Difference between revisions of "2010 AMC 12A Problems/Problem 8"
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− | == Problem | + | == Problem == |
Triangle <math>ABC</math> has <math>AB=2 \cdot AC</math>. Let <math>D</math> and <math>E</math> be on <math>\overline{AB}</math> and <math>\overline{BC}</math>, respectively, such that <math>\angle BAE = \angle ACD</math>. Let <math>F</math> be the intersection of segments <math>AE</math> and <math>CD</math>, and suppose that <math>\triangle CFE</math> is equilateral. What is <math>\angle ACB</math>? | Triangle <math>ABC</math> has <math>AB=2 \cdot AC</math>. Let <math>D</math> and <math>E</math> be on <math>\overline{AB}</math> and <math>\overline{BC}</math>, respectively, such that <math>\angle BAE = \angle ACD</math>. Let <math>F</math> be the intersection of segments <math>AE</math> and <math>CD</math>, and suppose that <math>\triangle CFE</math> is equilateral. What is <math>\angle ACB</math>? | ||
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Let <math>\angle BAE = \angle ACD = x</math>. | Let <math>\angle BAE = \angle ACD = x</math>. | ||
− | < | + | <cmath>\begin{align*}&ngle BCD = \angle AEC = 60^\circ\\ |
+ | &\angle EAC + \angle FCA + \angle ECF + \angle AEC = \angle EAC + x + 60^\circ + 60^\circ = 180^\circ\\ | ||
+ | &\angle EAC = 60^\circ - x\\ | ||
+ | &\angle BAC = \angle EAC + \angle BAE = 60^\circ - x + x = 60^\circ\end{align*}</cmath> | ||
− | <math>\ | + | Since <math>\frac{AC}{AB} = \frac{1}{2}</math>, <math>\angle BCA = \boxed{90^\circ\ \textbf{(C)}}</math> |
− | + | == See also == | |
+ | {{AMC12 box|year=2010|num-b=7|num-a=9|ab=A}} | ||
− | + | [[Category:Introductory Geometry Problems]] | |
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Revision as of 23:00, 25 February 2010
Problem
Triangle has
. Let
and
be on
and
, respectively, such that
. Let
be the intersection of segments
and
, and suppose that
is equilateral. What is
?
Solution
Let .
Since ,
See also
2010 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 7 |
Followed by Problem 9 |
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All AMC 12 Problems and Solutions |