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

 
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==Problem 21==
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[[Category: Introductory Geometry Problems]]
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==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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==Solution==
 
==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 <math>b=\sqrt{3a}</math>.
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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 <math>b=\sqrt{3}a</math>.
Drawing the points out, it is possible to have a diagram where <math>b=\sqrt{3a}</math>. It turns out that <math>a, 2a, and b</math> could be the lengths of a 30-60-90 triangle, and the other 3 <math>"a's"</math> can be the lengths of an equilateral triangle formed from connecting the dots.
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Drawing the points out, it is possible to have a diagram where <math>b=\sqrt{3}a</math>. It turns out that <math>a,</math> <math>2a,</math> and <math>b</math> could be the lengths of a 30-60-90 triangle, and the other 3 <math>a\text{'s}</math> can be the lengths of an equilateral triangle formed from connecting the dots.
So, <math>b=\sqrt{3a}</math>, so <math>b:a= \sqrt{3}=(A)</math>
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So, <math>b=\sqrt{3}a</math>, so <math>b:a= \boxed{\textbf{(A)} \: \sqrt{3}}</math>
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<asy>draw((0, 0)--(1/2, sqrt(3)/2)--(1, 0)--cycle);
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draw((1/2, sqrt(3)/2)--(2, 0)--(1,0));
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label("$a$", (0, 0)--(1, 0), S);
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label("$a$", (1, 0)--(2, 0), S);
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label("$a$", (0, 0)--(1/2, sqrt(3)/2), NW);
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label("$a$", (1, 0)--(1/2, sqrt(3)/2), NE);
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label("$b=\sqrt{3}a$", (1/2, sqrt(3)/2)--(2, 0), NE);
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</asy>
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== 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}=\sqrt{3} \boxed{\mathrm{(A)}}</math>
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~ Nafer
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== Solution (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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== See Also ==
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{{AMC10 box|year=2012|ab=B|num-b=20|num-a=22}}
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{{MAA Notice}}

Latest revision as of 20:38, 5 September 2020

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: $\sqrt{3}$.


Video Solution by Richard Rusczyk

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

~dolphin7

See Also

2012 AMC 10B (ProblemsAnswer KeyResources)
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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