2006B AMC 10 真题（P版）

1 答案 C

共有 \(2006\) 项。把相邻项配对，得到 \((-1+1)+(-1+1)+\cdots\)。由于 \(2006\) 是偶数，每一项都能配对，因此总和为 \(0\)。

所以正确答案是 C。

English

There are \(2006\) terms. Pairing consecutive terms gives \((-1+1)+(-1+1)+\cdots.\) Since \(2006\) is even, every term pairs off and the sum is \(0.\)

Thus, the correct answer is C.

2 答案 A

因为 \(x \spadesuit y = (x+y)(x-y) = x^2 - y^2\)，所以 \(4 \spadesuit 5 = 16 - 25 = -9\)。

接着 \(3 \spadesuit (-9) = 9 - 81 = -72\)。

所以正确答案是 A。

English

Since \(x \spadesuit y = (x+y)(x-y) = x^2 - y^2,\) we have \(4 \spadesuit 5 = 16 - 25 = -9.\)

Then \(3 \spadesuit (-9) = 9 - 81 = -72.\)

Thus, the correct answer is A.

3 答案 A

设 \(c\) 和 \(p\) 分别为 Cougars 和 Panthers 的得分。则 \(c+p=34\)，且 \(c-p=14\)。两式相减得 \(2p=20\)，所以 \(p=10\)。

所以正确答案是 A。

English

Let \(c\) and \(p\) be the Cougars’ and Panthers’ scores. Then \(c+p=34\) and \(c-p=14.\) Subtracting gives \(2p=20,\) so \(p=10.\)

Thus, the correct answer is A.

4 答案 D

红色圆面积为 \(\pi(\tfrac12)^2 = \tfrac{\pi}{4}\)，大圆面积为 \(\pi(\tfrac32)^2 = \tfrac{9\pi}{4}\)。蓝色圆环面积为 \(\tfrac{9\pi}{4}-\tfrac{\pi}{4}=2\pi\)。

因此蓝色面积是红色面积的 \(2\pi \div \tfrac{\pi}{4} = 8\) 倍。

所以正确答案是 D。

English

The red circle has area \(\pi(\tfrac12)^2 = \tfrac{\pi}{4},\) and the large circle has area \(\pi(\tfrac32)^2 = \tfrac{9\pi}{4}.\) The blue ring is \(\tfrac{9\pi}{4}-\tfrac{\pi}{4}=2\pi.\)

The ratio is \(2\pi \div \tfrac{\pi}{4} = 8.\)

Thus, the correct answer is D.

5 答案 B

将两个矩形并排放置，使长度为 \(3\) 的边竖直。它们宽度相加为 \(2+3=5\)，高度 \(3\) 和 \(4\) 都能放进边长 \(5\) 的正方形。

因为两个矩形都与正方形的边平行且内部不重叠，所以它们的水平投影或竖直投影必定互不重叠。在任一方向上，第一个矩形至少占据 \(2\) 的长度，第二个矩形至少占据 \(3\) 的长度，因此正方形边长至少为 \(2+3=5\)。所以最小面积为 \(5^2=25\)。

所以正确答案是 B。

English

Place the rectangles side by side with their \(3\)-length sides vertical. Their widths add to \(2+3=5,\) and the heights \(3\) and \(4\) both fit within \(5.\)

Because the rectangles are axis-aligned and their interiors do not overlap, their horizontal projections or their vertical projections must be disjoint. In either direction, the first rectangle spans at least \(2\) and the second spans at least \(3,\) so the square’s side is at least \(2+3=5.\) The smallest area is therefore \(5^2=25.\)

Thus, the correct answer is B.

6 答案 D

每条边长为 \(\tfrac{2}{\pi}\)，也是一个半圆弧的直径，所以每条弧长为 \(\tfrac12\pi\cdot\tfrac{2}{\pi}=1\)。

边界由四条这样的弧组成，所以周长为 \(4\cdot1=4\)。

所以正确答案是 D。

English

Each side has length \(\tfrac{2}{\pi},\) the diameter of a semicircular arc, so each arc has length \(\tfrac12\pi\cdot\tfrac{2}{\pi}=1.\)

The boundary consists of four such arcs, so the perimeter is \(4\cdot1=4.\)

Thus, the correct answer is D.

