2012 AMC 12A Problem 21

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

\begin{aligned} &a^2 + 3b^2 + 3c^2 \\ &\quad {}- 3ab - 2ac - 2bc = -1997. \end{aligned}

What is

a?

249

250

251

252

253

Add the two equations to get a clean symmetric expression

The sum simplifies to

(a-b)^2

+ (b-c)^2

+ (c-a)^2 = 14,

and

14 = 3^2 + 2^2 + 1^2

uniquely

Adding the two equations gives

2a^2 + 2b^2 + 2c^2

- 2ab - 2bc - 2ca = 14,

that is,

\begin{aligned} &(a-b)^2 + (b-c)^2 \\ &\quad {}+ (c-a)^2 = 14. \end{aligned}

The only way to write

14

as a sum of three squares is

9 + 4 + 1.

Since

a \ge b \ge c,

we get

a - c = 3,

with either

(a-b, b-c) = (2,1)

or

(1,2).

Substituting

(a, b, c) = (c+3, c+1, c)

into the first equation gives

3(2c+3) + 2(c+1) = 2011,

so

c = 250

and

(a, b, c) = (253, 251, 250).

The other case has no integer solution.

Thus, the correct answer is E.

Problem 21 in other years