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Difference between revisions of "2011 AMC 10B Problems/Problem 17"

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[[Category: Introductory Geometry Problems]]
 
[[Category: Introductory Geometry Problems]]
  
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== Solution ==
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== Solution 1==
  
 
We can let <math>\angle AEB</math> be <math>4x</math> and <math>\angle ABE</math> be <math>5x</math> because they are in the ratio <math>4 : 5</math>. When an [[inscribed angle]] contains the [[diameter]], the inscribed angle is a [[right angle]]. Therefore by triangle sum theorem, <math>4x+5x+90=180 \longrightarrow x=10</math> and <math>\angle ABE = 50</math>.
 
We can let <math>\angle AEB</math> be <math>4x</math> and <math>\angle ABE</math> be <math>5x</math> because they are in the ratio <math>4 : 5</math>. When an [[inscribed angle]] contains the [[diameter]], the inscribed angle is a [[right angle]]. Therefore by triangle sum theorem, <math>4x+5x+90=180 \longrightarrow x=10</math> and <math>\angle ABE = 50</math>.
  
 
<math>\angle ABE = \angle BED</math> because they are alternate interior angles and <math>\overline{AB} \parallel \overline{ED}</math>. Opposite angles in a [[cyclic]] quadrilateral are [[supplementary]], so <math>\angle BED + \angle BCD = 180</math>. Use substitution to get <math>\angle ABE + \angle BCD = 180 \longrightarrow 50 + \angle BCD = 180 \longrightarrow \angle BCD = \boxed{\textbf{(C)} 130}</math>
 
<math>\angle ABE = \angle BED</math> because they are alternate interior angles and <math>\overline{AB} \parallel \overline{ED}</math>. Opposite angles in a [[cyclic]] quadrilateral are [[supplementary]], so <math>\angle BED + \angle BCD = 180</math>. Use substitution to get <math>\angle ABE + \angle BCD = 180 \longrightarrow 50 + \angle BCD = 180 \longrightarrow \angle BCD = \boxed{\textbf{(C)} 130}</math>
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==Note:==
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We could also tell that quadrilateral <math>BEDC</math> is an isosceles trapezoid because for <math>\overline{EB}</math> and <math>\overline{DC}</math> to be parallel, the line going through the center of the circle and perpendicular to <math>\overline{DC}</math> must fall through the center of <math>\overline{DC}</math>.
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==Solution 2==
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Note <math>\angle ABE = \angle BED=50</math> as before. The sum of the interior angles for quadrilateral <math>EBCD</math> is <math>360</math>. Denote the center of the circle as <math>P</math>. <math>\angle PDE = \angle PED = 50</math>. Denote <math>\angle PDC = \angle PCD = x</math> and <math>\angle PBC = \angle PCB = y</math>. We wish to find <math>\angle BCD = x+y</math>. Our equation is <math>(\angle PDE +\angle PED)+(\angle PDC+\angle PCD)+(\angle PBC + \angle PCB) = 360 \longrightarrow 2(50) + 2x +2y = 360</math>. Our final equation becomes <math>2(x+y)+100 = 360</math>. After subtracting <math>100</math> and dividing by <math>2</math>, our answer becomes <math>x+y=\boxed{\textbf{(C)} 130}</math>
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==Solution 3 (Circle Geometry)==
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Note that <math>\overset{\Large\frown} {BE}</math> intercepts <math>\angle BAE</math>. Since, <math>\overset{\Large\frown} {BE}=180</math>, thus <math>\angle BAE=90°</math> (courtesy of the Inscribed Angles Theorem).
 +
 +
Since we know that <math>\angle BAE=90°</math>, then <math>\angle AEB + \angle ABE = 90°</math>, (courtesy of the Triangle Sum Theorem) and also <math>5\angle AEB = 4\angle ABE</math>. By solving this variation, <math>\angle AEB = 40</math> and <math>\angle ABE = 50</math>. After that, due to the Alternate Interior Angles Theorem, <math>\angle ABE \cong \angle BED</math>, which means <math>\angle BED = 50</math>.
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 +
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After doing some angle chasing, then these following facts should be true,
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<math>\overset{\Large\frown} {AB}=80</math> <math>\overset{\Large\frown} {BD}=100</math> <math>\overset{\Large\frown} {AE}=100</math>.
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Note that the arcs have to equal 360, so, <math>360=\overset{\Large\frown} {AB}+\overset{\Large\frown} {BD}+\overset{\Large\frown} {DE}+\overset{\Large\frown} {AE}</math>
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<math>360=80+100+100+\overset{\Large\frown} {DE}</math>
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<math>\overset{\Large\frown} {DE}=80</math>
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Notice how <math>\overset{\Large\frown} {DB}</math> intercepts <math>\angle BCD</math> and that <math>\overset{\Large\frown} {DB}=\overset{\Large\frown} {DE}+\overset{\Large\frown} {AE}+\overset{\Large\frown} {AB}</math>.
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<math>\overset{\Large\frown} {DB}=80+100+80</math>
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<math>\overset{\Large\frown} {DB}=260</math>
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According to the Inscribed Angles Theorem, <math>2\angle BCD=\overset{\Large\frown} {DB}</math>, therefore the answer is <math>\frac{260}{2}= \boxed{\textbf{(C)} 130}</math>
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~ghfhgvghj10
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== Video Solution by OmegaLearn==
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https://youtu.be/NsQbhYfGh1Q?t=4155
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 +
~ pi_is_3.14
  
 
== See Also==
 
== See Also==

Latest revision as of 03:53, 21 January 2023

Problem

In the given circle, the diameter $\overline{EB}$ is parallel to $\overline{DC}$, and $\overline{AB}$ is parallel to $\overline{ED}$. The angles $AEB$ and $ABE$ are in the ratio $4 : 5$. What is the degree measure of angle $BCD$?

