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

Revision as of 21:57, 15 May 2021

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

~ Nafer

Solution (using the answer choices)

We know that $a-b-2a$ form a triangle. From triangle inequality, we see that $b>a$ . Then, we also see that there is an isosceles triangle with lengths $a-a-b$ . From triangle inequality: $b<2a$ . The only answer choice that holds these two inequalities is: $\sqrt{3}$ .

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2012amc10b/271

~dolphin7

@@ Line 28: / Line 28: @@
 If exactly <math>3</math> points are collinear, the only possibility that works is when a <math>30^{\circ}-90^{\circ}-60^{\circ}</math> triangle is formed.
-Thus <math>b=\sqrt{3}a</math>, or <math>\frac{b}{a}=\boxed{\mathrm{(A)}}\sqrt{3}</math>
+Thus <math>b=\sqrt{3}a</math>, or <math>\frac{b}{a}=\boxed{\mathrm{(A)}\sqrt{3}}</math>
 ~ Nafer

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