2016B AMC 10 真题（P版）

原式等于 \[2a^{-2}+\frac{a^{-2}}{2} = 2.5 (a^{-1})^2\text{。}\] 此时 \(a^{-1} \) 等于 \(\dfrac{1}{\frac 12} = 2\)，所以原式为 \(2.5\cdot 2^2 = 10\)。

所以正确答案是 D。

English

The expression is equivalent to \[2a^{-2}+\frac{a^{-2}}{2} = 2.5 (a^{-1})^2.\] Then \(a^{-1} \) is equal to \(\dfrac{1}{\frac 12} = 2,\) so our expression is equal to \(2.5\cdot 2^2 = 10.\)

Thus, the correct answer is D .

对任意非零 \(a,b\)，有 \[\dfrac{a\heartsuit b}{b\heartsuit a} = \dfrac{a^3b^2}{b^3 a^2} = \dfrac ab \text{。}\] 令 \(a=2,b=4\)，得到 \(\frac{ 2}{4} = \frac 12\)。

所以正确答案是 B。

English

For any nonzero \(a,b,\) we have \[\dfrac{a\heartsuit b}{b\heartsuit a} = \dfrac{a^3b^2}{b^3 a^2} = \dfrac ab .\] Using \(a=2,b=4,\) we get that \(\frac{ 2}{4} = \frac 12.\)

Thus, the correct answer is B .

由 \(x=-2016<0\) 可知 \(|x|=2016\)，所以 \[ \begin{aligned} |x|-x&=2016-(-2016) \\ &=4032 \end{aligned}\text{，}\] 最内层的绝对值为 \(4032\)。因此整个表达式为 \[|4032-2016|+2016=4032\text{。}\]

所以正确答案是 D。

English

Since \(x=-2016<0,\) we have \(|x|=2016.\) Thus \[ \begin{aligned} |x|-x&=2016-(-2016) \\ &=4032, \end{aligned} \] so the innermost absolute value is \(4032.\) Therefore the full expression is \[|4032-2016|+2016=4032.\]

Thus, the correct answer is D .

读完 \(15\) 本书共需要 \(\dfrac{ 15\cdot 16}{2} =120\) 天。因此第十五本读完是在第一本读完后的 \(120-1=119\) 天。因为这是 \(7\) 的倍数，所以第 \(15\) 本是在与第 \(1\) 本相同的星期几读完的，也就是星期一。

所以正确答案是 B。

English

The number of days it takes to read \(15\) books is \(\dfrac{ 15\cdot 16}{2} =120.\) Therefore, it is \(120-1=119 \) days after the first book is read. This is a multiple of \(7,\) so the \(15\)th book was finished on the same day as the \(1\)st book. Therefore, it was finished on a Monday.

Thus, the correct answer is B .

四个年龄的总和是 \(4\cdot8=32\)。由于中位数是 \(5\)，中间两个年龄的平均数是 \(5\)，所以它们的和是 \(10\)。

因此最小年龄与最大年龄的和为 \(32-10=22\)。

所以正确答案是 D。

English

The total of the four cousin ages is \(4\cdot8=32\). Since the median is \(5\), the two middle ages have average \(5\), so their sum is \(10\).

Therefore the youngest and oldest ages sum to \(32-10=22\).

Thus, the correct answer is D.

为使两个加数的和尽量小，百位取两个最小的非零数字 \(1\) 和 \(2\)，剩下的四个数位取余下最小的四个数字 \(0,3,4,5\)。把较小的数字放在十位上可知，任何可能的和都至少是 \(104+235=339\)。另一方面，数字和小于 \(4\) 的三位数最大也只有 \(300\)。因此 \(S\) 的数字和至少为 \(4\)。

例子 \(157+243=400\) 使用六个互不相同的数字，且数字和为 \(4\)。因此最小可能值是 \(4\)。

所以正确答案是 B。

English

To minimize the sum of the addends, use the two smallest nonzero hundreds digits, \(1\) and \(2\), and then the four smallest remaining digits, \(0,3,4,5\). Placing the smaller digits in the tens places shows that every possible sum is at least \(104+235=339\). On the other hand, any three-digit number with digit sum less than \(4\) is at most \(300\). Therefore the digit sum of \(S\) is at least \(4\).

