Question

如图所示排列二十个正方体积木。首先将 $10$ 个积木排成三角形；再将由 $6$ 个积木排成的三角形层居中放在这 $10$ 个积木上；然后将由 $3$ 个积木排成的三角形层居中放在这 $6$ 个积木上；最后在第三层顶端中央放一个积木。底层积木以某种顺序编号为 $1$ 到 $10$ 。第 $2$ 、 $3$ 、 $4$ 层的每个积木所标的数，等于支撑它的三个积木所标数之和。求顶层积木可能标上的最小数。

Twenty cubical blocks are arranged as shown. First, $10$ are arranged in a triangular pattern; then a layer of $6,$ arranged in a triangular pattern, is centered on the $10;$ then a layer of $3,$ arranged in a triangular pattern, is centered on the $6;$ and finally one block is centered on top of the third layer. The blocks in the bottom layer are numbered $1$ through $10$ in some order. Each block in layers $2,$ $3$ and $4$ is assigned the number which is the sum of the numbers assigned to the three blocks on which it rests. Find the smallest possible number which could be assigned to the top block.

$55$

$83$

$114$

$137$

$144$

答案：C

知识点：weighted sum rearrangement principle triangular stack

难度评级：1960

小提示：

确定每个底层积木对顶层数值贡献了多少次。

Determine how many times each bottom block contributes to the top

大提示：

中心位置、六个边缘位置和三个角位置的系数分别为 $6$ 、 $3$ 和 $1$ 。

The center, six edge positions, and three corner positions have coefficients $6,$ $3,$ and $1$

解答：

逐层展开各个和，顶层数值为 $6c+3(e_1+e_2+e_3$ ${}+e_4+e_5+e_6)$ ${}+(v_1+v_2+v_3)$ ，其中 $c$ 是底层中心积木， $e_i$ 是六个非角落的边界积木， $v_i$ 是三个角落积木。为使这个加权和最小，将 $1$ 放在系数为 $6$ 的位置，将 $2$ 、 $\ldots$ 、 $7$ 放在系数为 $3$ 的位置，并将 $8$ 、 $9$ 、 $10$ 放在系数为 $1$ 的位置。最小值为 $6+3(2+3+4$ ${}+5+6+7)$ ${}+(8+9+10)=114$ 。因此正确答案是 C。

Expanding the sums layer by layer, the top value is $6c+3(e_1+e_2+e_3$ ${}+e_4+e_5+e_6)$ ${}+(v_1+v_2+v_3),$ where $c$ is the center bottom block, the $e_i$ are the six non-corner boundary blocks, and the $v_i$ are the three corner blocks. To minimize this weighted sum, assign $1$ to the coefficient- $6$ position, $2,$ $\ldots,$ $7$ to the coefficient- $3$ positions, and $8,$ $9,$ $10$ to the coefficient- $1$ positions. The minimum is $6+3(2+3+4$ ${}+5+6+7)$ ${}+(8+9+10)=114.$ Thus the correct answer is C.

1993 AMC 12 第 22 题

答案：C

难度评级：1960

小提示：

大提示：

解答：

其他年份的第 22 题