Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 5 (Similarity & Circle Geometry))
(→‎Solution 5 (Similarity and Circle Geometry))
 
(39 intermediate revisions by 16 users not shown)
Line 4: Line 4:
 
on <math>\overline{BE}</math> such that <math>\overline{AF}</math> is perpendicular to <math>\overline{BE}</math>. What is the degree measure of <math>\angle BFC</math>?
 
on <math>\overline{BE}</math> such that <math>\overline{AF}</math> is perpendicular to <math>\overline{BE}</math>. What is the degree measure of <math>\angle BFC</math>?
  
−
==Solution (Law of Sines and Law of Cosines)==
+
<math>\textbf{(A)}\ 110 \qquad\textbf{(B)}\ 111 \qquad\textbf{(C)}\ 112 \qquad\textbf{(D)}\ 113 \qquad\textbf{(E)}\ 114</math>
  
−
Without loss of generality, we assume the length of each side of <math>ABCD</math> is 2.
+
==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 <math>ABCD</math> is <math>2</math>.
 
Because <math>E</math> is the midpoint of <math>CD</math>, <math>CE = 1</math>.
 
Because <math>E</math> is the midpoint of <math>CD</math>, <math>CE = 1</math>.
  
Line 18: Line 44:
 
</cmath>
 
</cmath>
  
−
We have <math>\angle BCE = 180^\circ - \angle FBC - \angle BCE = 46^\circ - \angle FBC</math>.
+
We have <math>\angle BEC = 180^\circ - \angle FBC - \angle BCE = 46^\circ - \angle FBC</math>.
  
 
Hence,
 
Hence,
Line 89: Line 115:
  
 
~mathfan2020
 
~mathfan2020
 +
 +
A little bit faster: <math>AOFB</math> is cyclic <math>\implies \angle OFE = \angle BAO</math>.
 +
 +
<math>AB \parallel CD \implies \angle BAO = \angle OCE</math>.
 +
 +
Therefore <math>\angle OFE=\angle OCE \implies OECF</math> is cyclic.
 +
 +
Hence <math>\angle CFE=\angle COE=\angle CAD = 67^\circ</math>.
 +
 +
~asops
  
 
==Solution 4==
 
==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>.  
 
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>
+
Substituting, <math>\angle BFC = 90 + \frac{46}{2} = \boxed{\textbf{(D)} \ 113}</math>.
  
 
~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>.  
+
This solution refers to the <b>Diagram</b> section.
  
 +
We extend <math>AD</math> and <math>BE</math> to point <math>G</math>, as shown below:
 
<asy>
 
<asy>
−
pair A = (0,0);
+
/*
−
label("$D$", A, NW);
+
Made by ghfhgvghj10
−
pair B = (2.25,3);
+
Edited by MRENTHUSIASM
−
label("$A$", B, NW);
+
*/
−
pair C = (6,3);
+
size(300);
−
label("$B$", C, NE);
+
pair A, B, C, D, E, F, G;
−
pair D = (3.75,0);
+
D = origin;
−
label("$C$", D, SE);
+
A = 6*dir(46);
−
pair E = (1.875,0);
+
C = (6,0);
−
label("$E$", E, S);
+
B = C + (A-D);
−
draw(C--E);
+
E = midpoint(C--D);
−
draw(A--B--C--D--E--cycle);
+
F = foot(A,B,E);
−
label("$F$", (3.3,1), S);
+
G = 6*dir(226);
−
draw(B--(3.5,1.2));
+
dot("$A$",A,1.5*NW,linewidth(5));
−
draw(rightanglemark(B,(3.5,1.2),E));
+
dot("$B$",B,1.5*NE,linewidth(5));
−
draw((3.5,1.2)--D);
+
dot("$C$",C,1.5*SE,linewidth(5));
−
label("$G$", (-2.25, -3), SW);
+
dot("$D$",D,1.5*NW,linewidth(5));
−
draw(A--(-2.25, -3));
+
dot("$E$",E,1.5*S,linewidth(5));
−
draw(E--(-2.25, -3));
+
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>
 
</asy>
 +
We know that <math>AB=AD=2</math> and <math>CE=DE=1</math>.
  
−
We know that <math>AB=2, AD=2, DE=1</math>, and <math>CE=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>.
−
 
 
−
By SAS Similarity, <math>\triangle ABG \sim \triangle DEG</math> with a ratio of <math>2:1</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.  
 
