2022B AMC 10 真题（P版）

1 答案 A

从最内层的运算开始，依次向外计算可得 \[ \begin{aligned} 1\mathbin\diamondsuit(2\mathbin\diamondsuit3)&=|1-|2-3||=0,\\ (1\mathbin\diamondsuit2)\mathbin\diamondsuit3&=||1-2|-3|=2 \end{aligned}\text{。}\]因此，题中所求的差为 \(0-2=-2\)。

所以正确答案是 A。

English

Working from the innermost operations outward gives \[ \begin{aligned} 1\mathbin\diamondsuit(2\mathbin\diamondsuit3)&=|1-|2-3||=0,\\ (1\mathbin\diamondsuit2)\mathbin\diamondsuit3&=||1-2|-3|=2. \end{aligned} \] Therefore, the given difference is \(0-2=-2.\)

Thus, the answer is A .

2 答案 D

因为 \(ABCD\) 是菱形，所以 \[AB=AD=AP+PD=5\text{。}\]在直角三角形 \(ABP\) 中，\[\begin{aligned}BP&=\sqrt{AB^2-AP^2}\\&=\sqrt{25-9}=4\end{aligned}\text{。}\]因此，菱形的底 \(AD=5\)，高 \(BP=4\)，面积为 \(5\cdot4=20\)。

所以正确答案是 D。

English

Since \(ABCD\) is a rhombus, \[AB=AD=AP+PD=5.\] Right triangle \(ABP\) then gives \[\begin{aligned}BP&=\sqrt{AB^2-AP^2}\\&=\sqrt{25-9}=4.\end{aligned}\] Thus the rhombus has base \(AD=5\) and height \(BP=4,\) so its area is \(5\cdot4=20.\)

Thus, the answer is D .

3 答案 D

百位和十位共有 \(9\cdot10=90\) 种选择。固定这两位后，个位的奇偶性必须使三个数字中偶数数字的总数为奇数。所需奇偶性的数字总有 \(5\) 个，其中偶数数字包括 \(0\)。

因此，符合条件的整数共有 \(90\cdot5=450\) 个。

所以正确答案是 D。

English

There are \(9\cdot10=90\) choices for the hundreds and tens digits. Once those two digits are fixed, the units digit must have whichever parity makes the total number of even digits odd. There are always \(5\) digits of the required parity (including \(0\) among the even digits).

Therefore, the number of integers is \(90\cdot5=450.\)

Thus, the answer is D .

4 答案 A

要求第 \(700\) 次打嗝，就要看它比第一次晚了 \(699\) 个间隔。

这相当于 \(699\cdot 5 = 3495\) 秒。注意 \(3495 = 60\cdot 58+15\)，所以这个时刻在第一次打嗝之后 \(58\) 分 \(15\) 秒。因此它是 \(4:58\) 再过 \(15\) 秒。

所以答案是 A。

English

Since we want to look at the \(700\)th hiccup, we need to look at time that is \(699\) hiccups after the first one.

This would be \(699\cdot 5 = 3495\) seconds. Note that \(3495 = 60\cdot 58+15,\) so the time would be \(58\) minutes and \(15\) seconds after the first hiccup. This would therefore be \(4:58\) and \(15\) seconds.

Thus, the answer is A .

5 答案 B

设题中正数表达式为 \(E\)。将它平方，并利用 \(1-\frac{1}{p^2}=(1-\frac{1}{p})(1+\frac{1}{p})\)，可得 \[ \begin{aligned} E^2&=\prod_{p\in\{3,5,7\}}\frac{1+\frac{1}{p}}{1-\frac{1}{p}}\\ &=\frac42\cdot\frac64\cdot\frac86=4 \end{aligned}\text{。}\]因为这个表达式为正数，所以 \(E=2\)。

所以正确答案是 B。

English

Let the given positive expression be \(E.\) Squaring it and using \(1-\frac{1}{p^2}=(1-\frac{1}{p})(1+\frac{1}{p})\) gives \[ \begin{aligned} E^2&=\prod_{p\in\{3,5,7\}}\frac{1+\frac{1}{p}}{1-\frac{1}{p}}\\ &=\frac42\cdot\frac64\cdot\frac86=4. \end{aligned} \] Hence \(E=2.\)

Thus, the answer is B .

6 答案 A

我们断言这些数都不可能是质数。

第 \(n\) 项可写成 \[\sum_{k=0}^{2n} 10^k + 10^n = \sum_{k=0}^{n} 10^k + \] \[ \sum_{k=n}^{2n} 10^k =\sum_{k=0}^{n} 10^k + \sum_{k=0}^{n} 10^k \cdot 10^n \] \[= (10^n+1)(\sum_{k=0}^{n} 10^k)\text{。}\]这说明每一项都能写成两个大于 \(1\) 的整数的乘积，所以前十项中没有质数。

所以正确答案是 A。

English

We claim that none of these numbers can ever be prime.

