2024A AMC 10 真题（P版）

1 答案 A

直接计算：\(9901 \cdot 101 = 990100 + 9901\) \(= 1000001\)，而 \(99 \cdot 10101 = 999999\)。因此差为 \(1000001 - 999999 = 2\)，正确答案是 A。

English

Just compute each piece. We have \(9901 \cdot 101 = 990100 + 9901\) \(= 1000001,\) and \(99 \cdot 10101 = 999999.\) Subtracting, \(1000001 - 999999 = 2.\) Thus, A is the correct answer.

2 答案 B

两个方程为 \(1.5a + 800b = 69\) 和 \(1.2a + 1100b = 69\)。相减消去 \(69\)，得到 \(0.3a - 300b = 0\)，所以 \(a = 1000b\)。代入第一式得 \(1500b + 800b = 2300b = 69\)，因此 \(b = 0.03\)、\(a = 30\)。所求时间为 \(T = 30(4.2) + 0.03(4000)\) \(= 126 + 120\) \(= 246\)。所以正确答案是 B。

English

Subtract the two equations \(1.5a + 800b = 69\) and \(1.2a + 1100b = 69\) to kill the \(69.\) That leaves \(0.3a - 300b = 0,\) so \(a = 1000b.\) Now substitute: \(1500b + 800b = 2300b = 69,\) so \(b = 0.03\) and \(a = 30.\) Then \(T = 30(4.2) + 0.03(4000)\) \(= 126 + 120\) \(= 246.\) Therefore, the answer is B.

3 答案 B

如果 \(2\) 是五个质数之一，则总和是大于 \(2\) 的偶数，因此是合数。所以五个质数都必须是奇数。五个最小奇质数之和为 \(3 + 5 + 7 + 11 + 13 = 39\) \(= 3 \cdot 13\)，不是质数。下一个可能的和是把 \(13\) 换成下一个质数 \(17\)；若改动更靠前的项，为保持五个质数互不相同，增加量至少同样大。这样得到 \(3 + 5 + 7 + 11 + 17 = 43\)，它是质数，其各位数字之和为 \(4 + 3 = 7\)，正确答案是 B。

English

Suppose \(2\) is one of the five primes. Then the total is even and bigger than \(2,\) so it is composite. Thus all five primes must be odd. The five smallest odd primes give \(3 + 5 + 7 + 11 + 13 = 39\) \(= 3 \cdot 13,\) which is not prime. The next possible sum is obtained by replacing \(13\) with the next prime, \(17;\) changing any earlier term forces at least as large an increase to keep the primes distinct. This gives \(3 + 5 + 7 + 11 + 17 = 43,\) which is prime. Its digit sum is \(4 + 3 = 7.\) Thus, B is the correct answer.

4 答案 B

每个两位数至多为 \(99\)，所以 \(k\) 个两位数的和至多为 \(99k\)。需要 \(99k \ge 2024\)，因此 \(k \ge 20.4\)，所以 \(k \ge 21\)。\(21\) 个确实可行：二十个 \(99\) 加上一个 \(44\)，和为 \(1980 + 44 = 2024\)。所以正确答案是 B。

English

Each two-digit number is at most \(99,\) so \(k\) of them sum to at most \(99k.\) We need \(99k \ge 2024,\) which forces \(k \ge 20.4,\) so \(k \ge 21.\) And \(21\) really works: twenty \(99\)s plus one \(44\) give \(1980 + 44 = 2024.\) Therefore, the answer is B.

5 答案 D

分解得 \(2024 = 2^3 \cdot 11 \cdot 23\)。质数 \(23\) 是关键：要使 \(n!\) 含有因子 \(23\)，必须有 \(n \ge 23\)。当 \(n = 23\) 时，\(23!\) 已经含有因子 \(23\)、\(11\)，以及足够多的因子 \(2\)，所以 \(23!\) 能被 \(2024\) 整除，最小值为 \(23\)。因此正确答案是 D。

English

Factor \(2024 = 2^3 \cdot 11 \cdot 23.\) The prime \(23\) is the bottleneck: for \(23\) to divide \(n!,\) we need \(n \ge 23.\) At \(n = 23,\) the product \(23!\) already has \(23,\) \(11,\) and plenty of factors of \(2,\) so \(23!\) is divisible by \(2024.\) The least value is \(23.\) Thus, D is the correct answer.

