Difference between revisions of "2015 USAMO Problems/Problem 2"

Revision as of 17:39, 28 July 2019

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 $\triangle PAX \sim \triangle XAP$ 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

Fake Solution

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. By MSmathlete1018

@@ Line 56: / Line 56: @@
 To begin with, we connect <math>\overline{AT}</math> and we construct the nine-point circle of <math>\triangle AST</math> centered at <math>N_9</math>.
-Lemma <math>1</math>: <math>AX \cdot AS = AP^2</math>
+Lemma <math>1</math>: <math>AX \cdot AS = AP^2</math>.
 We proceed on a directed angle chase. We get <math>\measuredangle ASP = \measuredangle AQP = \measuredangle QPA</math>, so <math>\triangle PAX \sim \triangle XAP</math> and the desired result follows by side length ratios.

Page

Toolbox

Search