7 答案 A

分母可化简为 \(1-\dfrac{x-1}{x}=\dfrac{x-(x-1)}{x}=\dfrac{1}{x}\)。

所以原式为 \(\sqrt{\dfrac{x}{\frac{1}{x}}}=\sqrt{x^2}=|x|\)。由于 \(x \lt 0\)，它等于 \(-x\)。

所以正确答案是 A。

English

The denominator simplifies: \(1-\dfrac{x-1}{x}=\dfrac{x-(x-1)}{x}=\dfrac{1}{x}.\)

So the expression is \(\sqrt{\dfrac{x}{\frac{1}{x}}}=\sqrt{x^2}=|x|.\) Since \(x \lt 0,\) this equals \(-x.\)

Thus, the correct answer is A.

8 答案 B

设正方形边长为 \(s\)，则 \(s^2=40\)。它的底边在直径上居中，而上方的一个顶点 \(\left(\tfrac{s}{2},s\right)\) 位于圆上。

因此 \(r^2=\left(\tfrac{s}{2}\right)^2+s^2=\tfrac{40}{4}+40=50\)，半圆面积为 \(\tfrac12\pi r^2=\tfrac12\pi(50)=25\pi\)。

所以正确答案是 B。

English

Let the square have side \(s,\) so \(s^2=40.\) Its base lies centered on the diameter, and a top corner at \(\left(\tfrac{s}{2},s\right)\) lies on the circle.

Then \(r^2=\left(\tfrac{s}{2}\right)^2+s^2=\tfrac{40}{4}+40=50.\) The semicircle area is \(\tfrac12\pi r^2=\tfrac12\pi(50)=25\pi.\)

Thus, the correct answer is B.

9 答案 B

柠檬水总重 \(100+100+400=600\) 克，含 \(25+386=411\) 卡路里。

\(200\) 克中有 \(411\cdot\tfrac{200}{600}=\tfrac{411}{3}=137\) 卡路里。

所以正确答案是 B。

English

The lemonade totals \(100+100+400=600\) grams containing \(25+386=411\) calories.

In \(200\) grams there are \(411\cdot\tfrac{200}{600}=\tfrac{411}{3}=137\) calories.

Thus, the correct answer is B.

10 答案 A

设三边为 \(x\)、\(3x\) 和 \(15\)。三角形不等式 \(x+15 \gt 3x\) 给出 \(x \lt 7.5\)。

最大整数为 \(x=7\)，三边为 \(7\)、\(21\)、\(15\)，周长为 \(7+21+15=43\)。

所以正确答案是 A。

English

Let the sides be \(x,\) \(3x,\) and \(15.\) The triangle inequality \(x+15 \gt 3x\) gives \(x \lt 7.5.\)

The largest integer is \(x=7,\) giving sides \(7,\) \(21,\) \(15\) and perimeter \(7+21+15=43.\)

Thus, the correct answer is A.

11 答案 C

当 \(n\ge 10\) 时，\(n!\) 能被 \(100\) 整除，所以不影响最后两位。

十位数字来自 \(7!+8!+9!\) \(=5040+40320+362880\) \(=408240\)，其十位数字为 \(4\)。

所以正确答案是 C。

English

For \(n\ge 10,\) \(n!\) is divisible by \(100,\) so it does not affect the last two digits.

The tens digit comes from \(7!+8!+9!\) \(=5040+40320+362880\) \(=408240,\) whose tens digit is \(4.\)

Thus, the correct answer is C.

