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Difference between revisions of "2025 SSMO Speed Round Problems/Problem 4"

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==Problem==
 
==Problem==
  
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In rectangle <math>ABCD,</math> let <math>AB = 8,BC = 15,\omega</math> be the circumcircle of <math>ABCD</math>, <math>\ell</math> be the line through <math>B</math> parallel to <math>AC,</math> and <math>X \neq B</math> be the intersection of <math>\ell</math> and <math>\omega</math>. Suppose the value of <math>BX</math> can be expressed as <math>\frac{m}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.  
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In rectangle <math>ABCD,</math> let <math>AB = 8,BC = 15,\omega</math> be the circumcircle of <math>ABCD</math>, <math>\ell</math> be the line through <math>B</math> parallel to <math>AC,</math> and <math>X \neq B</math> be the intersection of <math>\ell</math> and <math>\omega</math>. Suppose the value of <math>BX</math> can be expressed as <math>\tfrac{m}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
  
 
==Solution==
 
==Solution==

Latest revision as of 15:13, 10 September 2025

Problem

In rectangle $ABCD,$ let $AB = 8,BC = 15,\omega$ be the circumcircle of $ABCD$, $\ell$ be the line through $B$ parallel to $AC,$ and $X \neq B$ be the intersection of $\ell$ and $\omega$. Suppose the value of $BX$ can be expressed as $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

[asy] import geometry; unitsize(2.7cm); point B=dir(aSin(0.47)); point C=dir(180-aSin(0.47)); point D=dir(180+aSin(0.47)); point A=dir(-aSin(0.47)); point X=dir(180-3*aSin(0.47)); point Y=intersectionpoint(line(C,D),line(B,X)); draw(A--B--C--D--cycle,p=black+0.3mm); draw(unitcircle,p=blue+0.3mm); draw(1.4*B-0.4*Y--1.2*Y-0.2*B,p=red+0.3mm); draw(C--Y,dashed); draw(A--C,dashed+red); dot(A,linewidth(4)); dot(B,linewidth(4)); dot(C,linewidth(4)); dot(D,linewidth(4)); dot(X,linewidth(4)); dot(Y,linewidth(4)); label("$A$",A,dir(-45)); label("$B$",B,dir(45)); label("$C$",C,dir(145)); label("$D$",D,dir(215)); label("$\omega$",(0,-0.9)); label("$\ell$",1.3*B-0.3*Y,dir(45)); label("$X$",X,dir(90)); label("$Y$",Y,dir(75)); [/asy]

Extend $CD$ to intersect $\ell$ at point $Y$. Note that $AB\parallel CY$ and $AC\parallel BY$, so $ABYC$ is a parallelogram. This implies that $BY = AC = \sqrt{8^2+15^2} = 17$ and $CY = AB = 8$. Let $t = BX$. By power of a point on $Y$ with respect to $\omega$, we have $XY\cdot BY = CY\cdot DY$. We know that $CY = 8$, $DY = 16$, $BY = 17$, and $XY = 17-t$. Thus, \[17(17-t) = 128.\] Solving this equation yields $t=\tfrac{161}{17}$, and we extract $161+17 = \boxed{178}$.

~Sedro

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