2024A AMC 12 真题（P版）

1 答案 A

直接计算，\(9901\cdot101=990100\) \(+9901=1000001\) 而 \(99\cdot10101=1010100\) \(-10101=999999\)。它们的差是 \(1000001-999999=2\)。因此正确答案是 A。

English

Directly, \(9901\cdot101=990100\) \(+9901=1000001\) and \(99\cdot10101=1010100\) \(-10101=999999.\) Their difference is \(1000001-999999=2.\) Thus, the correct answer is A.

2 答案 B

由 \(1.5a+800b=1.2a+1100b\) 可得 \(0.3a=300b\)，所以 \(a=1000b\)。于是 \(69=1.5a+800b\) \(=1500b+800b=2300b\)，得到 \(b=0.03\)，\(a=30\)。当 \(L=4.2,\ G=4000\) 时，\(T=30(4.2)\) \(+0.03(4000)\) \(=126+120=246\)。因此正确答案是 B。

English

From \(1.5a+800b=1.2a+1100b\) we get \(0.3a=300b,\) so \(a=1000b.\) Then \(69=1.5a+800b\) \(=1500b+800b=2300b,\) giving \(b=0.03\) and \(a=30.\) For \(L=4.2,\ G=4000,\) \(T=30(4.2)\) \(+0.03(4000)\) \(=126+120=246.\) Thus, the correct answer is B.

3 答案 B

每个数至多为 \(99\)，所以 \(k\) 个数的和至多为 \(99k\)。因为 \(99\cdot20=1980\lt2024\)，至少需要 \(21\) 个数。用 \(21\) 个数时，可以取二十个 \(99\) 和一个 \(44\)：\(20\cdot99+44=1980+44\) \(=2024\)。因此正确答案是 B。

English

Each number is at most \(99,\) so \(k\) numbers sum to at most \(99k.\) Since \(99\cdot20=1980\lt2024,\) at least \(21\) numbers are required. With \(21,\) we can use twenty \(99\)s and one \(44\): \(20\cdot99+44=1980+44\) \(=2024.\) Thus, the correct answer is B.

4 答案 D

分解得 \(2024=2^3\cdot11\cdot23\)。阶乘 \(n!\) 只有在 \(n\ge23\) 时才包含质因数 \(23\)。当 \(n=23\) 时，\(23!\) 已经包含 \(23,\ 11\)，以及足够多的因数 \(2\)，所以 \(23!\) 是 \(2024\) 的倍数。因此正确答案是 D。

English

Factoring, \(2024=2^3\cdot11\cdot23.\) The factorial \(n!\) contains the prime \(23\) only when \(n\ge23.\) At \(n=23,\) the product \(23!\) already includes \(23,\ 11,\) and plenty of factors of \(2,\) so \(23!\) is a multiple of \(2024.\) Thus, the correct answer is D.

5 答案 D

原数据集的总和是 \(20\cdot45=900\)。设有 \(k\) 个六，删去后剩下 \(20-k\) 个数，总和为 \(900-6k\)，平均数为 \(66\)，所以 \(900-6k=66(20-k)\) \(=1320-66k\)。因此 \(60k=420\)，得到 \(k=7\)。所以正确答案是 D。

English

The full set sums to \(20\cdot45=900.\) Removing \(k\) sixes leaves \(20-k\) numbers summing to \(900-6k,\) with mean \(66,\) so \(900-6k=66(20-k)\) \(=1320-66k.\) Then \(60k=420,\) giving \(k=7.\) Thus, the correct answer is D.

6 答案 B

要得到较小的正和，可以取两个负整数 \(-p,-q\) 和一个正整数 \(r\)，其中 \(pqr=60\)。不计顺序时，\(60\) 的正因数三元组为 \[ \begin{gathered} (1,1,60),(1,2,30),\\ (1,3,20),(1,4,15),\\ (1,5,12),(1,6,10),\\ (2,2,15),(2,3,10),\\ (2,5,6),(3,4,5) \end{gathered}\text{。}\] 把最大的因数取作 \(r\) 时，\(r-p-q\) 的正值最小。表中得到的正值为 \(58,27,16,10,6,3,11,5\)；最后两个三元组给出负值。因此最小的正和为 \(10-1-6=3\)。（三个正整数之和至少为 \(3\)，而三个负整数的乘积不可能为正。）因此正确答案是 B。

English

To obtain a small positive sum, use two negative integers \(-p,-q\) and one positive integer \(r,\) where \(pqr=60.\) Up to order, the positive factor triples of \(60\) are \[ \begin{gathered} (1,1,60),(1,2,30),\\ (1,3,20),(1,4,15),\\ (1,5,12),(1,6,10),\\ (2,2,15),(2,3,10),\\ (2,5,6),(3,4,5). \end{gathered} \] A positive value of \(r-p-q\) is smallest when the largest factor is chosen as \(r.\) The positive values from the list are \(58,27,16,10,6,3,11,5;\) the last two triples give negative values. Thus the least positive sum is \(10-1-6=3.\) (Three positive integers have sum at least \(3,\) and three negative integers cannot have positive product.) Thus, the correct answer is B.