  
 +
Note that as <math>\angle{AFG}</math> has an angle of 90 deg and <math>AG=2DG</math>, we can 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, as shown below:
 
<asy>
 
<asy>
−
pair A = (0,0);
+
/*
−
label("$D$", A, NW);
+
Made by ghfhgvghj10
−
pair B = (2.25,3);
+
Edited by MRENTHUSIASM
−
label("$A$", B, NW);
+
*/
−
pair C = (6,3);
+
size(300);
−
label("$B$", C, NE);
+
pair A, B, C, D, E, F, G;
−
pair D = (3.75,0);
+
D = origin;
−
label("$C$", D, SE);
+
A = 6*dir(46);
−
pair E = (1.875,0);
+
C = (6,0);
−
label("$E$", E, S);
+
B = C + (A-D);
−
draw(C--E);
+
E = midpoint(C--D);
−
draw(A--B--C--D--E--cycle);
+
F = foot(A,B,E);
−
label("$F$", (3.3,1), S);
+
G = 6*dir(226);
−
draw(B--(3.5,1.2));
+
dot("$A$",A,1.5*NE,linewidth(5));
−
draw(rightanglemark(B,(3.5,1.2),E));
+
dot("$B$",B,1.5*NE,linewidth(5));
−
draw((3.5,1.2)--D);
+
dot("$C$",C,1.5*SE,linewidth(5));
−
label("$G$", (-2.25, -3), SW);
+
dot("$D$",D,1.5*NW,linewidth(5));
−
draw(A--(-2.25, -3));
+
dot("$E$",E,1.5*S,linewidth(5));
−
draw(E--(-2.25, -3));
+
dot("$F$",F,1.5*dir(-20),linewidth(5));
−
pair O1 = (0,0);
+
dot("$G$",G,1.5*SW,linewidth(5));
−
draw(circle(O1,3.75));
+
markscalefactor=0.04;
−
label("$H$", (-3.75,0), SW);
+
draw(rightanglemark(A,F,B),red);
−
draw(D--(-3.75,0));
+
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>
 
</asy>
 +
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 arch <math>CG</math>.  
+
Let's call <math>\angle CFE = \theta</math>.
  
−
Lets call <math>\angle CFE = \theta</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 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>.  
+
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>.
  
−
<math>\angle CDG = 2\theta</math>.
+
~ghfhgvghj10 (If I make any minor mistakes, feel free to make minor fixes and edits).
 +
~mathboy282
  
−
Notice how <math>CDG</math> and <math>ADC</math> are supplementary to each other.
+
== Solution 6 (Simplification/Reduction) ==
  
−
This concludes that, <math>2\theta=180-\angle ADC</math>.  
+
If angle <math>ADC</math> was a right angle, it would be much easier. Thus, first pretend that <math>ADC</math> is a right angle. <math>ABCD</math> is now a square. WLOG, let each of the side lengths be 1. We can use the Pythagorean Theorem to find the length of line <math>AE</math>, which is <math>\sqrt{5}/2</math>. We want the measure of angle <math>BFC</math>, so to work closer to it, we should try finding the length of line <math>BF</math>. Angle <math>FAB</math> and angle <math>ABF</math> are complementary. Angle <math>ABF</math> and angle <math>FBC</math> are also complementary. Thus, <math>\sin FAB=\cos ABF=\sin FBC</math>. <math>\sin FAB=\sin FBC=(1/2)/(\sqrt{5}/2)=1/\sqrt{5}</math>. Since <math>\sin FAB=1\sqrt{5}</math>,and <math>AB=1</math>, <math>FB=\sin FAB</math>. It follows now that <math>FE=3*\sqrt{5}/10</math>.  
  
−
<math>2\theta=180-46</math>
+
Now, zoom in on triangle <math>BEC</math>. To use the Law of Cosines on triangle <math>FBC</math>, we need the length of <math>FC</math>. Use the Law of Cosines on triangle <math>EFC</math>. Cos <math>E=1/\sqrt{5}</math>. Thus, after using the Law of Cosines, <math>FC=\sqrt{2/5}</math>.
  
−
<math>2\theta=134</math>
+
Since we now have SSS on <math>BEC</math>, we can get use the Law of Cosines. <math>\cos BFC=1/-\sqrt{2}</math>. <math>\arccos 1/-\sqrt{2}</math> is 45, but if the cosine is negative that means that the angle is the supplement of the positive cosine value. <math>180-45=135</math>. Angle <math>BFC</math> is <math>135^\circ</math>.
  
−
<math>\theta=67°</math>.  
+
Realize that, around point F, there will always be 3 right angles, regardless of what angle <math>ADC</math> is. There are only two angles that change when <math>ADC</math> changes. Break up angle <math>BFC</math> into angle <math>BFB'</math>, which is always 90 degrees, and angle <math>B'FC</math>, which we have discovered to to be half of <math>ADC</math>. Thus, when angle <math>ADC</math> is 46 degrees, then <math>B'FC</math> will be 23. <math>23+90=113</math>. Angle <math>BFC</math> is <math>\boxed{\textbf{(D) }113}</math> degrees.
  
