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

Revision as of 21:13, 18 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

Video Solution

https://youtu.be/Ysb1EK_5B2g

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

@@ Line 66: / Line 66: @@
 Therefore, <math>\angle BFC =
-\boxed{\textbf{(D) 113}}</math>.
+\boxed{\textbf{(D)} \ 113}</math>.
 ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
@@ Line 81: / Line 81: @@
 So now, because <math>ABCD</math> is a rhombus, <math>FD = AD = CD</math>. This means that there exists a circle from <math>D</math> with radius <math>AD</math> that passes through <math>F</math>, <math>A</math>, and <math>C</math>.
-AG is a diameter of this circle because <math>\angle AFG=90^\circ</math>. This means that <math>\angle GFC = \angle GAC = \frac{1}{2} \angle GDC</math>, so <math>\angle GFC = \frac{1}{2}(180^\circ - 46^\circ)=67^\circ</math>, which means that <math>\angle BFC = \boxed{\textbf{(D) 113}}</math>
+AG is a diameter of this circle because <math>\angle AFG=90^\circ</math>. This means that <math>\angle GFC = \angle GAC = \frac{1}{2} \angle GDC</math>, so <math>\angle GFC = \frac{1}{2}(180^\circ - 46^\circ)=67^\circ</math>, which means that <math>\angle BFC = \boxed{\textbf{(D)} \ 113}</math>
 ~popop614
+==Solution 3==
+Let <math>\overline{AC}</math> meet <math>\overline{BD}</math> at <math>O</math>, then <math>AOFB</math> is cyclic and <math>\angle FBO = \angle FAO</math>. Also, <math>AC \cdot BO = [ABCD] = 2 \cdot [ABE] = AF \cdot BE</math>, so <math>\frac{AF}{BO} = \frac{AC}{BE}</math>, thus <math>\triangle AFC \sim \triangle BOE</math> by SAS, and <math>\angle OEB = \angle ACF</math>, then <math>\angle CFE = \angle EOC = \angle DAC = 67^\circ</math>, and <math>\angle BFC = \boxed{\textbf{(D)} \ 113}</math>
+~mathfan2020
+==Solution 4==
+Observe that all answer choices are close to <math>112.5 = 90+\frac{45}{2}</math>. A quick solve shows that having <math>\angle D = 90^\circ</math> yields <math>\angle BFC = 135^\circ = 90 + \frac{90}{2}</math>, meaning that <math>\angle BFC</math> increases with <math>\angle D</math>.
+Substituting, <math>\angle BFC = 90 + \frac{46}{2} = \boxed{\textbf{(D)} \ 113}</math>
+~mathfan2020
 ==Video Solution==

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