2005A AMC 10 真题（P版）

1 答案 D

Mike 的 \(\$2\) 小费是账单的 \(10\%\)，所以他的账单是 \(2 \times 10 = 20\) 美元。Joe 的 \(\$2\) 小费是账单的 \(20\%\)，所以他的账单是 \(2 \times 5 = 10\) 美元。差为 \(20 - 10 = 10\) 美元。

所以正确答案是 D。

English

Mike’s \(\$2\) tip is \(10\%\) of his bill, so his bill is \(2 \times 10 = 20\) dollars. Joe’s \(\$2\) tip is \(20\%\) of his bill, so his bill is \(2 \times 5 = 10\) dollars. The difference is \(20 - 10 = 10\) dollars.

Thus, the correct answer is D.

2 答案 C

首先，\((1 \star 2) = \dfrac{1+2}{1-2} = \dfrac{3}{-1} = -3\)。然后，\((-3 \star 3) = \dfrac{-3+3}{-3-3} = \dfrac{0}{-6} = 0\)。

所以正确答案是 C。

English

First \((1 \star 2) = \dfrac{1+2}{1-2} = \dfrac{3}{-1} = -3.\) Then \((-3 \star 3) = \dfrac{-3+3}{-3-3} = \dfrac{0}{-6} = 0.\)

Thus, the correct answer is C.

3 答案 B

由 \(2x + 7 = 3\) 得 \(x = -2\)。代入另一个方程，\(-2b - 10 = -2\)，所以 \(-2b = 8\)，\(b = -4\)。

所以正确答案是 B。

English

From \(2x + 7 = 3\) we get \(x = -2.\) Substituting, \(-2b - 10 = -2,\) so \(-2b = 8\) and \(b = -4.\)

Thus, the correct answer is B.

4 答案 B

设宽为 \(w\)，长为 \(2w\)。由勾股定理，\(x^2 = w^2 + (2w)^2 = 5w^2\)，所以 \(w^2 = \dfrac{x^2}{5}\)。长方形面积为 \(w \cdot 2w = 2w^2 = \dfrac{2}{5}x^2\)。

所以正确答案是 B。

English

Let the width be \(w,\) so the length is \(2w.\) Then \(x^2 = w^2 + (2w)^2 = 5w^2,\) giving \(w^2 = \dfrac{x^2}{5}.\) The area is \(w \cdot 2w = 2w^2 = \dfrac{2}{5}x^2.\)

Thus, the correct answer is B.

5 答案 A

Dave 单独购买时，付 \(6\) 扇的钱并得到一扇免费窗户，达到 \(7\) 扇，共 \(\$600\)；Doug 单独购买时，付 \(7\) 扇的钱并得到一扇免费窗户，达到 \(8\) 扇，共 \(\$700\)。分别购买总共 \(\$1300\)。一起购买需要 \(15\) 扇窗户，买 \(12\) 扇可获得 \(3\) 扇免费窗户，共 \(\$1200\)。节省 \(1300 - 1200 = 100\) 美元。

所以正确答案是 A。

English

Alone, Dave pays for \(6\) windows and receives one free to reach \(7,\) costing \(\$600;\) Doug pays for \(7\) and receives one free to reach \(8,\) costing \(\$700.\) Separately they pay \(\$1300.\) Together they need \(15\) windows: buying \(12\) yields \(3\) free, for \(\$1200.\) The savings are \(1300 - 1200 = 100\) dollars.

Thus, the correct answer is A.

6 答案 B

两组数的总和为 \(20 \cdot 30 + 30 \cdot 20 = 600 + 600\) \(= 1200\)。所以这 \(50\) 个数的平均数为 \(\dfrac{1200}{50} = 24\)。

所以正确答案是 B。

English

The combined sum is \(20 \cdot 30 + 30 \cdot 20 = 600 + 600\) \(= 1200.\) The average of all \(50\) numbers is \(\dfrac{1200}{50} = 24.\)

Thus, the correct answer is B.

7 答案 B

设 Mike 骑了 \(m\) 英里。Josh 的速度是 Mike 的 \(\frac{4}{5}\)，时间是 Mike 的 \(2\) 倍，所以 Josh 的距离为 \(\frac{8}{5}m\)。两人合计骑了 \(13\) 英里，因此 \(m + \frac{8}{5}m = \frac{13}{5}m = 13\)，得到 \(m = 5\)。

所以正确答案是 B。

English

Let Mike ride \(m\) miles. Josh rides \(\frac{4}{5}\) the rate for \(2\) times the time, so Josh’s distance is \(\frac{8}{5}m.\) Together they cover \(13,\) so \(m + \frac{8}{5}m = \frac{13}{5}m = 13,\) giving \(m = 5.\)

Thus, the correct answer is B.

