Difference between revisions of "2012 AMC 10B Problems/Problem 21"

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When you see a and 2a, you could think of 30-60-90 triangles. Since all of the other's lengths are a, you could think that b is root3a.
 
When you see a and 2a, you could think of 30-60-90 triangles. Since all of the other's lengths are a, you could think that b is root3a.
Drawing the points out, it is possible to have a diagram where b=root3a.
+
Drawing the points out, it is possible to have a diagram where b=<math>\sqrt{3a}</math>
 
So, b=<math>\sqrt{3a}</math>, so B:A= <math>\sqrt{3}</math>
 
So, b=<math>\sqrt{3a}</math>, so B:A= <math>\sqrt{3}</math>

Revision as of 18:34, 12 March 2012

When you see a and 2a, you could think of 30-60-90 triangles. Since all of the other's lengths are a, you could think that b is root3a. Drawing the points out, it is possible to have a diagram where b=$\sqrt{3a}$ So, b=$\sqrt{3a}$, so B:A= $\sqrt{3}$

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