The example \(157+243=400\) uses six distinct digits and has digit sum \(4\). Thus the smallest possible digit sum is \(4\).

Thus, the correct answer is B.

设较小角为 \(x\)。则较大角为 \(\frac 54 x\)，两角和为 \(\frac 94 x\)。

它们的余角分别是 \(90-x\) 与 \(90 - \frac 54x\)。由于 \(90-x\) 大于 \(90-\frac 54 x\)，所以 \[\begin{aligned}90 -x &= 2\left(90 - \frac 54 x\right) \\90-x&= 180-\frac 52 x\\x&=60\end{aligned}\text{。}\]

因此两角和为 \[\dfrac 94 \cdot 60=135\text{。}\]

所以正确答案是 C。

English

Let the smaller angle be \(x.\) Then, the larger angle is \(\frac 54 x.\) Note that their sum is \(\frac 94 x.\)

Their complements are then \(90-x\) and \(90 - \frac 54x.\) Since \(90-x\) is larger than \(90-\frac 54 x,\) we know \[\begin{aligned}90 -x &= 2\left(90 - \frac 54 x\right) \\90-x&= 180-\frac 52 x\\x&=60.\end{aligned}\]

Therefore, the sum is \[\dfrac 94 \cdot 60=135.\]

Thus, the correct answer is C .

为了求十位数字，先求该数模 \(100\) 的余数。先求 \(2015^{2016} \mod 100\)，可用中国剩余定理分别考虑 \(\mod 4\) 和 \(\mod 25\)。

有 \[2015^{2016} \equiv 0 \mod 25\] 因为它是 \(25\) 的倍数。

另一方面，底数除以四的余数为负一，所以 \[2015^{2016} \equiv (-1)^{2016}\] \[ \equiv 1 \mod 4\text{。}\]

由中国剩余定理可知，这个数同余于 \(25 \mod 100\)。

因此 \[2015^{2016}-2017 \]\[\equiv 25 - 17 \mod 100\]\[\equiv 8 \mod 100\text{。}\] 这说明十位数字是 \(0\)。

所以正确答案是 A。

English

To find the tens digit, we first need to find the number modulo \(100.\) First, we can find \(2015^{2016} \mod 100.\) We can do this with the Chinese Remainder Theorem by first getting the number \(\mod 4\) and then \(\mod 25.\)

The number \[2015^{2016} \equiv 0 \mod 25\] since it is a multiple of \(25.\)

Then, observe that \[2015^{2016} \equiv (-1)^{2016}\]\[ \equiv 1 \mod 4 .\]

By the Chinese remainder theorem, we can get that our number is congruent to \(25 \mod 100.\)

Therefore, \[2015^{2016}-2017 \]\[\equiv 25 - 17 \mod 100\]\[\equiv 8 \mod 100 .\] This means the tens digit is \(0.\)

Thus, the correct answer is A .

设三个顶点为 \[A=(0,0)\text{，}\] \[B=(x_1,y_1)\text{，}\] \[C=(x_2,y_2)\text{。}\]

因为 \(BC\) 平行于 \(x\) 轴，所以 \(y_1=y_2\)，记为 \(y\)。又两点都在抛物线上，\(x_1^2 = y=x_2^2\)，即 \(x_1^2=x_2^2 \)。因此 \(x_1=-x_2 \) 或 \(x_1 = x_2\)。后一种会让两点重合，面积为 \(0\)。所以令 \(x= -x_1=x_2\)，同时 \(y=x^2\)。

于是点为 \[A=(0,0)\text{，}\]\[B=(-x,x^2)\text{，}\]\[C=(x,x^2)\text{。}\] 以 \(BC\) 为底时，底长为 \(2x\)，高为 \(x^2\)，面积为 \(x^3 = 64\)。因此 \(x=4\)，所以 \(BC = 2\cdot 4 = 8\)。

所以正确答案是 C。

English

Let the points be \[A=(0,0),\]\[B=(x_1,y_1),\]\[C=(x_2,y_2).\]

Then, since \(BC\) is parallel with the \(x\)-axis, we know \(y_1=y_2,\) which we will let be \(y.\) Then, \(x_1^2 = y=x_2^2,\) so \(x_1^2=x_2^2 .\) This implies that either \(x_1=-x_2 \) or \(x_1 = x_2.\) The second option cannot happen since that would set two points as the same, which would create an area of \(0.\) As such, let \(x= -x_1=x_2.\) Then \(y=x^2\) also.