We prove this claim by noticing that the \(n\)th number is \[\sum_{k=0}^{2n} 10^k + 10^n = \sum_{k=0}^{n} 10^k + \] \[ \sum_{k=n}^{2n} 10^k =\sum_{k=0}^{n} 10^k + \sum_{k=0}^{n} 10^k \cdot 10^n \] \[= (10^n+1)(\sum_{k=0}^{n} 10^k).\] This shows that the number can be written as the product of two numbers greater than \(1,\) so there are no primes.

Thus, the answer is A .

7 答案 B

设两个根为 \(r,s\)。展开 \((x-r)(x-s)\) 得到 \(x^2-(r+s)x+rs\)。与题中多项式比较系数，可得 \(rs=36\) 且 \(r+s=-k\)。

因此，需要找出满足 \(rs = 36\) 的不同整数 \(r\) 和 \(s\)。所有可能的因数对为 \[ \pm\{1,36\},\pm\{2,18\},\pm\{3,12\}\] 以及 \[\pm\{4,9\}\text{。}\]

每个无序因数对都会给出一个不同的 \(k\) 值，所以共有 \(8\) 个可能的 \(k\) 值。

所以正确答案是 B。

English

Let the roots be \(r,s.\) Expanding \((x-r)(x-s)\) gives \(x^2-(r+s)x+rs.\) Comparing coefficients with the given polynomial yields \(rs=36\) and \(r+s=-k.\)

Therefore, we need \(r\) and \(s\) distinct such that \(rs = 36.\) All the possible factor pairs are \[ \pm\{1,36\},\pm\{2,18\},\pm\{3,12\}\] and \[\pm\{4,9\}. \]

Each of these unordered pairs produces a unique value for \(k,\) so there are \(8\) possible values for \(k.\)

Thus, B is the correct answer.

8 答案 B

一个由十个连续整数组成的集合恰好含有两个 \(7\) 的倍数，当且仅当其中第一个 \(7\) 的倍数位于前三个位置之一。因此，这个倍数的个位数字必须是 \(1\)、\(2\) 或 \(3\)。

这些个位数字所对应的 \(7\) 的倍数分别为 \[\begin{gathered}21+70j,\\42+70j,\\63+70j\end{gathered}\text{。}\]对每个表达式，\(j=0,1,\ldots,13\) 都会给出不超过 \(1000\) 的数，因此每一类对应 \(14\) 个集合。总数为 \(3\cdot14=42\)。

所以正确答案是 B。

English

A block of ten consecutive integers contains exactly two multiples of \(7\) precisely when its first multiple of \(7\) is in one of the first three positions. Thus that multiple must have units digit \(1,\) \(2,\) or \(3.\)

The multiples of \(7\) with those units digits are, respectively, \[\begin{gathered}21+70j,\\42+70j,\\63+70j.\end{gathered}\] For each expression, \(j=0,1,\ldots,13\) gives a value at most \(1000,\) so each class contributes \(14\) blocks. Therefore, the total is \(3\cdot14=42.\)

Thus, the answer is B .

9 答案 D

每一项都可以裂项相消，因为 \[\frac{k}{(k+1)!}=\frac{1}{k!}-\frac{1}{(k+1)!}\text{。}\]求和后，中间所有阶乘的倒数都相消，只剩下 \[1-\frac1{2022!}\text{。}\]因此，\(a=1, b=2022\)，所以 \(a+b=2023\)。

所以正确答案是 D。

English

Each term telescopes because \[\frac{k}{(k+1)!}=\frac{1}{k!}-\frac{1}{(k+1)!}.\] Summing makes every intermediate factorial reciprocal cancel, leaving \[1-\frac1{2022!}.\] Hence \(a=1, b=2022,\) and \(a+b=2023.\)

Thus, our answer is D .

10 答案 D

设五个数按从小到大为 \(a,b,c,d,e\)。中位数为 \(c\)，众数为 \(c+2\)。

因为众数大于中位数且唯一，最后两个数必须都是 \(c+2\)，所以数列为 \(a,b,c,c+2,c+2\)。

平均数为 \(c-2\)，所以 \[ \begin{aligned} &\frac{a+b+c+(c+2)+(c+2)}{5} \\ &\quad = c-2 \end{aligned}\text{。}\]化简得 \(a+b+3c+4=5c-10\)，所以 \(a+b=2c-14\)。

为了让众数唯一，\(a\) 和 \(b\) 必须是不同的正整数，且都小于 \(c\)。由于 \(a+b=2c-14\) 为偶数，最小可行和为 \(1+3=4\)，所以 \(2c-14\ge4\)，即 \(c\ge9\)。

因此最小可能众数为 \(c+2=11\)，并且 \(1,3,9,11,11\) 确实可行。

所以正确答案是 D。

English

Let the integers in increasing order be \(a,b,c,d,e.\) The median is \(c,\) and the unique mode is \(c+2.\)

