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Difference between revisions of "2021 Fall AMC 12B Problems/Problem 24"
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~Steven Chen (www.professorchenedu.com)
 
~Steven Chen (www.professorchenedu.com)
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==Solution 4==
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Denote <math>B=(0, 0)</math> and <math>C=(24, 0)</math>. Note that by Heron's formula the area of <math>\triangle ABC</math> is <math>\frac{165\sqrt{7}}{4}</math> so the <math>y</math>-coordinate of <math>A</math> (height of <math>A</math> above the <math>x</math>-axis) is easily computed by the base-height formula as <math>\frac{55\sqrt7}{16}</math>.
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Now, since <math>AB=11</math>, the <math>x</math>-coordinate of <math>A</math> satisfies <math>x^2+(\frac{55\sqrt7}{16})^2=11^2</math> and solving gives <math>x=\frac{99}{16}</math>.
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The circumcircle of <math>\triangle ABC</math> has radius <math>\frac{abc}{4A}=\frac{11\cdot 24\cdot 20}{165\sqrt7}=\frac{32}{\sqrt7}</math>. We know by the perpendicular bisector rule that the circumcenter <math>O</math> is located directly below the midpoint of <math>\overline{BC}</math> (<math>x</math>-coordinate <math>12</math>).
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So, the negative <math>y</math> coordinate of <math>O</math> satisfies <math>12^2+y^2=(\frac{32}{\sqrt7})^2</math> and solving gives <math>y=-\frac{4}{\sqrt7}</math>.
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It's also clear that point <math>E</math> is going to be located directly below <math>O</math> 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 <math>\frac{32}{\sqrt7}</math>, we have the coordinates of <math>E=(12, -\frac{36}{\sqrt7})</math>
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Solving for point <math>D</math> (the point on the <math>x</math>-axis between <math>A</math> and <math>E</math>), we get that <math>D=(\frac{264}{31}, 0)</math>.
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So now we know six of the critical points: <math>A=(\frac{99}{16}, \frac{55\sqrt7}{16})</math>; <math>B=(0, 0)</math>; <math>C=(24, 0)</math>; <math>D=(\frac{264}{31}, 0)</math>; <math>E=(12, -\frac{36}{\sqrt7})</math>; <math>O=(12, -\frac{4}{\sqrt7})</math>.
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We are now ready to add in the circumcircle of <math>\triangle BDE</math>, which has radius <math>\frac{BD\cdot DE\cdot BE}{4[BDE]}</math>. From the above information, <math>BD=\frac{264}{31}</math>, <math>DE=\sqrt{(\frac{108}{31})^2+(\frac{36}{\sqrt7})^2}</math>, and <math>BE=\sqrt{12^2+(\frac{36}{\sqrt7})^2}</math>.
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 +
After a bit of simplification we end up with <math>DE=\frac{1152}{31\sqrt7}</math> and <math>BE=\frac{48}{\sqrt7}</math>.
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 +
For the area of <math>\triangle BDE</math>, the altitude dropped from vertex <math>E</math> has height <math>\frac{36}{\sqrt7}</math>, and the base <math>\overline{BD}</math> has length <math>\frac{264}{31}</math>, so its area is <math>\frac12\cdot\frac{36}{\sqrt7}\cdot\frac{264}{31}=\frac{4752}{31\sqrt7}</math>.
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Thus, <math>\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}}</math> which after tons of cancellations becomes <math>\frac{768}{31\sqrt7}</math>.
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We know from the perpendicular bisector rule that the circumcenter <math>P</math> of <math>\triangle BDE</math> is located directly below the midpoint of <math>\overline{BD}</math> (<math>x</math>-coordinate <math>\frac{132}{31}</math>).
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So, the negative <math>y</math>-coordinate of <math>P</math> satisfies <math>(\frac{132}{31})^2+y^2=(\frac{768}{31\sqrt7})^2</math>, and solving gives <math>y=-\frac{684}{31\sqrt7}</math>. Thus, the equation of the circumcircle of <math>\triangle BDE</math> is <math>(x-\frac{132}{31})^2+(y+\frac{684}{31\sqrt7})^2=(\frac{768}{31\sqrt7})^2</math>.
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Point <math>F</math> is the intersection of this circle and the line <math>\overline{AB}</math>, which has equation <math>y=\frac{5\sqrt7}{9}x</math>. So, we substitute <math>y=\frac{5\sqrt7}{9}x</math> into the equation of the circle to get <math>(x-\frac{132}{31})^2+(\frac{5\sqrt7}{9}x+\frac{684}{31\sqrt7})^2=(\frac{768}{31\sqrt7})^2</math>.
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After simplifying, we have <math>\frac{256}{81}x^2+16x=0</math> (the <math>\frac{768}{31\sqrt7}</math>'s cancel out), whose solutions are <math>x=0</math> and <math>x=-\frac{81}{16}</math>. The first corresponds to the origin, and the second corresponds to point <math>F</math>. Thus the coordinates of <math>F</math> are <math>(-\frac{81}{16}, \frac{5\sqrt7}{9}\cdot\frac{-81}{16})=(-\frac{81}{16}, -\frac{45\sqrt7}{16})</math>.
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The coordinates of <math>C</math> are <math>(24, 0)</math>, so <cmath>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.</cmath>
  
 
==Video Solution by Power of Logic(Trig and Power of a point)==
 
==Video Solution by Power of Logic(Trig and Power of a point)==

Revision as of 18:05, 30 March 2022

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 $CF$?

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

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 \[\frac{AC}{AD} = \frac{AE}{AB} \to AB \cdot AC = AD \cdot AE = AB \cdot AF\] 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 and the definition of angle bisectors note that\[\angle ABD=\angle ABC=\angle AEC\ \text{and}\ \angle BAD=\angle DAC=\angle EAC\]so $\triangle ABD\sim\triangle AEC$. Therefore $\frac{AB}{AD}=\frac{AE}{AC}\rightarrow AB\cdot AC=AD\cdot AE$. By PoP, we can also express $AD\cdot AE$ as $AB\cdot AF,$ so $AB\cdot AC=AB\cdot AF\rightarrow AC=AF=20$ and $BF=20-AB=20-11=9$. Let $CF=x$. Applying Stewart’s theorem on $\triangle ACF$ with cevian $\overrightarrow{CB},$ we have \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

See Also

2021 Fall AMC 12B (Problems • Answer Key • Resources)
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

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