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Difference between revisions of "2022 AMC 10B Problems/Problem 20"

(→‎Solution 5 (Similarity & Circle Geometry))
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</asy>
 
</asy>
  
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Notice how <math>F</math> and <math>C</math> are on the circle and that <math>\angle CFE</math> intercepts with arch <math>CG</math>.  
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Notice how <math>F</math> and <math>C</math> are on the circle and that <math>\angle CFE</math> intercepts with <math>\overset{\Large\frown} {CG}</math>.  
  
 
Lets call <math>\angle CFE = \theta</math>.
 
Lets call <math>\angle CFE = \theta</math>.
  
−
<math>\angle CDG</math> also intercepts arch <math>CG</math>, but it's vertical angle (<math>\angle ADH</math>), also intercepts an arch congruent to arch <math>CG</math>.  So <math>\angle CDG = 2\angle CFE</math>.  
+
<math>\angle CDG</math> also intercepts <math>\overset{\Large\frown} {CG}</math>, but it's vertical angle (<math>\angle ADH</math>), also intercepts an arch congruent to <math>\overset{\Large\frown} {CG}</math>.  So <math>\angle CDG = 2\angle CFE</math>.  
  
 
<math>\angle CDG = 2\theta</math>.
 
<math>\angle CDG = 2\theta</math>.

Revision as of 23:00, 21 November 2022

Problem

Let $ABCD$ be a rhombus with $\angle ADC = 46^\circ$. Let $E$ be the midpoint of $\overline{CD}$, and let $F$ be the point on $\overline{BE}$ such that $\overline{AF}$ is perpendicular to $\overline{BE}$. What is the degree measure of $\angle BFC$?

Solution (Law of Sines and Law of Cosines)

Without loss of generality, we assume the length of each side of $ABCD$ is 2. Because $E$ is the midpoint of $CD$, $CE = 1$.

Because $ABCD$ is a rhombus, $\angle BCE = 180^\circ - \angle D$.

In $\triangle BCE$, following from the law of sines, \[ \frac{CE}{\sin \angle FBC} = \frac{BC}{\sin \angle BEC} . \]

We have $\angle BCE = 180^\circ - \angle FBC - \angle BCE = 46^\circ - \angle FBC$.

Hence, \[ \frac{1}{\sin \angle FBC} = \frac{2}{\sin \left( 46^\circ - \angle FBC \right)} . \]

By solving this equation, we get $\tan \angle FBC = \frac{\sin 46^\circ}{2 + \cos 46^\circ}$.

Because $AF \perp BF$, \begin{align*} BF & = AB \cos \angle ABF \\ & = 2 \cos \left( 46^\circ - \angle FBC \right) . \end{align*}

In $\triangle BFC$, following from the law of sines, \[ \frac{BF}{\sin \angle BCF} = \frac{BC}{\sin \angle BFC} . \]

Because $\angle BCF = 180^\circ - \angle BFC - \angle FBC$, the equation above can be converted as \[ \frac{BF}{\sin \left( \angle BFC + \angle FBC \right)} = \frac{BC}{\sin \angle BFC} . \]

Therefore, \begin{align*} \tan \angle BFC & = \frac{\sin \angle FBC}{\cos \left( 46^\circ - \angle FBC \right) - \cos \angle FBC} \\ & = \frac{1}{\sin 46^\circ - \left( 1 - \cos 46^\circ \right) \cot \angle FBC} \\ & = \frac{\sin 46^\circ}{\cos 46^\circ - 1} \\ & = - \frac{\sin 134^\circ}{1 + \cos 134^\circ} \\ & = - \tan \frac{134^\circ}{2} \\ & = - \tan 67^\circ \\ & = \tan \left( 180^\circ - 67^\circ \right) \\ & = \tan 113^\circ . \end{align*}

Therefore, $\angle BFC = \boxed{\textbf{(D)} \ 113}$.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 2

Extend segments $\overline{AD}$ and $\overline{BE}$ until they meet at point $G$.

Because $\overline{AB} \parallel \overline{ED}$, we have $\angle ABG = \angle DEG$ and $\angle GDE = \angle GAB$, so $\triangle ABG \sim \triangle DEG$ by AA.

Because $ABCD$ is a rhombus, $AB = CD = 2DE$, so $AG = 2GD$, meaning that $D$ is a midpoint of segment $\overline{AG}$.

Now, $\overline{AF} \perp \overline{BE}$, so $\triangle GFA$ is right and median $FD = AD$.

So now, because $ABCD$ is a rhombus, $FD = AD = CD$. This means that there exists a circle from $D$ with radius $AD$ that passes through $F$, $A$, and $C$.

