Difference between revisions of "2025 SSMO Accuracy Round Problems/Problem 5"

Latest revision as of 14:45, 17 September 2025

Problem

$ABC$ is an isosceles triangle with base $BC = 6$ and $AB = AC$ . Point $M$ is the midpoint of $BC$ such that $AM = 9$ . Circle $\omega_1$ is the circumcircle of $ABC$ with radius $R,$ and $\omega_2$ is a circle passing through $B$ and $C$ with radius $2R$ and center on the opposite side of $BC$ as $A$ . Segment $AM$ intersects $\omega_2$ at point $X$ and $\omega_1$ at point $Y,$ where $X$ lies between $A$ and $Y$ . The length $XY$ can be expressed as $m - \sqrt{n},$ where $m$ and $n$ are positive integers. Find $m+n$ .

Solution

$[asy] unitsize(0.3cm); import geometry; point A=(0,9); point B=(-3,0); point C=(3,0); point M=0.5*C+0.5*B; point Y=intersectionpoints(line(A,M),circle(A,B,C))[0]; point Z=(0,-19.54); point X=intersectionpoints(circle(B,C,Z),line(A,M))[1]; draw(A--B--C--cycle); draw(circle(A,B,C),blue); draw(circle(B,C,Z),blue); draw(A--Y,red); draw(Y--Z,red+dashed); dot(A); dot(B); dot(C); dot(M); dot(X); dot(Y); dot(Z); label("$A$",A,dir(90)); label("$B$",B+(-0.35,0.2),dir(180)); label("$C$",C+(0.35,0.2),dir(0)); label("$Z$",Z,dir(-90)); label("$\omega_1$",(-4,4),dir(90)); label("$\omega_2$",(9,-10),dir(90)); point S=(30,-5); point Ap=(0,27)+S; point Bp=(-9,0)+S; point Cp=(9,0)+S; point Mp=0.5*Bp+0.5*Cp; point Yp=intersectionpoints(line(Ap,Mp),circle(Ap,Bp,Cp))[0]; point Zp=(0,-61.62)+S; point Xp=intersectionpoints(circle(Bp,Cp,Zp),line(Ap,Mp))[1]; point P=circumcenter(Ap,Bp,Cp); point Q=circumcenter(Bp,Cp,Zp); real r=circumradius(Ap,Bp,Cp); real s=circumradius(Bp,Cp,Zp); draw(Bp--Cp); draw(Xp--Xp+(0,3),red); draw(Xp--Yp,red); draw(Yp--Yp+(0,-3),red+dashed); draw(arccircle(P+r*dir(320),Yp,P+r*dir(220)),blue); draw(arccircle(Q+s*dir(115),Xp,Q+s*dir(65)),blue); draw(Bp--0.85*Bp+0.15*Ap); draw(Cp--0.85*Cp+0.15*Ap); dot(Bp); dot(Cp); dot(Mp); dot(Xp); dot(Yp); label("$B$",Bp+(-0.8,0.3),dir(180)); label("$C$",Cp+(0.8,0.3),dir(0)); label("$M$",Mp+(0.1,-0.2),dir(215)); label("$X$",Xp+(-0.8,0),dir(90)); label("$Y$",Yp+(0.8,0),dir(-90)); [/asy]$

Suppose line $AM$ intersects $\omega_2$ again at $Z\ne X$ . Note that because $\triangle ABC$ is isosceles, $AY$ and $XZ$ are diameters of $\omega_1$ and $\omega_2$ , respectively, so $XZ = 2\cdot AY$ . Now, by power of a point on $M$ with respect to $\omega_1$ , we have $AM\cdot YM = BM\cdot CM$ . Since $AM = 9$ and $BM=CM = 3$ , we have $YM = 1$ . Therefore, $AY = 10$ and $XZ = 20$ .

Let $t = XM$ ; note that $ZM = 20-t$ . By power of a point on $M$ with respect to $\omega_2$ , we have $XM\cdot ZM = BM\cdot CM$ , or $t(20-t) = 9$ . This is a quadratic in $t$ that we can solve to find that $t = 10+\sqrt{91}$ or $t=10-\sqrt{91}$ . Because the radius of $\omega_2$ is greater than the radius of $\omega_1$ , it follows that $M$ lies between $X$ and the center of $\omega_2$ . Thus, $t < 10$ , so $t = 10 - \sqrt{91}$ . To finish, we have $XY = XM+YM = 11-\sqrt{91}$ , so the answer is $11+91 = \boxed{102}$ .

