Sample Question 24:

Triangle \(ABC\) has side lengths \(AB = 11, BC=24\), and \(CA = 20\). The bisector of \(\angle{BAC}\) intersects \(\overline{BC}\) in point \(D\), and intersects the circumcircle of \(\triangle{ABC}\) in point \(E \ne A\). The circumcircle of \(\triangle{BED}\) intersects the line \(AB\) in points \(B\) and \(F \ne B\). What is \(CF\)?

\(\textbf{(A) } 28 \qquad \textbf{(B) } 20\sqrt{2} \qquad \textbf{(C) } 30 \qquad \textbf{(D) } 32 \qquad \textbf{(E) } 20\sqrt{3}\)

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Answer Keys

Question 24: C

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Solutions

Question 24
Step 1: Determine the cos of angle A via the Law of Cosine. The computation gives us \(\cos A = -\frac18\).

Step 2: Notice that since ABEC is a cyclic quadrilateral, we have \(\angle CEA = \angle CBA\). Similarly, since BDEF is a cyclic quadrilateral, \(\angle CBA = \angle FEA\).

Step 3: By transitivity, since \(\angle CEA = \angle FEA\) and \(\angle CAE = \angle FAE\), we deduce that \(\triangle AFE \cong \triangle ACE\) via the Angle-Side-Angle congruence theorem, hence \(AF = AC = 20\).

Step 4: Use the Law of Cosine one more time in triangle ACF to get
\(CF^2 = AF^2+AC^2+2 \cdot AF \cdot AC \cdot \cos(\angle CAF)\). We can substitute \(AF = AC = 20\) and \(\cos(\angle CAF) = -\frac{1}{8}\) which we computed in step 1 to get \(CF^2 = 20^2 + 20^2 - 2 \cdot 20 \cdot 20 (-\frac18)= 900\), and taking the square root results in \(CF=30\).

So the answer is \(\textbf{(C) } 30\).