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Difference between revisions of "2017 AMC 12A Problems/Problem 24"
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Quadrilateral <math>ABCD</math> is inscribed in circle <math>O</math> and has side lengths <math>AB=3, BC=2, CD=6</math>, and <math>DA=8</math>. Let <math>X</math> and <math>Y</math> be points on <math>\overline{BD}</math> such that <math>\frac{DX}{BD} = \frac{1}{4}</math> and <math>\frac{BY}{BD} = \frac{11}{36}</math>.
 
Quadrilateral <math>ABCD</math> is inscribed in circle <math>O</math> and has side lengths <math>AB=3, BC=2, CD=6</math>, and <math>DA=8</math>. Let <math>X</math> and <math>Y</math> be points on <math>\overline{BD}</math> such that <math>\frac{DX}{BD} = \frac{1}{4}</math> and <math>\frac{BY}{BD} = \frac{11}{36}</math>.
 +
Let <math>E</math> be the intersection of line <math>AX</math> and the line through <math>Y</math> parallel to <math>\overline{AD}</math>. Let <math>F</math> be the intersection of line <math>CX</math> and the line through <math>E</math> parallel to <math>\overline{AC}</math>. Let <math>G</math> be the point on circle <math>O</math> other than <math>C</math> that lies on line <math>CX</math>. What is <math>XF\cdot XG</math>?

Revision as of 17:13, 8 February 2017

Quadrilateral $ABCD$ is inscribed in circle $O$ and has side lengths $AB=3, BC=2, CD=6$, and $DA=8$. Let $X$ and $Y$ be points on $\overline{BD}$ such that $\frac{DX}{BD} = \frac{1}{4}$ and $\frac{BY}{BD} = \frac{11}{36}$. Let $E$ be the intersection of line $AX$ and the line through $Y$ parallel to $\overline{AD}$. Let $F$ be the intersection of line $CX$ and the line through $E$ parallel to $\overline{AC}$. Let $G$ be the point on circle $O$ other than $C$ that lies on line $CX$. What is $XF\cdot XG$?

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