8 答案 C

三角形 \(ABH\)、\(BCE\)、\(CDF\) 和 \(DAG\) 是全等直角三角形。在 \(\triangle BCE\) 中，斜边 \(BC = \sqrt{50}\)，且 \(BE = 1\)，所以 \(CE = \sqrt{50 - 1} = 7\)。因为 \(BH = CE = 7\)，且 \(E\) 在 \(BH\) 上、\(BE = 1\)，所以内正方形边长 \(EH = 7 - 1 = 6\)，面积为 \(6^2 = 36\)。

所以正确答案是 C。

English

The triangles \(ABH,\) \(BCE,\) \(CDF,\) and \(DAG\) are congruent right triangles. In \(\triangle BCE\) the hypotenuse is \(BC = \sqrt{50}\) and \(BE = 1,\) so \(CE = \sqrt{50 - 1} = 7.\) Since \(BH = CE = 7\) and \(E\) lies on \(BH\) with \(BE = 1,\) the inner square’s side is \(EH = 7 - 1 = 6,\) giving area \(6^2 = 36.\)

Thus, the correct answer is C.

9 答案 B

三个 X 的位置可以是 \(\binom{5}{3} = 10\) 种等可能选择，其中只有一种位置选择产生 XOXOX，所以概率为 \(\dfrac{1}{10}\)。

所以正确答案是 B。

English

The three X positions can be any of \(\binom{5}{3} = 10\) equally likely choices, and exactly one of them produces XOXOX. So the probability is \(\dfrac{1}{10}.\)

Thus, the correct answer is B.

10 答案 A

方程可写为 \(4x^2 + (a+8)x + 9 = 0\)，它只有一个解当且仅当判别式 \((a+8)^2 - 144 = 0\)。所以 \(a + 8 = \pm 12\)，即 \(a = 4\) 或 \(a = -20\)，二者之和为 \(-16\)。

所以正确答案是 A。

English

Writing the equation as \(4x^2 + (a+8)x + 9 = 0,\) there is one solution exactly when the discriminant \((a+8)^2 - 144 = 0.\) Then \(a + 8 = \pm 12,\) so \(a = 4\) or \(a = -20,\) and their sum is \(-16.\)

Thus, the correct answer is A.

11 答案 B

小立方体共有 \(6n^3\) 个面，其中来自原立方体外表面的红色面共有 \(6n^2\) 个。因此 \(\dfrac{6n^2}{6n^3} = \dfrac{1}{n} = \dfrac{1}{4}\)，所以 \(n = 4\)。

所以正确答案是 B。

English

The unit cubes have \(6n^3\) faces total, of which the original surface accounts for \(6n^2\) red faces. Then \(\dfrac{6n^2}{6n^3} = \dfrac{1}{n} = \dfrac{1}{4},\) so \(n = 4.\)

Thus, the correct answer is B.

12 答案 B

底边长 \(2\) 等于两个半径，所以半径为 \(1\)。这个三叶形由四个等边三角形和四个圆弓形组成，它们可以重新拼成四个半径为 \(1\)、圆心角为 \(60^\circ\) 的扇形，总面积为 \(4 \cdot \dfrac{60}{360}\pi (1)^2 = \dfrac{2}{3}\pi\)。

所以正确答案是 B。

English

Since the base \(2\) equals two radii, the radius is \(1.\) The trefoil is made of four equilateral triangles and four circular segments, which reassemble into four \(60^\circ\) sectors of a circle of radius \(1.\) Their total area is \(4 \cdot \dfrac{60}{360}\pi (1)^2 = \dfrac{2}{3}\pi.\)

Thus, the correct answer is B.

13 答案 E

取第 \(50\) 次方根，条件变为 \(130n \gt n^2 \gt 2^4 = 16\)。由 \(n^2 \gt 16\) 得 \(n \gt 4\)，由 \(130n \gt n^2\) 得 \(n \lt 130\)。所以 \(n\) 可以是整数 \(5, 6, \ldots, 129\)，共有 \(125\) 个。

所以正确答案是 E。

English

Taking \(50\)th roots, the condition becomes \(130n \gt n^2 \gt 2^4 = 16.\) From \(n^2 \gt 16\) we get \(n \gt 4,\) and from \(130n \gt n^2\) we get \(n \lt 130.\) So \(n\) ranges over the integers \(5, 6, \ldots, 129,\) which is \(125\) values.

