2015 USAMO Problems/Problem 2

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, $\tan{A}=t$ , 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 $(u-\frac{1}/{2})^2+v^2=a+\frac{1}/{4}$ , which is a circle centered at $(\frac{1}{2},0)$ with radius $r=\sqrt{a+\frac{1}{4}}$ .

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2015 USAMO Problems/Problem 2

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