2023B AMC 12 真题（P版）

1 答案 C

橙汁总量是 \(3+\tfrac13=\tfrac{10}{3}\) 杯。平均分成四份后，每个杯子应有 \(\tfrac{10}{3}\div 4=\tfrac{5}{6}\) 杯。因此前三个杯子中的每一个都必须倒出 \(1-\tfrac{5}{6}=\tfrac{1}{6}\)。

因此，正确答案是 C。

English

The total juice is \(3+\tfrac13=\tfrac{10}{3}\) glasses. Split evenly, each glass should hold \(\tfrac{10}{3}\div 4=\tfrac{5}{6}\) of a glass. Each of the first three glasses must therefore give up \(1-\tfrac{5}{6}=\tfrac{1}{6}.\)

Thus, the correct answer is C.

2 答案 B

原价为 \(p\) 的一双鞋最终费用是 \(0.8\times 1.075\times p=0.86p\)。由 \(0.86p\le 43\) 得 \(p\le 50\)，所以他能买得起的最贵跑鞋原价为 \(50\) 美元。

因此，正确答案是 B。

English

The final cost of a pair with original price \(p\) is \(0.8\times 1.075\times p=0.86p.\) Setting \(0.86p\le 43\) gives \(p\le 50,\) so the most expensive affordable pair originally cost \(50\) dollars.

Thus, the correct answer is B.

3 答案 D

内接直角三角形的斜边是直径，所以圆 \(A\) 的直径为 \(5\)，圆 \(B\) 的直径为 \(13\)。面积之比为 \(\left(\tfrac{5}{13}\right)^2= \tfrac{25}{169}\)。

因此，正确答案是 D。

English

The hypotenuse of an inscribed right triangle is a diameter, so circle \(A\) has diameter \(5\) and circle \(B\) has diameter \(13.\) The ratio of areas is \(\left(\tfrac{5}{13}\right)^2= \tfrac{25}{169}.\)

Thus, the correct answer is D.

4 答案 C

换算单位后，这条窄条宽 \(0.65\) 厘米，长 \(2500\) 厘米，所以面积为 \(0.65\times 2500=1625\) 平方厘米。

因此，正确答案是 C。

English

Converting units, the strip is \(0.65\) cm wide and \(2500\) cm long, so its area is \(0.65\times 2500=1625\) square centimeters.

Thus, the correct answer is C.

5 答案 C

一组猜过的小正方形能保证碰到这个骨牌，当且仅当未猜的小正方形中没有两个相邻；否则骨牌 可以藏在那一对相邻小正方形上。\(3\times 3\) 方格中两两不相邻的小正方形集合最大是 \(5\) 个小正方形的棋盘格集合（四个角加中心）。所以最多能留下 \(5\) 个小正方形不猜，必须猜 \(9-5=4\)。

因此，正确答案是 C。

English

A set of guessed squares is guaranteed to hit the domino if and only if the un-guessed squares contain no two adjacent squares, since otherwise the domino could hide on that adjacent pair. The largest set of pairwise non-adjacent squares in the \(3\times 3\) grid is the \(5\)-square checkerboard (four corners plus the center). So at most \(5\) squares can be left unguessed, and you must guess \(9-5=4.\)

Thus, the correct answer is C.

6 答案 C

因子 \((x-k)\) 的指数是 \(k\)，所以 \(P\) 的符号只在 \(x=k\) 且 \(k\) 为奇数时改变，也就是在 \(1,3,5,7,9\) 处改变。当 \(x\gt 10\) 时每个因子都为正，所以 \(P\gt 0\)。向左扫描并在每个奇数根处翻转符号，正区间为 \((10,\infty)\)、\((9,10)\)、\((6,7)\)、\((5,6)\)、\((2,3)\) 和 \((1,2)\)，共六个区间。

因此，正确答案是 C。

English

The exponent of the factor \((x-k)\) is \(k,\) so the sign of \(P\) changes at \(x=k\) only when \(k\) is odd, i.e. at \(1,3,5,7,9.\) For \(x\gt 10\) every factor is positive, so \(P\gt 0.\) Sweeping left and flipping at each odd root, the positive intervals are \((10,\infty),\) \((9,10),\) \((6,7),\) \((5,6),\) \((2,3),\) and \((1,2)\) — six intervals in all.

Thus, the correct answer is C.

