AMC12 2005 B

AMC12 2005 B · Q20

AMC12 2005 B · Q20. It mainly tests Averages (mean), Optimization (basic).

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 $(a+b+c+d)^{2}+(e+f+g+h)^{2}?$

设$a,b,c,d,e,f,g$和$h$是集合$\{-7,-5,-3,-2,2,4,6,13\}.$中不同的元素。 $(a+b+c+d)^{2}+(e+f+g+h)^{2}$的最小可能值是多少？

(A) 30 30

(B) 32 32

(D) 40 40

(E) 50 50

Answer

Correct choice: (C)

正确答案：（C）

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 $3^2+5^2=34\Rightarrow\boxed{\mathrm{C}}$.

该集合元素之和为$-7-5-3-2+2+4+6+13=8$，因此若能使每个括号内四项之和都为$4$，则最小值为$2(4^2)=32$。考虑包含$13$的那组四个数：其和为偶数当且仅当其中含有两个或四个奇数。若含四个奇数，则和为$13-7-5-3=-2$；若含两个奇数，则其最小可能和为$13-7+2-2=6$，因此无法得到两个和都为$4$。最接近$4$与$4$的情况是$3$与$5$，可由$13-7-5+2$与$6-3-2+4$实现。因此最小可能值为$3^2+5^2=34\Rightarrow\boxed{\mathrm{C}}$。

Topics