Because the mode is larger than the median and is unique, the last two entries must both be \(c+2,\) so the list is \(a,b,c,c+2,c+2.\)

The mean is \(c-2,\) so \[ \begin{aligned} &\frac{a+b+c+(c+2)+(c+2)}{5} \\ &\quad = c-2. \end{aligned} \] Hence \(a+b+3c+4=5c-10,\) so \(a+b=2c-14.\)

To keep the mode unique, \(a\) and \(b\) must be distinct positive integers, both less than \(c.\) Since \(a+b=2c-14\) is even, the smallest such sum is \(1+3=4,\) so \(2c-14\ge4,\) giving \(c\ge9.\)

The smallest possible mode is therefore \(c+2=11,\) and it is attainable with \(1,3,9,11,11.\)

Thus, the answer is D .

11 答案 B

原命题可写为：如果一所学校比 Euclid HS 更大，那么它卖出的 T 恤就不比 Euclid HS 更多。其逆否命题是：如果一所学校卖出的 T 恤比 Euclid HS 更多，那么它就不比 Euclid HS 更大。这正是选项 B。选项 C 比原命题更强，因为它还排除了较大的学校卖出同样多 T 恤的情形。

所以正确答案是 B。

English

The statement says: if a school is bigger than Euclid HS, then it did not sell more T-shirts than Euclid HS. Its contrapositive is: if a school sold more T-shirts than Euclid HS, then it is not bigger than Euclid HS. This is exactly choice B . Choice C is stronger than the original statement because it rules out a bigger school selling the same number of T-shirts.

Thus, the answer is B .

12 答案 C

可以改求使一次也没有掷出点数和为 \(7\) 的概率小于 \(\dfrac 12\) 的最小 \(n\)。每次掷出点数和为 \(7\) 的概率是 \(\dfrac 16\)，所以没有掷出点数和 \(7\) 的概率是 \(\dfrac 56\)。

因此，所有投掷都没有出现点数和 \(7\) 的概率是 \(\left(\dfrac 56\right)^n\)。我们要找使 \(\left(\dfrac 56\right)^n < \dfrac 12\) 的最小 \(n\)。

当 \(n=3\) 时，概率为 \(\dfrac{125}{216}\)，大于 \(\dfrac 12\)。

当 \(n=4\) 时，概率为 \(\dfrac{625}{1296}\)，小于 \(\dfrac 12\)。因此答案是 \(4\)。

所以正确答案是 C。

English

To compute this, we can also find the least \(n\) such that the probability of not rolling a \(7\) is less than \(\dfrac 12.\) Each roll has an independent probability of \(\dfrac 16\) of getting \(7,\) so it has a \(\dfrac 56\) probability of not landing on \(7.\)

Thus, the probability of none of the rolls being \(7\) is \(\left(\dfrac 56\right)^n.\) We must find the least \(n\) such that \(\left(\dfrac 56\right)^n < \dfrac 12. \)

If \(n=3,\) then the probability is \(\dfrac{125}{216},\) which is greater than \(\dfrac 12.\)

If \(n=4,\) then the probability is \(\dfrac{625}{1296},\) which is less than \(\dfrac 12.\) This makes the answer \(4.\)

Thus, the answer is C .

13 答案 E

由于两个质数相差 \(2\)，可以将它们写成 \(m-1,m+1\)，其中 \(m\) 是它们的平均数。

于是 \[(m+1)^3-(m-1)^3 = 31106\text{，}\]即 \[m^3+3m^2+3m+1\] \[-(m^3-3m^2+3m-1) \] \[= 6m^2+2 = 31106\text{。}\]

因此 \(m^2= 5184\)，所以 \(m=72\)。

因此两个质数为 \(71,73\)。大于它们的最小质数是 \(79\)，其数字和为 \(16\)。

所以答案是 E。

English

Since the primes are \(2\) away from each other, we can make them equal to \(m-1,m+1,\) where \(m\) is their average.

Then, \[(m+1)^3-(m-1)^3 = 31106 ,\] making \[m^3+3m^2+3m+1\]\[-(m^3-3m^2+3m-1) \]\[= 6m^2+2 = 31106.\]

Therefore, \(m^2= 5184,\) so \(m=72.\)

The primes are therefore \(71,73.\) The least prime greater than both of those is \(79,\) and its digit sum is \(16.\)

Thus, the answer is E .