Then, the points are \[A=(0,0),\]\[B=(-x,x^2),\]\[C=(x,x^2).\] With a base of \(BC,\) the length is \(2x\) and the height is \(x^2.\) This would make the area \(x^3 = 64.\) Therefore, \(x=4,\) so \(BC = 2\cdot 4 = 8.\)

Thus, the correct answer is C .

面积按 \(\frac 53 ^2 \) 倍增加，因为边长按 \(\frac 53\) 倍增加。厚度不变，所以体积也按同样倍数增加。又因为木材密度相同，重量也按 \(\frac 53^2 \) 倍增加，于是新重量为 \[12 \cdot \dfrac {25}{9} = \dfrac{100}3\text{。}\] 约为 \(33.3\)。

所以正确答案是 D。

English

The surface area is increased by a factor of \(\frac 53 ^2 \) since the side lengths are increased by a factor of \(\frac 53.\) Also since the thickness is constant, the volume is scaled up by that much. Then, the wood having the same density makes the weight increase by a factor of \(\frac 53^2 \) as well, so the new weight is \[12 \cdot \dfrac {25}{9} = \dfrac{100}3.\] This is approximately \(33.3.\)

Thus, the correct answer is D .

设长边上有 \(l\) 根柱子，短边上有 \(s\) 根柱子，均包括两个角。总柱子数为 \[2l+2s-4=20\text{，}\] 其中 \(-4\) 是为了避免四个角被重复计算。

又长边柱子数是短边的两倍，所以 \(l=2s\)。代入得 \(2(2s)+2s-4=20\)，因此 \(s=4\)，\(l=8\)。

短边有 \(3\) 个间隔，长边有 \(7\) 个间隔，每个间隔 \(4\) 码，所以边长分别为 \(12\) 和 \(28\)，面积为 \(12\cdot28=336\)。

所以正确答案是 B。

English

Let \(l\) be the number of posts on a long side, including corners, and let \(s\) be the number of posts on a short side, including corners. The total number of posts is \[2l+2s-4=20,\] where the \(-4\) avoids double-counting the corners.

Since a long side has twice as many posts as a short side, \(l=2s\). Thus \(2(2s)+2s-4=20\), so \(s=4\) and \(l=8\).

There are \(3\) equal gaps on a short side and \(7\) equal gaps on a long side. With \(4\) yards between neighboring posts, the side lengths are \(12\) and \(28\), so the area is \(12\cdot28=336\).

Thus, the correct answer is B.

乘积为奇数当且仅当两个数都是奇数。有 \(\binom 32=3\) 种奇数对，总共有 \(\binom 52=10\) 种数对。因此乘积为奇数的概率为 \(\frac 3{10} = 0.3\)，乘积为偶数的概率为 \(1-0.3=0.7\)。

所以正确答案是 D。

English

The product is odd if and only if both numbers are odd. There are \(\binom 32=3\) ways to do this out of a possible \(\binom 52=10 \) ways. This makes the probability of it being odd equal to \(\frac 3{10} = 0.3.\) This means the probability it is even is \(1-0.3=0.7.\)

Thus, the correct answer is D .

设双胞胎、三胞胎、四胞胎的组数分别为 \(w,r,q\)。婴儿总数给出 \[2w+3r+4q=1000\text{。}\] 条件给出 \(r=4q\)，且 \(w=3r=12q\)。

代入得 \(2(12q)+3(4q)+4q\) \(=40q=1000\)，所以 \(q=25\)。四胞胎中的婴儿数为 \(4q=100\)。

所以正确答案是 D。

English

Let \(w,r,q\) be the numbers of sets of twins, triplets, and quadruplets. Then \[2w+3r+4q=1000.\] The statement gives \(r=4q\) and \(w=3r=12q\).

Substituting gives \(2(12q)+3(4q)+4q\) \(=40q=1000\), so \(q=25\). The number of babies in sets of quadruplets is \(4q=100\).

Thus, the correct answer is D.