Thus, the correct answer is E.

14 答案 E

首位和末位数字必须奇偶性相同，平均数才是一个数字。若二者都为奇数，有 \(5 \cdot 5 = 25\) 对。若二者都为偶数，且首位非零，有 \(4 \cdot 5 = 20\) 对。每一对首末数字唯一确定中间数字，所以共有 \(25 + 20 = 45\) 个三位数。

所以正确答案是 E。

English

The first and last digits must have the same parity so their average is a digit. Both odd gives \(5 \cdot 5 = 25\) pairs. Both even, with a nonzero leading digit, gives \(4 \cdot 5 = 20\) pairs. Each pair fixes the middle digit, for a total of \(25 + 20 = 45\) numbers.

Thus, the correct answer is E.

15 答案 E

质因数分解为 \(3! \cdot 5! \cdot 7! = 2^8 \cdot 3^4 \cdot 5^2 \cdot 7\)。立方数因子的质数指数必须是 \(3\) 的倍数：\(2\) 的指数可为 \(0\)、\(3\) 或 \(6\)，有 \(3\) 种；\(3\) 的指数可为 \(0\) 或 \(3\)，有 \(2\) 种；\(5\) 和 \(7\) 的指数只能为 \(0\)。因此共有 \(3 \cdot 2 \cdot 1 \cdot 1 = 6\) 个立方数因子。

所以正确答案是 E。

English

As a product of primes, \(3! \cdot 5! \cdot 7! = 2^8 \cdot 3^4 \cdot 5^2 \cdot 7.\) A cube divisor uses exponents that are multiples of \(3:\) the exponent of \(2\) can be \(0,\) \(3,\) or \(6\) (\(3\) choices), the exponent of \(3\) can be \(0\) or \(3\) (\(2\) choices), and the exponents of \(5\) and \(7\) must be \(0.\) That gives \(3 \cdot 2 \cdot 1 \cdot 1 = 6\) cubes.

Thus, the correct answer is E.

16 答案 D

若这个两位数为 \(10a + b\)，则 \((10a + b) - (a + b) = 9a\)。\(9a\) 的个位数字为 \(6\) 只在 \(a = 4\) 时发生，因为 \(9 \cdot 4 = 36\)。此时 \(b\) 可以是 \(0\) 到 \(9\) 中任意数字，所以从 \(40\) 到 \(49\) 共有十个数。

所以正确答案是 D。

English

If the number is \(10a + b,\) then \((10a + b) - (a + b) = 9a.\) The units digit of \(9a\) is \(6\) only when \(a = 4,\) since \(9 \cdot 4 = 36.\) The digit \(b\) can then be anything from \(0\) to \(9,\) giving the ten numbers \(40\) through \(49.\)

Thus, the correct answer is D.

17 答案 D

每个数字都是两条线段的端点，所以五个线段和的总和为 \(2(3 + 5 + 6 + 7 + 9) = 60\)。五项等差数列的中间项等于平均数，因此为 \(\dfrac{60}{5} = 12\)。

所以正确答案是 D。

English

Every number is an endpoint of two segments, so the five segment sums total \(2(3 + 5 + 6 + 7 + 9) = 60.\) The middle term of a five-term arithmetic sequence equals its mean, which is \(\dfrac{60}{5} = 12.\)

Thus, the correct answer is D.

18 答案 A

假设五场都进行，则所有五场结果序列等可能。要求 B 赢第 \(2\) 场且 A 最终以三场胜利赢得系列赛，得到以下等可能序列：

\[ \begin{gathered} \text{BBAAA}, \quad \text{ABBAA}, \\ \text{ABABA}, \quad \text{ABAAB}, \\ \text{ABAAA} \end{gathered}\text{。}\]

其中只有 BBAAA 的第一场是 B 赢，所以概率为 \(\dfrac{1}{5}\)。

所以正确答案是 A。

English

Suppose all five games are played, so every sequence of five results is equally likely. Requiring that B wins game \(2\) and A ends up with the series (three wins) leaves the equally likely sequences

\[ \begin{gathered} \text{BBAAA}, \quad \text{ABBAA}, \\ \text{ABABA}, \quad \text{ABAAB}, \\ \text{ABAAA}. \end{gathered} \]

Only in BBAAA does team B win the first game, so the probability is \(\dfrac{1}{5}.\)

Thus, the correct answer is A.

