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2005 AMC 12B Problems/Problem 20

Problem

Let $a,b,c,d,e,f,g$ and $h$ be distinct elements in the set $\{-7,-5,-3,-2,2,4,6,13\}.$

What is the minimum possible value of $(a+b+c+d)^{2}+(e+f+g+h)^{2}?$

$\mathrm{(A)}\ 30 \qquad \mathrm{(B)}\ 32 \qquad \mathrm{(C)}\ 34 \qquad \mathrm{(D)}\ 40 \qquad \mathrm{(E)}\ 50$

Solution

The sum of the set is $-7-5-3-2+2+4+6+13=8$, so if we could have the sum in each set of parenthesis be $4$ then the minimum value would be $2(4^2)=32$. Considering the set of four terms containing $13$, this sum could only be even if it had two or four odd terms. If it had all four odd terms then it would be $13-7-5-3=-2$, and with two odd terms then its minimum value is $13-7+2-2=6$, so we cannot achieve two sums of $4$. The closest we could have to $4$ and $4$ is $3$ and $5$, which can be achieved through $13-7-5+2$ and $6-3-2+4$. So the minimum possible value is $3^2+5^2=34\Rightarrow\boxed{\mathrm{C}}$.

Solution 2 (better explanation of 1st)

Trying out the values for a bit leads to us getting $34$ with $(-3, -2, 2, 6)$ in 1 set for a total of 9 and the other numbers giving a total of 25.

We start off with trying to get a $2(4^2) = 32$ solution, so we only need to find one set with a sum of 4 (the other will automatically have a sum of 4).

We can add 7 to each number. We are now trying to get a set with 32.

\[(0, 2, 4, 5, 9, 11, 13, 20)\]

Observe that there are 4 odd and 4 even values. To get 32 from 4 numbers $a,b,c,d$, the numbers either have to be 4 odds, 4 evens, or 2 odds and evens. The 4 odd numbers do not work (38) nor the 4 even numbers (26).

If $a,b,c,d$ has 2 odd numbers and 2 even numbers we can divide them into two cases: whether $a,b,c,d$ has 20 or doesn't have 20.

Case 1: $a,b,c,d$ doesn't have 20.

The maximum sum of the 2 even numbers are 2 + 4 = 6 and the maximum sum of the 2 odd numbers are 11 + 13 = 24. This is not enough to reach the 32 we need.

Case 2: $a,b,c,d$ has 20.

The minimum sum of the 2 even numbers are 20 + 0 = 20 and the minimum sum of the 2 odd numbers are 5 + 9 = 14. This is already greater than the 32 we need.

We have proven that we cannot partition the 8 numbers into two groups of 4 with the same sum, so the smallest value is $\boxed{C}$.

Solution 3

Apply Cauchy's inequality to bound the sum, yielding $(1+1)\left((a+b+c+d)^{2}+(e+f+g+h)^{2}\right) \ge (a+b+c+d+e+f+g+h)^2$ and rearrange to get $(a+b+c+d)^{2}+(e+f+g+h)^{2} \ge 32$, where equality is attained iff $\frac{a+b+c+d}{1}=\frac{e+f+g+h}{1}=4$, which is impossible from the numbers given upon observing. Therefore, we can eliminate choices $\mathrm{(A)}$ and $\mathrm{(B)}$. Now experiment to get that $13-7-5+2$ and $6-3-2+4$ give $34$, which must be the minimum since we've eliminated the smaller answer choices. $\implies \boxed{\mathrm{C}}$

~bomberdoodles

See also

2005 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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