6 答案 D

反转六个字母会颠倒每一对字母的相对顺序，所以共有 \(\binom{6}{2} = 15\) 对字母变成逆序。每次相邻交换只改变一个逆序对数，所以至少需要 \(15\) 次。逐个把字母冒泡到正确位置可以正好用 \(15\) 次完成，因此这个下界可以达到，正确答案是 D。

English

Reversing all six letters flips the relative order of every pair, so all \(\binom{6}{2} = 15\) pairs end up inverted. Each adjacent swap fixes exactly one inversion. So we need at least \(15\) swaps, and bubbling each letter into place hits \(15\) exactly. Therefore, the answer is D.

7 答案 B

正乘积来自三个正整数，或一个正整数与两个负整数。三个正整数的和至少为 \(3\)。在第二种情形中，把三个数写成 \(-x,-y,z\)，其中 \(x,y,z\) 均为正数且 \(xyz=60\)。若三数之和为正，则 \(z \gt x+y \ge 2\sqrt{xy}\)，所以 \(\frac{60}{xy} \gt 2\sqrt{xy}\)，从而 \(xy \lt 10\)。检查 \(60\) 的因数中满足 \(xy\) 的因数对，\(1,2,3,4,5,6\) 的最小正值在 \(\frac{60}{xy}-x-y\) 时取得，等于 \(3\)。因此 \((-1)(-6)(10)=60\)，最小正和为 \(10-1-6=3\)，正确答案是 B。

English

A positive product comes from three positive integers or one positive and two negative integers. Three positive integers have sum at least \(3.\) In the second case write the numbers as \(-x,-y,z,\) where \(x,y,z\) are positive and \(xyz=60.\) A positive sum requires \(z \gt x+y \ge 2\sqrt{xy},\) so \(\frac{60}{xy} \gt 2\sqrt{xy}\) and hence \(xy \lt 10.\) The possible products \(xy\) that divide \(60\) are \(1,2,3,4,5,6.\) Checking their factor pairs, the smallest positive value of \(\frac{60}{xy}-x-y\) is \(10-1-6=3.\) Thus \((-1)(-6)(10)=60,\) and no positive sum below \(3\) is possible. Therefore, the answer is B.

8 答案 A

从下午 \(1{:}00\) 到 \(2{:}45\) 是 \(105\) 分钟。Amy、Bomani 和 Charlie 合计每 \(3\) 分钟包装 \(4 + 3 + 3 = 10\) 包，即每分钟 \(\tfrac{10}{3}\) 包，也就是共包装 \(\tfrac{10}{3} \cdot 105 = 350\) 包。剩下 \(450 - 350 = 100\) 包由 Daria 完成。Daria 每分钟包装 \(\tfrac54\) 包，因此需要 \(\frac{100}{\tfrac54} = 80\) 分钟。她工作了最后 \(80\) 分钟，也就是在下午 \(1{:}00\) 过后 \(105 - 80 = 25\) 分钟加入，即下午 \(1{:}25\)。因此正确答案是 A。

English

From \(1{:}00\) to \(2{:}45\) is \(105\) minutes. Amy, Bomani, and Charlie pack \(4 + 3 + 3 = 10\) packages every \(3\) minutes, so \(\tfrac{10}{3}\) per minute, which is \(\tfrac{10}{3} \cdot 105 = 350\) packages. That leaves \(450 - 350 = 100\) for Daria, who packs \(\tfrac54\) per minute and so needs \(\frac{100}{\tfrac54} = 80\) minutes. She worked the last \(80\) minutes, joining \(105 - 80 = 25\) minutes after \(1{:}00.\) That’s \(1{:}25\) PM. Therefore, the answer is A.

