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Difference between revisions of "2015 USAJMO Problems/Problem 3"

(Created page with "===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 ...")
 
(→‎Solution)
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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
 
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).
 
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).
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===Solution 2===
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Note that each point <math>X</math> on <math>PQ</math> corresponds to exactly one point on arc <math>PBQ</math>. Also notice that since <math>AB</math> is the diameter of <math>\omega</math>, <math>\angle ASB</math> is always a right angle; therefore, point <math>T</math> is always <math>B</math>. WLOG, assume that <math>\omega</math> is on the coordinate plane, and <math>B</math> corresponds to the origin. The locus of <math>M</math>, since the locus of <math>S</math> is arc <math>PBQ</math>, is the arc that is produced when arc <math>PBQ</math> is dilated by <math>\frac {1} {2}</math> with respect to the origin, which resides on the circle <math>\psi</math>, which is produced when <math>\omega</math> is dilated by <math>\frac {1} {2}</math> with respect to the origin.

Revision as of 23:05, 6 March 2016

Problem

Quadrilateral $APBQ$ is inscribed in circle $\omega$ with $\angle P = \angle Q = 90^{\circ}$ and $AP = AQ < BP$. Let $X$ be a variable point on segment $\overline{PQ}$. Line $AX$ meets $\omega$ again at $S$ (other than $A$). Point $T$ lies on arc $AQB$ of $\omega$ such that $\overline{XT}$ is perpendicular to $\overline{AX}$. Let $M$ denote the midpoint of chord $\overline{ST}$. As $X$ varies on segment $\overline{PQ}$, show that $M$ 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).

Solution 2

Note that each point $X$ on $PQ$ corresponds to exactly one point on arc $PBQ$. Also notice that since $AB$ is the diameter of $\omega$, $\angle ASB$ is always a right angle; therefore, point $T$ is always $B$. WLOG, assume that $\omega$ is on the coordinate plane, and $B$ corresponds to the origin. The locus of $M$, since the locus of $S$ is arc $PBQ$, is the arc that is produced when arc $PBQ$ is dilated by $\frac {1} {2}$ with respect to the origin, which resides on the circle $\psi$, which is produced when $\omega$ is dilated by $\frac {1} {2}$ with respect to the origin.

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