Question

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 $a?$

$249$

$250$

$251$

$252$

$253$

Solution:

Adding together the equations gives us $2(a^2 + b^2 + c^2) - 2(ab + ac + bc)$ $= 14.$

We can group terms and factor this to get $(a - b)^2 + (a - c)^2 + (b - c)^2$ $= 14.$

Note that every term on the left hand side is a positive square integer. The only triple of squares that add to $14$ is $9, 4,$ and $1.$

We have that $a - c$ is the biggest difference among the three pairs. Therefore, $a - c = 3.$

We cannot discern which of the other terms we can match with the other squares. Let us try $a - b = 1$ and $b - c = 2.$

Plugging in these values into the first equation gives us $a^2 - (a - 1)^2 - (a - 3)^2$ $+ a(a - 1) = 2011.$ Simplifying yields $7a = 2021.$ Since $2021$ is not divisible by $7,$ we have that $a - b = 2$ and $b - c = 1.$

Plugging these in again and solving gives us $a = 253.$

Thus, E is the correct answer.

2012 AMC 10A Problem 24

Solution:

Problem 24 in Other Years