9 答案 B

将 \(6\) 名十一年级生分成三个无序对，有 \(\frac{6!}{2!^3 3!} = 15\) 种方法。十二年级生同样有 \(15\) 种分法。每个队伍由一个十一年级生对子和一个十二年级生对子组成，所以还要将三个十一年级生对子与三个十二年级生对子配对，有 \(3! = 6\) 种方式。总数为 \(15 \cdot 15 \cdot 6 = 1350\)，正确答案是 B。

English

Split the \(6\) juniors into three unordered pairs. There are \(\frac{6!}{2!^3 3!} = 15\) ways, and the same \(15\) for the seniors. Each team is one junior-pair paired with one senior-pair, so we match the three junior-pairs to the three senior-pairs in \(3! = 6\) ways. That’s \(15 \cdot 15 \cdot 6 = 1350\) sets of teams. Thus, B is the correct answer.

10 答案 C

从 \(100\) 开始直接运行：\(100 \to 110 \to 120 \to 40\) \(\to 50 \to 60 \to 20\) \(\to 30 \to 10 \to 20\) \(\to 30 \to 10 \to \cdots\)。第 \(8\) 步后到达 \(10\)，之后按 \(10, 20, 30\) 循环，周期为 \(3\)。因此第 \(8 + k\) 步是循环中的第 \(k\) 项。对第 \(100\) 步，有 \(k = 92\)，且 \(92 \equiv 2 \pmod 3\)，所以落在 \(30\)。因此正确答案是 C。

English

Just run it from \(100:\) \(100 \to 110 \to 120 \to 40\) \(\to 50 \to 60 \to 20\) \(\to 30 \to 10 \to 20\) \(\to 30 \to 10 \to \cdots.\) After the \(8\)th step we’re at \(10,\) and from there it cycles \(10, 20, 30\) with period \(3.\) So step \(8 + k\) is the \(k\)th entry of the cycle. For step \(100,\) \(k = 92,\) and \(92 \equiv 2 \pmod 3,\) which lands on \(30.\) Therefore, the answer is C.

11 答案 D

因为 \(m = \sqrt{n^2 - 49} \ge 0\) 且为整数，所以 \(n^2 - 49 = m^2\)，即 \((n - m)(n + m) = 49\)。分解 \(49\) 可得 \(|n| = 25, m = 24\)，或 \(|n| = 7, m = 0\)。因此有序对 \((m, n)\) 为 \((24, 25)\)、\((24, -25)\)、\((0, 7)\)、\((0, -7)\)，共 \(4\) 个，正确答案是 D。

English

Note \(m = \sqrt{n^2 - 49} \ge 0\) has to be an integer, so \(n^2 - 49 = m^2,\) which means \((n - m)(n + m) = 49.\) The factorizations of \(49\) give \(|n| = 25, m = 24\) or \(|n| = 7, m = 0.\) So the ordered pairs \((m, n)\) are \((24, 25),\) \((24, -25),\) \((0, 7),\) \((0, -7).\) That’s \(4\) of them. Thus, D is the correct answer.

12 答案 E

从 \(1700\) 开始依次应用每日变化 \(+80, -90, -10, +60, -40\)。六天得分为 \(1700, 1780, 1690\)、\(1680, 1740, 1700\)。总和为 \(10290\)，平均分为 \(\frac{10290}{6} = 1715\)。所以正确答案是 E。

English

Apply the daily changes \(+80, -90, -10, +60, -40\) to the starting \(1700.\) The six scores are \(1700, 1780, 1690,\) \(1680, 1740, 1700.\) They add to \(10290,\) so the average is \(\frac{10290}{6} = 1715.\) Therefore, the answer is E.

