2012 AMC 10B Problems/Problem 21

Revision as of 16:08, 15 January 2013 by N1000 (talk | contribs) (Solution)

Problem 21

Four distinct points are arranged on a plane so that the segments connecting them have lengths $a$, $a$, $a$, $a$, $2a$, and $b$. What is the ratio of $b$ to $a$?

$\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \sqrt{5}\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ \pi$


When you see that there are lengths a and 2a, one could think of 30-60-90 triangles. Since all of the other's lengths are a, you could think that $b=\sqrt{3a}$. Drawing the points out, it is possible to have a diagram where $b=\sqrt{3a}$. It turns out that $a,$ $2a,$ and $b$ could be the lengths of a 30-60-90 triangle, and the other 3 $"a's"$ can be the lengths of an equilateral triangle formed from connecting the dots. So, $b=\sqrt{3a}$, so $b:a= \sqrt{3}=(A)$

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