2005A AMC 12 真题（P版）

1 答案 D

由 \(0.1x = 2\) 得 \(x = 20\)，由 \(0.2y = 2\) 得 \(y = 10\)。

因此 \(x - y = 20 - 10 = 10\)。

所以正确答案是 D。

English

From \(0.1x = 2\) we get \(x = 20,\) and from \(0.2y = 2\) we get \(y = 10.\)

Therefore \(x - y = 20 - 10 = 10.\)

Thus, the correct answer is D.

2 答案 B

解 \(2x + 7 = 3\) 得 \(x = -2\)。

代入第二个方程，得到 \(-2b - 10 = -2\)，所以 \(-2b = 8\)，从而 \(b = -4\)。

所以正确答案是 B。

English

Solving \(2x + 7 = 3\) gives \(x = -2.\)

Substituting into the second equation, \(-2b - 10 = -2,\) so \(-2b = 8\) and \(b = -4.\)

Thus, the correct answer is B.

3 答案 B

设宽为 \(w\)，则长为 \(2w\)。由勾股定理，\[ x^2 = w^2 + (2w)^2 = 5w^2\text{。}\]

面积为 \(w \cdot 2w = 2w^2 = \dfrac{2}{5}x^2\)。

所以正确答案是 B。

English

Let the width be \(w.\) Then the length is \(2w,\) and the diagonal gives \[ x^2 = w^2 + (2w)^2 = 5w^2. \]

The area is \(w \cdot 2w = 2w^2 = \dfrac{2}{5}x^2.\)

Thus, the correct answer is B.

4 答案 A

分别购买时，Dave 买 \(7\) 扇需付 \(6\) 扇的钱，即 \(\$600\)；Doug 买 \(8\) 扇需付 \(7\) 扇的钱，即 \(\$700\)，合计 \(\$1300\)。

合起来买时共需 \(15\) 扇，付 \(12\) 扇的钱可得 \(3\) 扇免费，费用为 \(\$1200\)。

因此节省 \(\$1300 - \$1200 = \$100\)。

所以正确答案是 A。

English

Buying separately, Dave gets \(7\) windows by paying for \(6\) (\(\$600\)), and Doug gets \(8\) by paying for \(7\) (\(\$700\)), for a total of \(\$1300.\)

Buying together, they need \(15\) windows: paying for \(12\) yields \(3\) free, for a cost of \(\$1200.\)

The savings are \(\$1300 - \$1200 = \$100.\)

Thus, the correct answer is A.

5 答案 B

全部 \(50\) 个数的总和为 \(20 \cdot 30 + 30 \cdot 20\) \(= 600 + 600\) \(= 1200\)。

平均数为 \(1200 \div 50 = 24\)。

所以正确答案是 B。

English

The total of all \(50\) numbers is \(20 \cdot 30 + 30 \cdot 20\) \(= 600 + 600\) \(= 1200.\)

The average is \(1200 \div 50 = 24.\)

Thus, the correct answer is B.

6 答案 B

因为路程等于速度乘以时间，Josh 骑的距离是 Mike 的 \(\dfrac{4}{5} \cdot 2 = \dfrac{8}{5}\)。

设 Mike 骑了 \(m\) 英里，则 \[ 13 = m + \dfrac{8}{5}m = \dfrac{13}{5}m\text{，} \] 所以 \(m = 5\)。

所以正确答案是 B。

English

Since distance is rate times time, Josh rode \(\dfrac{4}{5} \cdot 2 = \dfrac{8}{5}\) as far as Mike.

Let \(m\) be the miles Mike rode. Then \[ 13 = m + \dfrac{8}{5}m = \dfrac{13}{5}m, \] so \(m = 5.\)

Thus, the correct answer is B.