正方形必须在 \(y=-0.1\) 上方、\(x=5.1\) 左方、且在 \(y=\pi x\) 下方。因为 \(\pi\) 略大于 \(3\)，在 \(x=1,2,3,4\) 处可用的整数高度分别是 \(3,6,9,12\)，边长大于 \(3\) 的正方形不能完全放入。

按左上顶点和边长计数。边长为 \(1\) 时有 \(3+6+9+12=30\) 个；边长为 \(2\) 时有 \(2+5+8=15\) 个；边长为 \(3\) 时有 \(1+4=5\) 个。

总数为 \(30+15+5=50\)。

所以正确答案是 D。

English

A square must lie above \(y=-0.1\), to the left of \(x=5.1\), and below \(y=\pi x\). Since \(\pi\) is a little more than \(3\), the lattice heights available at \(x=1,2,3,4\) are \(3,6,9,12\), and side lengths larger than \(3\) cannot fit.

Count by side length using the top-left lattice point. For side length \(1\), there are \(3+6+9+12=30\) choices. For side length \(2\), there are \(2+5+8=15\) choices. For side length \(3\), there are \(1+4=5\) choices.

The total is \(30+15+5=50\).

Thus, the correct answer is D.

把方阵像棋盘一样染色，使中心和四个角是一种颜色，四个边格是另一种颜色。相邻整数占据相邻格，所以路径 \(1,2,\ldots,9\) 的颜色交替。因此五个奇数全在一个颜色类中，四个偶数全在另一个颜色类中。

由中心和四个角组成的颜色类有五个格子，所以其中放的是五个奇数。因此四角与中心之和为 \[1+3+5+7+9=25\text{。}\]

由于四个角的和为 \(18\)，中心的数为 \(25-18=7\)。

所以正确答案是 C。

English

Color the array like a checkerboard, with the center and four corners one color and the four edge squares the other color. Consecutive numbers occupy adjacent squares, so the path \(1,2,\ldots,9\) alternates colors. Therefore all five odd numbers occupy one color class and all four even numbers occupy the other.

The color class consisting of the center and four corners has five squares, so it contains the five odd numbers. Hence the sum of the corners and center is \[1+3+5+7+9=25.\]

Since the corners have a sum of \(18,\) the center has a value of \(25-18=7.\)

Thus, the correct answer is C .

设首项为 \(a\)，公比为 \(r\)。由第二项为一可得 \[\begin{aligned}S&=\dfrac{a}{1-r}\\ &= \dfrac{ar}{r(1-r)} \\&= \dfrac{1}{r(1-r)}\end{aligned}\text{。}\] 因此要使级数和最小，就要找使下式最大的 \(r\)：\[r(1-r)= 0.25-(r-0.5)^2\text{。}\] 这个最大值在 \(r=0.5\) 时取到。

所以 \(S = \dfrac{1}{0.5(0.5)}= 4\)。

所以正确答案是 E。

English

Let the first value of the series be \(a,\) and let the ratio be \(r.\) Thus, \[\begin{aligned}S&=\dfrac{a}{1-r}\\ &= \dfrac{ar}{r(1-r)} \\&= \dfrac{1}{r(1-r)}.\end{aligned}\] This means we have to find \(r\) that maximizes \[r(1-r)= 0.25-(r-0.5)^2.\] This maximization will happen with \(r=0.5.\)

Therefore, \(S = \dfrac{1}{0.5(0.5)}= 4.\)

Thus, the correct answer is E .

把三对相对面上的数记为 \((a_1,a_2)\)、\((b_1,b_2)\)、\((c_1,c_2)\)。每个顶点的乘积从每一对中各取一个数，所以八个顶点乘积之和为 \[(a_1+a_2)(b_1+b_2)(c_1+c_2)\text{。}\]

六个面上数字之和为 \(2+3+4+5+6+7=27\)，所以三对相对面之和的总和为 \(27\)。它们的乘积在尽量相等时最大，即 \(9,9,9\)，最大为 \(9^3=729\)。

这个最大值可以通过把 \(2\) 与 \(7\)、\(3\) 与 \(6\)、\(4\) 与 \(5\) 配成相对面来达到。因此最大可能和为 \(729\)。

所以正确答案是 D。

English

Pair opposite faces as \((a_1,a_2)\), \((b_1,b_2)\), and \((c_1,c_2)\). Each vertex product uses one number from each pair, so the sum of all eight vertex products is \[(a_1+a_2)(b_1+b_2)(c_1+c_2).\]

The six face labels sum to \(2+3+4+5+6+7=27\), so the three opposite-pair sums have total \(27\). Their product is maximized when the sums are as equal as possible, namely \(9,9,9\), giving at most \(9^3=729\).