7 答案 D

点 \(P_k\) 关于 \(AC\) 的中点 \(M\) 对称，所以把 \(P_k\) 和它的对称点配对可得 \(\vec{BP_k}+\vec{BP_{2025-k}}=2\,\vec{BM}\)。因此整个和为 \(2024\,\vec{BM}\)。在直角三角形中，斜边上的中线长度等于斜边的一半；此处 \(AC=2\)，所以 \(BM=1\)。向量和的长度为 \(2024\cdot1=2024\)。因此正确答案是 D。

English

The points \(P_k\) are symmetric about the midpoint \(M\) of \(AC,\) so pairing \(P_k\) with its mirror gives \(\vec{BP_k}+\vec{BP_{2025-k}}=2\,\vec{BM}.\) Hence the whole sum is \(2024\,\vec{BM}.\) In a right triangle the median to the hypotenuse has length half the hypotenuse; here \(AC=2,\) so \(BM=1.\) The length of the sum is \(2024\cdot1=2024.\) Thus, the correct answer is D.

8 答案 A

方程表示 \(\sin(3\theta)\cos(2\theta)=1\)，且两个因子都必须为正（对数才有定义）。因为 \(\sin(3\theta)\le1\) 且 \(\cos(2\theta)\le1\)，它们的乘积为 \(1\) 只能在 \(\sin(3\theta)=1\) 且 \(\cos(2\theta)=1\) 同时成立时发生。但 \(\cos(2\theta)=1\) 强制 \(\theta\in\{0,\pi,2\pi\}\)，此时 \(\sin(3\theta)=0\ne1\)。没有角满足条件。因此正确答案是 A。

English

The equation means \(\sin(3\theta)\cos(2\theta)=1\) with both factors positive (for the logs to be defined). Since \(\sin(3\theta)\le1\) and \(\cos(2\theta)\le1,\) their product is \(1\) only if \(\sin(3\theta)=1\) and \(\cos(2\theta)=1\) simultaneously. But \(\cos(2\theta)=1\) forces \(\theta\in\{0,\pi,2\pi\},\) where \(\sin(3\theta)=0\ne1.\) No angle works. Thus, the correct answer is A.

9 答案 E

写成 \(M+1213=a^2\)、\(M+3773=b^2\)，则 \(b^2-a^2=2560\)，即 \((b-a)(b+a)=2560\)。两个因数同奇偶，所以都为偶数。要最大化 \(a\)（也就是最大化 \(M\)），应最小化 \(b-a\)：取 \(b-a=2,\ b+a=1280\)，得 \(a=639\)。于是 \(M=639^2-1213\) \(=408321-1213\) \(=407108\)，个位数字为 \(8\)。因此正确答案是 E。

English

Write \(M+1213=a^2\) and \(M+3773=b^2,\) so \(b^2-a^2=2560,\) i.e. \((b-a)(b+a)=2560.\) Both factors have the same parity, hence both even. To maximize \(a\) (and thus \(M\)), minimize \(b-a:\) take \(b-a=2,\ b+a=1280,\) so \(a=639.\) Then \(M=639^2-1213\) \(=408321-1213\) \(=407108,\) whose units digit is \(8.\) Thus, the correct answer is E.

10 答案 C

\(3\text{-}4\text{-}5\) 三角形的最小角满足 \(\tan\alpha=\tfrac34\)。因此 \[ \begin{aligned} &\tan2\alpha=\frac{2\cdot\frac34}{1-\frac{9}{16}} \\ &=\frac{\frac{3}{2}}{\frac{7}{16}}=\frac{24}{7} \end{aligned}\text{。}\] \(7\text{-}24\text{-}25\) 三角形的最小角满足 \(\tan\beta=\tfrac{7}{24}=\cot2\alpha\) \(=\tan\!\left(\tfrac{\pi}{2}-2\alpha\right)\)。所以 \(\beta=\tfrac{\pi}{2}-2\alpha\)。因此正确答案是 C。

English

The smallest angle of the \(3\text{-}4\text{-}5\) triangle has \(\tan\alpha=\tfrac34.\) Then \[ \begin{aligned} &\tan2\alpha=\frac{2\cdot\frac34}{1-\frac{9}{16}} \\ &=\frac{\frac{3}{2}}{\frac{7}{16}}=\frac{24}{7}. \end{aligned} \] The smallest angle of the \(7\text{-}24\text{-}25\) triangle has \(\tan\beta=\tfrac{7}{24}=\cot2\alpha\) \(=\tan\!\left(\tfrac{\pi}{2}-2\alpha\right).\) Hence \(\beta=\tfrac{\pi}{2}-2\alpha.\) Thus, the correct answer is C.

