Difference between revisions of "2009 AMC 10B Problems/Problem 9"
(New page: == Problem == Segment <math>BD</math> and <math>AE</math> intersect at <math>C</math>, as shown, <math>AB=BC=CD=CE</math>, and <math>\angle A = \frac 52 \angle B</math>. What is the degre...) |
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Revision as of 11:53, 4 July 2013
Problem
Segment 
and 
intersect at 
, as shown, 
, and 
. What is the degree measure of 
?
![[asy] unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=4; pair C=(0,0), Ep=dir(35), D=dir(-35), B=dir(145); pair A=intersectionpoints(Circle(B,1),C--(-1*Ep))[0]; pair[] ds={A,B,C,D,Ep}; dot(ds); draw(A--Ep--D--B--cycle); label("$A$",A,SW); label("$B$",B,NW); label("$C$",C,N); label("$E$",Ep,E); label("$D$",D,E); [/asy]](http://latex.artofproblemsolving.com/2/6/4/2648b032deb3bc4fba6f53bcb3b07b32bcfad484.png)
Solution
is isosceles, hence 
.
The sum of internal angles of can now be expressed as 
, hence 
, and each of the other two angles is 
.
Now we know that 
.
Finally, 
is isosceles, hence each of the two remaining angles ( and 
) is equal to 
.
See Also
| 2009 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
