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

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Quadrilateral <math>APBQ</math> is inscribed in circle <math>\omega</math> with <math>\angle P = \angle Q = 90^{\circ}</math> and <math>AP = AQ < BP</math>. Let <math>X</math> be a variable point on segment <math>\overline{PQ}</math>. Line <math>AX</math> meets <math>\omega</math> again at <math>S</math> (other than <math>A</math>). Point <math>T</math> lies on arc <math>AQB</math> of <math>\omega</math> such that <math>\overline{XT}</math> is perpendicular to <math>\overline{AX}</math>. Let <math>M</math> denote the midpoint of chord <math>\overline{ST}</math>. As <math>X</math> varies on segment <math>\overline{PQ}</math>, show that <math>M</math> moves along a circle.
 
Quadrilateral <math>APBQ</math> is inscribed in circle <math>\omega</math> with <math>\angle P = \angle Q = 90^{\circ}</math> and <math>AP = AQ < BP</math>. Let <math>X</math> be a variable point on segment <math>\overline{PQ}</math>. Line <math>AX</math> meets <math>\omega</math> again at <math>S</math> (other than <math>A</math>). Point <math>T</math> lies on arc <math>AQB</math> of <math>\omega</math> such that <math>\overline{XT}</math> is perpendicular to <math>\overline{AX}</math>. Let <math>M</math> denote the midpoint of chord <math>\overline{ST}</math>. As <math>X</math> varies on segment <math>\overline{PQ}</math>, show that <math>M</math> moves along a circle.
  
−
===Solution===
+
===Solution 1===
−
WLOG, let the circle be the unit circle centered at the origin, <math>A=(1,0) P=(1-a,b), Q=(1-a,-b)</math>, where <math>(1-a)^2+b^2=1</math>. Let angle <XAB=A, which is an acute angle, <math>\tan{A}=t</math>, then <math>X=(1-a,at)</math>.
 
  
−
Angle <BOS=2A, S=(-cos2A,sin2A).
+
We will use coordinate geometry.
−
Let M=(u,v), then T=(2u+cos2A, 2v-sin2A)
 
  
−
The condition TX perpendicular to AX yields (2v-sin2A-at)/(2u+cos2A+a-1)=cotA.    (E1)
+
Without loss of generality,
−
Use identities (cosA)^2=1/(1+t^2), cos2A=2(cosA)^2-1= 2/(1+t^2) -1, sin2A=2sinAcosA=2t^2/(1+t^2), we obtain 2vt-at^2=2u+a.   (E1')
+
let the circle be the unit circle centered at the origin,
 +
<cmath>A=(1,0) P=(1-a,b), Q=(1-a,-b)</cmath>,
 +
where <math>(1-a)^2+b^2=1</math>.
  
−
The condition that T is on the circle yields (2u+cos2A)^2+ (2v-sin2A)^2=1, namely vsin2A-ucos2A=u^2+v^2.  (E2)
+
Let angle <math>\angle XAB=A</math>, which is an acute angle, <math>\tan{A}=t</math>, then <math>X=(1-a,at)</math>.
  
−
M is the mid-point on the hypotenuse of triangle STX, hence MS=MX, yielding (u+cos2A)^2+(v-sin2A)^2=(u+a-1)^2+(v-at)^2.   (E3)
+
Angle <math>\angle BOS=2A</math>, <math>S=(-\cos(2A),\sin(2A))</math>.
 +
Let <math>M=(u,v)</math>, then <math>T=(2u+\cos(2A), 2v-\sin(2A))</math>.
  
−
Expand (E3), using (E2) to replace 2(vsin2A-ucos2A) with 2(u^2+v^2), and using (E1') to replace a(-2vt+at^2) with -a(2u+a), and we obtain
+
The condition <math>TX \perp AX</math> yields: <math>(2v-\sin(2A)-at)/(2u+\cos(2A)+a-1)=\cot A. </math>    (E1)
−
u^2-u-a+v^2=0, namely (u-1/2)^2+v^2=a+1/4, which is a circle centered at (1/2,0) with radius r=sqrt(a+1/4).
+
 
 +
Use identities <math>(\cos A)^2=1/(1+t^2)</math>,  <math>\cos(2A)=2(\cos A)^2-1= 2/(1+t^2) -1</math>, <math>\sin(2A)=2\sin A\cos A=2t^2/(1+t^2)</math>, we obtain <math>2vt-at^2=2u+a</math>.  (E1')
 +
 