19 答案 D

旋转后的正方形向下移动时，它的两条下边落在相邻正方形的内侧上角，这些角的高度为 \(1\)。下顶点位于两角正中，所以每个角与它的水平距离为 \(\tfrac12\)。下边斜率的绝对值为 \(1\)，因此从下顶点到角会升高 \(\tfrac12\)。所以下顶点的高度是 \(1-\tfrac12=\tfrac12\)。

点 \(B\) 是对角顶点，比下顶点高一条长度为 \(\sqrt2\) 的完整竖直对角线。因此它的高度是 \(\tfrac12+\sqrt2\)。

所以正确答案是 D。

English

When lowered, the rotated square’s two lower edges rest on the inner top corners of the adjoining squares, which are at height \(1.\) The bottom vertex is centered between those corners, so each corner is horizontally \(\tfrac12\) unit from it. A lower edge has slope \(1\) in magnitude, so it rises \(\tfrac12\) unit on the way to a corner. Therefore the bottom vertex is at height \(1-\tfrac12=\tfrac12.\)

Point \(B\) is the opposite vertex, a full vertical diagonal of length \(\sqrt2\) higher. Its height is therefore \(\tfrac12+\sqrt2.\)

Thus, the correct answer is D.

20 答案 A

延长四条长为 \(1\) 的边可形成一个正方形。每条短边 \(\dfrac{\sqrt{2}}{2}\) 是等腰直角三角形的斜边，对应直角边为 \(\dfrac{1}{2}\)，因此八边形可看作从边长 \(1 + 2 \cdot \frac{1}{2} = 2\) 的正方形四角切去四个这样的三角形，面积为 \(2^2 - 4 \cdot \frac{1}{2}\left(\frac{1}{2}\right)^2 = 4 - \frac{1}{2} = \dfrac{7}{2}\)。

所以正确答案是 A。

English

Extend the four sides of length \(1\) to form a square. Each short side \(\dfrac{\sqrt{2}}{2}\) is the hypotenuse of an isosceles right triangle with legs \(\dfrac{1}{2},\) and cutting these four corners from a square of side \(1 + 2 \cdot \frac{1}{2} = 2\) gives the octagon. Its area is \(2^2 - 4 \cdot \frac{1}{2}\left(\frac{1}{2}\right)^2 = 4 - \frac{1}{2} = \dfrac{7}{2}.\)

Thus, the correct answer is A.

21 答案 B

由 \(1 + 2 + \cdots + n = \dfrac{n(n+1)}{2}\) 可知，商为 \(\dfrac{6n}{\frac{n(n+1)}{2}} = \dfrac{12}{n+1}\)。它是整数当且仅当 \(n + 1\) 整除 \(12\)。\(12\) 的因数中不小于 \(2\) 的有 \(2\)、\(3\)、\(4\)、\(6\)、\(12\)，对应 \(n = 1\)、\(2\)、\(3\)、\(5\)、\(11\)，共五个值。

所以正确答案是 B。

English

Since \(1 + 2 + \cdots + n = \dfrac{n(n+1)}{2},\) the quotient is \(\dfrac{6n}{\frac{n(n+1)}{2}} = \dfrac{12}{n+1},\) which is an integer exactly when \(n + 1\) divides \(12.\) The divisors of \(12\) that are at least \(2\) are \(2,\) \(3,\) \(4,\) \(6,\) \(12,\) giving \(n = 1,\) \(2,\) \(3,\) \(5,\) \(11\) — five values.

Thus, the correct answer is B.

22 答案 D

\(S\) 和 \(T\) 的共同元素是 \(\operatorname{lcm}(4,6) = 12\) 的倍数。集合 \(S\) 包含不超过 \(4 \cdot 2005 = 8020\) 的 \(4\) 的倍数，而 \(T\) 最大到 \(6 \cdot 2005 = 12{,}030\)，所以上界由 \(8020\) 决定。共同元素是不超过该上界的 \(12\) 的倍数，共有 \(\left\lfloor \dfrac{8020}{12} \right\rfloor = 668\) 个。

所以正确答案是 D。

English

The elements common to \(S\) and \(T\) are the multiples of \(\operatorname{lcm}(4,6) = 12.\) Now \(S\) contains multiples of \(4\) up to \(4 \cdot 2005 = 8020,\) while \(T\) reaches up to \(6 \cdot 2005 = 12{,}030,\) so the common elements are the multiples of \(12\) not exceeding \(8020.\) There are \(\left\lfloor \dfrac{8020}{12} \right\rfloor = 668\) of them.