[asy] unitsize(7mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; real r=3; pair A=(-3cos(80),-3sin(80)); pair D=(3cos(80),3sin(80)), C=(-3cos(80),3sin(80)); pair O=(0,0), E=(-3,0), B=(3,0); path outer=Circle(O,r); draw(outer); draw(E--B); draw(E--A); draw(B--A); draw(E--D); draw(C--D); draw(B--C); pair[] ps={A,B,C,D,E,O}; dot(ps); label("$A$",A,N); label("$B$",B,NE); label("$C$",C,S); label("$D$",D,S); label("$E$",E,NW); label("$$",O,N); [/asy]

$\textbf{(A)}\ 120 \qquad\textbf{(B)}\ 125 \qquad\textbf{(C)}\ 130 \qquad\textbf{(D)}\ 135 \qquad\textbf{(E)}\ 140$

Solution 1

We can let $\angle AEB$ be $4x$ and $\angle ABE$ be $5x$ because they are in the ratio $4 : 5$. When an inscribed angle contains the diameter, the inscribed angle is a right angle. Therefore by triangle sum theorem, $4x+5x+90=180 \longrightarrow x=10$ and $\angle ABE = 50$.

$\angle ABE = \angle BED$ because they are alternate interior angles and $\overline{AB} \parallel \overline{ED}$. Opposite angles in a cyclic quadrilateral are supplementary, so $\angle BED + \angle BCD = 180$. Use substitution to get $\angle ABE + \angle BCD = 180 \longrightarrow 50 + \angle BCD = 180 \longrightarrow \angle BCD = \boxed{\textbf{(C)} 130}$

Note:

We could also tell that quadrilateral $BEDC$ is an isosceles trapezoid because for $\overline{EB}$ and $\overline{DC}$ to be parallel, the line going through the center of the circle and perpendicular to $\overline{DC}$ must fall through the center of $\overline{DC}$.

Solution 2

Note $\angle ABE = \angle BED=50$ as before. The sum of the interior angles for quadrilateral $EBCD$ is $360$. Denote the center of the circle as $P$. $\angle PDE = \angle PED = 50$. Denote $\angle PDC = \angle PCD = x$ and $\angle PBC = \angle PCB = y$. We wish to find $\angle BCD = x+y$. Our equation is $(\angle PDE +\angle PED)+(\angle PDC+\angle PCD)+(\angle PBC + \angle PCB) = 360 \longrightarrow 2(50) + 2x +2y = 360$. Our final equation becomes $2(x+y)+100 = 360$. After subtracting $100$ and dividing by $2$, our answer becomes $x+y=\boxed{\textbf{(C)} 130}$

Solution 3 (Circle Geometry)

Note that $\overset{\Large\frown} {BE}$ intercepts $\angle BAE$. Since, $\overset{\Large\frown} {BE}=180$, thus $\angle BAE=90°$ (courtesy of the Inscribed Angles Theorem).

Since we know that $\angle BAE=90°$, then $\angle AEB + \angle ABE = 90°$, (courtesy of the Triangle Sum Theorem) and also $5\angle AEB = 4\angle ABE$. By solving this variation, $\angle AEB = 40$ and $\angle ABE = 50$. After that, due to the Alternate Interior Angles Theorem, $\angle ABE \cong \angle BED$, which means $\angle BED = 50$.


After doing some angle chasing, then these following facts should be true, $\overset{\Large\frown} {AB}=80$ $\overset{\Large\frown} {BD}=100$ $\overset{\Large\frown} {AE}=100$.

Note that the arcs have to equal 360, so, $360=\overset{\Large\frown} {AB}+\overset{\Large\frown} {BD}+\overset{\Large\frown} {DE}+\overset{\Large\frown} {AE}$

$360=80+100+100+\overset{\Large\frown} {DE}$

$\overset{\Large\frown} {DE}=80$

Notice how $\overset{\Large\frown} {DB}$ intercepts $\angle BCD$ and that $\overset{\Large\frown} {DB}=\overset{\Large\frown} {DE}+\overset{\Large\frown} {AE}+\overset{\Large\frown} {AB}$.

$\overset{\Large\frown} {DB}=80+100+80$

$\overset{\Large\frown} {DB}=260$

According to the Inscribed Angles Theorem, $2\angle BCD=\overset{\Large\frown} {DB}$, therefore the answer is $\frac{260}{2}= \boxed{\textbf{(C)} 130}$

~ghfhgvghj10

Video Solution by OmegaLearn

https://youtu.be/NsQbhYfGh1Q?t=4155

~ pi_is_3.14

See Also

2011 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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All AMC 10 Problems and Solutions

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