2012 AMC 10A Problem 24

1. Add the equations. The two left sides look unrelated, but adding them gives
$$2a^2 + 2b^2 + 2c^2 - 2ab - 2ac - 2bc = 2011 - 1997 = 14.$$
2. Recognise squares. Expand three squares:
$$(a - b)^2 + (b - c)^2 + (a - c)^2 = (a^2 - 2ab + b^2) + (b^2 - 2bc + c^2) + (a^2 - 2ac + c^2).$$
This is exactly the left side from step 1. So the three gaps between $a$, $b$, $c$ have squares that add to $14$.
3. Find the gaps. Let $x = a - b$ and $y = b - c$. Both are whole numbers, $0$ or more, since $a \ge b \ge c$. Then $a - c = x + y$, and
$$x^2 + y^2 + (x + y)^2 = 14.$$
Each square is at most $14$, so $x + y \le 3$. Try the options: $x + y = 3$ with gaps $1$ and $2$ gives $1 + 4 + 9 = 14$. The others fail ($0, 1, 2$ for $x + y$ give totals of at most $8$, and $0$ and $3$ gives $0 + 9 + 9 = 18$). So the gaps are $1$ and $2$ in some order.
4. Use the first equation. Write $b = a - x$ and $c = a - x - y$, and put them into $a^2 - b^2 - c^2 + ab = 2011$. Try both orders.
- $x = 1$, $y = 2$: $b = a - 1$, $c = a - 3$. Expanding, $a^2 - (a^2 - 2a + 1) - (a^2 - 6a + 9) + (a^2 - a) = 7a - 10$. So $7a - 10 = 2011$, giving $7a = 2021$. But $2021$ is not a multiple of $7$ ($7 \times 288 = 2016$). No good.
- $x = 2$, $y = 1$: $b = a - 2$, $c = a - 3$. Expanding, $a^2 - (a^2 - 4a + 4) - (a^2 - 6a + 9) + (a^2 - 2a) = 8a - 13$. So $8a - 13 = 2011$, giving $8a = 2024$ and $a = 253$.
5. Check. $a = 253$, $b = 251$, $c = 250$. The first equation holds by step 4. The two equations add to $14$ (the gaps give $4 + 1 + 9 = 14$), so the second one is $14 - 2011 = -1997$, as required.

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

When two equations look messy, try adding or subtracting them to see if something neat appears. The expression $a^2 + b^2 + c^2 - ab - bc - ca$ is half the sum of the squared gaps, and a small total like $14$ leaves only a few possibilities. Writing $b$ and $c$ as $a$ minus small gaps then makes the $a^2$ terms cancel, leaving a simple equation.

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 $\boxed{\textbf{(E)}\ 253}$.