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 $\textbf{(C) } 30$.