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 \(\textbf{(A) }17\).