2013 AMC 12B Problem 17

Attempt Problem 17 of the 2013 AMC 12B below, then check your answer against the professionally curated solution 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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

17.

Let a,a, b,b, and cc be real numbers such that
a+b+c=2 a + b + c = 2 and a2+b2+c2=12. a^2 + b^2 + c^2 = 12.
What is the difference between the maximum and minimum possible values of c?c?
22
103\dfrac{10}{3}
44
163\dfrac{16}{3}
203\dfrac{20}{3}
Answer: D
Concepts:quadraticinequalityextremal argument
Small hint:
Write a+b=2−ca + b = 2 - c and a2+b2=12−c2a^2 + b^2 = 12 - c^2
Big hint:
Real a,ba, b with a given sum and sum of squares exist iff (a+b)2≤2(a2+b2)(a+b)^2 \le 2(a^2 + b^2)
Solution:
From the equations, a+b=2−ca + b = 2 - c and a2+b2=12−c2.a^2 + b^2 = 12 - c^2. Real numbers a,ba, b with a given sum and sum of squares exist iff (a+b)2≤2(a2+b2),(a + b)^2 \le 2(a^2 + b^2), i.e. (2−c)2≤2(12−c2).(2 - c)^2 \le 2(12 - c^2). This simplifies to (3c−10)(c+2)≤0,(3c - 10)(c + 2) \le 0, so −2≤c≤103.-2 \le c \le \tfrac{10}{3}. The difference is 103−(−2)=163.\tfrac{10}{3} - (-2) = \tfrac{16}{3}. Thus, the correct answer is D.
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