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

Revision as of 11:15, 9 October 2019

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

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: $\sqrt{3}$ .

@@ Line 30: / Line 30: @@
 ~ Nafer
+== Solution (using the answer choices) ==
+We know that <math>a-b-2a</math> form a triangle. From triangle inequality, we see that <math>b>a</math>. Then, we also see that there is an isosceles triangle with lengths <math>a-a-b</math>. From triangle inequality: <math>b<2a</math>. The only answer choice that holds these two inequalities is: <math>\sqrt{3}</math>.
 == See Also ==

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