Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 3)
(→‎Solution 1)
Line 3: Line 3:
  
 
===Solution 1===
 
===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.
 
We will use coordinate geometry.

Revision as of 17:44, 10 March 2019

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. $QED$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_USAJMO_Problems/Problem_3&oldid=104228"