2019B AMC 10 真题（P版）

1 答案 D

设第一个和第二个容器的体积分别为 \(F\) 和 \(S\)。水量既等于 \(\dfrac56F\)，也等于 \(\dfrac34S\)，所以 \(\dfrac56F=\dfrac34S\)。

因此 \(\dfrac{F}{S}=\dfrac{\frac{3}{4}}{\frac{5}{6}}=\dfrac9{10}\)。第一个容器较小，所以所求体积比为 \(\dfrac9{10}\)。所以正确答案是 D。

English

Let the volumes of the first and second containers be \(F\) and \(S\). The amount of water is both \(\dfrac56F\) and \(\dfrac34S\), so \(\dfrac56F=\dfrac34S\).

Thus \(\dfrac{F}{S}=\dfrac{\frac{3}{4}}{\frac{5}{6}}=\dfrac9{10}\). Since the first container is smaller, the ratio of the smaller container to the larger container is \(\dfrac9{10}\). Thus, D is the correct answer.

2 答案 E

反例必须满足 \(n\) 不是质数，所以 \(n\) 的候选值只有 \(15,21,27\)。同时 \(n-2\) 也必须不是质数，只有 \(27\) 符合。

所以正确答案是 E。

English

We need \(n\) to not be prime, so \(n\) can only be \(15,21,27.\) Then, \(n-2\) must be not prime, leaving just \(27.\)

Thus, the answer is E .

3 答案 B

设高年级学生人数为 \(s\)。则非高年级学生有 \(500-s\) 人。高年级学生中 \(60\%\) 不会演奏乐器。因此，不会演奏乐器的人数可表示为 \[ \begin{aligned} &0.3(500-s)+0.6(s) \\ &= 0.3s+150 \end{aligned} \] 另一方面， \[0.468\cdot 500=234\text{。}\] 所以 \[0.3s+150=234\] \[s=280\text{。}\] 非高年级学生有 \(220\) 人，其中 \(70\%\) 会演奏乐器，因此人数为 \[220\cdot 0.7=154\text{。}\]

所以答案是 B。

English

Let the number of seniors be \(s.\) Then, \(500-s\) people aren’t seniors. We know \(60\%\) of seniors don’t play an instrument. Then, the number of students who don’t play an instrument can be represented as \[ \begin{aligned} &0.3(500-s)+0.6(s) \\ &= 0.3s+150 \end{aligned} \] and \[0.468\cdot 500=234.\] Thus, \[0.3s+150=234\] \[s=280.\] This makes the number of non-seniors equal to \(220.\) Since \(70\%\) of non-seniors play instruments, we have the total number as \[220\cdot 0.7=154.\]

Thus, the answer is B .

4 答案 A

令 \(d=b-a\)。

于是 \[(a,b,c)= (a,a+d,a+2d)\text{。}\] 因此 \[ax+(a+d)y=a+2d\text{。}\] 分别比较 \(a\) 和 \(d\) 的部分，可得 \[ax+ay=a\] 以及 \[dy=2d\] 对所有 \(a,d\) 都成立。所以 \[y=2,x+y=1\] 从而 \[x=-1\text{。}\] 公共点为 \((-1,2)\)。

所以答案是 A。

English

Let \(d=b-a.\)

Then, we have \[(a,b,c)= (a,a+d,a+2d).\] Thus, \[ax+(a+d)y=a+2d.\] If we match the parts of \(a\) and \(d,\) we get \[ax+ay=a\] and \[dy=2d\] for all \(a,d.\) Therefore, we have \[y=2,x+y=1\] implying that \[x=-1.\] This makes the pair \((-1,2).\)

Thus, the answer is A .

5 答案 E

关于 \(y=x\) 的反射把 \((x,y)\) 映为 \((y,x)\)。因此它既保持第一象限，也保持面积，所以选项 A 和 B 必为真。

从 \((x,y)\) 指向它的像 \((y,x)\) 的方向向量是 \((y-x,x-y)\)，由于 \(x\ne y\)，它的斜率为 \(-1\)。因此选项 C 和 D 也必为真。

对于选项 E，取 \(A=(2,1)\)、\(B=(3,2)\) 和 \(C=(4,1)\)。这些点满足全部条件，但 \(AB\) 和 \(A'B'\) 的斜率都是 \(1\)，所以这两条直线平行而不是垂直。

所以正确答案是 E。

English

Reflection across \(y=x\) sends \((x,y)\) to \((y,x)\). It therefore preserves the first quadrant and preserves area, so A and B are always true.

