Difference between revisions of "2015 USAMO Problems/Problem 2"
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===Problem=== | ===Problem=== | ||
| − | + | Quadrilateral <math>APBQ</math> is inscribed in circle <math>\omega</math> with <math>\angle P = \angle Q = 90^{\circ}</math> and <math>AP = AQ < BP</math>. Let <math>X</math> be a variable point on segment <math>\overline{PQ}</math>. Line <math>AX</math> meets <math>\omega</math> again at <math>S</math> (other than <math>A</math>). Point <math>T</math> lies on arc <math>AQB</math> of <math>\omega</math> such that <math>\overline{XT}</math> is perpendicular to <math>\overline{AX}</math>. Let <math>M</math> denote the midpoint of chord <math>\overline{ST}</math>. As <math>X</math> varies on segment <math>\overline{PQ}</math>, show that <math>M</math> moves along a circle. | |
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===Solution=== | ===Solution=== | ||
| − | + | WLOG, let the circle be the unit circle centered at the origin, A=(1,0) P=(1-a,b), Q=(1-a,-b), where (1-a)^2+b^2=1. Let angle <XAB=A, which is an acute angle, tanA=t, then X=(1-a,at). | |
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| − | + | Angle <BOS=2A, S=(-cos2A,sin2A). | |
| + | Let M=(u,v), then T=(2u+cos2A, 2v-sin2A) | ||
| − | + | The condition TX perpendicular to AX yields (2v-sin2A-at)/(2u+cos2A+a-1)=cotA. (E1) | |
| + | Use identities (cosA)^2=1/(1+t^2), cos2A=2(cosA)^2-1= 2/(1+t^2) -1, sin2A=2sinAcosA=2t^2/(1+t^2), we obtain 2vt-at^2=2u+a. (E1') | ||
| − | + | The condition that T is on the circle yields (2u+cos2A)^2+ (2v-sin2A)^2=1, namely vsin2A-ucos2A=u^2+v^2. (E2) | |
| − | + | M is the mid-point on the hypotenuse of triangle STX, hence MS=MX, yielding (u+cos2A)^2+(v-sin2A)^2=(u+a-1)^2+(v-at)^2. (E3) | |
| − | + | Expand (E3), using (E2) to replace 2(vsin2A-ucos2A) with 2(u^2+v^2), and using (E1') to replace a(-2vt+at^2) with -a(2u+a), and we obtain | |
| + | u^2-u-a+v^2=0, namely (u-1/2)^2+v^2=a+1/4, which is a circle centered at (1/2,0) with radius r=sqrt(a+1/4). | ||
Revision as of 12:11, 12 May 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
WLOG, let the circle be the unit circle centered at the origin, A=(1,0) P=(1-a,b), Q=(1-a,-b), where (1-a)^2+b^2=1. Let angle <XAB=A, which is an acute angle, tanA=t, then X=(1-a,at).
Angle <BOS=2A, S=(-cos2A,sin2A). Let M=(u,v), then T=(2u+cos2A, 2v-sin2A)
The condition TX perpendicular to AX yields (2v-sin2A-at)/(2u+cos2A+a-1)=cotA. (E1) Use identities (cosA)^2=1/(1+t^2), cos2A=2(cosA)^2-1= 2/(1+t^2) -1, sin2A=2sinAcosA=2t^2/(1+t^2), we obtain 2vt-at^2=2u+a. (E1')
The condition that T is on the circle yields (2u+cos2A)^2+ (2v-sin2A)^2=1, namely vsin2A-ucos2A=u^2+v^2. (E2)
M is the mid-point on the hypotenuse of triangle STX, hence MS=MX, yielding (u+cos2A)^2+(v-sin2A)^2=(u+a-1)^2+(v-at)^2. (E3)
Expand (E3), using (E2) to replace 2(vsin2A-ucos2A) with 2(u^2+v^2), and using (E1') to replace a(-2vt+at^2) with -a(2u+a), and we obtain u^2-u-a+v^2=0, namely (u-1/2)^2+v^2=a+1/4, which is a circle centered at (1/2,0) with radius r=sqrt(a+1/4).
