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

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[[Category: Introductory Geometry Problems]]
 
==Problem==
 
==Problem==
 
Four distinct points are arranged on a plane so that the segments connecting them have lengths <math>a</math>, <math>a</math>, <math>a</math>, <math>a</math>, <math>2a</math>, and <math>b</math>. What is the ratio of <math>b</math> to <math>a</math>?
 
Four distinct points are arranged on a plane so that the segments connecting them have lengths <math>a</math>, <math>a</math>, <math>a</math>, <math>a</math>, <math>2a</math>, and <math>b</math>. What is the ratio of <math>b</math> to <math>a</math>?
  
 
<math>\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \sqrt{5}\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ \pi</math>
 
<math>\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \sqrt{5}\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ \pi</math>
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[[Category: Introductory Geometry Problems]]
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==Solution==
 
==Solution==
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== Solution 2 ==
 
== Solution 2 ==
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For any <math>4</math> non-collinear points with the given requirement, notice that there must be a triangle with side lengths <math>a</math>, <math>a</math>, <math>2a</math>, which is not possible as <math>a+a=2a</math>. Thus at least <math>3</math> of the <math>4</math> points must be collinear.
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If all <math>4</math> points are collinear, then there would only be <math>3</math> lines of length <math>a</math>, which wouldn't work.
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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.
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Thus <math>b=\sqrt{3}a</math>, or <math>\frac{b}{a}=\boxed{\mathrm{(A)}\sqrt{3}}</math>
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~ Nafer
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== Solution 3 (using the answer choices) ==
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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>.
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==Video Solution by Richard Rusczyk==
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https://artofproblemsolving.com/videos/amc/2012amc10b/271
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~dolphin7
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(Direct Youtube Link)
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https://www.youtube.com/watch?v=6sL7bhkVivo
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~lukiebear
  
 
== See Also ==
 
== See Also ==

Latest revision as of 19:18, 23 October 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 3 (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

(Direct Youtube Link) https://www.youtube.com/watch?v=6sL7bhkVivo

~lukiebear

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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