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

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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>.
 
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>.
  
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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\in \{10+\sqrt{91}, 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>.
+
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
 
~Sedro

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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