Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(added box+MAA notice)
m (→‎Solution)
Line 11: Line 11:
 
==Solution==
 
==Solution==
  
−
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>.
+
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>.
  
 
(Solution by AwesomeToad)
 
(Solution by AwesomeToad)

Revision as of 20:15, 29 June 2014

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

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 AwesomeToad)

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_12A_Problems/Problem_14&oldid=62418"