Difference between revisions of "2011 AMC 10B Problems/Problem 18"
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Revision as of 13:12, 4 July 2013
Problem
Rectangle has
and
. Point
is chosen on side
so that
. What is the degree measure of
?
Solution
![[asy] unitsize(10mm); defaultpen(linewidth(.5pt)+fontsize(10pt)); dotfactor=3; pair A=(0,3), B=(6,3), C=(6,0), D=(0,0); pair M=(0.80385,3); draw(A--B--C--D--cycle); draw(M--C); draw(M--D); draw(anglemark(A,M,D)); draw(anglemark(D,M,C)); draw(anglemark(C,D,M)); pair[] ps={A,B,C,D,M}; dot(ps); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$M$",M,N); label("$6$",midpoint(C--M),SW); label("$6$",midpoint(A--B),N); label("$3$",midpoint(B--C),E); [/asy]](http://latex.artofproblemsolving.com/6/e/5/6e52470a180b9bf804714a44c042322ee1ffdadd.png)
It is given that . Since
and
are alternate interior angles and
,
. Use the Base Angle Theorem to show
. We know that
is a rectangle, so it follows that
. We notice that
is a
triangle, and
. If we let
be the measure of
then
See Also
2011 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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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