Difference between revisions of "2013 AMC 12B Problems/Problem 24"
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| + | ==Problem== | ||
| + | Let <math>ABC</math> be a triangle where <math>M</math> is the midpoint of <math>\overline{AC}</math>, and <math>\overline{CN}</math> is the angle bisector of <math>\angle{ACB}</math> with <math>N</math> on <math>\overline{AB}</math>. Let <math>X</math> be the intersection of the median <math>\overline{BM}</math> and the bisector <math>\overline{CN}</math>. In addition <math>\triangle BXN</math> is equilateral with <math>AC=2</math>. What is <math>BN^2</math>? | ||
| + | <math>\textbf{(A)}\ \frac{10-6\sqrt{2}}{7} \qquad \textbf{(B)}\ \frac{2}{9} \qquad \textbf{(C)}\ \frac{5\sqrt{2}-3\sqrt{3}}{8} \qquad \textbf{(D)}\ \frac{\sqrt{2}}{6} \qquad \textbf{(E)}\ \frac{3\sqrt{3}-4}{5}</math> | ||
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| + | ==Solution== | ||
| + | |||
| + | == See also == | ||
| + | {{AMC12 box|year=2013|ab=B|num-b=23|num-a=25}} | ||
Revision as of 18:17, 22 February 2013
Problem
Let 
be a triangle where 
is the midpoint of 
, and 
is the angle bisector of 
with 
on 
. Let 
be the intersection of the median 
and the bisector . In addition 
is equilateral with 
. What is 
?
Solution
See also
| 2013 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 23 |
Followed by Problem 25 |
| 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 | |
