2021 Fall AMC 12B Problems/Problem 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 $CF$?

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

Solution 1 (Reflection)


By the Law of Cosine $\cos A = \frac{AC^2 + AB^2 - BC^2}{ 2 \cdot AC \cdot AB} = \frac{20^2 + 11^2 - 24^2}{2\cdot20\cdot11} = -\frac18$

As $ABEC$ is a cyclic quadrilateral, $\angle CEA = \angle CBA$. As $BDEF$ is a cyclic quadrilateral, $\angle CBA = \angle FEA$.

$\because \quad \angle CEA = \angle FEA \quad \text{and} \quad \angle CAE = \angle FAE$

$\therefore \quad \triangle AFE \cong \triangle ACE$ by $ASA$

Hence, $AF = AC = 20$

By the Law of Cosine $CF = \sqrt{20^2 + 20^2 - 2 \cdot 20 \cdot 20 (-\frac18)} = \sqrt{900} = \boxed{\textbf{C}~\text{30}}$

Note that $F$ is $C$'s reflection over line $AE$, quadrilateral $ACEF$ is a kite symmetrical by line $AE$, $AE \perp CF$.


Solution 2 (Olympiad Solution using Spiral Similarity)

Construct the $E$-antipode, $E^{\prime}\in(ABC)$. Notice $\triangle CE^{\prime}A\stackrel{+}{\sim}\triangle CBF$ by spiral similarity at $C$, thus $CF=\dfrac{CB\cdot CA}{CE^{\prime}}=\frac{480}{CE^{\prime}}$. Let $CE^{\prime}=x$; by symmetry $BE^{\prime}=x$ as well and $\cos\angle BE^{\prime}C=\cos\angle A=\tfrac{11^{2}+20^{2}-24^{2}}{2\cdot 11\cdot 20}=-\tfrac{1}{8}$ from Law of Cosines in $\triangle ABC$, so by Law of Cosines in $\triangle BE^{\prime}C$ we have \[x^{2}+x^{2}+\left(2x^{2}\right)\left(-\dfrac{1}{8}\right)=24^{2}\] from which $x=16$. Now, $CF=\dfrac{480}{16}=\boxed{\textbf{C}~\text{30}}$.

AMC 12 2021B Fall-24 Geogebra Diagram.png

Solution 3

Claim: $\triangle ADC \sim \triangle ABE.$

Proof: Note that $\angle CAD = \angle CAE = \angle EAB$ and $\angle DCA = \angle BCA = \angle BEA$ meaning that our claim is true by AA similarity.

Because of this similarity, we have that \[\frac{AC}{AD} = \frac{AE}{AB} \Longrightarrow AB \cdot AC = AD \cdot AE = AB \cdot AF\] by Power of a Point. Thus, $AC=AF=20.$

Two solution methods follow from here.

Solution 3.1 (Stewart's theorem)

Applying Stewart's theorem on $\triangle ABC$ with cevian $\overline{CF}$ using the directed lengths $AF = AC = 20$ and $FB = 11-20 = -9$, we obtain \begin{align*} (20)(-9)(11) + (CF)(11)(CF) &= (24)(20)(24) + (20)(-9)(20) \\ 11CF^{2} - 1980 &= 11520 - 3600\end{align*} so $CF=\sqrt{\frac{11520 - 3600 + 1980}{11}}=\sqrt{\frac{9900}{11}}=\sqrt{900}=\boxed{\textbf{(C) }30}$.

Solution 3.2 (Double Cosine Law)

Note that $\angle CAF = \angle CAB$ so we may plug into Law of Cosines to find the angle's cosine: \[AB^2+AC^2-2\cdot AB \cdot AC \cdot \cos(\angle CAB) = BC^2 \to \cos(\angle CAB) = -\frac{1}{8}.\]

So, we observe that we can use Law of Cosines again to find $CF$: \[CF^2 = AF^2+AC^2-2 \cdot AF \cdot AC \cdot \cos(\angle CAF) = 900 \to CF=\boxed{\textbf{(C) }30}\] both ways.

- Kevinmathz

Solution 4

This solution is based on this figure: 2021 AMC 12B (Nov) Problem 24, sol.png

Denote by $O$ the circumcenter of $\triangle BED$. Denote by $R$ the circumradius of $\triangle BED$.

