2015 USAMO Problems/Problem 2

Revision as of 16:33, 23 May 2015 by Sujaykazi (talk | contribs) (Solution)

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 $r=\sqrt{a+\frac{1}{4}}$.