12 答案 E

代入 \((1,2)\)：由 \(1=\tfrac14(2)+a\) 得 \(a=\tfrac12\)，由 \(2=\tfrac14(1)+b\) 得 \(b=\tfrac74\)。

因此 \(a+b=\tfrac12+\tfrac74=\tfrac94\)。

所以正确答案是 E。

English

Substituting \((1,2)\): from \(1=\tfrac14(2)+a\) we get \(a=\tfrac12,\) and from \(2=\tfrac14(1)+b\) we get \(b=\tfrac74.\)

Then \(a+b=\tfrac12+\tfrac74=\tfrac94.\)

Thus, the correct answer is E.

13 答案 E

Joe 加入 \(2\) 盎司奶油后没有再喝，所以他有 \(2\) 盎司奶油。

JoAnn 有 \(12\) 盎司咖啡加 \(2\) 盎司奶油，共 \(14\) 盎司均匀混合液。喝掉 \(2\) 盎司后，她保留了奶油的 \(\tfrac{12}{14}=\tfrac67\)，即 \(\tfrac67\cdot2=\tfrac{12}{7}\) 盎司。

所求比为 \(2\div\tfrac{12}{7}=\tfrac{7}{6}\)。

所以正确答案是 E。

English

Joe adds \(2\) ounces of cream and drinks nothing afterward, so he has \(2\) ounces of cream.

JoAnn has \(12\) ounces of coffee plus \(2\) ounces of cream, making \(14\) ounces of uniform mixture. After drinking \(2\) ounces she keeps \(\tfrac{12}{14}=\tfrac67\) of her cream, which is \(\tfrac67\cdot2=\tfrac{12}{7}\) ounces.

The ratio is \(2\div\tfrac{12}{7}=\tfrac{7}{6}.\)

Thus, the correct answer is E.

14 答案 D

因为 \(a\) 和 \(b\) 是 \(x^2-mx+2\) 的根，所以 \(ab=2\)。

\(q\) 是新方程两根的乘积：\[ \begin{aligned} q&=\left(a+\tfrac1b\right)\left(b+\tfrac1a\right)\\ &=ab+1+1+\tfrac{1}{ab}\\ &=2+2+\tfrac12=\tfrac92 \end{aligned}\text{。}\]

所以正确答案是 D。

English

Since \(a\) and \(b\) are roots of \(x^2-mx+2,\) we have \(ab=2.\)

The value \(q\) is the product of the new roots: \[ \begin{aligned} q&=\left(a+\tfrac1b\right)\left(b+\tfrac1a\right)\\ &=ab+1+1+\tfrac{1}{ab}\\ &=2+2+\tfrac12=\tfrac92. \end{aligned} \]

Thus, the correct answer is D.

15 答案 C

因为 \(AB=AD\) 且 \(\angle BAD=60^\circ\)，三角形 \(ABD\) 是等边三角形，三角形 \(CBD\) 也是等边三角形。

点 \(E\) 和 \(F\) 将这个菱形分成六个全等三角形，每个面积为 \(\tfrac{24}{6}=4\)。

菱形 \(BFDE\) 由三角形 \(BED\) 和 \(BFD\) 组成，所以面积为 \(2\cdot4=8\)。

所以正确答案是 C。

English

Because \(AB=AD\) and \(\angle BAD=60^\circ,\) triangle \(ABD\) is equilateral, and so is triangle \(CBD.\)

Points \(E\) and \(F\) split the rhombus into six congruent triangles, each of area \(\tfrac{24}{6}=4.\)

Rhombus \(BFDE\) is the union of triangles \(BED\) and \(BFD,\) so its area is \(2\cdot4=8.\)

Thus, the correct answer is C.

16 答案 E

从一个闰日到下一个闰日共有 \(3\cdot365+366=1461\) 天，且 \(1461\equiv 5\pmod 7\)。

从 \(2004\) 年到 \(2020\) 年有四个这样的周期，星期数推进 \(4\cdot5=20\equiv 6\pmod 7\)，也就是从星期日向前推 \(6\) 天。

因此 \(2020\) 年的闰日是星期六。

所以正确答案是 E。

English

From one Leap Day to the next is \(3\cdot365+366=1461\) days, and \(1461\equiv 5\pmod 7.\)

Over the four cycles from \(2004\) to \(2020,\) the weekday advances \(4\cdot5=20\equiv 6\pmod 7,\) that is, \(6\) days forward, which is one day back from Sunday.

