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

Revision as of 23:18, 28 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 $\angle BFC$ ?

Diagram

$[asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, E, F; D = origin; A = 6*dir(46); C = (6,0); B = C + (A-D); E = midpoint(C--D); F = foot(A,B,E); dot("$A$",A,1.5*NW,linewidth(5)); dot("$B$",B,1.5*NE,linewidth(5)); dot("$C$",C,1.5*SE,linewidth(5)); dot("$D$",D,1.5*SW,linewidth(5)); dot("$E$",E,1.5*S,linewidth(5)); dot("$F$",F,1.5*dir(-20),linewidth(5)); markscalefactor=0.04; draw(rightanglemark(A,F,B),red); draw(A--B--C--D--cycle^^A--F--C^^B--E); label("$46^{\circ}$",D,3*dir(26),red); [/asy]$ ~MRENTHUSIASM

Solution 1 (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 and Circle Geometry)

Let's make a diagram, but extend $AD$ and $BE$ to point $G$ . $[asy] /* Made by ghfhgvghj10 Edited by MRENTHUSIASM */ size(300); pair A, B, C, D, E, F, G; D = origin; A = 6*dir(46); C = (6,0); B = C + (A-D); E = midpoint(C--D); F = foot(A,B,E); G = 6*dir(226); dot("$A$",A,1.5*NW,linewidth(5)); dot("$B$",B,1.5*NE,linewidth(5)); dot("$C$",C,1.5*SE,linewidth(5)); dot("$D$",D,1.5*NW,linewidth(5)); dot("$E$",E,1.5*S,linewidth(5)); dot("$F$",F,1.5*dir(-20),linewidth(5)); dot("$G$",G,1.5*SW,linewidth(5)); markscalefactor=0.04; draw(rightanglemark(A,F,B),red); draw(A--B--C--D--cycle^^A--F--C^^B--E^^D--G^^E--G); label("$46^{\circ}$",D,3*dir(26),red+fontsize(10)); [/asy]$ We know that $AB=AD=2$ and $CE=DE=1$ .

By AA Similarity, $\triangle ABG \sim \triangle DEG$ with a ratio of $2:1$ . This implies that $2AD=AG$ and $AD \cong DG$ , so $AG=2AD=2\cdot2=4$ . That is, $D$ 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. $[asy] /* Made by ghfhgvghj10 Edited by MRENTHUSIASM */ size(300); pair A, B, C, D, E, F, G; D = origin; A = 6*dir(46); C = (6,0); B = C + (A-D); E = midpoint(C--D); F = foot(A,B,E); G = 6*dir(226); dot("$A$",A,1.5*NE,linewidth(5)); dot("$B$",B,1.5*NE,linewidth(5)); dot("$C$",C,1.5*SE,linewidth(5)); dot("$D$",D,1.5*NW,linewidth(5)); dot("$E$",E,1.5*S,linewidth(5)); dot("$F$",F,1.5*dir(-20),linewidth(5)); dot("$G$",G,1.5*SW,linewidth(5)); markscalefactor=0.04; draw(rightanglemark(A,F,B),red); draw(A--B--C--D--cycle^^A--F--C^^B--E^^D--G^^E--G); label("$46^{\circ}$",D,3*dir(26),red+fontsize(10)); draw(Circle(D,6),dashed); [/asy]$ Notice how $F$ and $C$ are on the circle and that $\angle CFE$ intercepts with $\overset{\Large\frown} {CG}$ .

Let's call $\angle CFE = \theta$ .

Note that $\angle CDG$ also intercepts $\overset{\Large\frown} {CG}$ , So $\angle CDG = 2\angle CFE$ .

Let $\angle CDG = 2\theta$ . Notice how $\angle CDG$ and $\angle ADC$ are supplementary to each other. We conclude that $\begin{align*} 2\theta &= 180-\angle ADC \\ 2\theta &= 180-46 \\ 2\theta &= 134 \\ \theta &= 67. \end{align*}$ Since $\angle BFC=180-\theta$ , we have $\angle BFC=180-67=\boxed{\textbf{(D)} \ 113}$ .

