2013 AMC 12B Problem 17

Below is the professionally curated solution for Problem 17 of the 2013 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AMC 12B solutions, or check the answer key.

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Concepts:quadratic inequality extremal argument

Difficulty rating: 1960

17.

Let $a,$ $b,$ and $c$ be real numbers such that

$a + b + c = 2 \quad\text{and}\quad a^2 + b^2 + c^2 = 12.$

What is the difference between the maximum and minimum possible values of $c?$

$2$

$\dfrac{10}{3}$

$4$

$\dfrac{16}{3}$

$\dfrac{20}{3}$

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 $(3c - 10)(c + 2) \le 0,$ so $-2 \le c \le \tfrac{10}{3}.$ The difference is $\tfrac{10}{3} - (-2) = \tfrac{16}{3}.$ Thus, the correct answer is D.

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