13 答案 C

伸缩只是关于原点放大，所以它与绕原点旋转可交换，也与关于横轴反射可交换。这给出 \(2\) 对。向右平移与关于 \(x\) 轴反射也可交换，因为两种顺序都把 \((x, y) \to (x + 2, -y)\)。另外三对不交换：平移与旋转、平移与伸缩不交换，旋转与反射不交换。共有 \(3\) 对，正确答案是 C。

English

The dilation just scales about the origin, so it commutes with both the rotation and the reflection. That’s \(2\) pairs. The translation commutes with the reflection across the \(x\)-axis too, since either order sends \((x, y) \to (x + 2, -y).\) The other three pairs fail: the translation clashes with the rotation and with the dilation, and the rotation clashes with the reflection. So \(3\) pairs commute. Thus, C is the correct answer.

14 答案 D

等边三角形边长为 \(16\sqrt3\)。将 \(\ell\) 放在 \(x\) 轴上，并取底角顶点 \(V = (16\sqrt3, 0)\)；斜边所在直线为 \(\sqrt3\,x + y = 48\)。圆与 \(\ell\) 相切，圆心高度为 \(12\)，且从外侧与该边相切。设圆心为 \(O = (20\sqrt3, 12)\)。令 \(T = (20\sqrt3, 0)\) 为圆在 \(\ell\) 上的切点，\(P\) 为圆在斜边上的切点。所求区域由从 \(V\) 引出的两条切线段和靠近的圆弧围成。切线长为 \(VT = VP = 4\sqrt3\)，因此风筝 \(VTOP\) 的面积为 \(4\sqrt3 \cdot 12 = 48\sqrt3\)。顶点 \(V\) 处夹角为 \(120^\circ\)，所以要减去的圆心角为 \(60^\circ\) 的扇形，面积为 \(\tfrac16 \pi (12)^2 = 24\pi\)。区域面积为 \(48\sqrt3 - 24\pi\)，因此 \(a + b + c = 48 + 3 + 24 = 75\)，正确答案是 D。

English

The equilateral triangle has side \(16\sqrt3.\) Put \(\ell\) on the \(x\)-axis with base vertex \(V = (16\sqrt3, 0)\); the slanted side lies on \(\sqrt3\,x + y = 48.\) The circle sits on \(\ell,\) has radius \(12,\) and touches that side externally, so its center is \(O = (20\sqrt3, 12).\) Let \(T = (20\sqrt3, 0)\) be its tangency point on \(\ell,\) and let \(P\) be the tangency point on the slanted side. The two tangent lengths from \(V\) satisfy \(VT = VP = 4\sqrt3,\) so kite \(VTOP\) has area \(4\sqrt3 \cdot 12 = 48\sqrt3.\) The angle at \(V\) is \(120^\circ,\) so the removed sector has angle \(60^\circ\) and area \(\tfrac16 \pi (12)^2 = 24\pi.\) The region has area \(48\sqrt3 - 24\pi,\) giving \(a + b + c = 48 + 3 + 24 = 75.\) Therefore, the answer is D.

15 答案 E

设 \(M + 1213 = y^2\)、\(M + 3773 = x^2\)。相减得 \(x^2 - y^2 = 2560\)，即 \((x - y)(x + y) = 2560\)。所以两个因子同奇偶，而乘积为偶数，二者都为偶数。写 \(x - y = 2s\)、\(x + y = 2t\)，则 \(st = 640\)。要使 \(M\) 最大，即使 \(y = t - s\) 最大，需要 \(s\) 尽量小。取 \(s = 1, t = 640\)，得 \(y = 639\)。因此 \(M = 639^2 - 1213 = 407108\)，个位数字为 \(8\)，正确答案是 E。

English

Set \(M + 1213 = y^2\) and \(M + 3773 = x^2.\) Subtracting, \(x^2 - y^2 = 2560,\) so \((x - y)(x + y) = 2560.\) The two factors share a parity, and their product is even, so both are even: write \(x - y = 2s,\) \(x + y = 2t,\) with \(st = 640.\) To make \(M\) as large as possible we want \(y = t - s\) as large as possible, so \(s\) as small as possible. Take \(s = 1, t = 640,\) giving \(y = 639.\) Then \(M = 639^2 - 1213 = 407108,\) whose units digit is \(8.\) Thus, E is the correct answer.