So Leap Day \(2020\) falls on a Saturday.

Thus, the correct answer is E.

17 答案 D

Alice 将一个球移给 Bob 后，Bob 的袋子中有 \(6\) 个球，其中恰好一种颜色出现两次。

两个袋子最终相同，当且仅当 Bob 还回这两个同色球中的一个。六个球中有两个符合条件，所以概率为 \(\tfrac26=\tfrac13\)。

所以正确答案是 D。

English

Alice moves one ball to Bob, so Bob’s bag holds \(6\) balls with exactly one color appearing twice.

The two bags end up identical exactly when Bob returns one of that duplicated pair. Two of the six balls qualify, so the probability is \(\tfrac26=\tfrac13.\)

Thus, the correct answer is D.

18 答案 E

前几项为 \(2,\,3,\,\tfrac32,\,\tfrac12,\,\tfrac13,\,\tfrac23\)，接着又是 \(2,\,3,\ldots\)，周期为 \(6\)。

因为 \(2006=6\cdot334+2\)，所以 \(a_{2006}=a_2=3\)。

所以正确答案是 E。

English

The terms are \(2,\,3,\,\tfrac32,\,\tfrac12,\,\tfrac13,\,\tfrac23,\) then \(2,\,3,\ldots,\) a cycle of length \(6.\)

Since \(2006=6\cdot334+2,\) we have \(a_{2006}=a_2=3.\)

Thus, the correct answer is E.

19 答案 A

因为 \(OA=1\)、\(OD=2\)，且 \(D\) 在直线 \(x=1\) 上，所以 \(\angle AOD=60^\circ\)，\(\angle COE=60^\circ\)，\(\angle DOE=30^\circ\)。

因此扇形 \(DOE\) 的面积为 \(\tfrac{30}{360}\pi(2^2)=\tfrac{\pi}{3}\)。

阴影区域是这个扇形减去三角形 \(OBD\) 和 \(OBE\)。由于 \(BD=BE=\sqrt3-1\)，每个三角形面积为 \(\tfrac12(\sqrt3-1)(1)\)，合计为 \(\sqrt3-1\)。

所以阴影面积为 \(\tfrac{\pi}{3}-(\sqrt3-1)=\tfrac{\pi}{3}+1-\sqrt3\)。

所以正确答案是 A。

English

Since \(OA=1\) and \(OD=2\) with \(D\) on the line \(x=1,\) we get \(\angle AOD=60^\circ,\) and likewise \(\angle COE=60^\circ,\) so \(\angle DOE=30^\circ.\)

The sector \(DOE\) has area \(\tfrac{30}{360}\pi(2^2)=\tfrac{\pi}{3}.\)

The region is this sector minus triangles \(OBD\) and \(OBE.\) With \(BD=BE=\sqrt3-1,\) each triangle has area \(\tfrac12(\sqrt3-1)(1),\) totaling \(\sqrt3-1.\)

So the shaded area is \(\tfrac{\pi}{3}-(\sqrt3-1)=\tfrac{\pi}{3}+1-\sqrt3.\)

Thus, the correct answer is A.

20 答案 E

\(\overline{AB}\) 的斜率为 \(\tfrac{178-(-22)}{2006-6}=\tfrac{200}{2000}=\tfrac1{10}\)。由于 \(\overline{AD}\perp\overline{AB}\)，其斜率为 \(-10\)，所以 \(\tfrac{y+22}{8-6}=-10\)，得到 \(y=-42\)。

于是 \(AB=\sqrt{2000^2+200^2}\) \(=200\sqrt{101}\)，且 \(AD=\sqrt{2^2+20^2}=2\sqrt{101}\)。

