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 $\angle XAB=A$ , which is an acute angle, $\tan{A}=t$ , then $X=(1-a,at)$ .

Angle $\angle BOS=2A$ , $S=(-\cos(2A),\sin(2A))$ . Let $M=(u,v)$ , then $T=(2u+\cos(2A), 2v-\sin(2A))$ .

The condition $TX \perp AX$ yields: $(2v-\sin(2A)-at)/(2u+\cos(2A)+a-1)=\cot A.$ (E1)

Use identities $(\cos A)^2=1/(1+t^2)$ , $\cos(2A)=2(\cos A)^2-1= 2/(1+t^2) -1$ , $\sin(2A)=2\sin A\cos A=2t^2/(1+t^2)$ , we obtain $2vt-at^2=2u+a$ . (E1')

The condition that $T$ is on the circle yields $(2u+\cos(2A))^2+ (2v-\sin(2A))^2=1$ , namely $v\sin(2A)-u\cos(2A)=u^2+v^2$ . (E2)

$M$ is the mid-point on the hypotenuse of triangle $STX$ , hence $MS=MX$ , yielding $(u+\cos(2A))^2+(v-\sin(2A))^2=(u+a-1)^2+(v-at)^2$ . (E3)

Expand (E3), using (E2) to replace $2(v\sin(2A)-u\cos(2A))$ 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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