14 答案 B

集合 \(S = \{13,14 \cdots ,25\}\) 有 \(13\) 个元素，且任意两个元素之和都大于 \(25\)，所以这个大小可以达到。

反过来，设 \(m\) 是 \(S\) 的最大元素。对 \(S\) 中每个满足 \(i<m\) 的 \(i\)，数 \(m-i\) 不能也属于 \(S\)。

因此，在小于 \(m\) 的数中，和为 \(m\) 的每一对至多选一个；若 \(m\) 为偶数，中间的数也不能选。于是小于 \(m\) 的元素至多有 \(\lfloor \dfrac {m-1}2 \rfloor\) 个，计入 \(m\) 本身后总数至多为 \(\lfloor \dfrac{m-1}2 \rfloor +1\)。

这个上界在 \(m=25\) 时最大，等于 \(13\)。

所以正确答案是 B。

English

The set \(S = \{13,14 \cdots ,25\}\) has \(13\) elements, and every pair has sum greater than \(25,\) so this size is attainable.

Conversely, let \(m\) be the maximum element of \(S.\) For every element of \(S\) satisfying \(i<m,\) the number \(i\) and the number \(m-i\) cannot both belong to \(S.\)

Thus, among the numbers below \(m,\) at most one number can be chosen from each pair with sum \(m\); if \(m\) is even, the middle number cannot be chosen either. Hence at most \(\lfloor \dfrac {m-1}2 \rfloor\) elements lie below \(m,\) and including \(m\) gives at most \(\lfloor \dfrac{m-1}2 \rfloor +1\) elements.

The maximum value of this has \(m=25,\) yielding \(13.\)

Thus, the answer is B .

15 答案 D

将第 \(n\) 项写成 \(a+2n\)，于是第一项的前一项为 \(a\)。那么 \[\begin{aligned}S_n&=\sum_{i=1}^n(a+2i)\\&=n(a+n+1)\end{aligned}\text{。}\]

因此 \[ \begin{aligned} \frac{S_{3n}}{S_n}&=\frac{3(a+3n+1)}{a+n+1}\\ &=9-\frac{6(a+1)}{a+n+1} \end{aligned}\text{。}\]

要使这个表达式与 \(n\) 无关，最后一个分式的分子 \(6(a+1)\) 必须为 \(0\)。因此 \(a=-1\)。

所以 \[S_{20}=20(-1+20+1)=400\text{。}\]

所以答案是 D。

English

Write the \(n\)th term as \(a+2n,\) so the term before the first term is \(a.\) Then \[\begin{aligned}S_n&=\sum_{i=1}^n(a+2i)\\&=n(a+n+1).\end{aligned}\]

Hence \[ \begin{aligned} \frac{S_{3n}}{S_n}&=\frac{3(a+3n+1)}{a+n+1}\\ &=9-\frac{6(a+1)}{a+n+1}. \end{aligned} \]

For this expression to be independent of \(n,\) its numerator \(6(a+1)\) in the final fraction must be \(0.\) Thus \(a=-1.\)

Therefore, \[S_{20}=20(-1+20+1)=400.\]

Thus, the answer is D .

16 答案 D

按下图标记各点：

因为 \(AB=4\)，且正方形边长 \(BC=5\)，直角三角形 \(ABC\) 给出 \(AC=3\)。又因为 \(BC\perp CE\)，且 \(A,C,D\) 共线，所以 \(\angle ABC=\angle DCE\)。直角三角形 \(ABC\) 和 \(CDE\) 的斜边相等，都是 \(BC=CE=5\)，所以两三角形全等。因此 \(CD=4, DE=3\)，且 \(EF=4-DE=1\)。

直角三角形 \(EFG\) 与 \(CDE\) 相似，所以 \[\frac{EG}{EF}=\frac{EC}{CD}=\frac54\text{。}\]因此 \(EG=\frac{5}{4}\)。阴影区域 \(BCEG\) 是梯形，两条平行边为 \(BC=5\) 和 \(EG=\frac{5}{4}\)，高为垂直边 \(CE=5\)。它的面积为 \[\frac12\left(5+\frac54\right)5=\frac{125}{8}=15\frac58\text{。}\]

所以答案是 D。

English

Label the points as shown:

Because \(AB=4\) and the square side \(BC=5,\) right triangle \(ABC\) gives \(AC=3.\) Also \(BC\perp CE\) and \(A,C,D\) are collinear, so \(\angle ABC=\angle DCE.\) The right triangles \(ABC\) and \(CDE\) have equal hypotenuses \(BC=CE=5,\) so they are congruent. Thus \(CD=4, DE=3,\) and \(EF=4-DE=1.\)

Right triangles \(EFG\) and \(CDE\) are similar, so \[\frac{EG}{EF}=\frac{EC}{CD}=\frac54.\] Hence \(EG=\frac{5}{4}.\) The shaded region \(BCEG\) is a trapezoid whose parallel sides are \(BC=5\) and \(EG=\frac{5}{4},\) and whose height is the perpendicular side \(CE=5.\) Its area is \[\frac12\left(5+\frac54\right)5=\frac{125}{8}=15\frac58.\]

Thus, the answer is D .