7 答案 C

由图形对称性，三角形 \(ABH\)、三角形 \(BCE\)、三角形 \(CDF\)、三角形 \(DAG\) 是全等直角三角形。于是 \[ \begin{aligned} &BH = CE = \sqrt{BC^2 - BE^2} \\ &= \sqrt{50 - 1} = 7 \end{aligned}\text{。}\]

因为 \(E\) 在 \(B\) 与 \(H\) 之间，内正方形边长为 \(EH = BH - BE = 7 - 1 = 6\)。

\(EFGH\) 的面积为 \(6^2 = 36\)。

所以正确答案是 C。

English

By the symmetry of the figure, triangles \(ABH,\) \(BCE,\) \(CDF,\) and \(DAG\) are congruent right triangles. Hence \[ \begin{aligned} &BH = CE = \sqrt{BC^2 - BE^2} \\ &= \sqrt{50 - 1} = 7. \end{aligned} \]

Since \(E\) lies between \(B\) and \(H,\) the side of the inner square is \(EH = BH - BE = 7 - 1 = 6.\)

Therefore the area of \(EFGH\) is \(6^2 = 36.\)

Thus, the correct answer is C.

8 答案 D

因为 \(A + M + C \le 9 + 9 + 9 = 27\)，且 \(2005 = 5 \cdot 401\)，所以数字和只能是 \(1\) 或 \(5\)。它不能是 \(1\)，否则就有 \(100A+10M+C=2005>999\)。因此它只能是较小的那个非平凡因数：\[ \begin{aligned} &100A + 10M + C = 401, \\ &\quad A + M + C = 5 \end{aligned}\text{。}\]

由第一式可直接读出 \(A = 4\)、\(M = 0\)、\(C = 1\)。

所以正确答案是 D。

English

Since \(A + M + C \le 9 + 9 + 9 = 27,\) and \(2005 = 5 \cdot 401,\) the digit sum can only be \(1\) or \(5.\) It cannot be \(1,\) because then \(100A+10M+C=2005>999.\) Thus it must be the smaller nontrivial factor: \[ \begin{aligned} &100A + 10M + C = 401, \\ &\quad A + M + C = 5. \end{aligned} \]

Reading off the digits, \(A = 4,\) \(M = 0,\) and \(C = 1.\)

Thus, the correct answer is D.

9 答案 A

方程为 \(4x^2 + (a+8)x + 9 = 0\)。它只有一个解时判别式为零： \[ (a+8)^2 - 4 \cdot 4 \cdot 9 = 0\text{，} \] 所以 \((a+8)^2 = 144\)，\(a + 8 = \pm 12\)。

因此 \(a = 4\) 或 \(a = -20\)，和为 \(-16\)。

所以正确答案是 A。

English

The equation is \(4x^2 + (a+8)x + 9 = 0.\) It has one solution when the discriminant vanishes: \[ (a+8)^2 - 4 \cdot 4 \cdot 9 = 0, \] so \((a+8)^2 = 144\) and \(a + 8 = \pm 12.\)

Thus \(a = 4\) or \(a = -20,\) and their sum is \(-16.\)

Thus, the correct answer is A.

10 答案 B

\(n^3\) 个单位立方体共有 \(6n^3\) 个面，其中红色面正是原立方体的表面，共 \(6n^2\) 个。

令红色面所占比例为四分之一：\[ \dfrac{6n^2}{6n^3} = \dfrac{1}{n} = \dfrac{1}{4}\text{，} \]所以 \(n = 4\)。

所以正确答案是 B。

English

The \(n^3\) unit cubes have \(6n^3\) faces total. The red faces are exactly the surface of the original cube, \(6n^2\) of them.

Setting the red fraction to one-fourth, \[ \dfrac{6n^2}{6n^3} = \dfrac{1}{n} = \dfrac{1}{4}, \] so \(n = 4.\)

Thus, the correct answer is B.

11 答案 E

中间数字为整数，当且仅当首位数字与末位数字奇偶性相同。每一对这样的首末数字都唯一确定中间数字。

首末数字都是奇数的选择有 \(5 \cdot 5 = 25\) 种；首末数字都是偶数时，首位不能为 \(0\)，所以有 \(4 \cdot 5 = 20\) 种。

总数为 \(25 + 20 = 45\)。

所以正确答案是 E。

English

The middle digit is an integer only when the first and last digits are both odd or both even. Each such pair determines the middle digit uniquely.

