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Difference between revisions of "2014 AMC 12A Problems/Problem 14"

(Created page with "==Problem== Let <math>a<b<c</math> be three integers such that <math>a,b,c</math> is an arithmetic progression and <math>a,c,b</math> is a geometric progression. What is the sm...")
 
 
(15 intermediate revisions by 9 users not shown)
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\textbf{(E) }6\qquad</math>
 
\textbf{(E) }6\qquad</math>
  
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==Solution==
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==Solution 1==
  
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We have <math>b-a=c-b</math>, so <math>a=2b-c</math>. Since <math>a,c,b</math> is geometric, <math>c^2=ab=(2b-c)b \Rightarrow 2b^2-bc-c^2=(2b+c)(b-c)=0</math>. Since <math>a<b<c</math>, we can't have <math>b=c</math> and thus <math>c=-2b</math>. then our arithmetic progression is <math>4b,b,-2b</math>. Since <math>4b < b < -2b</math>, <math>b < 0</math>. The smallest possible value of <math>c=-2b</math> is <math>(-2)(-1)=2</math>, or <math>\boxed{\textbf{(C)}}</math>.
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We have <math>b-a=c-b</math>, so <math>a=2b-c</math>. Since <math>a,c,b</math> is geometric, <math>c^2=ab=(2b-c)b \Rightarrow 2b^2-bc-c^2=(2b+c)(b-c)=0</math>. Since <math>a<b<c</math>, we can't have <math>b=c</math> and thus <math>c=-2b</math>. Then our arithmetic progression is <math>4b,b,-2b</math>. Since <math>4b < b < -2b</math>, <math>b < 0</math>. The smallest possible value of <math>c=-2b</math> is <math>(-2)(-1)=2</math>, or <math>\boxed{\textbf{(C)}}</math>.
  
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(Solution by AwesomeToad)
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(Solution by AT)
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 +
==Solution 2==
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Taking the definition of an arithmetic progression, there must be a common difference between the terms, giving us <math>(b-a) = (c-b)</math>. From this, we can obtain the expression <math>a = 2b-c</math>. Again, by taking the definition of a geometric progression, we can obtain the expression, <math>c=ar</math> and <math>b=ar^2</math>, where r serves as a value for the ratio between two terms in the progression. By substituting <math>b</math> and <math>c</math> in the arithmetic progression expression with the obtained values from the geometric progression, we obtain the equation, <math>a=2ar^2-ar</math> which can be simplified to <math>(r-1)(2r+1)=0</math> giving us <math>r=1</math> or <math>r=-1/2</math>. Thus, from the geometric progression, <math>a=a</math>, <math>b=1/4a</math> and <math>c=-1/2a</math>. Looking at the initial conditions of <math>a<b<c</math> we can see that the lowest integer value that would satisfy the above expressions is if <math>a = -4</math>, thus making <math>c=2</math> or <math>\boxed{\textbf{(C)}}</math>
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(Solution by thatuser)
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==Solution 3==
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By the definition of an arithmetic progression, we can label the terms <math>a</math>, <math>b</math>, and <math>c</math>, as <math>a</math>, <math>a+d</math>, and <math>a+2d</math>. Now, we have that <math>a</math>, <math>a+2d</math>, and <math>a+d</math> form a geometric progression. Since a geometric ratio has a common ratio between terms, we have <math>(a+2d)/a = (a+d)/a+2d</math>. Cross multiplying, we obtain the equation <math>(a+2d)^2=a(a+d)</math>. Multiplying it out and cancelling terms, we are left with the quadratic equation <math>4d^2+3ad=0</math>. Solving for <math>d</math> in terms of <math>a</math>, we get that <math>d=-3a/4</math> or <math>d=0</math>. Looking at the problem, we know that the <math>d</math> cannot be 0, therefore the arithmetic progression is <math>a, a/4, -a/2</math>, so we need to find the minimum value of <math>-a/2</math> while <math>-a/2>a</math>. Looking at our progression, we realize that a must be a multiple of 4 so that every term is an integer. Substituting <math>a=-4</math>, since that would yield the smallest value of <math>-a/2</math> which satisfies the conditions, we figure out that the answer is <math>\boxed{\textbf{(C)}}</math>.
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(Solution by i8Pie)
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==Video Solution==
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https://youtu.be/sxBuSZGfmHg
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 +
~Lucas
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 +
==See Also==
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{{AMC12 box|year=2014|ab=A|num-b=13|num-a=15}}
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{{MAA Notice}}

Latest revision as of 16:25, 19 September 2022

Problem

Let $a<b<c$ be three integers such that $a,b,c$ is an arithmetic progression and $a,c,b$ is a geometric progression. What is the smallest possible value of $c$?

$\textbf{(A) }-2\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }4\qquad \textbf{(E) }6\qquad$

Solution 1

We have $b-a=c-b$, so $a=2b-c$. Since $a,c,b$ is geometric, $c^2=ab=(2b-c)b \Rightarrow 2b^2-bc-c^2=(2b+c)(b-c)=0$. Since $a<b<c$, we can't have $b=c$ and thus $c=-2b$. Then our arithmetic progression is $4b,b,-2b$. Since $4b < b < -2b$, $b < 0$. The smallest possible value of $c=-2b$ is $(-2)(-1)=2$, or $\boxed{\textbf{(C)}}$.

(Solution by AT)

Solution 2

Taking the definition of an arithmetic progression, there must be a common difference between the terms, giving us $(b-a) = (c-b)$. From this, we can obtain the expression $a = 2b-c$. Again, by taking the definition of a geometric progression, we can obtain the expression, $c=ar$ and $b=ar^2$, where r serves as a value for the ratio between two terms in the progression. By substituting $b$ and $c$ in the arithmetic progression expression with the obtained values from the geometric progression, we obtain the equation, $a=2ar^2-ar$ which can be simplified to $(r-1)(2r+1)=0$ giving us $r=1$ or $r=-1/2$. Thus, from the geometric progression, $a=a$, $b=1/4a$ and $c=-1/2a$. Looking at the initial conditions of $a<b<c$ we can see that the lowest integer value that would satisfy the above expressions is if $a = -4$, thus making $c=2$ or $\boxed{\textbf{(C)}}$ (Solution by thatuser)

Solution 3

By the definition of an arithmetic progression, we can label the terms $a$, $b$, and $c$, as $a$, $a+d$, and $a+2d$. Now, we have that $a$, $a+2d$, and $a+d$ form a geometric progression. Since a geometric ratio has a common ratio between terms, we have $(a+2d)/a = (a+d)/a+2d$. Cross multiplying, we obtain the equation $(a+2d)^2=a(a+d)$. Multiplying it out and cancelling terms, we are left with the quadratic equation $4d^2+3ad=0$. Solving for $d$ in terms of $a$, we get that $d=-3a/4$ or $d=0$. Looking at the problem, we know that the $d$ cannot be 0, therefore the arithmetic progression is $a, a/4, -a/2$, so we need to find the minimum value of $-a/2$ while $-a/2>a$. Looking at our progression, we realize that a must be a multiple of 4 so that every term is an integer. Substituting $a=-4$, since that would yield the smallest value of $-a/2$ which satisfies the conditions, we figure out that the answer is $\boxed{\textbf{(C)}}$. (Solution by i8Pie)

Video Solution

https://youtu.be/sxBuSZGfmHg

~Lucas

See Also

2014 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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 12 Problems and Solutions

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