AMC12 2016 B

AMC12 2016 B · Q15

AMC12 2016 B · Q15. It mainly tests Inequalities (AM-GM etc. basic), 3D geometry (volume).

All the numbers 2, 3, 4, 5, 6, 7 are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?

将数字 2、3、4、5、6、7 分别分配到一个立方体的六个面上，每个面一个数字。对于立方体的每一个顶点（共 8 个），计算一个三个数字的乘积，这三个数字分别是包含该顶点的三个面的数字。求这 8 个乘积之和的最大可能值。

(A) 312 312

(B) 343 343

(D) 729 729

(E) 1680 1680

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Suppose that one pair of opposite faces of the cube are assigned the numbers $a$ and $b$, a second pair of opposite faces are assigned the numbers $c$ and $d$, and the remaining pair of opposite faces are assigned the numbers $e$ and $f$. Then the needed sum of products is $ace + acf + ade + adf + bce + bcf + bde + bdf = (a + b)(c + d)(e + f)$. The sum of these three factors is $2 + 3 + 4 + 5 + 6 + 7 = 27$. A product of positive numbers whose sum is fixed is maximized when the factors are all equal. Thus the greatest possible value occurs when $a + b = c + d = e + f = 9$, as in $(a,b,c,d,e,f) = (2,7,3,6,4,5)$. This results in the value $9^3 = 729$.

答案（D）：假设立方体一对相对的面被赋值为 $a$ 和 $b$，第二对相对的面被赋值为 $c$ 和 $d$，剩下的一对相对的面被赋值为 $e$ 和 $f$。那么所需的乘积和为 $ace + acf + ade + adf + bce + bcf + bde + bdf = (a + b)(c + d)(e + f)$。这三个因子的和为 $2 + 3 + 4 + 5 + 6 + 7 = 27$。在和固定的情况下，正数的乘积在各因子相等时取得最大值。因此当 $a + b = c + d = e + f = 9$ 时可取得最大值，例如 $(a,b,c,d,e,f) = (2,7,3,6,4,5)$。由此得到的数值为 $9^3 = 729$。

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