This maximum is attainable by pairing \(2\) with \(7\), \(3\) with \(6\), and \(4\) with \(5\). Hence the greatest possible sum is \(729\).

Thus, the correct answer is D.

设这个数列的长度为 \(s\)，首项为 \(x\)。于是 \[345=s\left(x+\frac{s-1}{2}\right)\text{，}\]即 \[s(2x+s-1)=690\text{。}\]

因此 \(s\) 必须是 \(690\) 的因子，而且 \(x=\frac{\frac{690}{s}-s+1}{2}\) 要是正整数。逐一检查可能的因子长度，得到 \[s=2,3,5,6,10,15,23\text{。}\]只有这些长度才能让 \(x\) 保持为正整数。

因此共有 \(7\) 种表示。

所以正确答案是 E。

English

Suppose the sequence has length \(s\) and first term \(x\). Then \[345=s\left(x+\frac{s-1}{2}\right),\] or \[s(2x+s-1)=690.\]

Thus \(s\) must be a divisor of \(690\), with \(x=\frac{\frac{690}{s}-s+1}{2}\) a positive integer. Checking the possible divisor lengths gives \[s=2,3,5,6,10,15,23.\] These are the only lengths that keep \(x\) positive and integral.

Therefore there are \(7\) representations.

Thus, the correct answer is E.

我们有 \(AE=AB-EB=4\) 和 \(FC=DC-DF=3\)。因为 \(AE\parallel FC\)，所以三角形 \(AEP\) 与 \(CFP\) 相似。于是 \[\frac{PF}{PE}=\frac{FC}{AE}=\frac34\text{，}\]从而 \(\frac{PF}{EF}=\frac37\)。

延长 \(AG\)，与直线 \(CD\) 交于 \(X\)。因为 \(BG=3\)，由相似得 \[\frac{AD}{DX}=\frac{BG}{AB},\qquad \frac4{DX}=\frac35\text{，}\]于是 \(DX=\frac{20}{3}\)。因此 \(FX=DX-DF=\frac{14}{3}\)。又因为 \(AE\parallel FX\)，三角形 \(AEQ\) 与 \(XFQ\) 相似，所以 \[\frac{QF}{QE}=\frac{FX}{AE}=\frac76\text{。}\]从而 \(\frac{QF}{EF}=\frac7{13}\)。

因此 \[ \begin{aligned} \frac{PQ}{EF}&=\frac{QF}{EF}-\frac{PF}{EF} \\ &=\frac7{13}-\frac37=\frac{10}{91} \end{aligned}\text{。}\]

所以正确答案是 D。

English

We have \(AE=AB-EB=4\) and \(FC=DC-DF=3.\) Since \(AE\parallel FC,\) the triangles \(AEP\) and \(CFP\) are similar. Therefore \[\frac{PF}{PE}=\frac{FC}{AE}=\frac34,\] so \(\frac{PF}{EF}=\frac37.\)

Extend \(AG\) to meet line \(CD\) at \(X.\) Because \(BG=3,\) similarity gives \[\frac{AD}{DX}=\frac{BG}{AB},\qquad \frac4{DX}=\frac35,\] and hence \(DX=\frac{20}{3}\). Thus \(FX=DX-DF=\frac{14}{3}.\) Since \(AE\parallel FX,\) triangles \(AEQ\) and \(XFQ\) are similar, so \[\frac{QF}{QE}=\frac{FX}{AE}=\frac76.\] Consequently \(\frac{QF}{EF}=\frac7{13}.\)

It follows that \[ \begin{aligned} \frac{PQ}{EF}&=\frac{QF}{EF}-\frac{PF}{EF} \\ &=\frac7{13}-\frac37=\frac{10}{91}. \end{aligned} \]

Thus, the correct answer is D .

