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

m (Solution 1 (Law of Sines and Law of Cosines))
(Solution 5 (Similarity & Circle Geometry): Made the solution more concise, and I deleted Point H (unnecessary to mention).)
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~mathfan2020
 
~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>.  
 
Let's make a diagram, but extend <math>AD</math> and <math>BE</math> to point <math>G</math>.  
 
<asy>
 
<asy>
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label("$46^{\circ}$",D,3*dir(26),red+fontsize(10));
 
label("$46^{\circ}$",D,3*dir(26),red+fontsize(10));
 
</asy>
 
</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.  
 
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.  
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*/
 
*/
 
size(300);
 
size(300);
pair A, B, C, D, E, F, G, H;
+
pair A, B, C, D, E, F, G;
 
D = origin;
 
D = origin;
 
A = 6*dir(46);
 
A = 6*dir(46);
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F = foot(A,B,E);
 
F = foot(A,B,E);
 
G = 6*dir(226);
 
G = 6*dir(226);
H = (-6,0);
 
 
dot("$A$",A,1.5*NE,linewidth(5));
 
dot("$A$",A,1.5*NE,linewidth(5));
 
dot("$B$",B,1.5*NE,linewidth(5));
 
dot("$B$",B,1.5*NE,linewidth(5));
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dot("$F$",F,1.5*dir(-20),linewidth(5));
 
dot("$F$",F,1.5*dir(-20),linewidth(5));
 
dot("$G$",G,1.5*SW,linewidth(5));
 
dot("$G$",G,1.5*SW,linewidth(5));
dot("$H$",H,1.5*W,linewidth(5));
 
 
markscalefactor=0.04;
 
markscalefactor=0.04;
 
draw(rightanglemark(A,F,B),red);
 
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));
 
label("$46^{\circ}$",D,3*dir(26),red+fontsize(10));
 
draw(Circle(D,6),dashed);
 
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>.  
 
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*}
 +
2\theta &= 180-\angle ADC \
 +
2\theta &= 180-46 \
 +
2\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).
 
~ghfhgvghj10 (If I make any minor mistakes, feel free to make minor fixes and edits).

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


See Also

2022 AMC 10B (ProblemsAnswer KeyResources)
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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