2005 AMC 12B Problem 20

Below is the professionally curated solution for Problem 20 of the 2005 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2005 AMC 12B solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:optimizationcompleting the square

Difficulty rating: 1910

20.

Let a,b,c,d,e,f,ga, b, c, d, e, f, g and hh be distinct elements in the set {7,5,3,2,2,4,6,13}. \{-7, -5, -3, -2, 2, 4, 6, 13\}. What is the minimum possible value of (a+b+c+d)2+(e+f+g+h)2? (a + b + c + d)^2 + (e + f + g + h)^2?

3030

3232

3434

4040

5050

Solution:

The elements sum to 8.8. If a+b+c+d=x,a + b + c + d = x, then e+f+g+h=8x,e + f + g + h = 8 - x, so x2+(8x)2=2(x4)2+32. x^2 + (8 - x)^2 = 2(x - 4)^2 + 32.

This is minimized when x=4,x = 4, giving 32.32. But 1313 must lie in one group, and no three of the remaining elements add with 1313 to make 44 (that would need three of them summing to 9,-9, which is impossible here). So x=4x = 4 is unattainable and (x4)21.(x - 4)^2 \ge 1.

The minimum is 2(1)+32=34,2(1) + 32 = 34, achieved for instance by {7,5,2,13}\{-7, -5, 2, 13\} (sum 33) and {3,2,4,6}\{-3, -2, 4, 6\} (sum 55).

Thus, the correct answer is C.

Problem 20 in Other Years