~Sedro

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Difference between revisions of "2025 SSMO Accuracy Round Problems/Problem 5"

Latest revision as of 14:45, 17 September 2025

Problem

Solution

@@ Line 4: / Line 4: @@
 ==Solution==
+<asy>
+unitsize(0.3cm);
+import geometry;
+point A=(0,9);
+point B=(-3,0);
+point C=(3,0);
+point M=0.5*C+0.5*B;
+point Y=intersectionpoints(line(A,M),circle(A,B,C))[0];
+point Z=(0,-19.54);
+point X=intersectionpoints(circle(B,C,Z),line(A,M))[1];
+draw(A--B--C--cycle);
+draw(circle(A,B,C),blue);
+draw(circle(B,C,Z),blue);
+draw(A--Y,red);
+draw(Y--Z,red+dashed);
+dot(A);
+dot(B);
+dot(C);
+dot(M);
+dot(X);
+dot(Y);
+dot(Z);
+label("$A$",A,dir(90));
+label("$B$",B+(-0.35,0.2),dir(180));
+label("$C$",C+(0.35,0.2),dir(0));
+label("$Z$",Z,dir(-90));
+label("$\omega_1$",(-4,4),dir(90));
+label("$\omega_2$",(9,-10),dir(90));
+point S=(30,-5);
+point Ap=(0,27)+S;
+point Bp=(-9,0)+S;
+point Cp=(9,0)+S;
+point Mp=0.5*Bp+0.5*Cp;
+point Yp=intersectionpoints(line(Ap,Mp),circle(Ap,Bp,Cp))[0];
+point Zp=(0,-61.62)+S;
+point Xp=intersectionpoints(circle(Bp,Cp,Zp),line(Ap,Mp))[1];
+point P=circumcenter(Ap,Bp,Cp);
+point Q=circumcenter(Bp,Cp,Zp);
+real r=circumradius(Ap,Bp,Cp);
+real s=circumradius(Bp,Cp,Zp);
+draw(Bp--Cp);
+draw(Xp--Xp+(0,3),red);
+draw(Xp--Yp,red);
+draw(Yp--Yp+(0,-3),red+dashed);
+draw(arccircle(P+r*dir(320),Yp,P+r*dir(220)),blue);
+draw(arccircle(Q+s*dir(115),Xp,Q+s*dir(65)),blue);
+draw(Bp--0.85*Bp+0.15*Ap);
+draw(Cp--0.85*Cp+0.15*Ap);
+dot(Bp);
+dot(Cp);
+dot(Mp);
+dot(Xp);
+dot(Yp);
+label("$B$",Bp+(-0.8,0.3),dir(180));
+label("$C$",Cp+(0.8,0.3),dir(0));
+label("$M$",Mp+(0.1,-0.2),dir(215));
+label("$X$",Xp+(-0.8,0),dir(90));
+label("$Y$",Yp+(0.8,0),dir(-90));
+</asy>
+Suppose line <math>AM</math> intersects <math>\omega_2</math> again at <math>Z\ne X</math>. Note that because <math>\triangle ABC</math> is isosceles, <math>AY</math> and <math>XZ</math> are diameters of <math>\omega_1</math> and <math>\omega_2</math>, respectively, so <math>XZ = 2\cdot AY</math>. Now, by power of a point on <math>M</math> with respect to <math>\omega_1</math>, we have <math>AM\cdot YM = BM\cdot CM</math>. Since <math>AM = 9</math> and <math>BM=CM = 3</math>, we have <math>YM = 1</math>. Therefore, <math>AY = 10</math> and <math>XZ = 20</math>.
+Let <math>t  = XM</math>; note that <math>ZM = 20-t</math>. By power of a point on <math>M</math> with respect to <math>\omega_2</math>, we have <math>XM\cdot ZM = BM\cdot CM</math>, or <math>t(20-t) = 9</math>. This is a quadratic in <math>t</math> that we can solve to find that <math>t = 10+\sqrt{91}</math> or <math>t=10-\sqrt{91}</math>. Because the radius of <math>\omega_2</math> is greater than the radius of <math>\omega_1</math>, it follows that <math>M</math> lies between <math>X</math> and the center of <math>\omega_2</math>. Thus, <math>t < 10</math>, so <math>t = 10 - \sqrt{91}</math>. To finish, we have <math>XY = XM+YM = 11-\sqrt{91}</math>, so the answer is <math>11+91 = \boxed{102}</math>.
+~Sedro