7 答案 E

设 \(L=\log n\)，则 \(\log(n^2)-(\log n)^2\) \(=2L-L^2\) \(=L(2-L)\)，分式为 \(\dfrac{L(2-L)}{L-3}\)。作符号分析可知它 \(\ge 0\) 当且仅当 \(L\le 0\) 或 \(2\le L\lt 3\)。由于 \(n\) 是正整数，\(L\le 0\) 只给出 \(n=1\)，而 \(2\le L\lt 3\) 给出 \(100\le n\le 999\)，共 \(900\) 个值。总数为 \(1+900=901\)。

因此，正确答案是 E。

English

Write \(L=\log n.\) Then \(\log(n^2)-(\log n)^2\) \(=2L-L^2\) \(=L(2-L),\) and the fraction is \(\dfrac{L(2-L)}{L-3}.\) A sign chart shows this is \(\ge 0\) exactly when \(L\le 0\) or \(2\le L\lt 3.\) Since \(n\) is a positive integer, \(L\le 0\) forces \(n=1,\) while \(2\le L\lt 3\) gives \(100\le n\le 999,\) which is \(900\) values. In total \(1+900=901.\)

Thus, the correct answer is E.

8 答案 D

如果最小元素为 \(m\ (m\ge 1)\)，则 \(|B|=m\)，其余 \(m-1\) 个元素来自 \(\{m+1,\ldots,12\}\)，这个集合大小为 \(12-m\)。总数为 \[ \begin{aligned} \sum_{m\ge 1}\binom{12-m}{m-1} &=\binom{11}{0} \\ &\quad {}+\binom{10}{1} \\ &\quad {}+\binom{9}{2} \\ &\quad {}+\binom{8}{3} \\ &\quad {}+\binom{7}{4} \\ &\quad {}+\binom{6}{5}\text{，} \end{aligned} \] 等于 \(1+10+36+56+35+6\) \(=144\)。

因此，正确答案是 D。

English

If the least element is \(m\ (m\ge 1),\) then \(|B|=m\) and the remaining \(m-1\) elements come from \(\{m+1,\ldots,12\},\) a set of size \(12-m.\) The count is \[ \begin{aligned} \sum_{m\ge 1}\binom{12-m}{m-1} &=\binom{11}{0} \\ &\quad {}+\binom{10}{1} \\ &\quad {}+\binom{9}{2} \\ &\quad {}+\binom{8}{3} \\ &\quad {}+\binom{7}{4} \\ &\quad {}+\binom{6}{5}, \end{aligned} \] which equals \(1+10+36+56+35+6\) \(=144.\)

Thus, the correct answer is D.

9 答案 B

将 \(x,y\) 替换为 \(|x|,|y|\)，条件 \(|x-1|+|y-1|\le 1\) 描述的是以 \((1,1)\) 为中心、对角线长为 \(2\) 的菱形，因此面积为 \(2\)。它完全位于第一象限（只在单个点 接触坐标轴），所以关于两条坐标轴反射后得到 \(4\) 个互不重叠的副本。总面积为 \(4\times 2=8\)。

因此，正确答案是 B。

English

Replacing \(x,y\) by \(|x|,|y|,\) the condition \(|x-1|+|y-1|\le 1\) describes a diamond centered at \((1,1)\) with diagonals of length \(2,\) hence area \(2.\) It lies entirely in the first quadrant (touching the axes only at single points), so reflecting across the two axes produces \(4\) disjoint copies. The total area is \(4\times 2=8.\)

Thus, the correct answer is B.

10 答案 E

两个圆分别为 \((x-4)^2+y^2=16\) 和 \(x^2+(y-10)^2=100\)，即 \(x^2+y^2=8x\) 和 \(x^2+y^2=20y\)。相减得 \(8x=20y\)，所以交点都在 \(y=\tfrac{2}{5}x\) 上，该直线的斜率为 \(\tfrac{2}{5}\)。

因此，正确答案是 E。

English

The circles are \((x-4)^2+y^2=16\) and \(x^2+(y-10)^2=100,\) i.e. \(x^2+y^2=8x\) and \(x^2+y^2=20y.\) Subtracting gives \(8x=20y,\) so the intersection points lie on \(y=\tfrac{2}{5}x,\) which has slope \(\tfrac{2}{5}.\)

Thus, the correct answer is E.