半径从 \(2\) 变为 \(3\)，所以位似比例因子为 \(\frac32\)。位似中心 \(C\) 在 \(A(2,2)\) 与 \(A'(5,6)\) 所在的直线上。

从 \(A\) 到 \(A'\) 的向量为 \((3,4)\)。若 \(C\) 是位似中心，则由比例因子可得 \[A'-A=\frac12(A-C)\text{。}\]于是 \(A-C=2(3,4)\)，所以 \(C=(2,2)-2(3,4)=(-4,-6)\)。

以 \(C\) 为中心作比例因子为 \(\frac32\) 的位似时，一个点移动的距离是它到 \(C\) 距离的一半。由于 \(CO=\sqrt{(-4)^2+(-6)^2}=\sqrt{52}\)，原点移动的距离为 \(\frac12\sqrt{52}=\sqrt{13}\)。

所以正确答案是 C。

English

The dilation scale factor is \(\frac32\), since the radius changes from \(2\) to \(3\). The center of dilation \(C\) lies on the line through \(A(2,2)\) and \(A'(5,6)\).

The vector from \(A\) to \(A'\) is \((3,4)\). If \(C\) is the dilation center, then the scale factor gives \[A'-A=\frac12(A-C).\] Thus \(A-C=2(3,4)\), so \(C=(2,2)-2(3,4)=(-4,-6).\)

Under a scale factor \(\frac32\) dilation about \(C\), a point moves by half its distance from \(C\). Since \(CO=\sqrt{(-4)^2+(-6)^2}=\sqrt{52}\), the origin moves \(\frac12\sqrt{52}=\sqrt{13}\).

Thus, the correct answer is C.

这个方程在四个象限中都是对称的。在第一象限里它化为 \[x^2+y^2=x+y\text{，}\]即 \[(x-\tfrac12)^2+(y-\tfrac12)^2=\tfrac12\text{。}\]

在第一象限中，围成的区域是直线 \(x+y=1\) 下方的三角形，面积为 \(\frac12\)，再加上一个半径为 \(\sqrt{\frac{1}{2}}\) 的半圆，面积为 \(\frac\pi4\)。

乘以 \(4\)，总面积为 \[4\left(\frac12+\frac\pi4\right)=2+\pi\text{。}\]

所以正确答案是 B。

English

The equation is symmetric in all four quadrants. In the first quadrant it becomes \[x^2+y^2=x+y,\] or \[(x-\tfrac12)^2+(y-\tfrac12)^2=\tfrac12.\]

In the first quadrant, the enclosed region is the triangle under \(x+y=1\), with area \(\frac12\), plus a semicircle of radius \(\sqrt{\frac{1}{2}}\), with area \(\frac\pi4\).

Multiplying by \(4\), the total area is \[4\left(\frac12+\frac\pi4\right)=2+\pi.\]

Thus, the correct answer is B.

队伍总数为 \(10+10+1=21\)。因此三队集合总数为 \(\binom{21}{3} = 1330\)。

没有循环胜负的三队集合中，有一支队伍击败另外两支。选择这支队伍有 \(21\) 种方法，再从它击败的十支队伍中选两支，有 \(\binom{10}{2} = 45\) 种方法。因此非循环集合有 \(21\cdot 45=945\) 个，循环集合数为 \(1330-945=385\)。

所以正确答案是 A。

English

The total number of teams is \(10+10+1=21.\) The total number of sets is therefore \(\binom{21}{3} = 1330.\)

Now, we must subtract the total number of sets such that there is no cycle. This only happens if one team beats the other two teams. There are \(21\) choices for the team that beat the other two and \(\binom{10}{2} = 45\) ways to choose the teams they beat. Thus, the total of non-cycles is \(21\cdot 45=945.\) This means the total number of cycles is \(1330-945=385.\)

Thus, the correct answer is A .

