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

(Created page with '== Problem 17 == In the given circle, the diameter <math>\overline{EB}</math> is parallel to <math>\overline{DC}</math>, and <math>\overline{AB}</math> is parallel to <math>\ove…')
 
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In the given circle, the diameter <math>\overline{EB}</math> is parallel to <math>\overline{DC}</math>, and <math>\overline{AB}</math> is parallel to <math>\overline{ED}</math>. The angles <math>AEB</math> and <math>ABE</math> are in the ratio <math>4 : 5</math>. What is the degree measure of angle <math>BCD</math>?
 
In the given circle, the diameter <math>\overline{EB}</math> is parallel to <math>\overline{DC}</math>, and <math>\overline{AB}</math> is parallel to <math>\overline{ED}</math>. The angles <math>AEB</math> and <math>ABE</math> are in the ratio <math>4 : 5</math>. What is the degree measure of angle <math>BCD</math>?
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<math> \textbf{(A)}\ 120 \qquad\textbf{(B)}\ 125 \qquad\textbf{(C)}\ 130 \qquad\textbf{(D)}\ 135 \qquad\textbf{(E)}\ 140</math>
 
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−
== Solution ==
 
  
 
<center><asy>
 
<center><asy>
−
unitsize(10mm);
+
unitsize(7mm);
 
defaultpen(linewidth(.8pt)+fontsize(10pt));
 
defaultpen(linewidth(.8pt)+fontsize(10pt));
 
dotfactor=4;
 
dotfactor=4;
  
 
real r=3;
 
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);
 
pair O=(0,0), E=(-3,0), B=(3,0);
 
path outer=Circle(O,r);
 
path outer=Circle(O,r);
 
draw(outer);
 
draw(outer);
 
draw(E--B);
 
draw(E--B);
 +
draw(E--A);
 +
draw(B--A);
 +
draw(E--D);
 +
draw(C--D);
 +
draw(B--C);
  
−
pair[] ps={B,E,0};
+
pair[] ps={A,B,C,D,E,O};
 
dot(ps);
 
dot(ps);
 +
 +
label("$A$",A,N);
 
label("$B$",B,NE);
 
label("$B$",B,NE);
 +
label("$C$",C,S);
 +
label("$D$",D,S);
 
label("$E$",E,NW);
 
label("$E$",E,NW);
 
label("$$",O,N);
 
label("$$",O,N);
 
</asy></center>
 
</asy></center>
 +
 +
<math> \textbf{(A)}\ 120 \qquad\textbf{(B)}\ 125 \qquad\textbf{(C)}\ 130 \qquad\textbf{(D)}\ 135 \qquad\textbf{(E)}\ 140</math>
 +
 +
== Solution ==
  
 
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{(E)} 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{(E)} 130}</math>

Revision as of 18:35, 26 May 2011

Problem 17

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

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 $\angle ABE + \angle BCD = 180 \longrightarrow 50 + \angle BCD = 180 \longrightarrow \angle BCD = \boxed{\textbf{(E)} 130}$

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