11 答案 D

设两条底边为 \(a\) 和 \(2a\)。每条腰的水平偏移为 \(\tfrac{a}{2}\)，所以高为 \(\sqrt{1-\tfrac{a^2}{4}}\)，面积为 \(A=\tfrac{3a}{2}\sqrt{1-\tfrac{a^2}{4}}\)。于是 \(A^2=\tfrac94\left(a^2-\tfrac{a^4}{4}\right)\)，当 \(a^2=2\) 时最大。此时高为 \(\tfrac{1}{\sqrt2}\)，且 \(A=\tfrac{3\sqrt2}{2}\cdot\tfrac{1}{\sqrt2}=\tfrac{3}{2}\)。

因此，正确答案是 D。

English

Let the bases be \(a\) and \(2a.\) Each leg has horizontal offset \(\tfrac{a}{2},\) so the height is \(\sqrt{1-\tfrac{a^2}{4}}\) and the area is \(A=\tfrac{3a}{2}\sqrt{1-\tfrac{a^2}{4}}.\) Then \(A^2=\tfrac94\left(a^2-\tfrac{a^4}{4}\right),\) maximized when \(a^2=2.\) There the height is \(\tfrac{1}{\sqrt2}\) and \(A=\tfrac{3\sqrt2}{2}\cdot\tfrac{1}{\sqrt2}=\tfrac{3}{2}.\)

Thus, the correct answer is D.

12 答案 E

令 \(z=a+bi\)，则 \(z\otimes z=a^2+b^2 i\)，且 \(z^2+40=(a^2-b^2+40)+2abi\)。比较实部得 \(a^2=a^2-b^2+40\)，所以 \(b^2=40\)。比较虚部得 \(b^2=2ab\)，所以 \(b=2a\)，\(a^2=\tfrac{b^2}{4}=10\)。因此 \(|z|^2=a^2+b^2=50\)，所以 \(|z|=5\sqrt{2}\)。

因此，正确答案是 E。

English

With \(z=a+bi,\) we have \(z\otimes z=a^2+b^2 i\) and \(z^2+40=(a^2-b^2+40)+2abi.\) The real parts give \(a^2=a^2-b^2+40,\) so \(b^2=40.\) The imaginary parts give \(b^2=2ab,\) so \(b=2a\) and \(a^2=\tfrac{b^2}{4}=10.\) Then \(|z|^2=a^2+b^2=50,\) so \(|z|=5\sqrt{2}.\)

Thus, the correct answer is E.

13 答案 D

由棱长和，\(4(a+b+c)=13\)，所以 \(a+b+c=\tfrac{13}{4}\)。由表面积，\(2(ab+bc+ca)=\tfrac{11}{2}\)，所以 \(ab+bc+ca=\tfrac{11}{4}\)。因此 \[ \begin{aligned} a^2+b^2+c^2 &=\left(\tfrac{13}{4}\right)^2-2\cdot\tfrac{11}{4} \\ &=\tfrac{169}{16}-\tfrac{88}{16} \\ &=\tfrac{81}{16}\text{，} \end{aligned} \] 所以对角线长为 \(\sqrt{\tfrac{81}{16}}=\tfrac{9}{4}\)。

因此，正确答案是 D。

English

From the edges, \(4(a+b+c)=13,\) so \(a+b+c=\tfrac{13}{4}.\) From the faces, \(2(ab+bc+ca)=\tfrac{11}{2},\) so \(ab+bc+ca=\tfrac{11}{4}.\) Then \[ \begin{aligned} a^2+b^2+c^2 &=\left(\tfrac{13}{4}\right)^2-2\cdot\tfrac{11}{4} \\ &=\tfrac{169}{16}-\tfrac{88}{16} \\ &=\tfrac{81}{16}, \end{aligned} \] so the diagonal is \(\sqrt{\tfrac{81}{16}}=\tfrac{9}{4}.\)

Thus, the correct answer is D.

14 答案 A

由韦达定理，三个不同整数根的乘积为 \(-6\)。三个不同整数且乘积为 \(-6\) 的集合为 \(\{1,2,-3\}\)，\(\{1,-2,3\}\)，\(\{-1,2,3\}\)，\(\{-1,-2,-3\}\)，和 \(\{1,-1,6\}\)。每个集合决定 \(a=-(p+q+r)\) 和 \(b=pq+qr+rp\)，并且这五个集合给出 不同的有序对，所以共有 \(5\) 个有序对 \((a,b)\)。

因此，正确答案是 A。

English

By Vieta, the three distinct integer roots multiply to \(-6.\) The sets of three distinct integers with product \(-6\) are \(\{1,2,-3\},\) \(\{1,-2,3\},\) \(\{-1,2,3\},\) \(\{-1,-2,-3\},\) and \(\{1,-1,6\}.\) Each set determines \(a=-(p+q+r)\) and \(b=pq+qr+rp,\) and all five give different pairs, so there are \(5\) ordered pairs \((a,b).\)

Thus, the correct answer is A.