The direction from \((x,y)\) to its image \((y,x)\) is \((y-x,x-y)\), whose slope is \(-1\) because \(x\ne y\). Thus C and D are always true as well.

For E, take \(A=(2,1)\), \(B=(3,2)\), and \(C=(4,1)\). These points satisfy all the conditions, but \(AB\) and \(A'B'\) both have slope \(1\), so the two lines are parallel rather than perpendicular.

Thus, the answer is E .

6 答案 C

左边可改写为 \[(n+1)n!+(n+2)(n+1)n!\] \[=((n+2)^2-1)n!\text{，}\] 所以 \[((n+2)^2-1) n! = 440 n!\] 因此 \[(n+2)^2=441\text{，}\] 得 \(n=19\)。各位数字和为 \(10\)。

所以答案是 C。

English

We can rewrite the left side as \[(n+1)n!+(n+2)(n+1)n!\]\[=((n+2)^2-1)n!,\] so \[((n+2)^2-1) n! = 440 n!\] Therefore, \[(n+2)^2=441,\] so \(n=19.\) The sum of its digits is \(10.\)

Thus, the answer is C .

7 答案 B

设 Casper 有 \(c\) 美分。那么 \(c\) 是 \(12,14\) 和 \(15\) 的公倍数，所以它必为 \(420\) 的倍数。

令 \(c=420k\)，其中 \(k\) 为整数。又有 \(c=20n\)，所以 \(420k=20n\)，即 \(n=21k\)。因为 \(k\) 是正整数，\(n\) 的最小值是 \(21\)。

所以正确答案是 B。

English

Let the number of cents he has \(c.\) Then, \(c\) is a multiple of \(12,14,\) and \(15.\) Thus, it must be a multiple of \(420.\)

Let \(c=420k\) for some \(k.\) Also, \(c=20n,\) so \(420k=20n,\) making \(n=21k.\) Since \(k\) is a whole number, the minimum possible value of \(n\) is \(21.\)

Thus, the answer is B .

8 答案 B

每个等边三角形的边长为 \(2\)，所以它的高为 \(\sqrt3\)。这条高恰好从正方形的一条边伸到正方形中心，因此正方形的边长为 \(2\sqrt3\)，面积为 \(12\)。

四个等边三角形的面积都是 \(\frac{\sqrt3}{4}(2^2)=\sqrt3\)。因此涂色区域的面积为 \[12-4\sqrt3\text{。}\]

所以正确答案是 B。

English

Each equilateral triangle has side length \(2\), so its altitude is \(\sqrt3\). Because that altitude runs from a side of the square to its center, the square has side length \(2\sqrt3\) and area \(12\).

Each of the four equilateral triangles has area \(\frac{\sqrt3}{4}(2^2)=\sqrt3\). Therefore the shaded area is \[12-4\sqrt3.\]

Thus, the answer is B .

9 答案 A

若 \(x\ge0\)，则 \(\lfloor|x|\rfloor=\lfloor x\rfloor=|\lfloor x\rfloor|\)，所以 \(f(x)=0\)。

若 \(x\) 是负整数，则两项都等于 \(|x|\)，所以仍有 \(f(x)=0\)。

若 \(x\) 是负数且不是整数，可以写成 \(x=-k-t\)，其中 \(k\) 是非负整数，且 \(0<t<1\)。此时 \(\lfloor|x|\rfloor=k\)，而 \(|\lfloor x\rfloor|=|-k-1|=k+1\)，所以 \(f(x)=-1\)。

因此值域为 \(\{-1,0\}\)。

所以答案是 A。

English

If \(x\ge0\), then \(\lfloor|x|\rfloor=\lfloor x\rfloor=|\lfloor x\rfloor|\), so \(f(x)=0\).

If \(x\) is a negative integer, both terms equal \(|x|\), so again \(f(x)=0\).

If \(x\) is negative and not an integer, write \(x=-k-t\), where \(k\) is a nonnegative integer and \(0<t<1\). Then \(\lfloor|x|\rfloor=k\), while \(|\lfloor x\rfloor|=|-k-1|=k+1\), so \(f(x)=-1\).

Therefore, the range is \(\{-1,0\}.\)

Thus, the answer is A .