17 答案 C

使用事实：\(a^n-b^n\) 能被 \(a-b\) 整除。

选项 A 是 \(2^{606}-1=4^{303}-1\)，能被 \(4-1=3\) 整除。

选项 B 是 \(2^{606}+1=4^{303}-(-1)^{303}\)，能被 \(4-(-1)=5\) 整除。

选项 D 是 \(2^{607}+1\)。因为 \(2^{606}-1\) 能被 \(3\) 整除，乘以 \(2\) 得到 \(2^{607}-2\) 能被 \(3\) 整除，从而 \(2^{607}+1\) 也能被 \(3\) 整除。

选项 E 是 \(3^{607}+2^{607}=3^{607}-(-2)^{607}\)，能被 \(3-(-2)=5\) 整除。

对于选项 C，\(2^{607}-1\) 是奇数。并且 \(2^{607}\equiv2\pmod3\)、\(2^{607}\equiv3\pmod5\)、\(2^{607}\equiv2\pmod7\)，所以 \(2^{607}-1\) 不能被 \(3,5\) 或 \(7\) 整除。

所以答案是 C。

English

Use the fact that \(a^n-b^n\) is divisible by \(a-b.\)

Choice A is \(2^{606}-1=4^{303}-1,\) which is divisible by \(4-1=3.\)

Choice B is \(2^{606}+1=4^{303}-(-1)^{303},\) which is divisible by \(4-(-1)=5.\)

Choice D is \(2^{607}+1.\) Since \(2^{606}-1\) is divisible by \(3,\) multiplying by \(2\) gives \(2^{607}-2\) divisible by \(3,\) so \(2^{607}+1\) is divisible by \(3.\)

Choice E is \(3^{607}+2^{607}=3^{607}-(-2)^{607},\) which is divisible by \(3-(-2)=5.\)

For choice C, \(2^{607}-1\) is odd. Also \(2^{607}\equiv2\pmod3,\) \(2^{607}\equiv3\pmod5,\) and \(2^{607}\equiv2\pmod7,\) so \(2^{607}-1\) is not divisible by \(3,5,\) or \(7.\)

Thus, our answer is C .

18 答案 B

共有 \(2^9=512\) 个有序二进制系数矩阵。齐次方程组只有零解，当且仅当它的三个行向量线性无关，因此我们数出这些矩阵后再从总数中减去。

线性无关矩阵的三行必须是互不相同的非零向量。这样的行有 \[7\cdot6\cdot5=210\] 种有序选法。在三个互不相同的非零二进制向量中，线性相关恰好发生在一个向量是另外两个向量的普通和时；这两个加数的支撑必须非空且互不相交。

若这个和的支撑大小为 \(2\)，有 \(3\) 种方法选择它的两个坐标，而两个加数就是对应的两个单位向量。若这个和的支撑大小为 \(3\)，有 \(3\) 种方法选择一个加数所占的单个坐标，其余两个坐标组成另一个加数。因此，共有 \(3+3=6\) 组无序相关三元组，每一组又有 \(3!=6\) 种行的排列顺序。

所以线性无关矩阵有 \(210-6\cdot6=174\) 个。所求的奇异矩阵数，也就是有非零解的方程组数，为 \[512-174=338\text{。}\]

所以正确答案是 B。

English

There are \(2^9=512\) ordered binary coefficient matrices. A homogeneous system has only the zero solution exactly when its three row vectors are linearly independent, so we count those matrices and subtract.

An independent matrix must have three distinct nonzero rows. There are \[7\cdot6\cdot5=210\] ordered choices of such rows. Among three distinct nonzero binary vectors, dependence occurs exactly when one is the ordinary sum of the other two; the two summands must have disjoint nonempty supports.

If the sum has support of size \(2,\) choose its two coordinates in \(3\) ways; its summands are the two corresponding unit vectors. If the sum has support of size \(3,\) choose which one coordinate forms one summand in \(3\) ways, with the other two coordinates forming the other summand. Thus there are \(3+3=6\) unordered dependent triples, each with \(3!=6\) row orders.

Hence the number of independent matrices is \(210-6\cdot6=174.\) The desired number of singular matrices, and therefore of systems with a nonzero solution, is \[512-174=338.\]

Thus, the answer is B .

19 答案 C

先假设中心格初始已填充。它必须恰有 \(2\) 或 \(3\) 个已填充邻格才能保持填充。每个这样的邻格已经与中心格相邻，所以要在变换后消失，它不能再与其他已填充邻格相邻。检查这些两两不相邻的位置后，唯一不会同时使某个空格恰有 \(3\) 个已填充邻格的选择，是两个相对的角格。这样的配置有 \(2\) 种。

再假设中心格初始为空。它的八个邻格中必须恰有 \(3\) 个被填充。这三个格都必须在变换后消失，所以其中任何一个都不能同时与另外两个相邻。此外，除中心格以外，不能有任何空格同时与这三个填充格相邻。应用这两个条件，可得到以下四种代表性图案：

前三种图案各有 \(4\) 个不同的旋转。最后一种有 \(4\) 个旋转及其 \(4\) 个镜像，共 \(8\) 种配置。因此，中心格初始为空的情形共有 \(4+4+4+8=20\) 种，全部配置共有 \(20+2=22\) 种。

所以正确答案是 C。

English

First suppose the center is initially filled. It must have exactly \(2\) or \(3\) filled neighbors to survive. Every such neighbor already touches the center, so to disappear it cannot touch any other filled neighbor. Checking these pairwise nonadjacent positions, the only choices that do not also give some empty square exactly \(3\) filled neighbors are two opposite corners. There are \(2\) such configurations.