15 答案 E

两边乘以 \(210\) 得 \(c=15a+14b\)。因为 \(15\equiv 1\pmod{14}\)，得 \(c\equiv a\pmod{14}\)，所以 \(\gcd(c,14)=1\) 当且仅当 \(\gcd(a,14)=1\)。因为 \(14\equiv -1\pmod{15}\)，得 \(c\equiv -b\pmod{15}\)，所以 \(\gcd(c,15)=1\) 当且仅当 \(\gcd(b,15)=1\)。又因为 \(210=14\cdot 15\) 且 \(\gcd(14,15)=1\)，陈述 III 成立：\(\gcd(c,210)=1\) 当且仅当两者都成立。陈述 II 是 III 的正向推论，因此为真。陈述 I 为假：如果 \(\gcd(a,14)=1\) 但 \(\gcd(b,15)\ne 1\)，则 \(\gcd(c,15)\ne 1\)，因而 \(\gcd(c,210)\ne 1\)。只有 II 和 III 为真。

因此，正确答案是 E。

English

Multiplying by \(210\) gives \(c=15a+14b.\) Since \(15\equiv 1\pmod{14},\) we get \(c\equiv a\pmod{14},\) so \(\gcd(c,14)=1\) iff \(\gcd(a,14)=1.\) Since \(14\equiv -1\pmod{15},\) we get \(c\equiv -b\pmod{15},\) so \(\gcd(c,15)=1\) iff \(\gcd(b,15)=1.\) As \(210=14\cdot 15\) with \(\gcd(14,15)=1,\) statement III follows: \(\gcd(c,210)=1\) iff both hold. Statement II is the forward implication of III, hence true. Statement I is false: if \(\gcd(a,14)=1\) but \(\gcd(b,15)\ne 1,\) then \(\gcd(c,15)\ne 1,\) so \(\gcd(c,210)\ne 1.\) Only II and III are true.

Thus, the correct answer is E.

16 答案 D

金额 \(30,31,32,33,34,35\) 都能凑出（例如 \(30=6\cdot 5\)，\(31=6+10+15\)，\(32=6\cdot 2+10\cdot 2\)，\(33=6\cdot 3+15\)，\(34=6\cdot 4+10\)，\(35=10+10+15\)）。之后不断加上 \(6\) 元硬币，就能凑出所有更大的金额。检查更小的数，\(29\) 无法凑出：若不用 \(15\) 元硬币，若干 \(6\) 元和 \(10\) 元硬币的总额为偶数；若用一枚 \(15\) 元硬币，剩余的 \(14\) 不能写成 \(6a+10b\)；两枚 \(15\) 元硬币已经超过 \(29\)。所以最大无法凑出的金额是 \(29\)，其数位和为 \(2+9=11\)。

因此，正确答案是 D。

English

The amounts \(30,31,32,33,34,35\) are all attainable (for instance \(30=6\cdot 5,\) \(31=6+10+15,\) \(32=6\cdot 2+10\cdot 2,\) \(33=6\cdot 3+15,\) \(34=6\cdot 4+10,\) \(35=10+10+15\)). Adding \(6\)’s then reaches every larger amount. Checking below, \(29\) is impossible: without a \(15\)-coin a sum of \(6\)’s and \(10\)’s is even, while with one \(15\)-coin the remaining \(14\) cannot be written as \(6a+10b;\) two \(15\)-coins already exceed \(29.\) So the largest impossible amount is \(29,\) whose digit sum is \(2+9=11.\)

Thus, the correct answer is D.

17 答案 E

设边长为 \(6,\ 6+d,\ 6+2d\)。\(120^\circ\) 角对着最长边，所以 \((6+2d)^2=6^2+(6+d)^2\) \(-2\cdot 6\cdot(6+d)\cos 120^\circ\)。用 \(\cos 120^\circ=-\tfrac12\) 得 \(3d^2+6d-72=0\)，所以 \(d=4\)，边长为 \(6,10,14\)。面积为 \(\tfrac12\cdot 6\cdot 10\cdot\sin 120^\circ\) \(=30\cdot\tfrac{\sqrt3}{2}\) \(=15\sqrt3\)。

因此，正确答案是 E。

English

Let the sides be \(6,\ 6+d,\ 6+2d.\) The \(120^\circ\) angle faces the longest side, so \((6+2d)^2=6^2+(6+d)^2\) \(-2\cdot 6\cdot(6+d)\cos 120^\circ.\) Using \(\cos 120^\circ=-\tfrac12\) gives \(3d^2+6d-72=0,\) so \(d=4\) and the sides are \(6,10,14.\) The area is \(\tfrac12\cdot 6\cdot 10\cdot\sin 120^\circ\) \(=30\cdot\tfrac{\sqrt3}{2}\) \(=15\sqrt3.\)

Thus, the correct answer is E.

