Difference between revisions of "2011 AMC 10B Problems/Problem 17"
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==Solution 3== | ==Solution 3== | ||
+ | Solve for <math>\angle{AEB}</math> and <math>\angle{ABE}</math> as above, and find them 40 and 50 degrees, respectively. Then, AB is parallel to ED, <math>\angle{BED}</math>=50. Since CD is parallel to EB, we have <math>\angle{EDC}</math>=130, and also because CD is parallel to EB (and because they are in a circle), we have <math>\angle{EDC}=\angle{BCD}</math>, so then <math>\angle{BCD}=\boxed{130}</math>. -wendizeng714 | ||
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== See Also== | == See Also== | ||
{{AMC10 box|year=2011|ab=B|num-b=16|num-a=18}} | {{AMC10 box|year=2011|ab=B|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 17:19, 27 October 2021
Problem
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 1
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
Note:
We could also tell that quadrilateral is an isosceles trapezoid because for
and
to be parallel, the line going through the center of the circle and perpendicular to
must fall through the center of
.
Solution 2
Note as before. The sum of the interior angles for quadrilateral
is
. Denote the center of the circle as
.
. Denote
and
. We wish to find
. Our equation is
. Our final equation becomes
. After subtracting
and dividing by
, our answer becomes
Video Solution
https://youtu.be/NsQbhYfGh1Q?t=4155
~ pi_is_3.14
Solution 3
Solve for and
as above, and find them 40 and 50 degrees, respectively. Then, AB is parallel to ED,
=50. Since CD is parallel to EB, we have
=130, and also because CD is parallel to EB (and because they are in a circle), we have
, so then
. -wendizeng714
See Also
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 |
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