Difference between revisions of "2015 USAMO Problems/Problem 2"
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===Solution=== | ===Solution=== | ||
− | + | Solution 1: Complex bash | |
WLOG, let the circle be the unit circle centered at the origin, <math>A=(1,0) P=(1-a,b), Q=(1-a,-b)</math>, where <math>(1-a)^2+b^2=1</math>. | WLOG, let the circle be the unit circle centered at the origin, <math>A=(1,0) P=(1-a,b), Q=(1-a,-b)</math>, where <math>(1-a)^2+b^2=1</math>. | ||
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<math>u^2-u-a+v^2=0</math>, namely <math>(u-\frac{1}{2})^2+v^2=a+\frac{1}{4}</math>, which is a circle centered at <math>(\frac{1}{2},0)</math> with radius <math>r=\sqrt{a+\frac{1}{4}}</math>. | <math>u^2-u-a+v^2=0</math>, namely <math>(u-\frac{1}{2})^2+v^2=a+\frac{1}{4}</math>, which is a circle centered at <math>(\frac{1}{2},0)</math> with radius <math>r=\sqrt{a+\frac{1}{4}}</math>. | ||
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+ | Solution 2: Mostly synthetic | ||
Let the midpoint of <math>AO</math> be <math>K</math>. We claim that <math>M</math> moves along a circle with radius <math>KP</math>. | Let the midpoint of <math>AO</math> be <math>K</math>. We claim that <math>M</math> moves along a circle with radius <math>KP</math>. |
Revision as of 23:13, 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.