11 答案 D

这里 \(2024_b=2b^3+2b+4\) \(=2(b^3+b+2)\)，所以 \(2024_b\) 是 \(16\) 的倍数当且仅当 \(b^3+b+2\) 是 \(8\) 的倍数。检查 \(\bmod 8\) 的剩余类可知，\(b^3+b+2\equiv0\) 恰好在 \(b\equiv3,6,7\pmod8\) 时成立。

统计 \(b\) 在 \([5,2024]\)：剩余 \(3\) 给出 \(11,\ldots,2019\)，共有 \(252\) 个；剩余 \(6\) 给出 \(6,\ldots,2022\)，共有 \(253\) 个；剩余 \(7\) 给出 \(7,\ldots,2023\)，共有 \(253\) 个。所以 \(K=252+253+253=758\)，各位数字之和为 \(7+5+8=20\)。

因此正确答案是 D。

English

Here \(2024_b=2b^3+2b+4\) \(=2(b^3+b+2),\) so \(2024_b\) is divisible by \(16\) exactly when \(b^3+b+2\) is divisible by \(8.\) Checking residues \(\bmod 8,\) \(b^3+b+2\equiv0\) precisely for \(b\equiv3,6,7\pmod8.\)

Counting \(b\) in \([5,2024]:\) residue \(3\) gives \(11,\ldots,2019\) (\(252\) values), residue \(6\) gives \(6,\ldots,2022\) (\(253\) values), and residue \(7\) gives \(7,\ldots,2023\) (\(253\) values). So \(K=252+253+253=758,\) and its digit sum is \(7+5+8=20.\)

Thus, the correct answer is D.

12 答案 E

因为各项成等比数列，\(720^2=ab\)，所以 \(ab=518400=2^8\cdot3^4\cdot5^2\)。因为 \(b=\frac{518400}{a}\)，要使 \(b\) 最小，就要找 \(518400\) 中大于 \(720\) 的最小因数。严格位于 \(720\) 与 \(768\) 之间的因数并不存在：如果它的 \(5\) 的指数为 \(2,1\) 或 \(0\)，那么分别除以 \(25,5\) 或 \(1\) 之后，形如 \(2^i3^j\) 的数必须位于 \((28.8,30.72),\ (144,153.6)\)，或 \((720,768)\)；而允许的幂 \(i\le8,j\le4\) 中没有符合者。因为 \(768=2^8\cdot3\) 是因数，所以它是最小可能的 \(b\)。与它配对的因数为 \(a=\frac{518400}{768}=675\)，而 \(b\) 的数位和为 \(7+6+8=21\)。因此，正确答案是 E。

English

Since the terms are geometric, \(720^2=ab,\) so \(ab=518400=2^8\cdot3^4\cdot5^2.\) Because \(b=\frac{518400}{a},\) minimizing \(b\) means finding the smallest divisor of \(518400\) greater than \(720.\) There is no divisor strictly between \(720\) and \(768:\) if its exponent of \(5\) is \(2,1,\) or \(0,\) then after dividing by \(25,5,\) or \(1,\) respectively, a number of the form \(2^i3^j\) would have to lie in \((28.8,30.72),\ (144,153.6),\) or \((720,768);\) the allowed powers \(i\le8,j\le4\) give none. Since \(768=2^8\cdot3\) is a divisor, it is the least possible \(b.\) Its paired divisor is \(a=\frac{518400}{768}=675,\) and the digit sum of \(b\) is \(7+6+8=21.\) Thus, the correct answer is E.

