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Difference between revisions of "2015 USAJMO Problems/Problem 3"

(Created page with "===Problem=== 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 ...")
 
(→‎Solution 2)
 
(8 intermediate revisions by 7 users not shown)
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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, A=(1,0) P=(1-a,b), Q=(1-a,-b), where (1-a)^2+b^2=1. Let angle <XAB=A, which is an acute angle, tanA=t, then X=(1-a,at).
+
<asy>
 +
size(8cm);
 +
pair A=(1,0);
 +
pair B=(-1,0);
 +
pair P=dir(70);
 +
pair Q=dir(-70);
 +
pair O=(0,0);
  
−
Angle <BOS=2A, S=(-cos2A,sin2A).
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pair X=0.3*P + 0.7*Q;
−
Let M=(u,v), then T=(2u+cos2A, 2v-sin2A)
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pair Y=5*X-4*A;
 +
pair S=intersectionpoints(A--Y,circle(O,1))[1];
 +
pair Z=(A-X)*dir(-90) + X;
 +
pair T=intersectionpoint(X--Z,circle(O,1));
 +
pair M=(S+T)/2;
  
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The condition TX perpendicular to AX yields (2v-sin2A-at)/(2u+cos2A+a-1)=cotA.    (E1)
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draw(circle(O,1));
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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')
+
draw(B--A--P--B--Q--A--S--T--X);
 +
draw(P--Q);
 +
dot("$A$",A,dir(A));
 +
dot("$B$",B,dir(B));
 +
dot("$P$",P,dir(P));
 +
dot("$Q$",Q,dir(Q));
 +
dot("$X$", X, SE);
 +
dot("$S$",S,dir(S));
 +
dot("$T$",T,dir(T));
 +
dot("$M$",M,dir(M));
 +
dot((0,0));
 +
</asy>
  
−
The condition that T is on the circle yields (2u+cos2A)^2+ (2v-sin2A)^2=1, namely vsin2A-ucos2A=u^2+v^2.  (E2)
 
  
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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)
+
We will use coordinate geometry.
  
−
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
+
Without loss of generality,
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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).
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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>.
 +
 
 +
Let angle <math>\angle XAB=A</math>, which is an acute angle, <math>\tan{A}=t</math>, then <math>X=(1-a,at)</math>.
 +
 
 +
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>.
 +
 
 +
The condition <math>TX \perp AX</math> yields: <math>(2v-\sin(2A)-at)/(2u+\cos(2A)+a-1)=\cot A. </math>    (E1)
 +
 
 +
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>
 +
 
 +
==More Solutions==
 +
https://artofproblemsolving.com/wiki/index.php/2015_USAMO_Problems/Problem_2

Latest revision as of 16:45, 29 April 2020

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

[asy] size(8cm); pair A=(1,0); pair B=(-1,0); pair P=dir(70); pair Q=dir(-70); pair O=(0,0); pair X=0.3*P + 0.7*Q; pair Y=5*X-4*A; pair S=intersectionpoints(A--Y,circle(O,1))[1]; pair Z=(A-X)*dir(-90) + X; pair T=intersectionpoint(X--Z,circle(O,1)); pair M=(S+T)/2; draw(circle(O,1)); draw(B--A--P--B--Q--A--S--T--X); draw(P--Q); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$P$",P,dir(P)); dot("$Q$",Q,dir(Q)); dot("$X$", X, SE); dot("$S$",S,dir(S)); dot("$T$",T,dir(T)); dot("$M$",M,dir(M)); dot((0,0)); [/asy]


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. $QED$

More Solutions

https://artofproblemsolving.com/wiki/index.php/2015_USAMO_Problems/Problem_2

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