference between revisions of "2015 USAMO Problems/Problem 2"

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Finally, we take a homothety of scale factor <math>\frac{3}{2}</math> about <math>A</math> mapping <math>G</math> to <math>M</math>. Hence, the locus of <math>M</math> is a circle, as desired. - Spacesam | Finally, we take a homothety of scale factor <math>\frac{3}{2}</math> about <math>A</math> mapping <math>G</math> to <math>M</math>. Hence, the locus of <math>M</math> is a circle, as desired. - Spacesam | ||

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## Latest revision as of 08:58, 20 August 2021

### 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