10 答案 A

面积条件确定了 \(C\) 到直线 \(AB\) 的距离。若该距离为 \(h\)，则 \[\frac{10h}{2}=100\text{，}\] 所以 \(h=20\)。因此 \(C\) 必须位于与 \(AB\) 平行且距离为 \(20\) 的直线上。

因为 \(C\) 到直线 \(AB\) 的垂直距离为 \(20\)，所以 \(AC\) 和 \(BC\) 都至少为 \(20\)。它们不能同时等于 \(20\)，否则从 \(C\) 向直线 \(AB\) 作的垂线就必须同时落在不同的点 \(A\) 和 \(B\) 上。因此 \(AC+BC>40\)，与周长条件 \(AC+BC=40\) 矛盾。所以不存在符合条件的点 \(C\)。

所以答案是 A。

English

The area condition fixes the distance from \(C\) to line \(AB.\) If that distance is \(h,\) then \[\frac{10h}{2}=100,\] so \(h=20.\) Thus \(C\) must lie on a line parallel to \(AB\) at distance \(20.\)

Since the perpendicular distance from \(C\) to line \(AB\) is \(20\), both \(AC\) and \(BC\) are at least \(20\). They cannot both equal \(20\), because that would require the perpendicular from \(C\) to meet line \(AB\) at both distinct points \(A\) and \(B\). Hence \(AC+BC>40\), contradicting the perimeter requirement \(AC+BC=40\). Therefore no point \(C\) works.

Thus, the answer is A .

11 答案 A

设绿珠数分别为 \(x\) 和 \(y\)，对应罐子为 \(1\) 和 \(2\)。那么两个罐子的弹珠总数分别为 \(10x\) 和 \(9y\)。

两个罐子的弹珠总数相同，所以 \(10x=9y\)。又有 \(x+y=95\)。解得 \(x=45\)，\(y=50\)。

罐子 \(1\) 有 \(9x=405\) 颗蓝珠，罐子 \(2\) 有 \(8y=400\) 颗蓝珠，相差 \(5\) 颗。所以正确答案是 A。

English

Let \(x\) and \(y\) be the numbers of green marbles in Jars \(1\) and \(2\), respectively. Then the total numbers of marbles in the jars are \(10x\) and \(9y\).

The jars contain the same number of marbles, so \(10x=9y\). Also \(x+y=95\). Solving gives \(x=45\) and \(y=50\).

Jar \(1\) has \(9x=405\) blue marbles, and Jar \(2\) has \(8y=400\) blue marbles. The difference is \(5\). Thus, A is the correct answer.

12 答案 C

首先，\(2019=5613_7\)。任何首位不超过 \(4\) 的数，其数位和至多为 \(4+6+6+6=22\)，而 \(4666_7<5613_7\) 达到了这个数位和。

若首位为 \(5\)，则第二位不超过 \(5\) 时，数位和至多为 \(5+5+6+6=22\)。若前两位是 \(56\)，与 \(5613_7\) 比较可知，末两位的数位和至多为 \(1+2\)，得到更小的总和。因此最大的数位和是 \(22\)。

所以答案是 C。

English

First, \(2019=5613_7\). Any number with leading digit at most \(4\) has digit sum at most \(4+6+6+6=22\), and \(4666_7<5613_7\) attains that sum.

If the leading digit is \(5\), then a second digit at most \(5\) gives digit sum at most \(5+5+6+6=22\). If the first two digits are \(56\), comparison with \(5613_7\) forces the last two digits to contribute at most \(1+2\), giving an even smaller sum. Therefore, the largest digit sum is \(22\).

Thus, the answer is C .

13 答案 A

平均数为 \(7+\frac{x}{5}\)。若 \(x<6\)，中位数为 \(6\)，所以 \(7+\frac{x}{5}=6\) 给出有效值 \(x=-5\)。

若 \(6\le x\le8\)，中位数为 \(x\)，但 \(7+\frac{x}{5}=x\) 给出 \(x=\frac{35}{4}\)，不在该区间内。若 \(x>8\)，中位数为 \(8\)，但 \(7+\frac{x}{5}=8\) 给出 \(x=5\)，仍不在所需范围内。

因此唯一可能的 \(x\) 是 \(-5\)，其和为 \(-5\)。

所以答案是 A。

English

The mean is \(7+\frac{x}{5}\). If \(x<6\), the median is \(6\), so \(7+\frac{x}{5}=6\) gives the valid value \(x=-5\).

If \(6\le x\le8\), the median is \(x\), but \(7+\frac{x}{5}=x\) gives \(x=\frac{35}{4}\), outside this interval. If \(x>8\), the median is \(8\), but \(7+\frac{x}{5}=8\) gives \(x=5\), again outside the required range.

Therefore, the only possible \(x\) is \(-5,\) making the sum \(-5.\)

Thus, the answer is A .

