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

2022 AMC 10B Problems/Problem 20

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

Note: You can also find that CFE is half of CDG via circle theorems. We know CDG = 180 - 46 = 134, therefore making CFE 67 and BFC 113.

~meikh_neiht

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.

Solution 7

Draw an auxiliary line from D to the midpoint of AB. Label it G. Then quadrilateral BGDE is a parallelogram. Hence AF and BD are perpendicular. Now, G being the midpoint of the hypontenuse of triangle ABF, it is the circumcenter of it. Thus, GF = GA and so DG is the perpendicular bisector of AF. Therefore, triangle AFD is also a isosceles triangle. Since AD = FD = CD, triangle CFD is also an isosceles triangle. Their one distinct angles' sum being 46 degrees, angle BFC = 113.

Jeongha Cho


Solution 8(lots and lots of tedious angle chasing)

It is a well known fact that connecting the midpoints of the sides of a rhombus gives us a rectangle. We let the midpoint of side $DC$ as $E$, the midpoint of side $AD$ as $F$, the midpoint of side $AB$ as $G$, and the midpoint of side $BC$ as $H$. We can connect $E, F, G, H$ to get rectangle $EFGH$. Note that the obtuse angles of the rhombus are each $134$ degrees. We can perform a little bit of angle chasing following this diagram(I cannot draw diagrams in LaTex so the rest of this solution will be diagram-free). Let the intersection point between line segment $BE$ and line segment $GH$ as $I$. Let angle $FEB$ to be $\theta$. Performing more angle chasing leads us to finding that angle $GIB$ is angle $\theta$ and angle $IBG$ is $113 - \theta$. Let point $J$ be on $BE$ such that $AJ$ is perpendicular to $BE$. Then, by performing yet more angle chasing leads us to finding that angle $JEC$ is $90 - \theta + 23 = 113 - \theta$. We can predict that triangle $JEC$ is similar to triangle $BIG$. This is because since $GB$ and $EC$ both bisect the sides of the rhombus and one angle is common(the $113 - \theta$ angle). Therefore, we can safely say that all angles in these two triangles must be the same and thus angle $CJE = IGB = 67$ degrees. Thus, our desired angle, $CJB$ is simply just $180 - 67 = \boxed{113}$.

~ilikemath247365

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_10B_Problems/Problem_20&oldid=256042"