2011 AMC 10B Problems/Problem 17: Difference between revisions

Revision as of 20:55, 26 November 2022

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

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

~ pi_is_3.14

@@ Line 48: / Line 48: @@
 ==Solution 3 (Circle Geometry)==
-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>.
+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 has to apply that <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>.
+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>.
-After doing some angle chasing, then these following facts should be true, (courtesy of the Inscribed Angles Theorem)
+After doing some angle chasing, then these following facts should be true,
 <math>\overset{\Large\frown} {AB}=80</math> <math>\overset{\Large\frown} {BD}=100</math> <math>\overset{\Large\frown} {AE}=100</math>.

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

Page

Toolbox

Search