There are \(5 \cdot 5 = 25\) odd-odd choices for the first and last digits. For even-even, the first digit cannot be \(0,\) giving \(4 \cdot 5 = 20\) choices.

The total is \(25 + 20 = 45.\)

Thus, the correct answer is E.

12 答案 D

斜率为 \[ \dfrac{1000 - 1}{100 - 1} = \dfrac{999}{99} = \dfrac{111}{11}\text{。}\]

直线上的点形如 \((1 + 11t,\ 1 + 111t)\)，并且恰好在 \(t\) 为整数时是格点。该点严格位于 \(A\)、\(B\) 之间时，\(0 \lt t \lt 9\)。

这样的整数 \(t\) 共有 \(8\) 个，因此符合条件的格点有 \(8\) 个。

所以正确答案是 D。

English

The slope is \[ \dfrac{1000 - 1}{100 - 1} = \dfrac{999}{99} = \dfrac{111}{11}. \]

So every point on the line has the form \((1 + 11t,\ 1 + 111t),\) which is a lattice point exactly when \(t\) is an integer. The point is strictly between \(A\) and \(B\) when \(0 \lt t \lt 9.\)

There are \(8\) such integers \(t,\) giving \(8\) lattice points.

Thus, the correct answer is D.

13 答案 D

每个数字恰好作为五条线段中两条的端点，所以五个端点和的总和为 \[ 2(3 + 5 + 6 + 7 + 9) = 60\text{。}\]

五项等差数列的中项等于平均数，即 \(60 \div 5 = 12\)。

所以正确答案是 D。

English

Every number appears as an endpoint of exactly two of the five segments, so the total of the five sums is \[ 2(3 + 5 + 6 + 7 + 9) = 60. \]

The middle term of a five-term arithmetic sequence is its mean, namely \(60 \div 5 = 12.\)

Thus, the correct answer is D.

14 答案 D

骰子共有 \(21\) 个点，所以去掉的点来自 \(n\) 点面的概率为 \(\dfrac{n}{21}\)。

若从奇数面去掉一点，该面变成偶数面，只剩两个奇数面，所以朝上一面为奇数点的概率是 \(\dfrac{2}{6}=\dfrac{1}{3}\)；若从偶数面去掉一点，该面变成奇数面，共有四个奇数面，所以概率是 \(\dfrac{4}{6}=\dfrac{2}{3}\)。去掉点来自奇数面的概率为 \(\dfrac{1 + 3 + 5}{21}\)，来自偶数面的概率为 \(\dfrac{2 + 4 + 6}{21}\)。

因此所求概率为 \[ \dfrac{1}{3} \cdot \dfrac{9}{21} + \dfrac{2}{3} \cdot \dfrac{12}{21} = \dfrac{33}{63} = \dfrac{11}{21}\text{。}\]

所以正确答案是 D。

English

The die has \(21\) dots, so a dot is removed from the face with \(n\) dots with probability \(\dfrac{n}{21}.\)

If a dot is removed from an odd face, that face becomes even, leaving two odd faces and hence probability \(\dfrac{2}{6}=\dfrac{1}{3}\) of an odd top. If a dot is removed from an even face, that face becomes odd, leaving four odd faces and hence probability \(\dfrac{4}{6}=\dfrac{2}{3}.\) The removed dot lies on an odd face with probability \(\dfrac{1 + 3 + 5}{21}\) and on an even face with probability \(\dfrac{2 + 4 + 6}{21}.\)

Hence the answer is \[ \dfrac{1}{3} \cdot \dfrac{9}{21} + \dfrac{2}{3} \cdot \dfrac{12}{21} = \dfrac{33}{63} = \dfrac{11}{21}. \]

Thus, the correct answer is D.

