Question

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 $c?$

$2$

$\dfrac{10}{3}$

$4$

$\dfrac{16}{3}$

$\dfrac{20}{3}$

Answer: D

Concepts:quadratic inequality extremal argument

Difficulty rating: 1960

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.

2013 AMC 12B Problem 17

Answer: D

Difficulty rating: 1960

Solution:

Problem 17 in Other Years