2012 AMC 10A Problem 24

The two left-hand sides look unrelated, but their sum is friendly:
$$2a^2 + 2b^2 + 2c^2 - 2ab - 2ac - 2bc = 2011 - 1997 = 14.$$
The left side is the well-known expression $(a-b)^2 + (b-c)^2 + (a-c)^2$, so the three pairwise differences have squares adding to $14$.

Let $x = a - b \ge 0$ and $y = b - c \ge 0$, so $a - c = x + y$. Then
$$x^2 + y^2 + (x+y)^2 = 14 \quad\Longrightarrow\quad x^2 + xy + y^2 = 7.$$
Each of $x, y$ is at most $2$, and checking the few options, only $\{x, y\} = \{1, 2\}$ works since $1 + 2 + 4 = 7$. In either case $c = a - 3$.

Now substitute $b = a - x$ and $c = a - x - y$ into the first equation. The $a^2$ terms cancel ($a^2 - a^2 - a^2 + a^2$), leaving
$$a(3x + 2y) - x^2 - (x+y)^2 = 2011.$$

• $x = 1, y = 2$: $7a - 1 - 9 = 2011$, so $7a = 2021$, not an integer.
• $x = 2, y = 1$: $8a - 4 - 9 = 2011$, so $8a = 2024$ and $a = 253$.

With $a = 253$, $b = 251$, $c = 250$: the first equation holds by construction, and since the sum of the two equations is $14$ (the differences satisfy $4 + 1 + 9 = 14$), the second holds as well.

The answer is $\boxed{\textbf{(E)}\ 253}$.

A system of quadratics with integer unknowns is usually cracked by finding one linear combination that becomes a sum of squares; a small sum of squares has only a handful of integer solutions, and each turns the remaining equation into a linear one. The pattern $a^2 + b^2 + c^2 - ab - bc - ca$ should trigger the identity immediately. Writing the unknowns as $a$ minus small gaps also keeps the arithmetic to linear terms.

• Trying to solve the system directly or subtracting the equations instead of adding them; the sum is the one combination that collapses to a sum of squares.
• Ignoring $a \ge b \ge c$ and admitting negative gaps such as $(x, y) = (1, -3)$, which also satisfy $x^2 + xy + y^2 = 7$ but violate the ordering.
• Expanding $(a-2)^2$ and $(a-3)^2$ numerically and slipping on signs; keep the quadratic terms symbolic so they cancel.