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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Expand (E3), using (E2) to replace <math>2(v\sin(2A)-u\cos(2A))</math> with <math>2(u^2+v^2)</math>, and using (E1') to replace <math>a(-2vt+at^2)</math> with <math>-a(2u+a)</math>, and we obtain | Expand (E3), using (E2) to replace <math>2(v\sin(2A)-u\cos(2A))</math> with <math>2(u^2+v^2)</math>, and using (E1') to replace <math>a(-2vt+at^2)</math> with <math>-a(2u+a)</math>, and we obtain | ||
<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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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 triangle <math>AMO</math>. | ||
+ | <math>KP^2 = \frac{AP^2+OP^2}{2}-\frac{AO^2}{4}</math> by the median formula on triangle <math>APO</math>. | ||
+ | <math>KM^2-KP^2 = \frac{1}{2}(AM^2+OM^2-AP^2-OP^2)</math>. | ||
+ | As <math>OP = OT</math>, <math>OP^2-OM^2 = MT^2</math> from right triangle <math>OMT</math>. <math>(1)</math> | ||
+ | By <math>(1)</math>, <math>KM^2-KP^2 = \frac{1}{2}(AM^2-MT^2-AP^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> | ||
+ | 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> | ||
+ | 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 00:11, 6 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.