Now suppose the center is initially empty. Exactly \(3\) of its eight neighbors must be filled. Each of those three must disappear, so none may be adjacent to both of the others. Also, no empty square besides the center may be adjacent to all three. Applying these two tests gives the following four representative patterns:

Each of the first three patterns has \(4\) distinct rotations. The last has \(4\) rotations and their \(4\) reflected images, for \(8\) configurations. Thus the center-empty case contributes \(4+4+4+8=20,\) and the total is \(20+2=22.\)

Thus, the answer is C .

20 答案 D

延长 \(\overline{BE}\)，与直线 \(AD\) 相交于 \(G\)。因为 \(AD\parallel BC\)，所以 \(\angle GDE=\angle ECB\)，而 \(\angle GED=\angle BEC\) 是对顶角。又有 \(DE=EC\)，所以 \(\triangle GDE\cong\triangle BCE\)。因此 \(DG=BC=AD\)。

所以 \(D\) 是 \(\overline{AG}\) 的中点。以 \(D\) 为圆心、经过 \(A\) 的圆也经过 \(C\) 和 \(G\)。因为 \(AF\perp FG\)，由泰勒斯定理，\(F\) 也在这个圆上。

因为 \(DG\) 与 \(DA\) 方向相反，\[\begin{aligned}\angle GDC&=180^\circ-\angle ADC\\&=134^\circ\end{aligned}\text{。}\]所对弧为 \(GC\) 的圆周角 \(\angle GFC\) 因此为 \(67^\circ\)。最后，\(B,F,G\) 共线，所以 \[\angle BFC=180^\circ-67^\circ=113^\circ\text{。}\]

所以答案是 D。

English

Extend \(\overline{BE}\) to meet line \(AD\) at \(G.\) Because \(AD\parallel BC,\) we have \(\angle GDE=\angle ECB,\) and \(\angle GED=\angle BEC\) are vertical angles. Also \(DE=EC,\) so \(\triangle GDE\cong\triangle BCE.\) Hence \(DG=BC=AD.\)

Thus \(D\) is the midpoint of \(\overline{AG}.\) The circle centered at \(D\) through \(A\) also passes through \(C\) and \(G.\) Since \(AF\perp FG,\) Thales’ theorem places \(F\) on this circle as well.

Because \(DG\) is opposite \(DA,\) \[\begin{aligned}\angle GDC&=180^\circ-\angle ADC\\&=134^\circ.\end{aligned}\] The inscribed angle \(\angle GFC\) subtending arc \(GC\) is therefore \(67^\circ.\) Finally, \(B,F,G\) are collinear, so \[\angle BFC=180^\circ-67^\circ=113^\circ.\]

Thus, the answer is D .

21 答案 E

第一个余式条件给出 \[\begin{aligned}P(x)&=(x^2+x+1)Q(x)\\&\quad+x+2\end{aligned}\text{。}\]其中 \(Q\) 是某个多项式。模 \(x^2+1\) 时有 \(x^2\equiv-1\)，所以 \[P(x)\equiv xQ(x)+x+2\text{。}\]

若 \(Q(x)=c\) 是常数，则这个余式为 \((c+1)x+2\)，不可能等于 \(2x+1\)。所以 \(Q\) 的次数至少为 \(1\)。

现在令 \(Q(x)=ax+b\)。模 \(x^2+1\) 化简可得 \[P(x)\equiv(b+1)x+(2-a)\text{。}\]与 \(2x+1\) 比较系数，得到 \(a=b=1\)。这构造出了一个 \(3\) 次多项式，而常数情形不成立又说明最低次数为 \(3\)。

因此 \[\begin{aligned}P(x)&=(x+1)(x^2+x+1)\\&\quad+x+2\\&=x^3+2x^2+3x+3\end{aligned}\text{。}\]它的系数平方和为 \(1^2+2^2+3^2+3^2=23\)。

所以正确答案是 E。

English

The first remainder condition gives \[\begin{aligned}P(x)&=(x^2+x+1)Q(x)\\&\quad+x+2.\end{aligned}\] for some polynomial \(Q.\) Modulo \(x^2+1,\) we have \(x^2\equiv-1,\) so \[P(x)\equiv xQ(x)+x+2.\]

If \(Q(x)=c\) is constant, this remainder is \((c+1)x+2,\) which cannot equal \(2x+1.\) Thus \(Q\) must have degree at least \(1.\)

Now let \(Q(x)=ax+b.\) Reducing modulo \(x^2+1\) gives \[P(x)\equiv(b+1)x+(2-a).\] Matching this with \(2x+1\) yields \(a=b=1.\) This constructs a degree-\(3\) polynomial, and the failed constant case proves that degree \(3\) is minimal.