14 答案 C

因为 \(19!\) 能被 \(1000\) 整除，它的末三位都是零，所以 \(H=0\)。

因为 \(19!\) 能被 \(9\) 整除，它的数位和 \(33+T+M\) 也能被 \(9\) 整除。因此 \(T+M\equiv3\pmod9\)，所以 \(T+M\) 是 \(3\) 或 \(12\)。

能被 \(11\) 整除说明交错数位和 \(T-M-7\) 能被 \(11\) 整除，所以 \(T-M\equiv7\pmod{11}\)。检查这两个同余式下的数字可能性，得到 \(T=4\)、\(M=8\)。因此 \(T+M+H=4+8+0=12\)。

所以答案是 C。

English

Because \(19!\) is divisible by \(1000\), its last three digits are zero, so \(H=0\).

Since \(19!\) is divisible by \(9\), its digit sum \(33+T+M\) is divisible by \(9\). Hence \(T+M\equiv3\pmod9\), so \(T+M\) is either \(3\) or \(12\).

Divisibility by \(11\) says the alternating digit sum \(T-M-7\) is divisible by \(11\), so \(T-M\equiv7\pmod{11}\). Checking the digit possibilities from these two congruences gives \(T=4\) and \(M=8\). Therefore \(T+M+H=4+8+0=12\).

Thus, the answer is C .

15 答案 A

设两条共有边的长度为 \(a\le b\)。由于两个三角形的面积不同，这两条共有边在两个三角形中不可能扮演相同角色。因此面积为 \(2\) 的三角形以 \(a\) 和 \(b\) 为两条直角边，而面积为 \(1\) 的三角形有一条直角边 \(a\)，另一条直角边为 \(\sqrt{b^2-a^2}\)，斜边为 \(b\)。

两条不共有的第三边之积为 \(\sqrt{a^2+b^2}\sqrt{b^2-a^2}\)，其平方为 \(b^4-a^4\)。

利用面积，\(\dfrac{ab}{2}=2\)，且 \(\dfrac{a\sqrt{b^2-a^2}}{2}=1\)。因此 \(a^2b^2=16\)，且 \(a^2(b^2-a^2)=4\)，所以 \(a^4=12\)。于是 \(b^4=\dfrac{256}{12}=\dfrac{64}{3}\)，且 \(b^4-a^4=\dfrac{28}{3}\)。所以正确答案是 A。

English

Let the two shared side lengths be \(a\le b\). Because the triangles have different areas, the shared sides cannot play the same roles in both triangles. Thus the area-\(2\) triangle has legs \(a\) and \(b\), while the area-\(1\) triangle has leg \(a\), other leg \(\sqrt{b^2-a^2}\), and hypotenuse \(b\).

The product of the two non-shared third sides is \(\sqrt{a^2+b^2}\sqrt{b^2-a^2}\), whose square is \(b^4-a^4\).

Using the areas, \(\dfrac{ab}{2}=2\) and \(\dfrac{a\sqrt{b^2-a^2}}{2}=1\). Hence \(a^2b^2=16\) and \(a^2(b^2-a^2)=4\), so \(a^4=12\). Then \(b^4=\dfrac{256}{12}=\dfrac{64}{3}\), and \(b^4-a^4=\dfrac{28}{3}\). Thus, A is the correct answer.

16 答案 A

缩放图形，使 \(AC=CD=4\)，且 \(DE=EB=3\)。设 \(A=\angle BAC\)、\(B=\angle ABC\)，所以 \(A+B=90^\circ\)。

等腰三角形 \(ACD\) 和 \(DEB\) 给出 \(\angle CDA=A\) 和 \(\angle EDB=B\)。因为 \(DA\) 与 \(DB\) 是反向射线，\[\angle CDE=180^\circ-A-B=90^\circ\text{。}\]因此 \(CE=\sqrt{4^2+3^2}=5\)，所以 \(BC=CE+EB=8\)，且 \(\tan A=\frac{BC}{AC}=2\)。

两个等腰三角形的底边长分别为 \(AD=8\cos A\) 和 \(BD=6\cos B=6\sin A\)。所以 \[\frac{AD}{BD}=\frac{8\cos A}{6\sin A}=\frac{4}{3\tan A}=\frac23\text{。}\]

所以答案是 A。

English

Scale the figure so that \(AC=CD=4\) and \(DE=EB=3\). Let \(A=\angle BAC\) and \(B=\angle ABC\), so \(A+B=90^\circ\).