13 答案 D

曲线 \(y=e^{x+1}+e^{-x}-2\) 关于其最小值所在的竖直直线对称。令导数 \(e^{x+1}-e^{-x}=0\) 得到 \(x+1=-x\)，所以 \(x=-\tfrac12\)。把 \(\left(-1,\tfrac12\right)\) 关于 \(x=-\tfrac12\) 反射，\(y\) 坐标不变，\(x=-1\) 变为 \(x=0\)。反射点是 \(\left(0,\tfrac12\right)\)。因此正确答案是 D。

English

The curve \(y=e^{x+1}+e^{-x}-2\) is symmetric about the vertical line through its minimum. Setting the derivative \(e^{x+1}-e^{-x}=0\) gives \(x+1=-x,\) so \(x=-\tfrac12.\) Reflecting \(\left(-1,\tfrac12\right)\) across \(x=-\tfrac12\) keeps the \(y\)-coordinate and sends \(x=-1\) to \(x=0.\) The image is \(\left(0,\tfrac12\right).\) Thus, the correct answer is D.

14 答案 C

行和列都为等差数列的网格，其条目可写成双线性形式 \(a(i,j)=\alpha+\beta i+\gamma j+\delta ij\)。四个已知值给出 \[ \alpha+5\beta+5\gamma+25\delta=0\text{，}\]\[ \alpha+2\beta+4\gamma+8\delta=48\text{，}\]\[ \alpha+4\beta+3\gamma+12\delta=16\text{，}\]\[ \alpha+3\beta+\gamma+3\delta=12\text{。}\]

解得 \(\delta=-5,\ \beta=5,\ \) \(\gamma=22,\ \alpha=-10\)。于是 \(a(1,2)=\alpha+\beta+2\gamma+2\delta\) \(=-10+5+44-10\) \(=29\)。

因此正确答案是 C。

English

A grid whose rows and columns are all arithmetic has entries of the bilinear form \(a(i,j)=\alpha+\beta i+\gamma j+\delta ij.\) The four givens yield \[ \alpha+5\beta+5\gamma+25\delta=0, \] \[ \alpha+2\beta+4\gamma+8\delta=48, \] \[ \alpha+4\beta+3\gamma+12\delta=16, \] \[ \alpha+3\beta+\gamma+3\delta=12. \]

Solving gives \(\delta=-5,\ \beta=5,\ \) \(\gamma=22,\ \alpha=-10.\) Then \(a(1,2)=\alpha+\beta+2\gamma+2\delta\) \(=-10+5+44-10\) \(=29.\)

Thus, the correct answer is C.

15 答案 D

因为 \(P(x)=(x-p)(x-q)(x-r)\)，将所有根上的 \(p^2+4=(p-2i)(p+2i)\) 分组可得 \[ \prod(p^2+4)=P(2i)\,P(-2i)\text{。} \] 计算得 \(P(2i)=-8i-8-2i+3\) \(=-5-10i\)，而 \(P(-2i)=8i-8+2i+3\) \(=-5+10i\)。它们的乘积为 \((-5)^2+10^2=25+100=125\)。因此正确答案是 D。

English

Since \(P(x)=(x-p)(x-q)(x-r),\) grouping \(p^2+4=(p-2i)(p+2i)\) over all roots gives \[ \prod(p^2+4)=P(2i)\,P(-2i). \] Compute \(P(2i)=-8i-8-2i+3\) \(=-5-10i\) and \(P(-2i)=8i-8+2i+3\) \(=-5+10i.\) Their product is \((-5)^2+10^2=25+100=125.\) Thus, the correct answer is D.

16 答案 C

把所有筹码视为可区分的；给每位玩家发 \(4\) 个的总方式数是 \(\binom{12}{4,4,4}=34650\)。对有利事件，选择哪位玩家得到红色、白色和蓝色有 \(3!=6\) 种方式。红色玩家还需 \(1\) 个筹码，白色玩家还需 \(2\) 个，蓝色玩家还需 \(3\) 个，全部为黑色；\(6\) 个黑色筹码按 \(1,2,3\) 分配有 \(\tfrac{6!}{1!\,2!\,3!}=60\) 种方式。概率为 \(\dfrac{6\cdot60}{34650}=\dfrac{360}{34650}=\dfrac{4}{385}\)。所以 \(m+n=4+385=389\)。因此正确答案是 C。

English

Treat all tokens as distinct; the total number of ways to deal \(4\) to each player is \(\binom{12}{4,4,4}=34650.\) For the favorable event, choose which player gets the reds, whites, and blue in \(3!=6\) ways. The red player needs \(1\) more token, the white player \(2\) more, and the blue player \(3\) more, all black; the \(6\) black tokens split as \(1,2,3\) in \(\tfrac{6!}{1!\,2!\,3!}=60\) ways. So the probability is \(\dfrac{6\cdot60}{34650}=\dfrac{360}{34650}=\dfrac{4}{385}.\) Then \(m+n=4+385=389.\) Thus, the correct answer is C.