18 答案 A

把 Zelda 第一学期的平均分从四个学期平均分中同时减去，并不影响比较。于是可以把学期平均分写成 Zelda 的 \(0,18\) 和 Yolanda 的 \(3,21\)。设 \(p\) 和 \(q\) 分别是 Yolanda 和 Zelda 在第二学期完成的小测所占的比例。她们的学年平均分之差为 \[ (3+18p)-18q=3+18(p-q)\text{。}\] 因为 \(0\lt p,q\lt 1\)，这个差小于 \(21\)，所以不可能等于 \(22\)。

其他陈述确实可能发生。取 \((p,q)=(\tfrac14,\tfrac34)\) 会使 Zelda 的平均分更高；取 \(p=q\) 会使 Yolanda 的平均分高 \(3\) 分（把 Zelda 的每次小测成绩都加上 \(3\) 分之后，这也验证了最后一个选项）；取 \((p,q)=(\tfrac13,\tfrac12)\) 会使两人的学年平均分相等。上面出现的每个分数都能用正整数的小测次数实现。

因此，正确答案是 A。

English

Subtracting Zelda’s first-semester average from all four semester averages does not affect the comparison. We may therefore write the semester averages as \(0,18\) for Zelda and \(3,21\) for Yolanda. Let \(p\) and \(q\) be the fractions of Yolanda’s and Zelda’s quizzes, respectively, that occurred in the second semester. Their yearly-average difference is \[ (3+18p)-18q=3+18(p-q). \] Because \(0\lt p,q\lt 1,\) this difference is less than \(21,\) so it cannot be \(22.\)

The other options really can occur. Taking \((p,q)=(\tfrac14,\tfrac34)\) makes Zelda’s average higher; taking \(p=q\) makes Yolanda’s average \(3\) points higher (and also verifies the last option after adding \(3\) to every Zelda score); and taking \((p,q)=(\tfrac13,\tfrac12)\) makes the yearly averages equal. Each displayed fraction can be realized by positive integer quiz counts.

Thus, the correct answer is A.

19 答案 E

用奇偶筛计数三个箱子全为奇数的分配数，对奇数 \(n\) 可得 \[ \scriptsize\frac{1}{8}\sum_{S\subseteq\{1,2,3\}}(-1)^{|S|}(3-2|S|)^n=\frac{3^n-3}{4} \] 再除以总分配数 \(3^n\)，概率为 \(\dfrac{3^n-3}{4\cdot 3^n}\)，当 \(n=2023\) 时极其接近 \(\tfrac14\)。

因此，正确答案是 E。

English

Counting assignments where all three bins are odd with the parity filter gives \[ \scriptsize\frac{1}{8}\sum_{S\subseteq\{1,2,3\}}(-1)^{|S|}(3-2|S|)^n=\frac{3^n-3}{4} \] for odd \(n.\) Dividing by the \(3^n\) total assignments, the probability is \(\dfrac{3^n-3}{4\cdot 3^n},\) which for \(n=2023\) is extremely close to \(\tfrac14.\)

Thus, the correct answer is E.

20 答案 E

取第一次跳跃为 \((2,0)\)，第二次为 \((2\cos\theta,2\sin\theta)\)，其中 \(\theta\) 在 \([0,2\pi)\) 上均匀分布。落点距离满足 \(R^2=(2+2\cos\theta)^2\) \(+(2\sin\theta)^2\) \(=8+8\cos\theta\)。需要 \(R\lt 1\)，即 \(\cos\theta\lt-\tfrac78\)。这样的角的测度为 \(2\arccos\tfrac78\)，所以概率为 \(\dfrac{2\arccos\tfrac78}{2\pi}= \dfrac{\arccos\tfrac78}{\pi}\)。利用 \(\arccos(1-2x^2)=2\arcsin x\)，并取 \(x=\tfrac14\)，得到 \(\arccos\tfrac78=2\arcsin\tfrac14\)，因此概率为 \(\dfrac{2\arcsin\tfrac14}{\pi}\)。

