2015 USAMO Problems/Problem 2


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 1

We will use coordinate geometry.

Without loss of generality, 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 $r=\sqrt{a+\frac{1}{4}}$.

Solution 2

Let the midpoint of $AO$ be $K$. We claim that $M$ moves along a circle with radius $KP$.

We will show that $KM^2 = KP^2$, which implies that $KM = KP$, and as $KP$ is fixed, this implies the claim.

$KM^2 = \frac{AM^2+OM^2}{2}-\frac{AO^2}{4}$ by the median formula on $\triangle AMO$.

$KP^2 = \frac{AP^2+OP^2}{2}-\frac{AO^2}{4}$ by the median formula on $\triangle APO$.

$KM^2-KP^2 = \frac{1}{2}(AM^2+OM^2-AP^2-OP^2)$.

As $OP = OT$, $OP^2-OM^2 = MT^2$ from right triangle $OMT$. $(1)$

By $(1)$, $KM^2-KP^2 = \frac{1}{2}(AM^2-MT^2-AP^2)$.

Since $M$ is the circumcenter of $\triangle XTS$, and $MT$ is the circumradius, the expression $AM^2-MT^2$ is the power of point $A$ with respect to $(XTS)$. However, as $AX*AS$ is also the power of point $A$ with respect to $(XTS)$, this implies that $AM^2-MT^2=AX*AS$. $(2)$

By $(2)$, $KM^2-KP^2 = \frac{1}{2}(AX*AS-AP^2)$

Finally, $\triangle APX \sim \triangle ASP$ by AA similarity ($\angle XAP = \angle SAP$ and $\angle APX = \angle AQP = \angle ASP$), so $AX*AS = AP^2$. $(3)$

By $(3)$, $KM^2-KP^2=0$, so $KM^2=KP^2$, as desired. $QED$

Solution 3(synthetic)

To begin with, we connect $\overline{AT}$ and we construct the nine-point circle of $\triangle AST$ centered at $N_9$.

Lemma $1$: $AX \cdot AS = AP^2$. We proceed on a directed angle chase. We get $\measuredangle ASP = \measuredangle AQP = \measuredangle QPA$, so $\triangle PAS \sim \triangle XAP$ and the desired result follows by side length ratios.

Lemma $2$: The locus of $N_9$ as $X$ moves along $\overline{PQ}$ is a circle centered about $A$. We add the midpoint of $\overline{AS}$, $N$, and let the circumradius of $\triangle AST$ be $R$. Taking the power of $A$ with respect to $(N_9)$, we get \[AN_9^2 - \left(\frac{1}{2} R\right)^2 = \text{Pow}_{(N_9)} A = AX \cdot AN = \frac{1}{2} AX \cdot AS = \frac{1}{2} AP^2.\] Hence, $AN_9 = \sqrt{\frac{1}{4}R^2 + \frac{1}{2}AP^2}$, which remains constant as $X$ moves.

Next, consider the homothety of scale factor $\frac{2}{3}$ about $O$ mapping $N_9$ to $G$. This means that the locus of $G$ is a circle as well.

Finally, we take a homothety of scale factor $\frac{3}{2}$ about $A$ mapping $G$ to $M$. Hence, the locus of $M$ is a circle, as desired. - Spacesam