2012 AMC 10A 第 24 题

Add the two equations first

The three squared differences must be $9,4,1$ in some order

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 nonnegative 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 $b - c = 1.$

In the other case, $b=a-2$ and $c=a-3.$ The first equation becomes $$\begin{aligned} a^2-(a-2)^2-(a-3)^2\\ {}+a(a-2)&=2011, \end{aligned}$$ or $8a-13=2011.$ Hence $a=253.$

Thus, E is the correct answer.