16 答案 D

十一个拼块的面积之和为 \(200\)。设外接矩形的高为 \(h\)，宽为 \(w=AB\)。相似图形的对应边长之比等于它们面积之比的平方根。由图可知，面积为 \(32\) 的拼块把它的短边贡献给左边，长度为 \(h\sqrt{\frac{32}{200}}=\tfrac25h\)。在它上方，面积为 \(36\) 的拼块贡献的是它的长边，长度为 \(w\sqrt{\frac{36}{200}}=\tfrac{3\sqrt2}{10}w\)。这两段合起来构成整个高，所以 \(h=\tfrac25h+\tfrac{3\sqrt2}{10}w\)，化简得 \(h=\frac{w}{\sqrt2}\)。由于 \(wh=200\)，我们得到 \(\frac{w^2}{\sqrt2}=200\)，从而 \(w=\sqrt{200\sqrt2}=10\sqrt[4]{8}\)。因此正确答案是 D。

English

The areas of the eleven pieces sum to \(200.\) Let the enclosing rectangle have height \(h\) and width \(w=AB.\) Similar figures have corresponding side lengths in the square-root ratio of their areas. From the diagram, the area-\(32\) piece contributes its short side to the left edge, of length \(h\sqrt{\frac{32}{200}}=\tfrac25h.\) Above it, the area-\(36\) piece contributes its long side, of length \(w\sqrt{\frac{36}{200}}=\tfrac{3\sqrt2}{10}w.\) These two segments make the full height, so \(h=\tfrac25h+\tfrac{3\sqrt2}{10}w,\) which simplifies to \(h=\frac{w}{\sqrt2}.\) Since \(wh=200,\) we get \(\frac{w^2}{\sqrt2}=200\) and \(w=\sqrt{200\sqrt2}=10\sqrt[4]{8}.\) Therefore, the answer is D.

17 答案 E

A 队第 \(1\) 场主场获胜概率为 \(\tfrac23\)，后两场客场每场获胜概率为 \(p\)。A 队赢得系列赛有三种互斥方式：赢第 \(1, 2\) 场；赢第 \(1\) 场、输第 \(2\) 场、赢第 \(3\) 场；输第 \(1\) 场、赢第 \(2, 3\) 场。因此 \(\tfrac23 p + \tfrac23(1 - p)p + \tfrac13 p^2 = \tfrac12\)。化简得 \(2p^2 - 8p + 3 = 0\)，合理根为 \(p = \tfrac{4 - \sqrt{10}}{2} = \tfrac12\!\left(4 - \sqrt{10}\right)\)。所以 \(m = 4\)、\(n = 10\)，\(m + n = 14\)，正确答案是 E。

English

Team A takes game \(1\) at home with probability \(\tfrac23,\) and each away game with probability \(p.\) It can win the playoff three disjoint ways: win games \(1, 2;\) win \(1,\) lose \(2,\) win \(3;\) lose \(1,\) win \(2, 3.\) Adding those, \(\tfrac23 p + \tfrac23(1 - p)p + \tfrac13 p^2 = \tfrac12.\) This cleans up to \(2p^2 - 8p + 3 = 0,\) so \(p = \tfrac{4 - \sqrt{10}}{2} = \tfrac12\!\left(4 - \sqrt{10}\right).\) Then \(m = 4,\) \(n = 10,\) and \(m + n = 14.\) Thus, E is the correct answer.

