Difference between revisions of "2015 USAMO Problems/Problem 2"
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We will show that <math>KM^2 = KP^2</math>, which implies that <math>KM = KP</math>, and as <math>KP</math> is fixed, this implies the claim. | We will show that <math>KM^2 = KP^2</math>, which implies that <math>KM = KP</math>, and as <math>KP</math> is fixed, this implies the claim. | ||
− | <math>KM^2 = \frac{AM^2+OM^2}{2}-\frac{AO^2}{4}</math> by the median formula on | + | <math>KM^2 = \frac{AM^2+OM^2}{2}-\frac{AO^2}{4}</math> by the median formula on<math>\triangle AMO</math>. |
− | <math>KP^2 = \frac{AP^2+OP^2}{2}-\frac{AO^2}{4}</math> by the median formula on | + | <math>KP^2 = \frac{AP^2+OP^2}{2}-\frac{AO^2}{4}</math> by the median formula on <math>\triangle APO</math>. |
<math>KM^2-KP^2 = \frac{1}{2}(AM^2+OM^2-AP^2-OP^2)</math>. | <math>KM^2-KP^2 = \frac{1}{2}(AM^2+OM^2-AP^2-OP^2)</math>. | ||
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Finally, <math>\triangle APX \sim \triangle ASP</math> by AA similarity (<math>\angle XAP = \angle SAP</math> and <math>\angle APX = \angle AQP = \angle ASP</math>), so <math>AX*AS = AP^2</math>. <math>(3)</math> | Finally, <math>\triangle APX \sim \triangle ASP</math> by AA similarity (<math>\angle XAP = \angle SAP</math> and <math>\angle APX = \angle AQP = \angle ASP</math>), so <math>AX*AS = AP^2</math>. <math>(3)</math> | ||
− | By <math>(3)</math>, <math>KM^2-KP^2=0</math>, so <math>KM^2=KP^2</math>, | + | By <math>(3)</math>, <math>KM^2-KP^2=0</math>, so <math>KM^2=KP^2</math>, as desired. <math>QED</math> |
Revision as of 17:35, 23 December 2015
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
Solution 1: Complex bash
WLOG, 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: Mostly synthetic
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.