In $\triangle BCF$, following from the law of cosines, we have \begin{align*} CF^2 & = BC^2 + BF^2 - 2 BC \cdot BF \cos \angle CBF \\ & = BC^2 + BF^2 + 2 BC \cdot BF \cos \angle ABC .  \hspace{1cm} (1) \end{align*} For $BF$, we have \begin{align*} BF & = 2 R \cos \angle FBO \\ & = 2 R \cos \left( 180^\circ - \angle ABC - \angle CBO \right) \\ & = 2 R \cos \left( 180^\circ - \angle ABC - \frac{180^\circ - \angle BOD}{2} \right) \\ & = 2 R \cos \left( 180^\circ - \angle ABC - \frac{180^\circ - 2 \angle BED}{2} \right) \\ & = 2 R \cos \left( 180^\circ - \angle ABC - \frac{180^\circ - 2 \angle BCA}{2} \right) \\ & = 2 R \cos \left( 90^\circ - \angle ABC + \angle BCA \right) \\ & = 2 R \sin \left( \angle ABC - \angle BCA \right) \\ & = \frac{BD}{\sin \angle BED} \sin \left( \angle ABC - \angle BCA \right) \\ & = \frac{BD}{\sin \angle BCA} \sin \left( \angle ABC - \angle BCA \right) \\ & = BD \left( \sin \angle ABC \cot \angle BCA - \cos \angle ABC \right) . \hspace{1cm} (2) \end{align*} The fourth equality follows from the property that $B$, $D$, $E$ are concyclic. The fifth and the ninth equalities follow from the property that $A$, $B$, $C$, $E$ are concyclic.

Because $AD$ bisects $\angle BAC$, following from the angle bisector theorem, we have \[ \frac{BD}{CD} = \frac{AB}{AC} . \] Hence, $BD = \frac{24 \cdot 11}{31}$.

In $\triangle ABC$, following from the law of cosines, we have \begin{align*} \cos \angle ABC & = \frac{AB^2 + BC^2 - AC^2}{2 AB \cdot BC} \\ & = \frac{9}{16} \end{align*} and \begin{align*} \cos \angle BCA & = \frac{AC^2 + BC^2 - AB^2}{2 AC \cdot BC} \\ & = \frac{57}{64} . \end{align*} Hence, $\sin \angle ABC = \frac{5 \sqrt{7}}{16}$ and $\sin \angle BCA = \frac{11 \sqrt{7}}{64}$. Hence, $\cot \angle BCA = \frac{57}{11 \sqrt{7}}$.

Now, we are ready to compute $BF$ whose expression is given in Equation (2). We get $BF = 9$.

Now, we can compute $CF$ whose expression is given in Equation (1). We have $CF = 30$.

Therefore, the answer is $\boxed{\textbf{(C) }30}$.

~Steven Chen (www.professorchenedu.com)

Solution 5

Denote $B=(0, 0)$ and $C=(24, 0)$. Note that by Heron's formula the area of $\triangle ABC$ is $\frac{165\sqrt{7}}{4}$ so the $y$-coordinate of $A$ (height of $A$ above the $x$-axis) is easily computed by the base-height formula as $\frac{55\sqrt7}{16}$.

Now, since $AB=11$, the $x$-coordinate of $A$ satisfies $x^2+(\frac{55\sqrt7}{16})^2=11^2$ and solving gives $x=\frac{99}{16}$.

The circumcircle of $\triangle ABC$ has radius $\frac{abc}{4A}=\frac{11\cdot 24\cdot 20}{165\sqrt7}=\frac{32}{\sqrt7}$. We know by the perpendicular bisector rule that the circumcenter $O$ is located directly below the midpoint of $\overline{BC}$ ($x$-coordinate $12$).

So, the negative $y$ coordinate of $O$ satisfies $12^2+y^2=(\frac{32}{\sqrt7})^2$ and solving gives $y=-\frac{4}{\sqrt7}$.