The isosceles triangles \(ACD\) and \(DEB\) give \(\angle CDA=A\) and \(\angle EDB=B\). Because \(DA\) and \(DB\) are opposite rays, \[\angle CDE=180^\circ-A-B=90^\circ.\] Therefore \(CE=\sqrt{4^2+3^2}=5\), so \(BC=CE+EB=8\) and \(\tan A=\frac{BC}{AC}=2\).

The bases of the two isosceles triangles have lengths \(AD=8\cos A\) and \(BD=6\cos B=6\sin A\). Hence \[\frac{AD}{BD}=\frac{8\cos A}{6\sin A}=\frac{4}{3\tan A}=\frac23.\]

Thus, the answer is A .

17 答案 C

已知两个球落在不同箱子中时，较大编号箱子中的球为红球的概率为 \(\frac 12\)。因此只需求出 \[\frac{P(\text{两球落在不同箱子})}2 \]\[= \frac{1-P(\text{两球落在同一箱子})}2\] 这个值，这里用了补集计数。

两个球都落在第 \(k\) 号箱子的概率为 \[2^{-k} \cdot 2^{-k} = 4^{-k}\text{。}\]

所以两个球落在同一个箱子的概率为 \[\sum_{k=1}^\infty 4^{-k}\text{。}\] 用等比数列求和公式，得到它等于 \[\frac 14 \cdot \dfrac{1}{1-\frac 14} = \frac 13\text{。}\]

故所求概率为 \[\frac{1-\frac 13}2 = \frac 13\text{。}\]

所以答案是 C。

English

Given that the two balls were tossed into separate bins, the probability that the ball in the higher-numbered bin is red is \(\frac 12.\) Thus we must find \[\frac{P(\text{Balls in different bins})}2 \]\[= \frac{1-P(\text{Balls in same bins})}2\] by complementary counting.

The probability that both balls are in bin \(k\) is \[2^{-k} \cdot 2^{-k} = 4^{-k}. \]

The probability that they are both in the same bin is therefore \[\sum_{k=1}^\infty 4^{-k}.\] Using the geometric sequence formula, we get this to be \[\frac 14 \cdot \dfrac{1}{1-\frac 14} = \frac 13.\]

Therefore, our answer is \[\frac{1-\frac 13}2 = \frac 13 .\]

Thus, the answer is C .

18 答案 C

设某个向家转向的极限点离家 \(x\) 千米。当 Henry 朝健身房走完这一段后，他的位置为 \[2-\frac{2-x}{4}=\frac32+\frac{x}{4}\text{。}\]

接着向家走该距离的四分之三，新位置离家为 \[\frac14\left(\frac32+\frac{x}{4}\right)=\frac38+\frac{x}{16}\text{。}\]

在极限点处，这个两步过程不改变位置，所以 \[x=\frac38+\frac{x}{16}\text{，}\] 由此得 \(x=\frac25\)。

另一个极限点就是朝健身房走完后到达的位置： \[2-\frac{2-\frac25}{4}=\frac85\text{。}\] 因此 \[|A-B|=\frac85-\frac25=\frac65=1\frac15\text{。}\]

所以正确答案是 C。

English

Suppose a limiting homeward turnaround point is \(x\) kilometers from home. After Henry walks toward the gym, his position is \[2-\frac{2-x}{4}=\frac32+\frac{x}{4}.\]

After he turns back toward home, one quarter of that distance remains, so his next homeward turnaround point is \[\frac14\left(\frac32+\frac{x}{4}\right)=\frac38+\frac{x}{16}.\]

At the limiting point this two-step map leaves the position unchanged, so \[x=\frac38+\frac{x}{16},\] which gives \(x=\frac25\).

The other limiting point is the position reached after walking toward the gym: \[2-\frac{2-\frac25}{4}=\frac85.\] Therefore \[|A-B|=\frac85-\frac25=\frac65=1\frac15.\]

Thus, the correct answer is C .

