AMC12 2012 A

AMC12 2012 A · Q21

AMC12 2012 A · Q21. It mainly tests Linear equations, Quadratic equations.

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 $a$?

设 $a$、$b$ 和 $c$ 是正整数且 $a\ge$ $b\ge$ $c$，使得 $a^2-b^2-c^2+ab=2011$ 且 $a^2+3b^2+3c^2-3ab-2ac-2bc=-1997$。求 $a$。

(A) 249 249

(B) 250 250

(D) 252 252

(E) 253 253

Answer

Correct choice: (E)

正确答案：（E）

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. We see that the only is possibility is $14 = 9 + 4 + 1$. $(a-c)^2 = 9 \Rightarrow a-c = 3$, 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 $\boxed{\textbf{(E)}\ 253}$.

将两式相加。 $2a^2 + 2b^2 + 2c^2 - 2ab - 2ac - 2bc = 14$。将其整理并因式分解。 $(a^2 - 2ab + b^2) + (a^2 - 2ac + c^2) + (b^2 - 2bc + c^2) = 14$ $(a - b)^2 + (a - c)^2 + (b - c)^2 = 14$ $a$、$b$、$c$ 都是整数，因此左边三项都必须是完全平方数。可见唯一可能是 $14 = 9 + 4 + 1$。 $(a-c)^2 = 9 \Rightarrow a-c = 3$，因为 $a-c$ 是最大的差。仅凭观察无法确定 $a-b$ 与 $b-c$ 分别是 $1$ 还是 $2$。我们要求 $a$，因此分两种情况分别用 $a$ 表示。两种情况为 $(a, b, c) = (a, a-1, a-3)$ 或 $(a, a-2, a-3)$。将这些值代入原方程之一，检验能否得到整数 $a$。当 $(a, b, c) = (a, a-1, a-3)$ 时，代入 $a^2 - (a-1)^2 - (a-3)^2 + a(a-1) = 2011$，化简得 $7a = 2021$。$2021$ 不能被 $7$ 整除，因此 $a$ 不是整数。另一种情况代入 $a^2 - (a-2)^2 - (a-3)^2 + a(a-2) = 2011$，化简得 $8a = 2024$。因此 $a = 253$，答案为 $\boxed{\textbf{(E)}\ 253}$。

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