2015 USAMO Problems/Problem 2
Contents
[hide]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