~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

@@ Line 120: / Line 120: @@
 ~mathfan2020
-==Solution 5 (Similarity & Circle Geometry)==
+==Solution 5 (Similarity and Circle Geometry)==
 Let's make a diagram, but extend <math>AD</math> and <math>BE</math> to point <math>G</math>.
 <asy>
@@ Line 148: / Line 148: @@
 label("$46^{\circ}$",D,3*dir(26),red+fontsize(10));
 </asy>
-We know that <math>AB=2, AD=2, DE=1</math>, and <math>CE=1</math>.
+We know that <math>AB=AD=2</math> and <math>CE=DE=1</math>.
-By SAS Similarity, <math>\triangle ABG \sim \triangle DEG</math> with a ratio of <math>2:1</math>.
+By AA Similarity, <math>\triangle ABG \sim \triangle DEG</math> with a ratio of <math>2:1</math>. This implies that <math>2AD=AG</math> and <math>AD \cong DG</math>, so <math>AG=2AD=2\cdot2=4</math>. That is, <math>D</math> is the midpoint of <math>AG</math>.
-This means that, <math>2AD=AG</math> and <math>AD \cong DG</math>.
-<math>AG=2AD=2(2)=4</math>.
-This also can prove that <math>D</math> is the midpoint of <math>AG</math>.
 Now, let's redraw our previous diagram, but construct a circle with radius <math>AD</math> or <math>2</math> centered at <math>D</math> and by extending <math>CD</math> to point <math>H</math>, which is on the circle.
@@ Line 165: / Line 159: @@
 */
 size(300);
-pair A, B, C, D, E, F, G, H;
+pair A, B, C, D, E, F, G;
 D = origin;
 A = 6*dir(46);
@@ Line 173: / Line 167: @@
 F = foot(A,B,E);
 G = 6*dir(226);
-H = (-6,0);
 dot("$A$",A,1.5*NE,linewidth(5));
 dot("$B$",B,1.5*NE,linewidth(5));
@@ Line 181: / Line 174: @@
 dot("$F$",F,1.5*dir(-20),linewidth(5));
 dot("$G$",G,1.5*SW,linewidth(5));
-dot("$H$",H,1.5*W,linewidth(5));
 markscalefactor=0.04;
 draw(rightanglemark(A,F,B),red);
-draw(A--B--C--D--cycle^^A--F--C^^B--E^^D--G^^E--G^^D--H);
+draw(A--B--C--D--cycle^^A--F--C^^B--E^^D--G^^E--G);
 label("$46^{\circ}$",D,3*dir(26),red+fontsize(10));
 draw(Circle(D,6),dashed);
@@ Line 190: / Line 182: @@
 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>.
+Let's call <math>\angle CFE = \theta</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>.
-Notice how <math>\angle CDG</math> and <math>\angle ADC</math> are supplementary to each other.
-This concludes that, <math>2\theta=180-\angle ADC</math>.
-<math>2\theta=180-46</math>
-<math>2\theta=134</math>
-<math>\theta=67°</math>.
-Realize how <math>\angle BFC=180-\theta</math>
-<math>\angle BFC=180-67</math>
+Note that <math>\angle CDG</math> also intercepts <math>\overset{\Large\frown} {CG}</math>, So <math>\angle CDG = 2\angle CFE</math>.
-<math>\angle BFC= 113.</math> Which means the answer is <math>\boxed{\textbf{(D)} \ 113}</math>.
+Let <math>\angle CDG = 2\theta</math>. Notice how <math>\angle CDG</math> and <math>\angle ADC</math> are supplementary to each other. We conclude that <cmath>\begin{align*}
+\theta &= 180-\angle ADC \\
+\theta &= 180-46 \\
+\theta &= 134 \\
+\theta &= 67.
+\end{align*}</cmath>
+Since <math>\angle BFC=180-\theta</math>, we have <math>\angle BFC=180-67=\boxed{\textbf{(D)} \ 113}</math>.
 ~ghfhgvghj10 (If I make any minor mistakes, feel free to make minor fixes and edits).

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
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