AG is a diameter of this circle because $\angle AFG=90^\circ$. This means that $\angle GFC = \angle GAC = \frac{1}{2} \angle GDC$, so $\angle GFC = \frac{1}{2}(180^\circ - 46^\circ)=67^\circ$, which means that $\angle BFC = \boxed{\textbf{(D)} \ 113}$

~popop614

Solution 3

Let $\overline{AC}$ meet $\overline{BD}$ at $O$, then $AOFB$ is cyclic and $\angle FBO = \angle FAO$. Also, $AC \cdot BO = [ABCD] = 2 \cdot [ABE] = AF \cdot BE$, so $\frac{AF}{BO} = \frac{AC}{BE}$, thus $\triangle AFC \sim \triangle BOE$ by SAS, and $\angle OEB = \angle ACF$, then $\angle CFE = \angle EOC = \angle DAC = 67^\circ$, and $\angle BFC = \boxed{\textbf{(D)} \ 113}$

~mathfan2020

Solution 4

Observe that all answer choices are close to $112.5 = 90+\frac{45}{2}$. A quick solve shows that having $\angle D = 90^\circ$ yields $\angle BFC = 135^\circ = 90 + \frac{90}{2}$, meaning that $\angle BFC$ increases with $\angle D$. Substituting, $\angle BFC = 90 + \frac{46}{2} = \boxed{\textbf{(D)} \ 113}$

~mathfan2020

Solution 5 (Similarity & Circle Geometry)

Let's make a diagram, but extend $AD$ and $BE$ to point $G$.

[asy] pair A = (0,0); label("$D$", A, NW); pair B = (2.25,3); label("$A$", B, NW); pair C = (6,3); label("$B$", C, NE); pair D = (3.75,0); label("$C$", D, SE); pair E = (1.875,0); label("$E$", E, S); draw(C--E); draw(A--B--C--D--E--cycle); label("$F$", (3.3,1), S); draw(B--(3.5,1.2)); draw(rightanglemark(B,(3.5,1.2),E)); draw((3.5,1.2)--D); label("$G$", (-2.25, -3), SW); draw(A--(-2.25, -3)); draw(E--(-2.25, -3)); [/asy]

We know that $AB=2, AD=2, DE=1$, and $CE=1$.

By SAS Similarity, $\triangle ABG \sim \triangle DEG$ with a ratio of $2:1$.

This means that, $2AD=AG$ and $AD \cong DG$.

$AG=2AD=2(2)=4$.

This also can prove that $D$ is the midpoint of $AG$.

Now, let's redraw our previous diagram, but construct a circle with radius $AD$ or $2$ centered at $D$ and by extending $CD$ to point $H$, which is on the circle.

[asy] pair A = (0,0); label("$D$", A, NW); pair B = (2.25,3); label("$A$", B, NW); pair C = (6,3); label("$B$", C, NE); pair D = (3.75,0); label("$C$", D, SE); pair E = (1.875,0); label("$E$", E, S); draw(C--E); draw(A--B--C--D--E--cycle); label("$F$", (3.3,1), S); draw(B--(3.5,1.2)); draw(rightanglemark(B,(3.5,1.2),E)); draw((3.5,1.2)--D); label("$G$", (-2.25, -3), SW); draw(A--(-2.25, -3)); draw(E--(-2.25, -3)); pair O1 = (0,0); draw(circle(O1,3.75)); label("$H$", (-3.75,0), SW); draw(D--(-3.75,0)); [/asy]

Notice how $F$ and $C$ are on the circle and that $\angle CFE$ intercepts with $\overset{\Large\frown} {CG}$.

Lets call $\angle CFE = \theta$.

$\angle CDG$ also intercepts $\overset{\Large\frown} {CG}$, but it's vertical angle ($\angle ADH$), also intercepts an arch congruent to $\overset{\Large\frown} {CG}$. So $\angle CDG = 2\angle CFE$.

$\angle CDG = 2\theta$.

Notice how $CDG$ and $ADC$ are supplementary to each other.

This concludes that, $2\theta=180-\angle ADC$.

$2\theta=180-46$

$2\theta=134$

$\theta=67°$.

Realize how $\angle BFC=180-\theta$

$\angle BFC=180-67$

$\angle BFC= 113.$ Which means the answer is $\boxed{\textbf{(D)} \ 113}$.

~ghfhgvghj10 (If I make any minor mistakes, feel free to make minor fixes and edits).

Video Solution

https://youtu.be/Ysb1EK_5B2g

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution by OmegaLearn Using Clever Similar Triangles and Angle Chasing

https://youtu.be/lEmCprb20n4

~ pi_is_3.14


See Also

2022 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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 10 Problems and Solutions

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