18 答案 D

在 \(b\) 进制中，\(2024_b = 2b^3 + 2b + 4\) \(= 2(b^3 + b + 2)\)，因此 \(2024_b\) 能被 \(16\) 整除，当且仅当 \(b^3 + b + 2\) 能被 \(8\) 整除。检查 \(b\) 模 \(8\) 的余数，可知恰当余数为 \(b \equiv 3, 6, 7 \pmod 8\)。统计 \(5 \le b \le 2024\) 中这三类数，得 \(K = 758\)，其各位数字之和为 \(7 + 5 + 8 = 20\)，正确答案是 D。

English

In base \(b,\) \(2024_b = 2b^3 + 2b + 4\) \(= 2(b^3 + b + 2),\) so \(2024_b\) is divisible by \(16\) exactly when \(b^3 + b + 2\) is divisible by \(8.\) Test the residues modulo \(8:\) this holds precisely for \(b \equiv 3, 6, 7 \pmod 8.\) Counting the \(b\) with \(5 \le b \le 2024\) in those three classes gives \(K = 758,\) whose digit sum is \(7 + 5 + 8 = 20.\) Therefore, the answer is D.

19 答案 E

因为 \(720^2 = ab\)，公比 \(r = \tfrac{720}{a} = \tfrac{b}{720}\) 是有理数。设 \(r = \tfrac{p}{q}\) 为最简分数，且 \(p \gt q\)。则 \(a = \tfrac{720q}{p}\) 和 \(b = \tfrac{720p}{q}\) 都为整数，所以 \(720\) 能被 \(p\) 整除，也能被 \(q\) 整除。为了使 \(b\) 最小，需要选择使 \(720\) 同时能被 \(p\) 和 \(q\) 整除的最小比值 \(\tfrac{p}{q} \gt 1\)，即 \(\tfrac{16}{15}\)。因此 \(b = 720 \cdot \tfrac{16}{15} = 768\)，\(a = 675\)，各位数字之和为 \(7 + 6 + 8 = 21\)，正确答案是 E。

English

Since \(720^2 = ab,\) the common ratio \(r = \tfrac{720}{a} = \tfrac{b}{720}\) is rational. Write \(r = \tfrac{p}{q}\) in lowest terms with \(p \gt q.\) Then \(a = \tfrac{720q}{p}\) and \(b = \tfrac{720p}{q}\) are integers, which forces \(720\) to be divisible by \(p\) and by \(q.\) To make \(b\) smallest, we want the smallest ratio \(\tfrac{p}{q} \gt 1\) with \(720\) divisible by both \(p\) and \(q,\) which is \(\tfrac{16}{15}.\) That gives \(b = 720 \cdot \tfrac{16}{15} = 768\) (and \(a = 675\)). The digit sum is \(7 + 6 + 8 = 21.\) Thus, E is the correct answer.

20 答案 C

两个条件说明选中数至少相差 \(3\)，而选中的奇数至少相差 \(7\)。在任意长度为 \(10\) 的区块中，若取四个数，就需要三个至少为 \(3\) 的间隔，因此它们只能占据位置 \(r,r+3,r+6,r+9\)。这样其中两个奇数会相差 \(6\)，与条件矛盾。所以每个完整的长度为 \(10\) 的区块最多含 \(3\) 个选中数，最后剩下的四个位置最多含 \(2\) 个，上界为 \(202\cdot3+2=608\)。这个上界可以达到：取 \(1,4,8,11,14,18,\ldots\)，即模十余 \(1,4,8\pmod{10}\) 的数，再加上 \(2021,2024\)。相邻选中数至少相差 \(3\)，而选中的奇数彼此相差 \(10\)，正确答案是 C。

English

The two conditions say chosen numbers are at least \(3\) apart, and chosen odd numbers at least \(7\) apart. Any four numbers in a block of \(10\) would need three gaps of at least \(3,\) so they would have to occupy positions \(r,r+3,r+6,r+9.\) Two of the odd entries would then differ by \(6,\) which is forbidden. Thus each full block of \(10\) contains at most \(3\) choices, and the last four positions contain at most \(2.\) This gives the upper bound \(202\cdot3+2=608.\) It is attained by the pattern \(1,4,8,11,14,18,\ldots\) (residues \(1,4,8\pmod{10}\)), together with \(2021,2024.\) Adjacent selected values differ by at least \(3,\) and the selected odd values are \(10\) apart. Therefore, the answer is C.