15 答案 C

设 \(O\) 为圆心。由 \(2 \cdot AC = BC\) 可知 \(AC = \dfrac{AB}{3}\)，又有 \(AO = \dfrac{AB}{2}\)，所以 \[ \begin{aligned} &CO = AO - AC \\ &= \dfrac{AB}{2} - \dfrac{AB}{3} \\ &= \dfrac{AB}{6} \end{aligned}\text{。}\]

三角形 \(DCO\) 与三角形 \(DAB\) 共享从 \(D\) 到直线 \(AB\) 的高，因此 \[ \dfrac{[DCO]}{[DAB]} = \dfrac{CO}{AB} = \dfrac{1}{6}\text{。}\]

因为 \(O\) 是 \(DE\) 的中点，三角形 \(DCO\) 与 \(ECO\) 面积相等，所以 \[ \begin{aligned} &[DCE] = 2\,[DCO] \\ &= \dfrac{2}{6}[DAB] = \dfrac{1}{3}[DAB] \end{aligned}\text{。}\]

所以正确答案是 C。

English

Let \(O\) be the center. Since \(2 \cdot AC = BC,\) we have \(AC = \dfrac{AB}{3},\) and \(AO = \dfrac{AB}{2},\) so \[ \begin{aligned} &CO = AO - AC \\ &= \dfrac{AB}{2} - \dfrac{AB}{3} \\ &= \dfrac{AB}{6}. \end{aligned} \]

Triangles \(DCO\) and \(DAB\) share the same altitude from \(D\) to line \(AB,\) so \[ \dfrac{[DCO]}{[DAB]} = \dfrac{CO}{AB} = \dfrac{1}{6}. \]

Because \(O\) is the midpoint of \(DE,\) triangles \(DCO\) and \(ECO\) have equal areas, so \[ \begin{aligned} &[DCE] = 2\,[DCO] \\ &= \dfrac{2}{6}[DAB] = \dfrac{1}{3}[DAB]. \end{aligned} \]

Thus, the correct answer is C.

16 答案 D

大圆圆心为 \((r, r)\)，第二个小圆圆心为 \((3s, s)\)。两圆外切，所以圆心距为 \(r + s\)。

两个圆心之间的水平距离为 \(r - 3s\)，竖直距离为 \(r - s\)，因此 \[ (r + s)^2 = (r - 3s)^2 + (r - s)^2\text{。}\]

展开并整理得 \(0 = r^2 - 10rs + 9s^2\) \(= (r - 9s)(r - s)\)。因为 \(r \ne s\)，所以 \(r = 9s\)，故 \(\frac{r}{s} = 9\)。

所以正确答案是 D。

English

Put the big circle’s center at \((r, r)\) and the second small circle’s center at \((3s, s).\) They are externally tangent, so the distance between centers is \(r + s.\)

The horizontal and vertical gaps are \(r - 3s\) and \(r - s,\) so \[ (r + s)^2 = (r - 3s)^2 + (r - s)^2. \]

Expanding gives \(0 = r^2 - 10rs + 9s^2\) \(= (r - 9s)(r - s).\) Since \(r \ne s,\) we get \(r = 9s,\) so \(\frac{r}{s} = 9.\)

Thus, the correct answer is D.

17 答案 A

两组互相垂直的切割都从上底面的棱一直切到下底面的中线。在 \(W\) 附近，它们切出一个正四棱锥：它的底面是下底面中与 \(W\) 相邻的那四分之一，顶点位于下底面中心正上方一个单位处。

它的底面是边长 \(\dfrac{1}{2}\) 的正方形，即下底面的四分之一，高为 \(1\)。因此体积为 \[ \dfrac{1}{3} \left(\dfrac{1}{2}\right)^2 (1) = \dfrac{1}{12}\text{。}\]

所以正确答案是 A。

English

The two perpendicular sets of cuts each run from top edges to midlines of the bottom face. Near \(W\) they carve out a square pyramid. Its base is the quarter of the bottom face adjacent to \(W,\) and its apex lies one unit above the center of the bottom face.

Thus its base is a square of side \(\dfrac{1}{2}\) and its altitude is the full height \(1.\) Therefore the volume is \[ \dfrac{1}{3} \left(\dfrac{1}{2}\right)^2 (1) = \dfrac{1}{12}. \]

Thus, the correct answer is A.

