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

Latest revision as of 15:45, 29 April 2020

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. $QED$

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

Latest revision as of 15:45, 29 April 2020

Contents

Problem

Solution 1

Solution 2

More Solutions

@@ Line 3: / Line 3: @@
 ===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.
@@ Line 52: / Line 81: @@
 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===
+==More Solutions==
+https://artofproblemsolving.com/wiki/index.php/2015_USAMO_Problems/Problem_2
-The below solution is false.
-Note that each point <math>X</math> on <math>PQ</math> corresponds to exactly one point on arc <math>PBQ</math>. Also notice that since <math>AB</math> is the diameter of <math>\omega</math>, <math>\angle ASB</math> is always a right angle; therefore, point <math>T</math> is always <math>B</math>. WLOG, assume that <math>\omega</math> is on the coordinate plane, and <math>B</math> corresponds to the origin. The locus of <math>M</math>, since the locus of <math>S</math> is arc <math>PBQ</math>, is the arc that is produced when arc <math>PBQ</math> is dilated by <math>\frac {1} {2}</math> with respect to the origin, which resides on the circle <math>\psi</math>, which is produced when <math>\omega</math> is dilated by <math>\frac {1} {2}</math> with respect to the origin. By MSmathlete1018
-The above solution is false; it gives the wrong locus. The step where they mess up is in stating that <math>T=B</math>, however this is not true because <math>\angle TXA</math> is right, not <math>\angle TSA</math>.