Therefore, \[\begin{aligned}P(x)&=(x+1)(x^2+x+1)\\&\quad+x+2\\&=x^3+2x^2+3x+3.\end{aligned}\] The sum of the squares of its coefficients is \(1^2+2^2+3^2+3^2=23.\)

Thus, the answer is E .

22 答案 E

将半径为 \(2\) 和 \(8\) 的两个同心圆分别称为内圆和外圆。设所求圆的半径为 \(r\)，圆心到原点的距离为 \(d\)。所求圆必与外圆内切，因此 \(d+r=8\)。

若所求圆与内圆外切，则 \(d-r=2\)，从而 \((r,d)=(3,5)\)。若所求圆包含内圆，则 \(r-d=2\)，从而 \((r,d)=(5,3)\)。因此，所求圆的半径必为 \(3\) 或 \(5\)。

第三个已知圆的半径为 \(\sqrt3\)，圆心为 \((5,0)\)。对 \(r\) 的两个可能值，所求圆都可以与第三个圆内切或外切，所以两圆心之间的距离为 \(r-\sqrt3\) 或 \(r+\sqrt3\)。在这四种情形中，若该距离为 \(q\)，则都有 \(|d-q|<5<d+q\)。因此，以原点为圆心、\(d\) 为半径的圆与所求圆圆心的轨迹交于关于 \(x\) 轴对称的两点，如图所示。于是，每一种半径都有 \(4\) 个所求圆。

所以总面积为 \[4(5^2\pi+3^2\pi)=136\pi\text{。}\]因此，正确答案是 E。

English

Call the concentric circles of radii \(2\) and \(8\) the inner and outer circles. Let a desired circle have radius \(r\) and let its center be distance \(d\) from the origin. It must be internally tangent to the outer circle, so \(d+r=8.\)

If it is externally tangent to the inner circle, then \(d-r=2,\) giving \((r,d)=(3,5).\) If it contains the inner circle, then \(r-d=2,\) giving \((r,d)=(5,3).\) Thus every desired circle has radius \(3\) or \(5.\)

The third given circle has radius \(\sqrt3\) and center \((5,0).\) For either value of \(r,\) a desired circle may be internally or externally tangent to it, so the distance between their centers is \(r-\sqrt3\) or \(r+\sqrt3.\) In all four cases, if this distance is \(q,\) then \(|d-q|<5<d+q,\) so the circle of possible centers intersects the circle of radius \(d\) about the origin in two points, symmetric across the \(x\)-axis, as shown. Hence there are \(4\) desired circles of each radius.

The total area is therefore \[4(5^2\pi+3^2\pi)=136\pi.\] Thus, E is the correct answer.

23 答案 C

停止时间只依赖时间变量，而最终位置只依赖距离变量，所以相应概率可以相乘。

两个独立 \((0,1)\) 数之和小于 \(1\) 的概率是单位正方形中一个直角三角形的面积，即 \(\frac12\)。

三个独立 \((0,1)\) 数之和小于 \(1\) 的概率是截距为 \(1\) 的四面体体积，即 \(\frac16\)。

若 \(t_1+t_2>1\)，Amelia 走两步后停止。其概率为 \(\frac12\)，且独立地有 \(x_1+x_2>1\) 的概率为 \(\frac12\)，贡献 \(\frac14\)。

若 \(t_1+t_2<1\)，Amelia 会走第三步。其概率为 \(\frac12\)，且独立地有 \(x_1+x_2+x_3>1\) 的概率为 \(1-\frac16=\frac56\)，贡献 \(\frac5{12}\)。

总概率为 \(\frac14+\frac5{12}=\frac23\)。

所以正确答案是 C。

English

The stopping time depends only on the time variables, while the final position depends only on the distance variables, so the corresponding probabilities multiply.

For two independent numbers in \((0,1),\) the probability that their sum is less than \(1\) is the area of a right triangle, namely \(\frac12.\)

For three independent numbers in \((0,1),\) the probability that their sum is less than \(1\) is the volume of a tetrahedron with side intercepts \(1,\) namely \(\frac16.\)

If \(t_1+t_2>1,\) Amelia stops after two steps. This has probability \(\frac12,\) and independently \(x_1+x_2>1\) has probability \(\frac12,\) contributing \(\frac14.\)

If \(t_1+t_2<1,\) Amelia takes the third step. This has probability \(\frac12,\) and independently \(x_1+x_2+x_3>1\) has probability \(1-\frac16=\frac56,\) contributing \(\frac5{12}.\)

The total probability is \(\frac14+\frac5{12}=\frac23.\)

Thus, the answer is C .