18 答案 A

在 \(1\) 到 \(999\) 这 \(999\) 个数中，由容斥原理可得 \[ \begin{aligned} &499 + 333 + 199 - 166 \\ &\quad {}- 99 - 66 + 33 = 733 \end{aligned} \] 个数能被 \(2, 3\) 或 \(5\) 中至少一个整除。

因此与 \(2, 3, 5\) 都互质的数共有 \(999 - 733 = 266\) 个；其中有 \(165\) 个质数（\(168\) 个质数去掉 \(2, 3, 5\)），另有 \(1\) 既不是质数也不是合数。

剩下 \(266 - 165 - 1 = 100\) 个就是看似质数。

所以正确答案是 A。

English

Among the \(999\) numbers from \(1\) to \(999,\) inclusion-exclusion gives \[ \begin{aligned} &499 + 333 + 199 - 166 \\ &\quad {}- 99 - 66 + 33 = 733 \end{aligned} \] that are divisible by \(2, 3,\) or \(5.\)

That leaves \(999 - 733 = 266\) numbers coprime to \(2, 3, 5.\) Of these, \(165\) are primes (the \(168\) primes minus \(2, 3, 5\)), and \(1\) is neither prime nor composite.

The remaining \(266 - 165 - 1 = 100\) numbers are prime-looking.

Thus, the correct answer is A.

19 答案 B

因为里程表从不显示 \(4\)，它只用 \(9\) 个符号计数，实际上相当于采用 \(9\) 进制；显示的 \(5, 6, 7, 8, 9\) 分别代表 \(9\) 进制数字 \(4, 5, 6, 7, 8\)。

读数 \(002005\) 对应 \(9\) 进制的 \(2004\)，其值为 \[ 2 \cdot 9^3 + 4 = 2 \cdot 729 + 4 = 1462\text{。}\]

所以正确答案是 B。

English

Because the odometer never displays a \(4,\) it uses only \(9\) symbols and counts in base \(9,\) where its digits \(5, 6, 7, 8, 9\) represent the base-\(9\) digits \(4, 5, 6, 7, 8.\)

The reading \(002005\) therefore corresponds to \(2004\) in base \(9,\) which equals \[ 2 \cdot 9^3 + 4 = 2 \cdot 729 + 4 = 1462. \]

Thus, the correct answer is B.

20 答案 E

设 \(g(n)\) 表示方程 \(f^{[n]}(x) = \tfrac{1}{2}\) 在 \([0, 1]\) 中的解数。由于 \(f\) 把两个半区间 \([0, \tfrac12]\) 与 \([\tfrac12, 1]\) 都映到整个 \([0, 1]\)，所以 \(f^{[n-1]}(y) = \tfrac12\) 的每个解都来自两个 \(x\) 值。

边界值 \(x = \tfrac12\) 满足 \(f^{[n]}(\tfrac12) = f^{[n-1]}(1) = 0 \ne \tfrac12\)，不会造成解的重合或丢失，因此 \(g(n) = 2\,g(n-1)\)。

由于 \(g(1) = 2\)，可得 \(g(2005) = 2^{2005}\)。

所以正确答案是 E。

English

Let \(g(n)\) count the solutions of \(f^{[n]}(x) = \tfrac{1}{2}\) in \([0, 1].\) Since \(f\) maps each of the two halves \([0, \tfrac12]\) and \([\tfrac12, 1]\) onto all of \([0, 1],\) every solution of \(f^{[n-1]}(y) = \tfrac12\) comes from two values of \(x\) (one in each half).

The boundary value \(x = \tfrac12\) satisfies \(f^{[n]}(\tfrac12) = f^{[n-1]}(1) = 0 \ne \tfrac12,\) so no solutions are lost, giving \(g(n) = 2\,g(n-1).\)

Since \(g(1) = 2,\) we conclude \(g(2005) = 2^{2005}.\)

Thus, the correct answer is E.

