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"

(→‎Problem 24)
Line 25: Line 25:
 
<math>a</math>, <math>b</math>, and <math>c</math> are all integers, so the three terms on the left side of the equation must all be perfect squares. Recognize that <math>14 = 9 + 4 + 1</math>.  
 
<math>a</math>, <math>b</math>, and <math>c</math> are all integers, so the three terms on the left side of the equation must all be perfect squares. Recognize that <math>14 = 9 + 4 + 1</math>.  
  
−
<math>(a-c)^2 = 9 -> a-c = 3</math>, since <math>a-c</math> is the biggest difference. It is impossible to determine by inspection whether <math>a-b = 2</math> or <math>1</math>, or whether <math>b-c = 1</math> or <math>2</math>.
+
<math>(a-c)^2 = 9 \rightarrow a-c = 3</math>, since <math>a-c</math> is the biggest difference. It is impossible to determine by inspection whether <math>a-b = 2</math> or <math>1</math>, or whether <math>b-c = 1</math> or <math>2</math>.
  
 
We want to solve for <math>a</math>, so take the two cases and solve them each for an expression in terms of <math>a</math>. Our two cases are <math>(a, b, c) = (a, a-1, a-3)</math> or <math>(a, a-2, a-3)</math>. Plug these values into one of the original equations to see if we can get an integer for <math>a</math>.
 
We want to solve for <math>a</math>, so take the two cases and solve them each for an expression in terms of <math>a</math>. Our two cases are <math>(a, b, c) = (a, a-1, a-3)</math> or <math>(a, a-2, a-3)</math>. Plug these values into one of the original equations to see if we can get an integer for <math>a</math>.

Revision as of 22:41, 9 February 2012

Problem 24

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:

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

and factored:

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

$a$, $b$, 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 $a-b = 2$ or $1$, 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 $\qquad\textbf{(E)}$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_24&oldid=44652"