21 答案 C

若每行和每列都是等差数列，则第 \(i\) 行第 \(j\) 列可写成双线性形式 \(f(i, j) = A + Bi + Cj + Dij\)。代入 \(f(5, 5) = 0\)、\(f(2, 4) = 48\)、\(f(4, 3) = 16\)、\(f(3, 1) = 12\) 并求解，得到 \(A = -10\)、\(B = 5\)、\(C = 22\)、\(D = -5\)。因此位置 \((1, 2)\) 的数为 \(-10 + 5 + 2 \cdot 22 - 2 \cdot 5 = 29\)，正确答案是 C。

English

If every row and every column is an arithmetic progression, the entry at row \(i,\) column \(j\) must take the bilinear form \(f(i, j) = A + Bi + Cj + Dij.\) Plug in \(f(5, 5) = 0,\) \(f(2, 4) = 48,\) \(f(4, 3) = 16,\) \(f(3, 1) = 12\) and solve: \(A = -10,\) \(B = 5,\) \(C = 22,\) \(D = -5.\) So position \((1, 2)\) is \(-10 + 5 + 2 \cdot 22 - 2 \cdot 5 = 29.\) Thus, C is the correct answer.

22 答案 B

每个风筝的一半都是 \(30\)-\(60\)-\(90\) 三角形，所以各边具有图示的长度和方向。取 \(A=(0,0)\)，并令 \(AB\) 水平。图形的水平跨度为六个单位长度，所以 \(B=(6,0)\)。沿着从 \(A\) 到 \(C\) 的外边界，三条边的向量分别为 \(\sqrt3(\cos30^\circ,\sin30^\circ)\)，\(\sqrt3(\cos90^\circ,\sin90^\circ)\)，和 \((1,0)\)。它们的和为 \((\tfrac52,\tfrac{3\sqrt3}{2})\)，所以 \(C=(\tfrac52,\tfrac{3\sqrt3}{2})\)。因此 \(AB=6\)，从 \(C\) 作出的高为 \(\tfrac{3\sqrt3}{2}\)，面积为 \(\tfrac12\cdot6\cdot\tfrac{3\sqrt3}{2}=\tfrac{9\sqrt3}{2}\)。所以答案是 B。

English

Each half of a kite is a \(30\)-\(60\)-\(90\) triangle, so its edges have the shown lengths and directions. Take \(A=(0,0)\) and \(AB\) horizontal. The horizontal span in the figure is six unit lengths, so \(B=(6,0).\) Along the outer boundary from \(A\) to \(C,\) the three edges have vectors \(\sqrt3(\cos30^\circ,\sin30^\circ),\) \(\sqrt3(\cos90^\circ,\sin90^\circ),\) and \((1,0).\) Their sum is \((\tfrac52,\tfrac{3\sqrt3}{2}),\) so \(C=(\tfrac52,\tfrac{3\sqrt3}{2}).\) Thus \(AB=6\) and the altitude from \(C\) is \(\tfrac{3\sqrt3}{2},\) giving area \(\tfrac12\cdot6\cdot\tfrac{3\sqrt3}{2}=\tfrac{9\sqrt3}{2}.\) Therefore, the answer is B.