因此，正确答案是 E。

English

Take the first jump as \((2,0)\) and the second as \((2\cos\theta,2\sin\theta)\) with \(\theta\) uniform on \([0,2\pi).\) The landing distance satisfies \(R^2=(2+2\cos\theta)^2\) \(+(2\sin\theta)^2\) \(=8+8\cos\theta.\) We need \(R\lt 1,\) i.e. \(\cos\theta\lt-\tfrac78.\) The measure of such angles is \(2\arccos\tfrac78,\) so the probability is \(\dfrac{2\arccos\tfrac78}{2\pi}= \dfrac{\arccos\tfrac78}{\pi}.\) Using \(\arccos(1-2x^2)=2\arcsin x\) with \(x=\tfrac14\) gives \(\arccos\tfrac78=2\arcsin\tfrac14,\) so the probability is \(\dfrac{2\arcsin\tfrac14}{\pi}.\)

Thus, the correct answer is E.

21 答案 E

将圆锥台延伸成完整圆锥。由于半径为 \(3\) 和 \(6\)，斜高带宽为 \(6\)，顶点到上边缘的斜距为 \(6\)，到下边缘的斜距为 \(12\)。下底周长 \(12\pi\) 展开成半径 \(12\)、角度 \(\dfrac{12\pi}{12}=\pi\) 的扇形。在这个展开图中把虫子放在 \((12,0)\)；蜂蜜位于沿底边走半圈的位置，在展开图中半径为 \(6\)、角度为 \(\tfrac{\pi}{2}\)。连接它们的直弦会进入半径 \(6\) 以内（不在表面上），所以测地线与半径 \(6\) 的圆相切：切线长为 \(\sqrt{12^2-6^2}=6\sqrt3\)，切点角度为 \(\tfrac{\pi}{3}\)，之后沿半径 \(6\) 的圆弧走过角度 \(\tfrac{\pi}{6}\)，弧长为 \(6\cdot\tfrac{\pi}{6}=\pi\)。最短路径为 \(6\sqrt3+\pi\)。

因此，正确答案是 E。

English

Extend the frustum to a full cone. Since the radii are \(3\) and \(6\) with slant band \(6,\) the apex is slant distance \(6\) from the top rim and \(12\) from the bottom rim. The bottom circumference \(12\pi\) unrolls to a sector of radius \(12\) and angle \(\dfrac{12\pi}{12}=\pi.\) Place the bug at \((12,0)\) in this pattern; the honey, halfway around the base, is at radius \(6\) and angle \(\tfrac{\pi}{2}.\) The straight chord between them passes within radius \(6\) (off the surface), so the geodesic goes tangent to the circle of radius \(6:\) the tangent has length \(\sqrt{12^2-6^2}=6\sqrt3\) and touches at angle \(\tfrac{\pi}{3},\) after which the path follows the arc of angle \(\tfrac{\pi}{6}\) on radius \(6,\) of length \(6\cdot\tfrac{\pi}{6}=\pi.\) The shortest path is \(6\sqrt3+\pi.\)

Thus, the correct answer is E.

22 答案 E

令 \(a=b=0\) 得 \(2f(0)=2f(0)^2\)，所以 \(f(0)=0\) 或 \(f(0)=1\)。如果 \(f(0)=0\)，则令 \(b=0\) 可推出 \(f\equiv 0\)，此时 \(f(1)=0\)。否则 \(f(0)=1\)，令 \(a=b\) 得 \(f(2a)=2f(a)^2-1\ge -1\) 对每个 \(a\) 都成立。特别地，取 \(a=\tfrac12\)，得 \(f(1)=2f\!\left(\tfrac12\right)^2-1\ge -1\)。所以 \(f(1)\ge -1\)，而且 \([-1,\infty)\) 中的每个值都可达到（例如 \(f(x)=\cos(kx)\) 或 \(\cosh(kx)\)）。因此 \(-2\) 不可能。

因此，正确答案是 E。

English

Setting \(a=b=0\) gives \(2f(0)=2f(0)^2,\) so \(f(0)=0\) or \(f(0)=1.\) If \(f(0)=0,\) then setting \(b=0\) forces \(f\equiv 0,\) giving \(f(1)=0.\) Otherwise \(f(0)=1,\) and setting \(a=b\) gives \(f(2a)=2f(a)^2-1\ge -1\) for every \(a.\) In particular, with \(a=\tfrac12,\) \(f(1)=2f\!\left(\tfrac12\right)^2-1\ge -1.\) So \(f(1)\ge -1,\) and indeed every value in \([-1,\infty)\) is attainable (e.g. \(f(x)=\cos(kx)\) or \(\cosh(kx)\)). Hence \(-2\) is impossible.