延长 \(\overline{FE}\) 和 \(\overline{CD}\)，直到它们相交于点 \(P\)。

设 \(d\) 为 \(ZW\) 与 \(FC\) 之间的距离。因为题中的四条直线彼此等距，且 \(FC\) 恰好位于 \(ZW\) 与 \(YX\) 的正中间，所以 \(ED\) 到 \(ZW\) 的距离是 \(2d\)。设从 \(P\) 到 \(ED\) 的高为 \(h\)。等边三角形 \(PED\) 与 \(PFC\) 的边长之比为 \(1:2\)，所以它们的高满足 \[h+3d=2h\text{，}\]从而 \(h=3d\)。因此 \(PZW\) 与 \(PED\) 的边长之比为 \[\frac{h+2d}{h}=\frac53\text{。}\]

取 \([PED]=1\)，由相似可得 \([PZW]=\frac{25}{9}\) 和 \([PFC]=4\)。于是 \[ \begin{aligned} [ZWCF]&=4-\frac{25}{9}=\frac{11}{9}, \\ [EDCF]&=4-1=3 \end{aligned}\text{。}\]所求的六边形由两块全等的 \(ZWCF\) 拼成，而正六边形由两块全等的 \(EDCF\) 拼成。因此所求的比值是 \(\frac{\frac{11}{9}}{3}=\frac{11}{27}\)。

所以正确答案是 C。

English

Extend \(\overline{FE}\) and \(\overline{CD}\) until they meet at \(P.\)

Let \(d\) be the distance between \(ZW\) and \(FC.\) Because the four given lines are equally spaced and \(FC\) lies halfway between \(ZW\) and \(YX,\) the distance from \(ED\) to \(ZW\) is \(2d.\) Let the altitude from \(P\) to \(ED\) be \(h.\) The equilateral triangles \(PED\) and \(PFC\) have side lengths in the ratio \(1:2,\) so their altitudes satisfy \[h+3d=2h,\] giving \(h=3d.\) Therefore the side-length ratio of \(PZW\) to \(PED\) is \[\frac{h+2d}{h}=\frac53.\]

Taking \([PED]=1,\) similarity gives \([PZW]=\frac{25}{9}\) and \([PFC]=4.\) Hence \[ \begin{aligned} [ZWCF]&=4-\frac{25}{9}=\frac{11}{9}, \\ [EDCF]&=4-1=3. \end{aligned} \] The desired hexagon consists of two congruent copies of \(ZWCF,\) while the regular hexagon consists of two congruent copies of \(EDCF.\) Thus the requested ratio is \(\frac{\frac{11}{9}}{3}=\frac{11}{27}.\)

Thus, the correct answer is C .

由 \(ab<bc<cd\)，可得 \(a\le b\le c\)。等差数列条件为 \[ \begin{aligned} 2(10b+c)&=(10a+b) \\ &\quad +(10c+d) \end{aligned}\text{，}\]整理得 \[10(a-2b+c)=-b+2c-d\text{。}\]右边是 \(10\) 的倍数。因为 \(b\le c\)，四个符号都是数字，且 \(b\ge1\)，所以右边介于 \(-9\) 和 \(17\) 之间。因此它只能是 \(0\) 或 \(10\)。

情况 \(1\)： \[ \begin{aligned} -b+2c-d&=10, \\ a&=2b-c+1 \end{aligned}\text{。}\]

下面逐一考察 \(c\) 的可能值。

\(c=6\)：\(b+d=2, 2b-5=a\)。由第一个方程有 \(b\leq 2\)，但这无法满足第二个方程。

\(c=7\)：\(b+d=4, 2b-6=a\)。由第一个方程有 \(b\leq 4\)，由第二个方程有 \(b > 3\)，所以只有 \(b=4\) 这一种情况。

\(c=8\)：\(b+d=6, 2b-7=a\)。由第一个方程有 \(b\leq 6\)，由第二个方程有 \(b > 3\)，所以 \(b=4,5,6\)。给出三种情况。

\(c=9\)：\(b+d=8, 2b-8=a\)。由第一个方程有 \(b\leq 8\)，由第二个方程有 \(b > 4\)，所以 \(b=5,6,7,8\)。给出四种情况。本情况的解为 \[ \begin{gathered} 2470,1482,3581,5680, \\ 2593,4692,6791,8890 \end{gathered}\text{，}\]共 \(8\) 个。