17 答案 D

第一个方程减去第二个方程得到 \((a-c)(b-1)=13\)。因此 \[ \begin{gathered} (a-c,b-1)\in\{(1,13),(13,1),\\ (-1,-13),(-13,-1)\} \end{gathered}\text{。}\] 相应的 \(b\) 值为 \(14,2,-12,0\)。把 \(a=c+(a-c)\) 代入 \(ab+c=100\)，前两种情形被排除，因为它们分别要求 \(15c=86\) 或 \(3c=74\)。\(b=-12\) 的情形给出 \(c=-8\) 和 \(a=-9\)，它满足全部三个方程。\(b=0\) 的情形给出 \((a,c)=(87,100)\)，但不满足 \(ca+b=60\)。因此唯一的整数解是 \((-9,-12,-8)\)，从而 \(ab+bc+ca\) \(=108+96+72=276\)。因此正确答案是 D。

English

Subtracting the second equation from the first gives \((a-c)(b-1)=13.\) Hence \[ \begin{gathered} (a-c,b-1)\in\{(1,13),(13,1),\\ (-1,-13),(-13,-1)\}. \end{gathered} \] The corresponding values of \(b\) are \(14,2,-12,0.\) Substituting \(a=c+(a-c)\) into \(ab+c=100\) eliminates the first two cases because they would require \(15c=86\) or \(3c=74.\) The case \(b=-12\) gives \(c=-8\) and \(a=-9,\) which satisfies all three equations. The case \(b=0\) gives \((a,c)=(87,100),\) which fails \(ca+b=60.\) Thus the unique integer solution is \((-9,-12,-8),\) and \(ab+bc+ca\) \(=108+96+72=276.\) Thus, the correct answer is D.

18 答案 A

卡片的对角线与长边所成角 \(\theta\) 满足 \(\tan\theta=\dfrac{1}{2+\sqrt3}\) \(=2-\sqrt3=\tan15^\circ\)，所以 \(2\theta=30^\circ\)。

每张新卡片与前一张共用一条对角线，而它的另一条对角线正是再顺时针转 \(30^\circ\) 得到的下一条直线。这些等长的对角线有共同的中点，都是同一个圆的直径。原卡片另一条对角线所在的直线经过 \(B\)，它相对 \(AC\) 逆时针转了 \(30^\circ\)；作为不带方向的直线，这等同于相对 \(AC\) 顺时针转了 \(150^\circ\)。每加一张卡片，未被共用的对角线就前进一步，加五张后共前进 \(5\cdot30^\circ=150^\circ\)，所以第六张卡片是第一张有顶点落在 \(B\) 上的新卡片。

因此正确答案是 A。

English

A diagonal makes an angle \(\theta\) with a long side, where \(\tan\theta=\dfrac{1}{2+\sqrt3}\) \(=2-\sqrt3=\tan15^\circ.\) Thus the acute angle between the two diagonals of a card is \(2\theta=30^\circ.\)

Each new card shares one diagonal with the previous card, and its other diagonal is the next line obtained by turning \(30^\circ\) clockwise. All these equal diagonals have the same midpoint and are diameters of one common circle. The line through the original card’s other diagonal, which contains \(B,\) is \(30^\circ\) counterclockwise from \(AC.\) As an unoriented line, this is the same as \(150^\circ\) clockwise from \(AC.\) Five additions advance the unused diagonal by \(5\cdot30^\circ=150^\circ,\) so the sixth card is the first new card with a vertex at \(B.\)

Thus, the correct answer is A.

19 答案 D

在 \(\triangle ACD\) 中，由余弦定理得 \(AC^2=9+25-2(15)\cos120^\circ\) \(=34+15=49\)，所以 \(AC=7\)。

因为 \(ABCD\) 是圆内接四边形，\(\angle ABC=180^\circ-120^\circ=60^\circ\)。在 \(BC=3\)、\(AC=7\) 的 \(\triangle ABC\) 中，由余弦定理得 \(49=AB^2+9-3AB\)，所以 \(AB=8\)。由托勒密定理，\(AC\cdot BD=AB\cdot CD+BC\cdot DA\) \(=8\cdot3+3\cdot5=39\)，所以 \(BD=\tfrac{39}{7}\)。这比 \(AC=7\) 短。

因此正确答案是 D。

English

In \(\triangle ACD,\) the law of cosines gives \(AC^2=9+25-2(15)\cos120^\circ\) \(=34+15=49,\) so \(AC=7.\)

Since \(ABCD\) is cyclic, \(\angle ABC=180^\circ-120^\circ=60^\circ.\) In \(\triangle ABC\) with \(BC=3\) and \(AC=7,\) the law of cosines gives \(49=AB^2+9-3AB,\) so \(AB=8.\) By Ptolemy, \(AC\cdot BD=AB\cdot CD+BC\cdot DA\) \(=8\cdot3+3\cdot5=39,\) hence \(BD=\tfrac{39}{7}.\) This is shorter than \(AC=7.\)

Thus, the correct answer is D.