Thus, the correct answer is E.

23 答案 A

每个骰子在质数 \(2,3,5\) 上贡献一个指数向量（点数 \(1\to(0,0,0)\)，\(2\to(1,0,0)\)，\(3\to(0,1,0)\)，\(4\to(2,0,0)\)，\(5\to(0,0,1)\)，\(6\to(1,1,0)\)），而乘积由这些向量之和决定。

要求恰有 \(k\) 枚骰子显示 \(5\)，从而固定 \(5\) 的指数，并令 \(m=n-k\)。对 \(3\) 的指数 \(b\)，其中 \(0\le b\le m\)，\(2\) 的可能指数 \(a\) 恰好是 \(0,1,\ldots,2m-b\)。若 \(a\le b\)，就用 \(a\) 枚显示 \(6\) 的骰子和 \(b-a\) 枚显示 \(3\) 的骰子。若 \(a\gt b\)，就用 \(b\) 枚显示 \(6\) 的骰子，再用显示 \(4\) 的骰子（必要时加一枚显示 \(2\) 的骰子）凑出余下的 \(a-b\) 个因子 \(2\)。由 \(a\le 2m-b\) 可知这样至多用 \(m\) 枚骰子；其余骰子都显示 \(1\)。因此指数对的个数为 \[ \begin{gathered} \sum_{b=0}^{m}(2m-b+1)\\ {}=\frac{(m+1)(3m+2)}2 \end{gathered}\text{。}\]

对 \(k=0,1,\ldots,n\) 求和，也就是对 \(m=0,1,\ldots,n\) 求和，不同乘积的个数为 \[ \begin{gathered} \sum_{m=0}^{n}\frac{(m+1)(3m+2)}2\\ {}=\frac{(n+1)^2(n+2)}2 \end{gathered}\text{。}\] 当 \(n=11\) 时它等于 \(\dfrac{12^2\cdot13}{2}=936\)，所以 \(n=11\)。

因此，正确答案是 A。

English

Each die contributes an exponent vector in the primes \(2,3,5\) (face \(1\to(0,0,0),\) \(2\to(1,0,0),\) \(3\to(0,1,0),\) \(4\to(2,0,0),\) \(5\to(0,0,1),\) \(6\to(1,1,0)\)), and a product is determined by the sum of these vectors.

Fix the exponent of \(5\) by requiring exactly \(k\) dice to show \(5,\) and put \(m=n-k.\) For an exponent \(b\) of \(3,\) where \(0\le b\le m,\) the possible exponents \(a\) of \(2\) are precisely \(0,1,\ldots,2m-b.\) If \(a\le b,\) use \(a\) faces showing \(6\) and \(b-a\) showing \(3.\) If \(a\gt b,\) use \(b\) faces showing \(6\) and make the remaining \(a-b\) factors of \(2\) with faces \(4\) and, if needed, one face \(2.\) The bound \(a\le 2m-b\) says this uses at most \(m\) dice; fill unused dice with \(1\)’s. Thus the number of exponent pairs is \[ \begin{gathered} \sum_{b=0}^{m}(2m-b+1)\\ {}=\frac{(m+1)(3m+2)}2. \end{gathered} \]

Summing over \(k=0,1,\ldots,n\), equivalently over \(m=0,1,\ldots,n,\) the number of distinct products is \[ \begin{gathered} \sum_{m=0}^{n}\frac{(m+1)(3m+2)}2\\ {}=\frac{(n+1)^2(n+2)}2. \end{gathered} \] For \(n=11\) this is \(\dfrac{12^2\cdot13}{2}=936,\) so \(n=11.\)

Thus, the correct answer is A.