情况 \(2\)： \[-b+2c-d=0\text{，}\]\[2b-a-c=0\text{，}\]这表示四个数字构成等差数列。

若公差为 \(1\)，则 \(1 \leq a \leq 6\)，给出 \(6\) 个解。

若公差为 \(2\)，则 \(1 \leq a \leq 3\)，给出 \(3\) 个解。若公差至少为 \(3\)，就会迫使 \(d=a+3r>9\)。因此本情况给出 \(9\) 个解。

总解数为 \(8+9 = 17\)。

所以正确答案是 D。

English

From \(ab<bc<cd\), we have \(a\le b\le c.\) The arithmetic-sequence condition is \[ \begin{aligned} 2(10b+c)&=(10a+b) \\ &\quad +(10c+d), \end{aligned} \] which rearranges to \[10(a-2b+c)=-b+2c-d.\] The right side is a multiple of \(10.\) Because \(b\le c\) and all four symbols are digits with \(b\ge1,\) it lies between \(-9\) and \(17\). Hence it is either \(0\) or \(10.\)

Case \(1:\) \[ \begin{aligned} -b+2c-d&=10, \\ a&=2b-c+1. \end{aligned} \]

We can look at the possible values of \(c.\)

\(c=6:\) \(b+d=2, 2b-5=a.\) Thus, \(b\leq 2\) from the first equation, but can’t work for the second equation.

\(c=7:\) \(b+d=4, 2b-6=a.\) Thus, \(b\leq 4\) from the first equation, and \(b > 3\) from the second equation. This makes one case for \(b=4.\)

\(c=8:\) \(b+d=6, 2b-7=a.\) Thus, \(b\leq 6\) from the first equation, and \(b > 3\) from the second equation. This makes three cases for \(b=4,5,6.\)

\(c=9:\) \(b+d=8, 2b-8=a.\) Thus, \(b\leq 8\) from the first equation, and \(b > 4\) from the second equation. This makes four cases for \(b=5,6,7,8.\) Altogether this case gives \[ \begin{gathered} 2470,1482,3581,5680, \\ 2593,4692,6791,8890, \end{gathered} \] for \(8\) solutions.

Case \(2:\) \[-b+2c-d=0,\]\[2b-a-c=0,\] which means the digits are an arithmetic sequence.

If the difference is \(1,\) then \(1 \leq a \leq 6\) makes \(6\) solutions.

If the difference is \(2,\) then \(1 \leq a \leq 3\) makes \(3\) solutions. A difference of at least \(3\) would force \(d=a+3r>9.\) This case therefore gives \(9\) solutions.

In total, the number of solutions is \(8+9 = 17.\)

Thus, the correct answer is D .

写 \(x=\lfloor x\rfloor+t\)，其中 \(0\le t\lt1\)。则 \[\lfloor kx\rfloor-k\lfloor x\rfloor=\lfloor kt\rfloor\text{，}\] 所以 \(f(x)\) 只取决于小数部分 \(t\)。

\(f\) 的值只会在 \(t\) 跨过某个分数 \(\frac{i}{k}\) 时改变，其中 \(2\le k\le10\)，\(1\le i\lt k\)。在 \((0,1)\) 内这些不同分数的个数是 \[ \begin{aligned} &\varphi(2)+\varphi(3) \\ &\quad {}+\cdots+\varphi(10) \\ &=1+2+2+4+2 \\ &\quad {}+6+4+6+4 \\ &=31 \end{aligned}\text{。}\]

加上第一个断点之前的初始取值，\(f\) 共有 \(31+1=32\) 个不同的值。

所以正确答案是 A。

English

Write \(x=\lfloor x\rfloor+t\), where \(0\le t\lt1\). Then \[\lfloor kx\rfloor-k\lfloor x\rfloor=\lfloor kt\rfloor,\] so \(f(x)\) depends only on the fractional part \(t\).

The value of \(f\) changes only when \(t\) crosses a fraction \(\frac{i}{k}\), where \(2\le k\le10\) and \(1\le i\lt k\). The number of distinct such fractions in \((0,1)\) is \[ \begin{aligned} &\varphi(2)+\varphi(3) \\ &\quad {}+\cdots+\varphi(10) \\ &=1+2+2+4+2 \\ &\quad {}+6+4+6+4 \\ &=31. \end{aligned} \]

Including the initial value before the first breakpoint, \(f\) assumes \(31+1=32\) distinct values.

Thus, the correct answer is A.