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

Revision as of 12:45, 27 July 2018

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, $\overline{DC}$ must fall in the center of the line going through the center of the circle and perpendicular to $\overline{DC}$ .

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 $x+y=\boxed{\textbf{(C)} 130}$

@@ Line 42: / Line 42: @@
 ==Note:==
-We could also tell that quadrilateral <math>BEDC</math> is an isosceles trapezoid because for <math>\overline{BE}</math> and <math>\overline{DC}</math>
+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, <math>\overline{DC}</math> must fall in the center of the line going through the center of the circle and perpendicular to <math>\overline{DC}</math>.
 ==Solution 2==

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