2012 AMC 10B Problems/Problem 21

Problem

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 $b=\sqrt{3}a$ . Drawing the points out, it is possible to have a diagram where $b=\sqrt{3}a$ . It turns out that $a,$ $2a,$ and $b$ could be the lengths of a 30-60-90 triangle, and the other 3 $a\text{'s}$ can be the lengths of an equilateral triangle formed from connecting the dots. So, $b=\sqrt{3}a$ , so $b:a= \boxed{\textbf{(A)} \: \sqrt{3}}$ $[asy]draw((0, 0)--(1/2, sqrt(3)/2)--(1, 0)--cycle); draw((1/2, sqrt(3)/2)--(2, 0)--(1,0)); label("$a$", (0, 0)--(1, 0), S); label("$a$", (1, 0)--(2, 0), S); label("$a$", (0, 0)--(1/2, sqrt(3)/2), NW); label("$a$", (1, 0)--(1/2, sqrt(3)/2), NE); label("$b=\sqrt{3}a$", (1/2, sqrt(3)/2)--(2, 0), NE); [/asy]$

Solution 2

For any $4$ distinct points with the given requirement, notice that there must be a triangle with side lengths $a$ , $a$ , $2a$ , which is not possible as $a+a=2a$ . Thus at least 3 of the 4 points must be collinear.

If all 4 points are collinear, then

2012 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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2012 AMC 10B Problems/Problem 21

Contents

Problem

Solution

Solution 2

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