Thus, the correct answer is D.

23 答案 C

设 \(O\) 是圆心。由 \(2 \cdot AC = BC\) 且 \(AC + BC = AB\)，得 \(AC = \dfrac{AB}{3}\)，因此 \(CO = \dfrac{AB}{2} - \dfrac{AB}{3} = \dfrac{AB}{6}\)。三角形 \(DCO\) 和 \(DAB\) 的顶点 \(D\) 相同，底边 \(CO\) 与 \(AB\) 在同一直线上，所以 \([\triangle DCO] = \dfrac{CO}{AB}[\triangle DAB]\) \(= \dfrac{1}{6}[\triangle DAB]\)。又因为 \(O\) 是 \(DE\) 的中点，\([\triangle DCE] = 2\,[\triangle DCO]\) \(= \dfrac{1}{3}[\triangle DAB]\)。

所以正确答案是 C。

English

Let \(O\) be the center. From \(2 \cdot AC = BC\) and \(AC + BC = AB,\) we get \(AC = \dfrac{AB}{3},\) so \(CO = \dfrac{AB}{2} - \dfrac{AB}{3} = \dfrac{AB}{6}.\) Triangles \(DCO\) and \(DAB\) share the apex \(D\) with bases \(CO\) and \(AB\) on the same line, so \([\triangle DCO] = \dfrac{CO}{AB}[\triangle DAB]\) \(= \dfrac{1}{6}[\triangle DAB].\) Because \(O\) is the midpoint of \(DE,\) \([\triangle DCE] = 2\,[\triangle DCO]\) \(= \dfrac{1}{3}[\triangle DAB].\)

Thus, the correct answer is C.

24 答案 B

条件 \(P(n) = \sqrt{n}\) 表示 \(n\) 是某个质数 \(q\) 的平方；同理 \(n + 48 = p^2\)，其中 \(p\) 是质数。因此 \(48 = p^2 - q^2 = (p - q)(p + q)\)。检查 \(48\) 的同奇偶因数分解，只有 \(p - q = 2,\ p + q = 24\) 给出质数，得到 \((p, q) = (13, 11)\)，所以 \(n = 121\)。因此恰好有一个这样的 \(n\)。

所以正确答案是 B。

English

The condition \(P(n) = \sqrt{n}\) means \(n\) is the square of a prime \(q,\) and likewise \(n + 48 = p^2\) for a prime \(p.\) Then \(48 = p^2 - q^2 = (p - q)(p + q).\) Checking the same-parity factorizations of \(48,\) only \(p - q = 2,\ p + q = 24\) yields primes, giving \((p, q) = (13, 11)\) and \(n = 121.\) So there is exactly one such \(n.\)

Thus, the correct answer is B.

25 答案 D

三角形 \(ADE\) 和 \(ABC\) 共用角 \(A\)，所以 \[ \begin{aligned} \dfrac{[ADE]}{[ABC]} &= \dfrac{AD \cdot AE}{AB \cdot AC} \\ &= \dfrac{19 \cdot 14}{25 \cdot 42} \\ &= \dfrac{266}{1050} \\ &= \dfrac{19}{75} \end{aligned}\text{。}\]又因为 \([BCED] = [ABC] - [ADE]\)，所以 \(\dfrac{[ADE]}{[BCED]} = \dfrac{19}{75 - 19} = \dfrac{19}{56}\)。

所以正确答案是 D。

English

Triangles \(ADE\) and \(ABC\) share angle \(A,\) so \[ \begin{aligned} \dfrac{[ADE]}{[ABC]} &= \dfrac{AD \cdot AE}{AB \cdot AC} \\ &= \dfrac{19 \cdot 14}{25 \cdot 42} \\ &= \dfrac{266}{1050} \\ &= \dfrac{19}{75}. \end{aligned} \] Since \([BCED] = [ABC] - [ADE],\) we get \(\dfrac{[ADE]}{[BCED]} = \dfrac{19}{75 - 19} = \dfrac{19}{56}.\)

Thus, the correct answer is D.