2021 Fall AMC 12B Problems/Problem 24: Difference between revisions

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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 $CF$ ?

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

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, $CF=\dfrac{480}{16}=\boxed{\textbf{C}~\text{30}}$ .

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Solution 1

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, $AC=AF=20.$

Now, note that $\angle CAF = \angle CAB$ and 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 $CF$ : $AF^2+AC^2-2 \cdot AF \cdot AC \cdot \cos(\angle CAF) = CF^2 \to CF = \boxed{\textbf{(C) }30}.$

- kevinmathz

Solution 2

By the Inscribed Angle Theorem note that $\angle ABD=\angle ABC=\angle AEC\ \text{and}\ \angle BAD=\angle DAC=\angle EAC$ since $\overline{ADE}$ bisects $\angle BAC$ . Thus, $\triangle ABD\sim\triangle AEC$ , and $\frac{AB}{AD}=\frac{AE}{AC}$ implying $AB\cdot AC=AD\cdot AE$ .

Note also that by Power of a Point, $AD\cdot AE=AB\cdot AF$ , so $AB\cdot AC=AB\cdot AF$ , giving $AC=AF=20$ and $BF=20-AB=20-11=9$ . Let $CF=x$ .

Applying Stewart's theorem on $\triangle ACF$ with cevian $\overline{CB}$ we obtain $\begin{align*} 11\cdot 9\cdot 20+24\cdot 20\cdot 24=11x^{2}+20\cdot 9\cdot 20 \\ 1980+11,520=11x^{2}+3600 \\ 13,500=11x^{2}+3600 \\ 11x^{2}=9900 \\ x^{2}=900 \\ x=\boxed{\textbf{(C)} ~30}\end{align*}$ ~Punxsutawney Phil

Solution 3

This solution is based on this figure: Image: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 4

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)

https://youtu.be/tEVbTtJlZjA

~math2718281828459

@@ Line 4: / Line 4: @@
 <math>\textbf{(A) } 28 \qquad \textbf{(B) } 20\sqrt{2} \qquad \textbf{(C) } 30 \qquad \textbf{(D) } 32 \qquad \textbf{(E) } 20\sqrt{3}</math>
+==Olympiad Solution using Spiral Similarity==
+Construct the <math>E</math>-antipode, <math>E^{\prime}\in(ABC)</math>. Notice <math>\triangle CE^{\prime}A\stackrel{+}{\sim}\triangle CBF</math> by spiral similarity at <math>C</math>, thus <math>CF=\dfrac{CB\cdot CA}{CE^{\prime}}=\frac{480}{CE^{\prime}}</math>. Let <math>CE^{\prime}=x</math>; by symmetry <math>BE^{\prime}=x</math> as well and <math>\cos\angle BE^{\prime}C=\cos\angle A=\tfrac{11^{2}+20^{2}-24^{2}}{2\cdot 11\cdot 20}=-\tfrac{1}{8}</math> from Law of Cosines in <math>\triangle ABC</math>, so by Law of Cosines in <math>\triangle BE^{\prime}C</math> we have <cmath>x^{2}+x^{2}+\left(2x^{2}\right)\left(-\dfrac{1}{8}\right)=24^{2}</cmath> from which <math>x=16</math>. Now, <math>CF=\dfrac{480}{16}=\boxed{\textbf{C}~\text{30}}</math>.
+[[File:AMC 12 2021B Fall-24 Geogebra Diagram.png|600px]]
 ==Solution 1==

2021 Fall AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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