It's also clear that point $E$ is going to be located directly below $O$ on the circle, because the angle bisector intersects the circumcircle at the midpoint of the arc (Fact 5). Since the radius of the circle is $\frac{32}{\sqrt7}$, we have the coordinates of $E=(12, -\frac{36}{\sqrt7})$

Solving for point $D$ (the point on the $x$-axis between $A$ and $E$), we get that $D=(\frac{264}{31}, 0)$.

So now we know six of the critical points: $A=(\frac{99}{16}, \frac{55\sqrt7}{16})$; $B=(0, 0)$; $C=(24, 0)$; $D=(\frac{264}{31}, 0)$; $E=(12, -\frac{36}{\sqrt7})$; $O=(12, -\frac{4}{\sqrt7})$.

We are now ready to add in the circumcircle of $\triangle BDE$, which has radius $\frac{BD\cdot DE\cdot BE}{4[BDE]}$. From the above information, $BD=\frac{264}{31}$, $DE=\sqrt{(\frac{108}{31})^2+(\frac{36}{\sqrt7})^2}$, and $BE=\sqrt{12^2+(\frac{36}{\sqrt7})^2}$.

After a bit of simplification we end up with $DE=\frac{1152}{31\sqrt7}$ and $BE=\frac{48}{\sqrt7}$.

For the area of $\triangle BDE$, the altitude dropped from vertex $E$ has height $\frac{36}{\sqrt7}$, and the base $\overline{BD}$ has length $\frac{264}{31}$, so its area is $\frac12\cdot\frac{36}{\sqrt7}\cdot\frac{264}{31}=\frac{4752}{31\sqrt7}$.

Thus, $\frac{BD\cdot DE\cdot BE}{4[BDE]}=\frac{\tfrac{264}{31}\cdot\tfrac{1152}{31\sqrt7}\cdot\tfrac{48}{\sqrt{7}}}{4\cdot\tfrac{4752}{31\sqrt7}}$ which after tons of cancellations becomes $\frac{768}{31\sqrt7}$.

We know from the perpendicular bisector rule that the circumcenter $P$ of $\triangle BDE$ is located directly below the midpoint of $\overline{BD}$ ($x$-coordinate $\frac{132}{31}$).

So, the negative $y$-coordinate of $P$ satisfies $(\frac{132}{31})^2+y^2=(\frac{768}{31\sqrt7})^2$, and solving gives $y=-\frac{684}{31\sqrt7}$. Thus, the equation of the circumcircle of $\triangle BDE$ is $(x-\frac{132}{31})^2+(y+\frac{684}{31\sqrt7})^2=(\frac{768}{31\sqrt7})^2$.

Point $F$ is the intersection of this circle and the line $\overline{AB}$, which has equation $y=\frac{5\sqrt7}{9}x$. So, we substitute $y=\frac{5\sqrt7}{9}x$ into the equation of the circle to get $(x-\frac{132}{31})^2+(\frac{5\sqrt7}{9}x+\frac{684}{31\sqrt7})^2=(\frac{768}{31\sqrt7})^2$.

After simplifying, we have $\frac{256}{81}x^2+16x=0$ (the $\frac{768}{31\sqrt7}$'s cancel out), whose solutions are $x=0$ and $x=-\frac{81}{16}$. The first corresponds to the origin, and the second corresponds to point $F$. Thus the coordinates of $F$ are $(-\frac{81}{16}, \frac{5\sqrt7}{9}\cdot\frac{-81}{16})=(-\frac{81}{16}, -\frac{45\sqrt7}{16})$.

The coordinates of $C$ are $(24, 0)$, so \[CF=\sqrt{(24+\frac{81}{16})^2+(\frac{45\sqrt7}{16})^2}=\sqrt{(\frac{465}{16})^2+(\frac{45\sqrt7}{16})^2}=\frac{\sqrt{465^2+(45\sqrt7)^2}}{16}=\frac{\sqrt{(15\cdot 31)^2+(15\cdot 3\sqrt7)^2}}{16}=\frac{15\sqrt{31^2+(3\sqrt7)^2}}{16}=\frac{15\sqrt{961+63}}{16}=\frac{15\sqrt{1024}}{16}=\frac{15}{16}\cdot 32=30.\]

Video Solution by Power of Logic(Trig and Power of a point)



See Also

2021 Fall AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png