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

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==Solution 2==
 
==Solution 2==
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Extend segments <math>AD</math> and <math>BE</math> until they meet at point <math>G</math>.
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Extend segments <math>\overline{AD}</math> and <math>\overline{BE}</math> until they meet at point <math>G</math>.
  
−
Because <math>AB \parallel ED</math>, we have <math>\angle ABG = \angle DEG</math> and <math>\angle GDE = \angle GAB</math>, so <math>\triangle ABG \sim \triangle DEG</math> by AA.
+
Because <math>\overline{AB} \parallel \overline{ED}</math>, we have <math>\angle ABG = \angle DEG</math> and <math>\angle GDE = \angle GAB</math>, so <math>\triangle ABG \sim \triangle DEG</math> by AA.
  
 
Because <math>ABCD</math> is a rhombus, <math>AB = CD = 2DE</math>, so <math>AG = 2GD</math>, meaning that <math>D</math> is a midpoint of segment <math>\overline{AG}</math>.
 
Because <math>ABCD</math> is a rhombus, <math>AB = CD = 2DE</math>, so <math>AG = 2GD</math>, meaning that <math>D</math> is a midpoint of segment <math>\overline{AG}</math>.
  
−
Now, <math>AF \perp BE</math>, so <math>\triangle GFA</math> is right and median <math>FD = AD</math>.
+
Now, <math>\overline{AF} \perp \overline{BE}</math>, so <math>\triangle GFA</math> is right and median <math>FD = AD</math>.
  
 
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>.
 
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>.

Revision as of 16:23, 17 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

Video Solution

https://youtu.be/Ysb1EK_5B2g

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

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