20 答案 D

令 \(x=\tfrac{AP}{AB}\)，\(y=\tfrac{AQ}{AC}\)，二者在 \([0,1]\) 上均匀分布，则面积比 \(\tfrac{[APQ]}{[ABC]}=xy\)。补事件 \(xy\ge\tfrac12\) 要求 \(x\ge\tfrac12\)，且 \(y\in[\tfrac{1}{2x},1]\)，其概率为 \[ \begin{aligned} &\int_{\frac{1}{2}}^{1}\left(1-\frac{1}{2x}\right)dx \\ &=\frac12-\frac{\ln2}{2}\approx0.153 \end{aligned}\text{。}\] 因此 \(P(xy\lt\tfrac12)\approx1-0.153=0.847\)，落在 \(\left(\tfrac34,\tfrac78\right]\) 中。因此正确答案是 D。

English

With \(x=\tfrac{AP}{AB}\) and \(y=\tfrac{AQ}{AC}\) uniform on \([0,1],\) the area ratio \(\tfrac{[APQ]}{[ABC]}=xy.\) The complementary event \(xy\ge\tfrac12\) requires \(x\ge\tfrac12\) and \(y\in[\tfrac{1}{2x},1],\) with probability \[ \begin{aligned} &\int_{\frac{1}{2}}^{1}\left(1-\frac{1}{2x}\right)dx \\ &=\frac12-\frac{\ln2}{2}\approx0.153. \end{aligned} \] Therefore \(P(xy\lt\tfrac12)\approx1-0.153=0.847,\) which lies in \(\left(\tfrac34,\tfrac78\right].\) Thus, the correct answer is D.

21 答案 B

递推式可化为 \(a_n=1+\tfrac{n-1}{n}(a_{n-1}+1)\)。计算前几项 \(2,\tfrac52,\tfrac{10}3,\tfrac{17}4,\ldots\)，可以猜出 \(a_n=n+\tfrac1n\)。这可以用归纳法证明：把 \(a_{n-1}=n-1+\tfrac1{n-1}\) 代入递推式得 \(a_n=1+\tfrac{n-1}{n}(n+\tfrac1{n-1})\) \(=n+\tfrac1n\)。于是 \(a_n^2=n^2+2+\tfrac1{n^2}\)，所以 \[ \begin{aligned} &\sum_{n=1}^{100}a_n^2 \\ &=\sum_{n=1}^{100}n^2+200 \\ &\quad {}+\sum_{n=1}^{100}\frac1{n^2} \\ &=338350+200+S \end{aligned}\text{，}\] 其中 \(S\gt1\) 且 \(S\lt1+\int_1^\infty x^{-2}\,dx=2\)。因此该和介于 \(338551\) 与 \(338552\) 之间，其向下取整为 \(338551\)。因此正确答案是 B。

English

The recurrence rearranges to \(a_n=1+\tfrac{n-1}{n}(a_{n-1}+1).\) Computing early terms \(2,\tfrac52,\tfrac{10}3,\tfrac{17}4,\ldots\) suggests \(a_n=n+\tfrac1n.\) This follows by induction: substituting \(a_{n-1}=n-1+\tfrac1{n-1}\) into the recurrence gives \(a_n=1+\tfrac{n-1}{n}(n+\tfrac1{n-1})\) \(=n+\tfrac1n.\) Then \(a_n^2=n^2+2+\tfrac1{n^2},\) so \[ \begin{aligned} &\sum_{n=1}^{100}a_n^2 \\ &=\sum_{n=1}^{100}n^2+200 \\ &\quad {}+\sum_{n=1}^{100}\frac1{n^2} \\ &=338350+200+S, \end{aligned} \] where \(S\gt1\) and \(S\lt1+\int_1^\infty x^{-2}\,dx=2.\) Hence the sum is between \(338551\) and \(338552,\) and its floor is \(338551.\) Thus, the correct answer is B.

