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

Revision as of 16:06, 9 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 $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Solution

$[asy] import geometry; unitsize(3cm); point B=dir(aSin(0.47)); point C=dir(180-aSin(0.47)); point D=dir(180+aSin(0.47)); point A=dir(-aSin(0.47)); point X=dir(180-3*aSin(0.47)); point Y=intersectionpoint(line(C,D),line(B,X)); draw(A--B--C--D--cycle,p=black+0.3mm); draw(unitcircle,p=blue+0.3mm); draw(1.4*B-0.4*Y--1.2*Y-0.2*B,p=red+0.3mm); draw(C--Y,dashed); draw(A--C,dashed+red); dot(A,linewidth(4)); dot(B,linewidth(4)); dot(C,linewidth(4)); dot(D,linewidth(4)); dot(X,linewidth(4)); dot(Y,linewidth(4)); label("$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 $161+17 = \boxed{178}$ .

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

Revision as of 16:06, 9 September 2025

Problem

Solution

@@ Line 10: / Line 10: @@
 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));
-<br /> point X=
+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("$A$",A,dir(-45));
-label("$B$",B,dir(135));
+label("$B$",B,dir(45));
-label("$C$",C,dir(215));
+label("$C$",C,dir(145));
-label("$D$",D,dir(-45));
+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 <math>CD</math> to intersect <math>\ell</math> at point <math>Y</math>. Note that <math>AB\parallel CY</math> and <math>AC\parallel BY</math>, so <math>ABYC</math> is a parallelogram. This implies that <math>BY = AC = \sqrt{8^2+15^2} = 17</math> and <math>CY = AB = 8</math>. Let <math>t = BX</math>. By power of a point on <math>Y</math> with respect to <math>\omega</math>, we have <math>XY\cdot BY = CY\cdot DY</math>. We know that <math>CY = 8</math>, <math>DY = 16</math>, <math>BY = 17</math>, and <math>XY = 17-t</math>. Thus,
+<cmath>17(17-t) = 128.</cmath>
+Solving this equation yields <math>t=\tfrac{161}{17}</math>, and we extract <math>161+17 = \boxed{178}</math>.