19 答案 C

首先注意到 \(100,000=2^5\cdot5^5\)。

因此，\(S\) 中任一元素都形如 \(2^a5^b\)，其中 \(0 \leq a,b \leq 5\)。

设 \(x,y \in S\) 是两个不同的元素，且 \[x = 2^a5^b\text{，}\]\[y=2^c5^d\text{。}\] 于是 \[xy = 2^{a+c}5^{b+d}\text{。}\] 因此 \[0 \leq a+c,b+d \leq 10\text{。}\] 这说明乘积的指数对至多有 \[(10+1)(10+1)=121\] 种。不过其中有些只能在 \(x=y\) 时出现，也就是 \((a,b)=(c,d)\) 的情形。

若 \(a+c=0\)，则必须有 \(a=0,c=0\)。

若 \(a+c=10\)，则必须有 \(a=5,c=5\)。

当 \(a+c\) 取其他值时，可以让 \(a \neq c\)。

对 \(b+d\) 也有类似的结构。因此，若 \[a+c,b+d \in \{0,10\}\text{，}\] 则 \[(a,b)=(c,d)\text{，}\] 从而 \(x=y\)。

因此必须排除 \(4\) 种选择，剩下 \[121-4=117\text{。}\]

所以答案是 C。

English

First, note that \(100,000=2^5\cdot5^5.\)

Therefore, any element of \(S\) must be of the form \(2^a5^b\) with \(0 \leq a,b \leq 5.\)

Suppose I have distinct \(x,y \in S\) with \[x = 2^a5^b,\]\[y=2^c5^d.\] Then, \[xy = 2^{a+c}5^{b+d}.\] Thus, \[0 \leq a+c,b+d \leq 10.\] This means that there are \[(10+1)(10+1)=121\] possible exponent pairs for a product. However, some products can arise only when \(x=y\), namely when \((a,b)=(c,d)\).

If \(a+c=0,\) then \(a=0,c=0\) must be true.

If \(a+c=10,\) then \(a=5,c=5\) must be true.

With any other value of \(a+c,\) we can have \(a \neq c.\)

Similar structure holds for \(b+d.\) Thus, if \[a+c,b+d \in \{0,10\},\] then \[(a,b)=(c,d),\] thus making \(x=y.\)

This means we have to eliminate \(4\) choices, leaving \[121-4=117.\]

Thus, the answer is C .

20 答案 E

直线 \(EG\) 经过半径为 \(2\) 的圆的圆心 \(F\)，所以阴影上半圆的面积为 \(2\pi\)。

弦 \(XZ\) 位于 \(AD\) 上，且它到 \(F\) 的距离为一个单位。因此 \[\angle XFZ=2\arccos\left(\frac12\right)=\frac{2\pi}{3}\text{。}\]\(AD\) 下方的阴影弓形面积为 \[ \begin{aligned} &\frac12(2^2)\left(\frac{2\pi}{3}\right)\\ &\quad-\frac12(2)(2)\sin\left(\frac{2\pi}{3}\right)\\ &=\frac{4\pi}{3}-\sqrt3 \end{aligned}\text{。}\]

\(EG\) 与 \(AD\) 之间的部分由四个如下的全等区域组成。

每个区域都是一个单位正方形去掉一个单位圆的四分之一，所以四个区域的总面积为 \[4\left(1-\frac\pi4\right)=4-\pi\text{。}\]

因此阴影总面积为 \[ \begin{aligned} &2\pi+\left(\frac{4\pi}{3}-\sqrt3\right)\\ &\quad+(4-\pi)\\ &=\frac{7\pi}{3}-\sqrt3+4 \end{aligned}\text{。}\]所以 \(a=7\)、\(b=3\)、\(c=3\)、\(d=4\)，从而 \(a+b+c+d=17\)。

所以答案是 E。

English

Line \(EG\) passes through the center \(F\) of the radius-\(2\) circle, so the shaded upper semicircle has area \(2\pi\).

The chord \(XZ\) lies on \(AD\), one unit from \(F\). Thus \[\angle XFZ=2\arccos\left(\frac12\right)=\frac{2\pi}{3}.\] The shaded circular segment below \(AD\) has area \[ \begin{aligned} &\frac12(2^2)\left(\frac{2\pi}{3}\right)\\ &\quad-\frac12(2)(2)\sin\left(\frac{2\pi}{3}\right)\\ &=\frac{4\pi}{3}-\sqrt3. \end{aligned} \]

The portion between \(EG\) and \(AD\) consists of four congruent pieces of the following form.

Each piece is a unit square with a quarter of a unit circle removed, so the four pieces have total area \[4\left(1-\frac\pi4\right)=4-\pi.\]

The total shaded area is therefore \[ \begin{aligned} &2\pi+\left(\frac{4\pi}{3}-\sqrt3\right)\\ &\quad+(4-\pi)\\ &=\frac{7\pi}{3}-\sqrt3+4. \end{aligned} \] Hence \(a=7\), \(b=3\), \(c=3\), and \(d=4\), giving \(a+b+c+d=17\).

Thus, the answer is E .