23 答案 D

用第一个方程减去第二个方程：\((a-c)(b-1)=13\)。令 \(u=a-c\)，\(v=b-1\)。四种可能 \((u,v)=(1,13)\)、\((13,1)\)、\((-1,-13)\) 和 \((-13,-1)\) 代入 \(bc+a=87\) 后，分别给出 \(c=\frac{86}{15}\)、\(c=\frac{74}{3}\)、\(c=-8\) 和 \(c=100\)。只有第三种给出三个原方程的整数解：\((a,b,c)=(-9,-12,-8)\)。（最后一个候选是 \((87,0,100)\)，它不满足 \(ca+b=60\)。）把原来的三个方程相加，得到 \(ab+bc+ca\) \({}+(a+b+c)=247\)。由于 \(a+b+c=-29\)，我们得到 \(ab+bc+ca\) \(=247-(-29)=276\)。因此正确答案是 D。

English

Subtract the second equation from the first: \((a-c)(b-1)=13.\) Put \(u=a-c\) and \(v=b-1.\) The four possibilities \((u,v)=(1,13),\) \((13,1),\) \((-1,-13),\) and \((-13,-1)\) give, after substitution into \(bc+a=87,\) respectively \(c=\frac{86}{15},\) \(c=\frac{74}{3},\) \(c=-8,\) and \(c=100.\) Only the third is an integer solution of all three original equations: \((a,b,c)=(-9,-12,-8).\) (The last candidate is \((87,0,100),\) which fails \(ca+b=60.\)) Adding the original equations gives \(ab+bc+ca\) \({}+(a+b+c)=247.\) Since \(a+b+c=-29,\) we obtain \(ab+bc+ca\) \(=247-(-29)=276.\) Thus, D is the correct answer.

24 答案 B

每次掷骰都会让蜜蜂沿 \(\pm x,\pm y\) 或 \(\pm z\) 方向移动一个单位，所以共有 \(6^4=1296\) 个等可能移动序列。有效路径分成两类。第一类走过某个正方形面的四条边：坐标方向对有 \(3\) 种选择，两个方向的正负号有 \(4\) 种选择，第一个方向有 \(2\) 种选择，共 \(24\) 条路径。第二类用到三个坐标方向，其中一个方向使用两次。重复方向有 \(3\) 种选择；它在四步方向序列中的两个位置不能相邻，这两个位置有 \(3\) 种选法，其余两个方向的顺序有 \(2\) 种。对每个这样的方向序列，恰有 \(2^3=8\) 种正负号选择能使四条不同的边位于同一个立方体上。因此第二类有 \(3\cdot3\cdot2\cdot8=144\) 条路径。总共有 \(24+144=168\) 个有利序列，所以概率为 \(\frac{168}{1296}=\frac{7}{54}\)，正确答案是 B。

English

Every roll moves the bee one unit along \(\pm x,\pm y,\) or \(\pm z,\) so there are \(6^4=1296\) equally likely sequences. There are two types of valid paths. A path around one square face has \(3\) choices of coordinate plane, \(4\) choices for the signs of its two axes, and \(2\) choices for which axis is used first, giving \(24.\) Otherwise all three coordinate directions are used, with one repeated: choose that axis in \(3\) ways, choose its two nonadjacent positions in \(3\) ways, order the other two axes in \(2\) ways, and choose their three initial signs in \(2^3=8\) ways. (The second step on the repeated axis must have the opposite sign.) This gives \(3\cdot3\cdot2\cdot8=144.\) Hence there are \(24+144=168\) favorable sequences, and the probability is \(\frac{168}{1296}=\frac{7}{54}.\) Therefore, the answer is B.

25 答案 C

中间一行的每个方格必须恰好接触一根牙签。先考虑从中间条带一侧穿到另一侧的回路。回路可以跨越全部 \(8\) 列、前 \(7\) 列、后 \(7\) 列，或中间 \(6\) 列；更窄的跨度会使外侧的一个中间方格没有接触牙签。固定两个端点后，每个内部的中间方格都可以独立选择把唯一一根牙签放在上边或下边，其余不自交回路随即唯一确定。因此四种情况分别给出 \(2^6,2^5,2^5\)，和 \(2^4\) 条回路。另有恰好两条不穿过中间条带的回路：完全沿顶部或完全沿底部延伸的水平长方形。所以总数为 \(2^6+2^5+2^5+2^4+2\) \(=64+32+32+16+2\) \(=146\)。所以正确答案是 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, C is the correct answer.