2013 AMC 12B Problem 17

Let

a,

b,

and

c

be real numbers such that

a + b + c = 2

and

a^2 + b^2 + c^2 = 12.

What is the difference between the maximum and minimum possible values of

c?

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Solution

From the equations,

a + b = 2 - c

and

a^2 + b^2 = 12 - c^2.

Real numbers

a, b

with a given sum and sum of squares exist iff

(a + b)^2 \le 2(a^2 + b^2),

i.e.

(2 - c)^2 \le 2(12 - c^2).

This simplifies to

(3c - 10)(c + 2) \le 0,

-2 \le c \le \tfrac{10}{3}.

The difference is

\tfrac{10}{3} - (-2) = \tfrac{16}{3}.

Thus, the correct answer is D.

Tagged: quadratic · inequality · extremal argument