 +
The condition that <math>T</math> is on the circle yields <math>(2u+\cos(2A))^2+ (2v-\sin(2A))^2=1</math>, namely <math>v\sin(2A)-u\cos(2A)=u^2+v^2</math>.  (E2)
 +
 
 +
<math>M</math> is the mid-point on the hypotenuse of triangle <math>STX</math>, hence <math>MS=MX</math>, yielding <math>(u+\cos(2A))^2+(v-\sin(2A))^2=(u+a-1)^2+(v-at)^2</math>.  (E3)
 +
 
 +
Expand (E3), using (E2) to replace <math>2(v\sin(2A)-u\cos(2A))</math> with <math>2(u^2+v^2)</math>, and using (E1') to replace <math>a(-2vt+at^2)</math> with <math>-a(2u+a)</math>, and we obtain
 +
<math>u^2-u-a+v^2=0</math>, namely <math>(u-\frac{1}{2})^2+v^2=a+\frac{1}{4}</math>, which is a circle centered at <math>(\frac{1}{2},0)</math> with radius <math>r=\sqrt{a+\frac{1}{4}}</math>.
 +
 
 +
 
 +
===Solution 2===
 +
 
 +
 
 +
Let the midpoint of <math>AO</math> be <math>K</math>. We claim that <math>M</math> moves along a circle with radius <math>KP</math>.
 +
 
 +
We will show that <math>KM^2 = KP^2</math>, which implies that <math>KM = KP</math>, and as <math>KP</math> is fixed, this implies the claim.
 +
 
 +
<math>KM^2 = \frac{AM^2+OM^2}{2}-\frac{AO^2}{4}</math> by the median formula on <math>\triangle AMO</math>.
 +
 
 +
<math>KP^2 = \frac{AP^2+OP^2}{2}-\frac{AO^2}{4}</math> by the median formula on <math>\triangle APO</math>.
 +
 
 +
<math>KM^2-KP^2 = \frac{1}{2}(AM^2+OM^2-AP^2-OP^2)</math>.
 +
 
 +
As <math>OP = OT</math>, <math>OP^2-OM^2 = MT^2</math> from right triangle <math>OMT</math>. <math>(1)</math>
 +
 
 +
By <math>(1)</math>, <math>KM^2-KP^2 = \frac{1}{2}(AM^2-MT^2-AP^2)</math>.
 +
 
 +
Since <math>M</math> is the circumcenter of <math>\triangle XTS</math>, and <math>MT</math> is the circumradius, the expression <math>AM^2-MT^2</math> is the power of point <math>A</math> with respect to <math>(XTS)</math>. However, as <math>AX*AS</math> is also the power of point <math>A</math> with respect to <math>(XTS)</math>, this implies that <math>AM^2-MT^2=AX*AS</math>. <math>(2)</math>
 +
 
 +
By <math>(2)</math>, <math>KM^2-KP^2 = \frac{1}{2}(AX*AS-AP^2)</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>, as desired. <math>QED</math>
 +
 
 +
== Solution 3(synthetic) ==
 +
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>.
 +
We proceed on a directed angle chase. We get <math>\measuredangle ASP = \measuredangle AQP = \measuredangle QPA</math>, so <math>\triangle PAS \sim \triangle XAP</math> and the desired result follows by side length ratios.
 +
 
 +
Lemma <math>2</math>: The locus of <math>N_9</math> as <math>X</math> moves along <math>\overline{PQ}</math> is a circle centered about <math>A</math>.
 +
We add the midpoint of <math>\overline{AS}</math>, <math>N</math>, and let the circumradius of <math>\triangle AST</math> be <math>R</math>. Taking the power of <math>A</math> with respect to <math>(N_9)</math>, we get <cmath>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.</cmath>
 +
Hence, <math>AN_9 = \sqrt{\frac{1}{4}R^2 + \frac{1}{2}AP^2}</math>, which remains constant as <math>X</math> moves.
 +
 
 +
Next, consider the homothety of scale factor <math>\frac{2}{3}</math> about <math>O</math> mapping <math>N_9</math> to <math>G</math>. This means that the locus of <math>G</math> is a circle as well.
 +
 
 +
Finally, we take a homothety of scale factor <math>\frac{3}{2}</math> about <math>A</math> mapping <math>G</math> to <math>M</math>. Hence, the locus of <math>M</math>  is a circle, as desired. - Spacesam

Latest revision as of 08:58, 20 August 2021

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 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

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