21 答案 C

条件 \(\log_a b = c^{2005}\) 等价于 \(b = a^{\left(c^{2005}\right)}\)。

若 \(c \ge 2\)，则 \(b = a^{\left(c^{2005}\right)} \ge 2^{\left(2^{2005}\right)}\)，远大于 \(2005\)，不可能满足 \(a + b + c = 2005\)。

当 \(c = 0\) 时，\(b = a^0 = 1\)，且 \(a + 1 + 0 = 2005\)，得到 \((a, b, c) = (2004, 1, 0)\)。当 \(c = 1\) 时，\(b = a^1 = a\)，且 \(2a + 1 = 2005\)，得到 \((a, b, c) = (1002, 1002, 1)\)。

共有 \(2\) 个三元组。

所以正确答案是 C。

English

The condition \(\log_a b = c^{2005}\) means \(b = a^{\left(c^{2005}\right)}.\)

If \(c \ge 2,\) then \(b = a^{\left(c^{2005}\right)} \ge 2^{\left(2^{2005}\right)},\) which vastly exceeds \(2005,\) so \(a + b + c = 2005\) is impossible.

For \(c = 0:\) \(b = a^0 = 1,\) so \(a + 1 + 0 = 2005\) gives \((a, b, c) = (2004, 1, 0).\) For \(c = 1:\) \(b = a^1 = a,\) so \(2a + 1 = 2005\) gives \((a, b, c) = (1002, 1002, 1).\)

There are \(2\) such triples.

Thus, the correct answer is C.

22 答案 B

设长方体的尺寸为 \(x, y, z\)。\(12\) 条棱给出 \(4(x + y + z) = 112\)，所以 \(x + y + z = 28\)；表面积给出 \(2xy + 2yz + 2xz = 384\)。

空间对角线是球的直径，因此 \[ \begin{aligned} &(2r)^2 = x^2 + y^2 + z^2 \\ &= (x + y + z)^2 \\ &\quad {}- (2xy + 2yz + 2xz) \\ &= 28^2 - 384 = 400 \end{aligned}\text{。}\]

因此 \(2r = 20\)，所以 \(r = 10\)。

所以正确答案是 B。

English

Let the dimensions be \(x, y, z.\) The \(12\) edges give \(4(x + y + z) = 112,\) so \(x + y + z = 28,\) and the surface area gives \(2xy + 2yz + 2xz = 384.\)

The space diagonal is a diameter of the sphere, so \[ \begin{aligned} &(2r)^2 = x^2 + y^2 + z^2 \\ &= (x + y + z)^2 \\ &\quad {}- (2xy + 2yz + 2xz) \\ &= 28^2 - 384 = 400. \end{aligned} \]

Thus \(2r = 20\) and \(r = 10.\)

Thus, the correct answer is B.

23 答案 B

设 \(a = 2^j\)，\(b = 2^k\)。则 \(\log_a b = \dfrac{k}{j}\)，它为整数当且仅当 \(k\) 是 \(j\) 的倍数。

对每个 \(j\)，满足 \(k \ne j\) 且位于 \(\{1, \ldots, 25\}\) 中的有效指数共有 \(\left\lfloor \tfrac{25}{j} \right\rfloor - 1\) 个。对 \(j\) 求和得到 \[ \begin{aligned} &24 + 11 + 7 + 5 + 4 + 3 + 2 \\ &\quad {}+ 2 + 4 \cdot 1 = 62 \end{aligned} \] 个有序对 \((a, b)\)。

总共有 \(25 \cdot 24 = 600\) 个有序不同对，所以概率为 \(\dfrac{62}{600} = \dfrac{31}{300}\)。

所以正确答案是 B。

English

Let \(a = 2^j\) and \(b = 2^k.\) Then \(\log_a b = \dfrac{k}{j},\) which is an integer exactly when \(k\) is a multiple of \(j.\)

For each \(j,\) the number of valid \(k \ne j\) in \(\{1, \ldots, 25\}\) is \(\left\lfloor \tfrac{25}{j} \right\rfloor - 1.\) Summing over \(j\) gives \[ \begin{aligned} &24 + 11 + 7 + 5 + 4 + 3 + 2 \\ &\quad {}+ 2 + 4 \cdot 1 = 62 \end{aligned} \] ordered pairs \((a, b).\)

Since there are \(25 \cdot 24 = 600\) ordered pairs of distinct elements, the probability is \(\dfrac{62}{600} = \dfrac{31}{300}.\)

Thus, the correct answer is B.

