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

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By <math>(2)</math>, <math>KM^2-KP^2 = \frac{1}{2}(AX*AS-AP^2)</math>
 
By <math>(2)</math>, <math>KM^2-KP^2 = \frac{1}{2}(AX*AS-AP^2)</math>
  
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Finally, <math>\triangle AXP \sim \triangle ASP</math> by AA similarity (<math>\angle XAP = \angle SAP</math> and <math>\angle APX = \angle AQP = \angle ASP</math>), so <math>AX*AS = AP^2</math>. <math>(3)</math>
+
Finally, <math>\triangle APX \sim \triangle ASP</math> by AA similarity (<math>\angle XAP = \angle SAP</math> and <math>\angle APX = \angle AQP = \angle ASP</math>), so <math>AX*AS = AP^2</math>. <math>(3)</math>
  
 
By <math>(3)</math>, <math>KM^2-KP^2=0</math>, so <math>KM^2=KP^2</math>, which implies that <math>KM=KP</math>. Finally, this implies the desired result, so we are done. <math>QED</math>
 
By <math>(3)</math>, <math>KM^2-KP^2=0</math>, so <math>KM^2=KP^2</math>, which implies that <math>KM=KP</math>. Finally, this implies the desired result, so we are done. <math>QED</math>

Revision as of 21:15, 27 September 2015

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

Solution 1: Complex bash

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


Solution 2: Mostly synthetic

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 $KM^2=KP^2$, which implies that $KM=KP$. Finally, this implies the desired result, so we are done. $QED$

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