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 "2012 AMC 10A Problems/Problem 24"

(→‎Video Solution by Richard Rusczyk)
(→‎Solution)
Line 17: Line 17:
 
<math>2a^2 + 2b^2 + 2c^2 - 2ab - 2ac - 2bc = 14</math>.
 
<math>2a^2 + 2b^2 + 2c^2 - 2ab - 2ac - 2bc = 14</math>.
  
−
Now, this can be rearranged:
+
Now, this can be rearranged and factored.
  
−
<math>(a^2 - 2ab + b^2) + (a^2 - 2ac + c^2) + (b^2 - 2bc + c^2) = 14</math>
+
<cmath>(a^2 - 2ab + b^2) + (a^2 - 2ac + c^2) + (b^2 - 2bc + c^2) = 14</cmath>
−
 
 
−
and factored:
 
  
 
<math>(a - b)^2 + (a - c)^2 + (b - c)^2 = 14</math>
 
<math>(a - b)^2 + (a - c)^2 + (b - c)^2 = 14</math>
Line 35: Line 33:
  
 
The other case gives <math>a^2 - (a-2)^2 - (a-3)^2 + a(a-2) = 2011</math>, which simplifies to <math>8a = 2024</math>. Thus, <math>a = 253</math> and the answer is <math>\boxed{\textbf{(E)}\ 253}</math>.
 
The other case gives <math>a^2 - (a-2)^2 - (a-3)^2 + a(a-2) = 2011</math>, which simplifies to <math>8a = 2024</math>. Thus, <math>a = 253</math> and the answer is <math>\boxed{\textbf{(E)}\ 253}</math>.
−
 
  
 
==Video Solution by Richard Rusczyk==
 
==Video Solution by Richard Rusczyk==

Revision as of 15:51, 20 January 2021

The following problem is from both the 2012 AMC 12A #21 and 2012 AMC 10A #24, so both problems redirect to this page.

Problem

Let $a$, $b$, and $c$ be positive integers with $a\ge$ $b\ge$ $c$ such that $a^2-b^2-c^2+ab=2011$ and $a^2+3b^2+3c^2-3ab-2ac-2bc=-1997$.

What is $a$?

$\textbf{(A)}\ 249\qquad\textbf{(B)}\ 250\qquad\textbf{(C)}\ 251\qquad\textbf{(D)}\ 252\qquad\textbf{(E)}\ 253$

Solution

Add the two equations.

$2a^2 + 2b^2 + 2c^2 - 2ab - 2ac - 2bc = 14$.

Now, this can be rearranged and factored.

\[(a^2 - 2ab + b^2) + (a^2 - 2ac + c^2) + (b^2 - 2bc + c^2) = 14\]

$(a - b)^2 + (a - c)^2 + (b - c)^2 = 14$

$a$, $b$, and $c$ are all integers, so the three terms on the left side of the equation must all be perfect squares. Recognize that $14 = 9 + 4 + 1$.

$(a-c)^2 = 9 \rightarrow a-c = 3$, since $a-c$ is the biggest difference. It is impossible to determine by inspection whether $a-b = 1$ or $2$, or whether $b-c = 1$ or $2$.

We want to solve for $a$, so take the two cases and solve them each for an expression in terms of $a$. Our two cases are $(a, b, c) = (a, a-1, a-3)$ or $(a, a-2, a-3)$. Plug these values into one of the original equations to see if we can get an integer for $a$.

$a^2 - (a-1)^2 - (a-3)^2 + a(a-1) = 2011$, after some algebra, simplifies to $7a = 2021$. $2021$ is not divisible by $7$, so $a$ is not an integer.

The other case gives $a^2 - (a-2)^2 - (a-3)^2 + a(a-2) = 2011$, which simplifies to $8a = 2024$. Thus, $a = 253$ and the answer is $\boxed{\textbf{(E)}\ 253}$.

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2012amc12a/250

~dolphin7

See Also

2012 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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
2012 AMC 12A (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 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=2012_AMC_10A_Problems/Problem_24&oldid=142869"