24 答案 B

因为 \(P(x)R(x)\) 次数为 \(6\)，而 \(P(Q(x))\) 次数为 \(3\deg Q\)，所以 \(\deg Q = 2\)。二次多项式 \(Q\) 由有序三元组 \((Q(1), Q(2), Q(3))\) 决定。

当 \(x = 1, 2, 3\) 时，右侧为零，所以 \(P(Q(x)) = 0\)，迫使每个 \(Q(1), Q(2), Q(3)\) 都属于 \(\{1, 2, 3\}\)，共有 \(27\) 个三元组。

其中 \(2\) 次以下的情形有五个：三个常数三元组 \((1,1,1), (2,2,2), (3,3,3)\)，以及线性函数 \(Q(x) = x\) 的 \((1,2,3)\) 和 \(Q(x) = 4 - x\) 的 \((3,2,1)\)。其余 \(27 - 5 = 22\) 个给出真正的二次多项式。

所以正确答案是 B。

English

Since \(P(x)R(x)\) has degree \(6\) and \(P(Q(x))\) has degree \(3\deg Q,\) we need \(\deg Q = 2.\) A quadratic \(Q\) is determined by the ordered triple \((Q(1), Q(2), Q(3)).\)

At \(x = 1, 2, 3\) the right side vanishes, so \(P(Q(x)) = 0,\) forcing each of \(Q(1), Q(2), Q(3)\) into \(\{1, 2, 3\}.\) That gives \(27\) triples.

Five of them give a polynomial of degree less than \(2:\) the constants from \((1,1,1), (2,2,2), (3,3,3)\) and the linear \(Q(x) = x\) from \((1,2,3)\) and \(Q(x) = 4 - x\) from \((3,2,1).\) The other \(27 - 5 = 22\) triples are non-collinear and yield genuine quadratics.

Thus, the correct answer is B.

25 答案 C

这样的三角形三条边长度相等。检查 \(3 \times 3 \times 3\) 网格中的可能平方边长，只有三类边会出现。

单位立方体的面对角线（长 \(\sqrt2\)）：\(8\) 个单位立方体中每个贡献 \(8\) 个三角形，共 \(8 \cdot 8 = 64\) 个。

\(2 \times 2 \times 2\) 大立方体的面对角线（长 \(2\sqrt2\)）：每个顶点相邻的三个面形成一个三角形，共 \(8\) 个。

边中点之间的线段（长 \(\sqrt6\)）：\(12\) 个边中点中的每一个，都是两个这种三角形的顶点，所以共有 \(\dfrac{12 \cdot 2}{3} = 8\) 个。

总数为 \(64 + 8 + 8 = 80\)。

所以正确答案是 C。

English

The three equal sides of such a triangle must all have the same length. Checking the possible squared lengths in the \(3 \times 3 \times 3\) grid, only three families of side occur.

Face diagonals of a unit cube (length \(\sqrt2\)): each of the \(8\) unit cubes contributes \(8\) triangles, one at each corner, for \(8 \cdot 8 = 64.\)

Face diagonals of the \(2 \times 2 \times 2\) cube (length \(2\sqrt2\)): the three faces meeting at a vertex form one triangle, giving \(8\) triangles.

Edge-midpoint segments (length \(\sqrt6,\) joining midpoints of two edges): each of the \(12\) edge midpoints is a vertex of two such triangles, for \(\dfrac{12 \cdot 2}{3} = 8.\)

The total is \(64 + 8 + 8 = 80.\)

Thus, the correct answer is C.