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

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Drawing the points out, it is possible to have a diagram where <math>b=\sqrt{3a}</math>. It turns out that <math>a,</math> <math>2a,</math> and <math>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.
 
Drawing the points out, it is possible to have a diagram where <math>b=\sqrt{3a}</math>. It turns out that <math>a,</math> <math>2a,</math> and <math>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>
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== See Also ==
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{{AMC10 box|year=2012|ab=B|num-b=23|num-a=25}}

Revision as of 20:18, 17 February 2013

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 $b:a= \sqrt{3}=(A)$

See Also

2012 AMC 10B (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 10 Problems and Solutions
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