Sample Question 24:

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 $XF\cdot XG$?

$\textbf{(A) }17\qquad\textbf{(B) }\frac{59 - 5\sqrt{2}}{3}\qquad\textbf{(C) }\frac{91 - 12\sqrt{3}}{4}\qquad\textbf{(D) }\frac{67 - 10\sqrt{2}}{3}\qquad\textbf{(E) }18$

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Answer Keys

Question 24: A

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Solutions

Question 24
Step 1: Determine that the ratio $XY/BD = 1 - 1/4 - 11/36 = 4/9$ based on the given ratios.

Step 2: Infer two sets of similar triangles, $triangle AXD \sim triangle EXY$ and $triangle ACX \sim triangle EFX$, using AA Similarity. This means $DX/XY = 9/16$ and $AX/XE = DX/XY = 9/16$. Hence, $XF/CX = 16/9$.

Step 3: Since quadrilateral ABCD is cyclic, use the Law of Cosines to find the length of $BD$. You will get:

- $BD^2=3^2+8^2-2*3*8*cos\angle BAD=2^2+6^2-2*2*6*cos (180-\angle BAD)$

- This leads to $\cos\angle BAD = 11/24$

- Hence, $BD=\sqrt{51}$.

Step 4: Use the Power of a Point theorem on point X. This gives $CX \cdot XG = DX \cdot XB = ( \sqrt{51}/4 ) * (3\sqrt{51}/4)$.

Step 5: Substitute $XF/CX$ into the equation $XF \cdot XG = XF/CX * CX \cdot XG$, giving the result $XF \cdot XG = 51/3 = 17$.

The answer is $\textbf{(A) }17$.