22 答案 C

中间一行的每个格子必须恰好接触一根牙签。先考虑从中间带的一侧穿到另一侧的回路。回路可以横跨全部 \(8\) 列、前 \(7\) 列、后 \(7\) 列或中间 \(6\) 列；更窄的跨度会使外侧某个中间格没有牙签。

一旦两端固定，每个内部中间格可以独立选择让其唯一牙签位于上边或下边，其余不自交回路都被迫确定。因此四种情形分别贡献 \(2^6,2^5,2^5\)，和 \(2^4\) 条回路。另外，恰有两条回路不穿过中间带：完全沿顶部或完全沿底部的水平长方形。因此总数为 \(2^6+2^5+2^5+2^4+2\) \(=64+32+32+16+2\) \(=146\)。

因此，正确答案是 C。

English

Each middle-row cell must touch exactly one toothpick. First consider loops that pass from one side of the middle strip to the other. The loop can span all \(8\) columns, the first \(7,\) the last \(7,\) or the middle \(6;\) a narrower span would leave an outer middle cell untouched.

Once the two ends are fixed, each interior middle cell independently has its one toothpick on its top or bottom side, and the rest of the non-self-intersecting loop is forced. The four cases therefore contribute \(2^6,2^5,2^5,\) and \(2^4\) loops. There are also exactly two loops that do not cross the middle strip: the horizontal rectangle running entirely along the top or entirely along the bottom. Hence the total is \(2^6+2^5+2^5+2^4+2\) \(=64+32+32+16+2\) \(=146.\)

Thus, the correct answer is C.

23 答案 B

令 \(a=\tan^2\tfrac{\pi}{16},\ \) \(b=\tan^2\tfrac{3\pi}{16},\ \) \(c=\tan^2\tfrac{5\pi}{16},\ \) \(d=\tan^2\tfrac{7\pi}{16}\)，原式为 \(ab+ac+bd+cd\) \(=(a+d)(b+c)\)。

因为 \(\tfrac{7\pi}{16}=\tfrac{\pi}{2}-\tfrac{\pi}{16}\)，所以 \(d=\cot^2\tfrac{\pi}{16}\)，从而 \(a+d=\tan^2\tfrac{\pi}{16}+\cot^2\tfrac{\pi}{16}\) \(=\tfrac{4}{\sin^2(\frac{\pi}{8})}-2\) \(=14+8\sqrt2\)。同理 \(b+c=\tfrac{4}{\sin^2(\frac{3\pi}{8})}-2\) \(=14-8\sqrt2\)。两者乘积为 \(14^2-(8\sqrt2)^2=196-128=68\)。

因此正确答案是 B。

English

With \(a=\tan^2\tfrac{\pi}{16},\ \) \(b=\tan^2\tfrac{3\pi}{16},\ \) \(c=\tan^2\tfrac{5\pi}{16},\ \) \(d=\tan^2\tfrac{7\pi}{16},\) the expression is \(ab+ac+bd+cd\) \(=(a+d)(b+c).\)

Since \(\tfrac{7\pi}{16}=\tfrac{\pi}{2}-\tfrac{\pi}{16},\) we have \(d=\cot^2\tfrac{\pi}{16},\) so \(a+d=\tan^2\tfrac{\pi}{16}+\cot^2\tfrac{\pi}{16}\) \(=\tfrac{4}{\sin^2(\frac{\pi}{8})}-2\) \(=14+8\sqrt2.\) Likewise \(b+c=\tfrac{4}{\sin^2(\frac{3\pi}{8})}-2\) \(=14-8\sqrt2.\) Their product is \(14^2-(8\sqrt2)^2=196-128=68.\)

Thus, the correct answer is B.

24 答案 D

等面四面体存在（可看作由一个长方体各面中心构成的四面体）当且仅当公共面三角形是锐角三角形，此时总表面积是单个面面积的 \(4\) 倍。把整数边长写成 \(u\lt v\lt w\)。若 \(u\le3\)，则 \(w\ge v+1\) 且 \(u^2+v^2\le(v+1)^2\)（只有 \((u,v,w)=(3,4,5)\) 时取等号），所以三角形不是锐角三角形。因此 \(u\ge4\)。

三角形 \((4,5,6)\) 是锐角三角形，因为 \(4^2+5^2\gt6^2\)。它的面积也是可能取到的最小值。任何锐角三角形的最大角至少为 \(60^\circ\)，所以满足 \((u,v)\ne(4,5)\) 的候选三角形面积至少为 \(\tfrac12uv\sin60^\circ\) \(\ge\tfrac12(4)(6)\tfrac{\sqrt3}{2}=6\sqrt3\)，这大于 \((4,5,6)\) 的面积 \(\tfrac{15\sqrt7}{4}\)。

