Cyclic Quattrocento (Fall 2021 AMC 12B #24)

Fall 2021 AMC 12B #24

Problem: 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 $CF$?

Solution: We can create a diagram to try to find unique relationships in this problem.

By angle chasing in the cyclic quadrilateral $ABEC$ given that $AE$ is the angle bisector of $\angle{BAC}$ we can find that $\angle{BAE} = \angle{CAE} = \angle{BCE} = \angle{CBE}$. This means that triangle $BCE$ is isosceles and $BE = CE$, which may be useful later on.

Now, we can focus on $CF$, which is the length we want to find. We can see that $ACF$ has a cevian $CB$, so we can use Stewart's Theorem (we will "derive" it with the law of cosines here) to solve for CF in terms of other side lengths.

We can let $\angle{ABC} = \theta$ which means $\angle{CBF} = 180^\circ - \theta$. Then, we can apply the law of cosines on triangles $ABC$ and $CBF$ and use the property that $\cos\theta = -\cos(180^\circ - \theta)$ to get the following equations \[ AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos{\theta} \] \[ CF^2 = BF^2 + BC^2 + 2 \cdot BF \cdot BC \cdot \cos{\theta} \] We can multiply the first equation by $BF$ and the second equation by $AB$ to eliminate the $2 \cdot AB \cdot BF \cdot BC \cdot \cos{\theta}$ term. So we have \[ AC^2 \cdot BF = AB^2 \cdot BF + BC^2 \cdot BF - 2 \cdot AB \cdot BF \cdot BC \cdot \cos{\theta} \] \[ CF^2 \cdot AB = BF^2 \cdot AB + BC^2 \cdot AB + 2 \cdot AB \cdot BF \cdot BC \cdot \cos{\theta} \] Adding the two we get \[ AC^2 \cdot BF + CF^2 \cdot AB = AB^2 \cdot BF + BF^2 \cdot AB + BC^2 \cdot BF + BC^2 \cdot AB \] We can simplify this to \[ AC^2 \cdot BF + CF^2 \cdot AB = (AB + BF) (AB \cdot BF) + (AB + BF) (BC^2) = (AB + BF) (AB \cdot BF + BC^2)\] Now, we can input the values of $AB, AC, BC$, which leaves us with \[ 20^2 BF + 11 CF^2 = (11 + BF) (11 BF + 24^2)\] Thus, we see that we only need to find $BF$ to solve for $CF$, so we can try to use similar triangles and $BE = CE$. 

We can first use power of a point with point A and the circumcircle of $BEF$ to get $AB \cdot AF = AD \cdot AE$ to find $AF = AB+BF = 11+BF$. Since we do not know $AD, AE$, we can try to express them in terms of other sides using similar triangles, trying to utilize $BE = CE$ and known side lengths. 

Since $\bigtriangleup ABE \cong \bigtriangleup BDE$ because $\angle AEB$ is shared and $\angle BAE = \angle DBE$, we can find $ \frac{AE}{AB} = \frac{BE}{BD}$. Similarly, $\bigtriangleup ABD \cong \bigtriangleup CED$ because $\angle BAD = \angle ECD$ and $\angle ADB = \angle CDE$, so $\frac{AD}{AB} = \frac{CD}{CE}$. Thus, we have $AD = \frac{AB \cdot CD}{CE}$ and $AE = \frac{AB \cdot BE}{BD}$.

Additionally, we can use the Angle Bisector Theorem to find that $\frac{CD}{BD} = \frac{AC}{AB}$. So, \[ AB \cdot AF = AD \cdot AE =  \frac{AB \cdot CD}{CE} \cdot \frac{AB \cdot BE}{BD} = AB^2 \cdot \frac{CD}{BD} = AB \cdot AC\] which means $AF = AC$.

Thus, $AF = 11 + BF = 20$, so $BF = 9$. We can substitute this in our previous result to find \[ 20^2 (9) + 11 CF^2 = (11 + 9) (11 (9) + 24^2)\] which gives us $CF^2 = 900$ or $CF = 30$.

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