21 答案 B

在最后一次重复结果出现之前，序列必须交替。若以 \(H\) 开始，第二个正面必定早于第二个反面出现，所以成功序列必须以 \(T\) 开始。为了在以 \(HH\) 结束之前看到第二个反面，开头必须是 \(THT\)。

因此成功序列为 \(THTHH,THTHTHH,\ldots\)：对每个不小于 \(5\) 的奇数长度，恰有一个序列。它们的总概率为 \[ \begin{gathered} \frac1{2^5}+\frac1{2^7}+\cdots\\ =\frac1{32}\cdot\frac1{1-\frac14}\\ =\frac1{24} \end{gathered}\text{。}\]

所以答案是 B。

English

Before the final repeated flip, the sequence must alternate. If it starts with \(H\), the second head necessarily occurs before the second tail, so a successful sequence must start with \(T\). To see a second tail before ending with \(HH\), it must begin \(THT\).

Thus the successful sequences are \(THTHH,THTHTHH,\ldots\): exactly one sequence of each odd length at least \(5\). Their total probability is \[ \begin{gathered} \frac1{2^5}+\frac1{2^7}+\cdots\\ =\frac1{32}\cdot\frac1{1-\frac14}\\ =\frac1{24}. \end{gathered} \]

Thus, the answer is B .

22 答案 B

不计顺序，唯一能达到的钱数状态是 \((1,1,1)\) 和 \((2,1,0)\)。玩家不可能在一轮结束时拥有全部 \(3\) 美元：每个轮初有钱的人都必须给别人一美元，而且不能给自己。从 \((1,1,1)\) 出发，下一状态仍为 \((1,1,1)\)，当且仅当三个人沿同一个循环方向传钱，其概率为 \(2\left(\dfrac12\right)^3=\dfrac14\)。

从 \((2,1,0)\) 出发，记三人的钱数为 \(A=2,B=1,C=0\)。下一状态恰为 \((1,1,1)\)，当且仅当 \(A\) 把钱给 \(B\)，且 \(B\) 把钱给 \(C\)，这是四个等可能选择组合中的一个。因此概率同样是 \(\dfrac14\)。

所以无论第 \(2018\) 次铃响后的状态如何，下一次铃响后状态为 \((1,1,1)\) 的概率都是 \(\dfrac14\)。所以正确答案是 B。

English

The only reachable money configurations up to order are \((1,1,1)\) and \((2,1,0)\). A player cannot finish a round with all \(3\) dollars: anyone who begins with money must give a dollar to someone else, and no one can give to themselves. From \((1,1,1)\), the next state is again \((1,1,1)\) exactly when all three players pass dollars in the same cyclic direction, which has probability \(2\left(\dfrac12\right)^3=\dfrac14\).

From \((2,1,0)\), label the players’ holdings \(A=2,B=1,C=0\). The next state is \((1,1,1)\) exactly when \(A\) gives to \(B\) and \(B\) gives to \(C\), one of the four equally likely pairs of choices. This also has probability \(\dfrac14\).

Therefore, regardless of the state after \(2018\) rings, the probability that the state after the next ring is \((1,1,1)\) is \(\dfrac14\). Thus, B is the correct answer.

23 答案 C

设两条切线交于 \(P\)。从同一点引出的两条切线长度相等，所以 \(PA=PB\)，因此 \(P\) 在 \(\overline{AB}\) 的垂直平分线上。

\(A(6,13)\) 与 \(B(12,11)\) 的中点为 \((9,12)\)，\(AB\) 的斜率为 \(-\dfrac13\)，所以垂直平分线为 \(y=3x-15\)。它与 \(x\) 轴交于 \(P=(5,0)\)。

经过 \(P\) 和 \(A\) 的切线斜率为 \(13\)，所以通向 \(A\) 的半径斜率为 \(-\dfrac1{13}\)。联立 \(y-13=-\dfrac1{13}(x-6)\) 与 \(y=3x-15\)，得到圆心 \(\left(\dfrac{37}{4},\dfrac{51}{4}\right)\)。

因此 \(r^2=\left(\dfrac{37}{4}-6\right)^2\) \(+\left(\dfrac{51}{4}-13\right)^2\) \(=\dfrac{85}{8}\)，所以圆面积为 \(\dfrac{85\pi}{8}\)。正确答案是 C。

English

Let \(P\) be the intersection point of the two tangents. Since tangent lengths from the same point are equal, \(PA=PB\), so \(P\) lies on the perpendicular bisector of \(\overline{AB}\).

