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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 <math>b=\sqrt{3a}</math>.
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==Problem 21==
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Drawing the points out, it is possible to have a diagram where <math>b=\sqrt{3a}</math>
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Four distinct points are arranged on a plane so that the segments connecting them have lengths <math>a</math>, <math>a</math>, <math>a</math>, <math>a</math>, <math>2a</math>, and <math>b</math>. What is the ratio of <math>b</math> to <math>a</math>?
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<math>\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \sqrt{5}\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ \pi</math>
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==Solution==
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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 <math>b=\sqrt{3a}</math>.
 +
Drawing the points out, it is possible to have a diagram where <math>b=\sqrt{3a}</math>. It turns out that <math>a, 2a, and b</math> could be the lengths of a 30-60-90 triangle, and the other 3 <math>"a's"</math> can be the lengths of an equilateral triangle formed from connecting the dots.
 
So, <math>b=\sqrt{3a}</math>, so <math>b:a= \sqrt{3}=(A)</math>
 
So, <math>b=\sqrt{3a}</math>, so <math>b:a= \sqrt{3}=(A)</math>

Revision as of 00:05, 14 March 2012

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 $a$?

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

Solution

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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