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=root3a.
So, b=root3a, so B:A=root 3a
+
So, b=root3a, so B:A=$\sqrt{3a}

Revision as of 18:31, 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=root3a. So, b=root3a, so B:A=$\sqrt{3a}

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