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. | ||
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<math>KM^2 = \frac{AM^2+OM^2}{2}-\frac{AO^2}{4}</math> by the median formula on triangle <math>AMO</math>. | <math>KM^2 = \frac{AM^2+OM^2}{2}-\frac{AO^2}{4}</math> by the median formula on triangle <math>AMO</math>. | ||
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<math>KP^2 = \frac{AP^2+OP^2}{2}-\frac{AO^2}{4}</math> by the median formula on triangle <math>APO</math>. | <math>KP^2 = \frac{AP^2+OP^2}{2}-\frac{AO^2}{4}</math> by the median formula on triangle <math>APO</math>. | ||
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<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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As <math>OP = OT</math>, <math>OP^2-OM^2 = MT^2</math> from right triangle <math>OMT</math>. <math>(1)</math> | As <math>OP = OT</math>, <math>OP^2-OM^2 = MT^2</math> from right triangle <math>OMT</math>. <math>(1)</math> | ||
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By <math>(1)</math>, <math>KM^2-KP^2 = \frac{1}{2}(AM^2-MT^2-AP^2)</math>. | By <math>(1)</math>, <math>KM^2-KP^2 = \frac{1}{2}(AM^2-MT^2-AP^2)</math>. | ||
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Since <math>M</math> is the circumcenter of <math>\triangle XTS</math>, and <math>MT</math> is the circumradius, the expression <math>AM^2-MT^2</math> is the power of point <math>A</math> with respect to <math>(XTS)</math>. However, as <math>AX*AS</math> is also the power of point <math>A</math> with respect to <math>(XTS)</math>, this implies that <math>AM^2-MT^2=AX*AS</math>. <math>(2)</math> | Since <math>M</math> is the circumcenter of <math>\triangle XTS</math>, and <math>MT</math> is the circumradius, the expression <math>AM^2-MT^2</math> is the power of point <math>A</math> with respect to <math>(XTS)</math>. However, as <math>AX*AS</math> is also the power of point <math>A</math> with respect to <math>(XTS)</math>, this implies that <math>AM^2-MT^2=AX*AS</math>. <math>(2)</math> | ||
By <math>(2)</math>, <math>KM^2-KP^2 = \frac{1}{2}(AX*AS-AP^2)</math> | By <math>(2)</math>, <math>KM^2-KP^2 = \frac{1}{2}(AX*AS-AP^2)</math> | ||
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Finally, <math>\triangle AXP \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 AXP \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> | ||
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By <math>(3)</math>, <math>KM^2-KP^2=0</math>, so <math>KM^2=KP^2</math>, which implies that <math>KM=KP</math>. Finally, this implies the desired result, so we are done. <math>QED</math> | By <math>(3)</math>, <math>KM^2-KP^2=0</math>, so <math>KM^2=KP^2</math>, which implies that <math>KM=KP</math>. Finally, this implies the desired result, so we are done. <math>QED</math> |
Revision as of 23:12, 5 June 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 triangle .
by the median formula on triangle .
.
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 , which implies that . Finally, this implies the desired result, so we are done.