−
Realize how <math>\angle BFC=180-\theta</math>
 
  
−
<math>\angle BFC=180-67</math>
+
==Video Solution (⚡️Just 1 min!⚡️)==
 +
https://youtu.be/CriWEtfD5GE
  
−
<math>\angle BFC= 113.</math> Which means the answer is <math>\boxed{\textbf{(D)} \ 113}</math>.
+
<i>~Education, the Study of Everything</i>
  
−
~ghfhgvghj10 (feel free to make minor edits)
+
==Video Solution==
 +
 
 +
https://www.youtube.com/watch?v=HWJe96s_ugs&list=PLmpPPbOoDfgj5BlPtEAGcB7BR_UA5FgFj&index=6
  
 
==Video Solution==
 
==Video Solution==
Line 196: Line 242:
 
~ pi_is_3.14
 
~ pi_is_3.14
  
 +
== Video Solution, best solution (not family friendly, no circles drawn) ==
 +
https://www.youtube.com/watch?v=vwI3I7dxw0Q
 +
 +
== Video Solution, by Challenge 25 ==
 +
https://youtu.be/W1jbMaO8BIQ (cyclic quads)
 +
==Video Solution by Interstigation==
 +
https://youtu.be/5Plt3mmZBC0
 +
 +
~Interstigation
 +
 +
==Video Solution (Cool Solution)==
 +
https://www.youtube.com/watch?v=cZcaeU9P25s&ab_channel=Chillin
  
 
== See Also ==
 
== See Also ==

Latest revision as of 08:43, 10 November 2024

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 $\angle BFC$?

$\textbf{(A)}\ 110 \qquad\textbf{(B)}\ 111 \qquad\textbf{(C)}\ 112 \qquad\textbf{(D)}\ 113 \qquad\textbf{(E)}\ 114$

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

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

A little bit faster: $AOFB$ is cyclic $\implies \angle OFE = \angle BAO$.

$AB \parallel CD \implies \angle BAO = \angle OCE$.

Therefore $\angle OFE=\angle OCE \implies OECF$ is cyclic.

Hence $\angle CFE=\angle COE=\angle CAD = 67^\circ$.

~asops

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)

This solution refers to the Diagram section.

We extend $AD$ and $BE$ to point $G$, as shown below: [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 $AG$.

Note that as $\angle{AFG}$ has an angle of 90 deg and $AG=2DG$, we can 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, as shown below: [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). ~mathboy282

Solution 6 (Simplification/Reduction)

If angle $ADC$ was a right angle, it would be much easier. Thus, first pretend that $ADC$ is a right angle. $ABCD$ is now a square. WLOG, let each of the side lengths be 1. We can use the Pythagorean Theorem to find the length of line $AE$, which is $\sqrt{5}/2$. We want the measure of angle $BFC$, so to work closer to it, we should try finding the length of line $BF$. Angle $FAB$ and angle $ABF$ are complementary. Angle $ABF$ and angle $FBC$ are also complementary. Thus, $\sin FAB=\cos ABF=\sin FBC$. $\sin FAB=\sin FBC=(1/2)/(\sqrt{5}/2)=1/\sqrt{5}$. Since $\sin FAB=1\sqrt{5}$,and $AB=1$, $FB=\sin FAB$. It follows now that $FE=3*\sqrt{5}/10$.

Now, zoom in on triangle $BEC$. To use the Law of Cosines on triangle $FBC$, we need the length of $FC$. Use the Law of Cosines on triangle $EFC$. Cos $E=1/\sqrt{5}$. Thus, after using the Law of Cosines, $FC=\sqrt{2/5}$.

Since we now have SSS on $BEC$, we can get use the Law of Cosines. $\cos BFC=1/-\sqrt{2}$. $\arccos 1/-\sqrt{2}$ is 45, but if the cosine is negative that means that the angle is the supplement of the positive cosine value. $180-45=135$. Angle $BFC$ is $135^\circ$.

Realize that, around point F, there will always be 3 right angles, regardless of what angle $ADC$ is. There are only two angles that change when $ADC$ changes. Break up angle $BFC$ into angle $BFB'$, which is always 90 degrees, and angle $B'FC$, which we have discovered to to be half of $ADC$. Thus, when angle $ADC$ is 46 degrees, then $B'FC$ will be 23. $23+90=113$. Angle $BFC$ is $\boxed{\textbf{(D) }113}$ degrees.


Video Solution (⚡️Just 1 min!⚡️)

https://youtu.be/CriWEtfD5GE

~Education, the Study of Everything

Video Solution

https://www.youtube.com/watch?v=HWJe96s_ugs&list=PLmpPPbOoDfgj5BlPtEAGcB7BR_UA5FgFj&index=6

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

Video Solution, best solution (not family friendly, no circles drawn)

https://www.youtube.com/watch?v=vwI3I7dxw0Q

Video Solution, by Challenge 25

https://youtu.be/W1jbMaO8BIQ (cyclic quads)

Video Solution by Interstigation

https://youtu.be/5Plt3mmZBC0

~Interstigation

Video Solution (Cool Solution)

https://www.youtube.com/watch?v=cZcaeU9P25s&ab_channel=Chillin

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_10B_Problems/Problem_20&oldid=232993"