24 答案 B

连续使用两次收缩不等式可得 \[ \begin{aligned} &|f(f(400))-f(f(300))|\\ &\quad\le\frac12|f(400)-f(300)|\\ &\quad\le25 \end{aligned}\text{，}\]同理，\[|f(f(800))-f(f(900))|\le25\text{。}\]

设 \(M=f(f(300))=f(f(900))\)。由三角不等式，\[ \begin{aligned} &|f(f(800))-f(f(400))|\\ &\quad\le|f(f(800))-M|\\ &\qquad+|M-f(f(400))|\\ &\quad\le50 \end{aligned}\text{。}\]

为取到这个上界，定义 \(f\) 为依次经过下列各点的分段线性函数：\[\begin{gathered}(300,600),(400,550),(550,575),\\ (650,625),(800,650),(900,600)\end{gathered}\text{。}\]并规定当 \(x\leq300\) 或 \(x\geq900\) 时 \(f(x)=600\)。每一段斜率的绝对值都不超过 \(\dfrac12\)，因此满足收缩条件。特别地，\(f(300)=f(900)=600\)、\(f(400)=550\)、\(f(800)=650\)。所以 \(f(f(400))=f(550)=575\)，而 \(f(f(800))=f(650)=625\)，两者之差为 \(50\)。

所以正确答案是 B。

English

Applying the contraction inequality twice gives \[ \begin{aligned} &|f(f(400))-f(f(300))|\\ &\quad\le\frac12|f(400)-f(300)|\\ &\quad\le25, \end{aligned} \] and similarly \[|f(f(800))-f(f(900))|\le25.\]

Set \(M=f(f(300))=f(f(900)).\) The triangle inequality now yields \[ \begin{aligned} &|f(f(800))-f(f(400))|\\ &\quad\le|f(f(800))-M|\\ &\qquad+|M-f(f(400))|\\ &\quad\le50. \end{aligned} \]

To attain the bound, define \(f\) by linear interpolation through the points \[\begin{gathered}(300,600),(400,550),(550,575),\\ (650,625),(800,650),(900,600).\end{gathered}\] and set \(f(x)=600\) for \(x\leq300\) or \(x\geq900.\) Every segment has slope with absolute value at most \(\dfrac12,\) so the contraction condition holds. In particular, \(f(300)=f(900)=600,\) \(f(400)=550,\) and \(f(800)=650.\) Hence \(f(f(400))=f(550)=575,\) while \(f(f(800))=f(650)=625,\) giving the difference \(50.\)

Thus, the answer is B .

25 答案 A

所求的和为 \[\frac{S_{2023}-S_{2019}}{2^{2019}}\text{。}\]此外，\(0\le S_n<2^n\)。

因此，存在唯一的 \(m_n\in\{0,1,\ldots,6\}\)，使得 \[7S_n=m_n2^n+1\text{。}\]模 \(7\) 化简得到 \(m_n2^n\equiv-1\pmod7\)。

因为 \(2^3\equiv1\pmod7\)，且 \(2019\equiv0\pmod3\)，所以 \(m_{2019}=6\)。又因为 \(2023\equiv1\pmod3\)，所以 \(2m_{2023}\equiv-1\pmod7\)，从而 \(m_{2023}=3\)。因此 \[\begin{gathered}S_{2019}=\frac{6\cdot2^{2019}+1}{7},\\ S_{2023}=\frac{3\cdot2^{2023}+1}{7}\end{gathered}\text{。}\]

最后，\[ \frac{S_{2023}-S_{2019}}{2^{2019}} =\frac{3\cdot2^4-6}{7}=6\text{。}\]

所以正确答案是 A。

English

The desired sum is \[\frac{S_{2023}-S_{2019}}{2^{2019}}.\] Also, \(0\le S_n<2^n.\)

Therefore, for a unique \(m_n\in\{0,1,\ldots,6\},\) \[7S_n=m_n2^n+1.\] Reducing modulo \(7\) gives \(m_n2^n\equiv-1\pmod7.\)

Since \(2^3\equiv1\pmod7\) and \(2019\equiv0\pmod3,\) we get \(m_{2019}=6.\) Since \(2023\equiv1\pmod3,\) we have \(2m_{2023}\equiv-1\pmod7,\) so \(m_{2023}=3.\) Hence \[\begin{gathered}S_{2019}=\frac{6\cdot2^{2019}+1}{7},\\ S_{2023}=\frac{3\cdot2^{2023}+1}{7}.\end{gathered}\]

Finally, \[ \frac{S_{2023}-S_{2019}}{2^{2019}} =\frac{3\cdot2^4-6}{7}=6. \]

Thus, the correct answer is A .