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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<center><asy> | <center><asy> | ||
| − | unitsize( | + | 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, | + | 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> | ||
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| + | == 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 
is parallel to 
, and 
is parallel to 
. The angles 
and 
are in the ratio 
. What is the degree measure of angle 
?
![[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]](http://latex.artofproblemsolving.com/b/a/d/bad25a60446625a8eac18a0209ed3ab5d7c02eff.png)
Solution
We can let 
be 
and 
be 
because they are in the ratio . When an inscribed angle contains the diameter, the inscribed angle is a right angle. Therefore by triangle sum theorem, 
and 
.
because they are alternate interior angles and 
. Opposite angles in a cyclic quadrilateral are supplementary, so 
. Use substitution to get 