24 答案 C

对每个质数分别处理 \(a,b,c,d\) 中的指数。

质数 \(2\)（总指数 \(6\)）：\(\max(b,c)=1\) 强制 \(a=3\)；然后 \(b+c+d=3\)，且 \(\max(b,d)=\max(c,d)=2\)，得 \(d=2\) 且 \(\{b,c\}=\{0,1\}\)，所以最小指数为 \(0\)。

质数 \(3\)（总指数 \(9\)）：\(\max(a,b)=2\)，而其他最小公倍数都等于 \(3\)，强制 \(c=d=3\)；然后 \(a+b=3\)，且 \(\max(a,b)=2\)，得 \(\{a,b\}=\{1,2\}\)，所以最小指数为 \(1\)。

质数 \(5\)（总指数 \(7\)）：\(\max(a,b)=3\)，且 \(\max(b,c)\)，\(\max(b,d),\max(c,d)=2\)，强制 \(a=3\)；然后 \(b+c+d=4\)，每个都 \(\le 2\)，且两两最大值为 \(2\)，所以其中两个等于 \(2\)，一个等于 \(0\)，最小指数为 \(0\)。

因此 \(\gcd(a,b,c,d)=2^0\cdot 3^1\cdot 5^0=3\)。

因此，正确答案是 C。

English

Handle each prime separately using the exponents of \(a,b,c,d.\)

Prime \(2\) (total \(6\)): \(\max(b,c)=1\) forces \(a=3;\) then \(b+c+d=3\) with \(\max(b,d)=\max(c,d)=2\) gives \(d=2\) and \(\{b,c\}=\{0,1\},\) so the minimum exponent is \(0.\)

Prime \(3\) (total \(9\)): \(\max(a,b)=2\) with the other lcms equal to \(3\) forces \(c=d=3;\) then \(a+b=3\) with \(\max(a,b)=2\) gives \(\{a,b\}=\{1,2\},\) so the minimum is \(1.\)

Prime \(5\) (total \(7\)): \(\max(a,b)=3\) with \(\max(b,c),\) \(\max(b,d),\max(c,d)=2\) forces \(a=3;\) then \(b+c+d=4\) with each \(\le 2\) and pairwise maxima \(2\) gives two of them equal to \(2\) and one equal to \(0,\) so the minimum is \(0.\)

Therefore \(\gcd(a,b,c,d)=2^0\cdot 3^1\cdot 5^0=3.\)

Thus, the correct answer is C.

25 答案 B

设原五边形的外接圆半径为 \(R\)。将一个顶点折到中心时，折痕是中心到该顶点线段的垂直平分线，即离中心距离为 \(\tfrac{R}{2}\) 的直线。五条折痕围成一个内切圆半径为 \(\tfrac{R}{2}\) 的正五边形，而原五边形的内切圆半径为 \(R\cos 36^\circ\)。面积按内切圆半径的平方成比例，所以比例为 \[ \frac{(\frac{R}{2})^2}{(R\cos 36^\circ)^2}=\frac{1}{4\cos^2 36^\circ}\text{。}\] 因为 \(\cos 36^\circ=\tfrac{1+\sqrt5}{4}\)，这个比例为 \(\tfrac{4}{(1+\sqrt5)^2}\) \(=\tfrac{4}{6+2\sqrt5}\) \(=\tfrac{2}{3+\sqrt5}\) \(=\tfrac{3-\sqrt5}{2}\)。乘以原面积 \(\sqrt5+1\) 得 \(\tfrac{(3-\sqrt5)(\sqrt5+1)}{2}\) \(=\tfrac{2\sqrt5-2}{2}\) \(=\sqrt5-1\)。

因此，正确答案是 B。

English

Let the original pentagon have circumradius \(R.\) Folding a vertex to the center creases along the perpendicular bisector of the segment from the center to that vertex, a line at distance \(\tfrac{R}{2}\) from the center. The five creases bound a regular pentagon with apothem \(\tfrac{R}{2},\) whereas the original has apothem \(R\cos 36^\circ.\) Areas scale as the square of the apothem, so the ratio is \[ \frac{(\frac{R}{2})^2}{(R\cos 36^\circ)^2}=\frac{1}{4\cos^2 36^\circ}. \] Since \(\cos 36^\circ=\tfrac{1+\sqrt5}{4},\) this ratio is \(\tfrac{4}{(1+\sqrt5)^2}\) \(=\tfrac{4}{6+2\sqrt5}\) \(=\tfrac{2}{3+\sqrt5}\) \(=\tfrac{3-\sqrt5}{2}.\) Multiplying by the original area \(\sqrt5+1\) gives \(\tfrac{(3-\sqrt5)(\sqrt5+1)}{2}\) \(=\tfrac{2\sqrt5-2}{2}\) \(=\sqrt5-1.\)

Thus, the correct answer is B.