# 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 $4$ non-collinear 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 there would only be $3$ lines of length $a$, which wouldn't work.

If exactly $3$ points are collinear, the only possibility that works is when a $30^{\circ}-90^{\circ}-60^{\circ}$ triangle is formed.

Thus $b=\sqrt{3}a$, or $\frac{b}{a}=\sqrt{3} \boxed{\mathrm{(A)}}$

~ Nafer