The midpoint of \(A(6,13)\) and \(B(12,11)\) is \((9,12)\), and the slope of \(AB\) is \(-\dfrac13\), so the perpendicular bisector is \(y=3x-15\). Its intersection with the \(x\)-axis is \(P=(5,0)\).

The tangent line through \(P\) and \(A\) has slope \(13\), so the radius to \(A\) has slope \(-\dfrac1{13}\). Intersecting \(y-13=-\dfrac1{13}(x-6)\) with \(y=3x-15\) gives center \(\left(\dfrac{37}{4},\dfrac{51}{4}\right)\).

Thus \(r^2=\left(\dfrac{37}{4}-6\right)^2\) \(+\left(\dfrac{51}{4}-13\right)^2\) \(=\dfrac{85}{8}\), so the area is \(\dfrac{85\pi}{8}\). Thus, C is the correct answer.

24 答案 C

令 \(a_n=x_n-4\)，则 \(a_0=1\)。化简递推式可得 \[a_{n+1}=\frac{a_n(a_n+9)}{a_n+10}\text{。}\] 只要 \(0<a_n\le1\)，就有 \[\frac9{10}a_n\le a_{n+1}\le\frac{10}{11}a_n\text{。}\]

由归纳法，\(\left(\dfrac9{10}\right)^n\le a_n\le\left(\dfrac{10}{11}\right)^n\)。当 \(n=80\) 时，\(\left(\dfrac{10}{9}\right)^{80}<2^{20}\)，所以 \(\left(\dfrac9{10}\right)^{80}>2^{-20}\)，从而 \(m>80\)。

另一方面，\(\left(\dfrac{11}{10}\right)^8>2\)，所以 \(\left(\dfrac{10}{11}\right)^{160}<2^{-20}\)，从而 \(m\le160\)。于是 \(81\le m\le160\)，所以 \(m\) 位于 \([81,242]\)。正确答案是 C。

English

Let \(a_n=x_n-4\). Then \(a_0=1\), and simplifying the recurrence gives \[a_{n+1}=\frac{a_n(a_n+9)}{a_n+10}.\] As long as \(0<a_n\le1\), this implies \[\frac9{10}a_n\le a_{n+1}\le\frac{10}{11}a_n.\]

By induction, \(\left(\dfrac9{10}\right)^n\le a_n\le\left(\dfrac{10}{11}\right)^n\). For \(n=80\), \(\left(\dfrac{10}{9}\right)^{80}<2^{20}\), so \(\left(\dfrac9{10}\right)^{80}>2^{-20}\), and therefore \(m>80\).

Also \(\left(\dfrac{11}{10}\right)^8>2\), so \(\left(\dfrac{10}{11}\right)^{160}<2^{-20}\), which gives \(m\le160\). Hence \(81\le m\le160\), so \(m\) lies in \([81,242]\). Thus, C is the correct answer.

25 答案 C

序列先以一个 \(0\) 开头，之后由若干个 \(110\) 块和 \(10\) 块依次组成，每个块都接在一个 \(0\) 后面。

设 \(110\) 块有 \(x\) 个，\(10\) 块有 \(y\) 个。于是序列的项数为 \[3x+2y+1=19\text{，}\] 即 \[3x+2y=18\text{。}\] 可能的有序对为 \[ \begin{aligned} (x,y)&=(6,0),(4,3),\\ &\quad(2,6),(0,9) \end{aligned}\text{。}\] 于是排列这 \(x+y\) 个块的方法数为 \[\binom{x+y}x\text{，}\] 因为只需选定 \(x+y\) 个位置中哪 \(x\) 个放 \(110\) 块。

因此总数为 \[\binom 66 + \binom 74 + \binom 82 + \binom 90\] \[=1+35+28+1\] \[=65\text{。}\]

所以正确答案是 C。

English

Our sequence starts with a \(0\) then has sequences of \(110\) and \(10\) in some order, where they each come after a \(0.\)

Let the number of \(110\) be \(x\) and let the number of \(10\) be \(y.\) Then the number of terms in the sequence is \[3x+2y+1=19,\] making \[3x+2y=18.\] The possible ordered pairs are \[ \begin{aligned} (x,y)&=(6,0),(4,3),\\ &\quad(2,6),(0,9). \end{aligned} \] Then, the number of ways to order the \(x+y\) blocks is \[\binom{x+y}x,\] since we choose which \(x\) of the \(x+y\) positions hold a \(110\) block.

Therefore, the total number of ways is \[\binom 66 + \binom 74 + \binom 82 + \binom 90\]\[=1+35+28+1\]\[=65.\]

Thus, the answer is C .