由海伦公式，半周长 \(s=\tfrac{15}2\)，面积为 \(\sqrt{\tfrac{15}2\cdot\tfrac72\cdot\tfrac52\cdot\tfrac32}=\tfrac{15\sqrt7}{4}\)。总表面积为 \(4\cdot\tfrac{15\sqrt7}{4}=15\sqrt7\)。

因此正确答案是 D。

English

A disphenoid exists (as the tetrahedron formed by the face-plane midpoints of a box) exactly when the common face triangle is acute, and its total surface area is \(4\) times one face’s area. Write the integer side lengths as \(u\lt v\lt w.\) If \(u\le3,\) then \(w\ge v+1\) and \(u^2+v^2\le(v+1)^2\) (with equality only for \((u,v,w)=(3,4,5)\)), so the triangle is not acute. Thus \(u\ge4.\)

The triangle \((4,5,6)\) is acute because \(4^2+5^2\gt6^2.\) It also has the least possible area. The largest angle of any acute triangle is at least \(60^\circ,\) so a candidate with \((u,v)\ne(4,5)\) has area at least \(\tfrac12uv\sin60^\circ\) \(\ge\tfrac12(4)(6)\tfrac{\sqrt3}{2}=6\sqrt3,\) which is greater than \(\tfrac{15\sqrt7}{4},\) the area of \((4,5,6).\)

By Heron’s formula with \(s=\tfrac{15}2,\) that area is \(\sqrt{\tfrac{15}2\cdot\tfrac72\cdot\tfrac52\cdot\tfrac32}=\tfrac{15\sqrt7}{4}.\) The total surface area is \(4\cdot\tfrac{15\sqrt7}{4}=15\sqrt7.\)

Thus, the correct answer is D.

25 答案 B

将 \(y=f(x)\) 的图像关于 \(y=x\) 反射，会得到其反函数的图像，所以图像关于 \(y=x\) 对称当且仅当 \(f\) 等于自己的反函数。对 \(f(x)=\tfrac{ax+b}{cx+d}\)，这有两种情况：\(a+d=0\) 且 \(ad-bc\ne0\)（真正的对合，包括 \(c=0\) 时斜率为 \(-1\) 的直线），或者 \(f\) 是恒等函数 \(y=x\)（\(b=c=0,\ a=d\ne0\)）。

对 \(a+d=0\)，令 \(d=-a\)；行列式 \(-a^2-bc\) 必须非零，所以需要 \(a^2+bc\ne0\)，同时 \((c,d)\ne(0,0)\)。当 \(a=0\) 时，\(b\) 和 \(c\) 都必须非零，给出 \(10^2=100\) 种选择。对每个非零的 \(a\)，先有 \((b,c)\) 的 \(11^2=121\) 种选择。若 \(|a|=1,3,4\) 或 \(5\)，恰有 \(2\) 组满足 \(bc=-a^2\)；若 \(|a|=2\)，则恰有 \(6\) 组。因此真正的对合共有 \[ \begin{gathered} 100+8(121-2)+2(121-6)\\ {}=1282 \end{gathered}\text{。}\] 恒等函数的情形再增加 \(10\) 个（\(a=d\in\{\pm1,\ldots,\pm5\}\)），总数为 \(1282+10=1292\)。

因此正确答案是 B。

English

Reflecting the graph of \(y=f(x)\) over \(y=x\) produces the graph of its inverse, so the graph is symmetric about \(y=x\) exactly when \(f\) equals its own inverse. For \(f(x)=\tfrac{ax+b}{cx+d}\) this happens in two ways: when \(a+d=0\) with \(ad-bc\ne0\) (a genuine involution, including the slope\(-1\) lines when \(c=0\)), or when \(f\) is the identity \(y=x\) (\(b=c=0,\ a=d\ne0\)).

For \(a+d=0,\) set \(d=-a;\) the determinant \(-a^2-bc\) must be nonzero, so we need \(a^2+bc\ne0,\) together with \((c,d)\ne(0,0).\) When \(a=0,\) both \(b\) and \(c\) must be nonzero, giving \(10^2=100\) choices. For each nonzero \(a,\) start with \(11^2=121\) choices of \((b,c).\) If \(|a|=1,3,4,\) or \(5,\) exactly \(2\) pairs satisfy \(bc=-a^2;\) if \(|a|=2,\) exactly \(6\) pairs do. Thus the genuine involutions number \[ \begin{gathered} 100+8(121-2)+2(121-6)\\ {}=1282. \end{gathered} \] The identity case adds \(10\) more (\(a=d\in\{\pm1,\ldots,\pm5\}\)), for a total of \(1282+10=1292.\)

Thus, the correct answer is B.