Difference between revisions of "2010 AMC 12A Problems/Problem 8"
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Since <math>\frac{AC}{AB} = \frac{1}{2}</math>, triangle <math>ABC</math> is a <math>30-60-90</math> triangle, so <math>\angle BCA = \boxed{90^\circ\,\textbf{(C)}}</math>. | Since <math>\frac{AC}{AB} = \frac{1}{2}</math>, triangle <math>ABC</math> is a <math>30-60-90</math> triangle, so <math>\angle BCA = \boxed{90^\circ\,\textbf{(C)}}</math>. | ||
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== See also == | == See also == |
Revision as of 14:16, 17 November 2019
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
![AMC 2010 12A Problem 8.png](https://wiki-images.artofproblemsolving.com//9/90/AMC_2010_12A_Problem_8.png)
Let .
Since , triangle
is a
triangle, so
.
See also
2010 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 7 |
Followed by Problem 9 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.