面积为 \(200\sqrt{101}\cdot2\sqrt{101}\) \(=400\cdot101\) \(=40{,}400\)。

所以正确答案是 E。

English

The slope of \(\overline{AB}\) is \(\tfrac{178-(-22)}{2006-6}=\tfrac{200}{2000}=\tfrac1{10}.\) Since \(\overline{AD}\perp\overline{AB},\) its slope is \(-10,\) so \(\tfrac{y+22}{8-6}=-10\) gives \(y=-42.\)

Then \(AB=\sqrt{2000^2+200^2}\) \(=200\sqrt{101}\) and \(AD=\sqrt{2^2+20^2}=2\sqrt{101}.\)

The area is \(200\sqrt{101}\cdot2\sqrt{101}\) \(=400\cdot101\) \(=40{,}400.\)

Thus, the correct answer is E.

21 答案 C

每个骰子掷出 \(k\) 的概率为 \(\tfrac{k}{1+2+\cdots+6}=\tfrac{k}{21}\)。

总和为 \(7\) 的有序数对 \((1,6),(2,5),\ldots,(6,1)\) 给出的概率为 \[ \begin{aligned} &\scriptsize\dfrac{1\cdot6+2\cdot5+3\cdot4+4\cdot3+5\cdot2+6\cdot1}{21^2}\\ &=\dfrac{56}{441}=\dfrac{8}{63} \end{aligned}\text{。}\]

所以正确答案是 C。

English

Each die shows \(k\) with probability \(\tfrac{k}{1+2+\cdots+6}=\tfrac{k}{21}.\)

For a total of \(7,\) the ordered pairs \((1,6),(2,5),\ldots,(6,1)\) contribute \[ \begin{aligned} &\scriptsize\dfrac{1\cdot6+2\cdot5+3\cdot4+4\cdot3+5\cdot2+6\cdot1}{21^2}\\ &=\dfrac{56}{441}=\dfrac{8}{63}. \end{aligned} \]

Thus, the correct answer is C.

22 答案 D

总成本为 \(N(4B+5J)=253\) 美分 \(=11\cdot23\)。因为 \(N \gt 1\)，所以 \(N\in\{11,23,253\}\)。

若 \(N=253\) 或 \(N=23\)，则 \(4B+5J\) 等于 \(1\) 或 \(11\)，对正整数不可能。

因此 \(N=11\)，且 \(4B+5J=23\)。唯一正整数解为 \(B=2\)、\(J=3\)。果酱成本为 \(11\cdot3\cdot5=165\) 美分，即 \(\$1.65\)。

所以正确答案是 D。

English

The total cost is \(N(4B+5J)=253\) cents \(=11\cdot23.\) Since \(N \gt 1,\) \(N\in\{11,23,253\}.\)

If \(N=253\) or \(N=23,\) then \(4B+5J\) equals \(1\) or \(11,\) impossible for positive integers.

So \(N=11\) and \(4B+5J=23,\) whose only positive solution is \(B=2,\) \(J=3.\) The jam costs \(11\cdot3\cdot5=165\) cents, or \(\$1.65.\)

Thus, the correct answer is D.

23 答案 D

将四边形分成两个面积为 \(R\) 和 \(S\) 的三角形，则阴影面积为 \(T=R+S\)。

比较共高三角形，由底边比得到 \(\tfrac{R}{3}=\tfrac{T+7}{10}\) 和 \(\tfrac{S}{7}=\tfrac{T+3}{14}\)。

因此 \[T=R+S=3\cdot\tfrac{T+7}{10}+7\cdot\tfrac{T+3}{14}\] 所以 \(10T=3(T+7)\) \(+5(T+3)\) \(=8T+36\)，得到 \(T=18\)。

所以正确答案是 D。

English

Split the quadrilateral into two triangles of areas \(R\) and \(S,\) so the shaded area is \(T=R+S.\)

Comparing triangles that share an altitude, base ratios give \(\tfrac{R}{3}=\tfrac{T+7}{10}\) and \(\tfrac{S}{7}=\tfrac{T+3}{14}.\)

Then \[T=R+S=3\cdot\tfrac{T+7}{10}+7\cdot\tfrac{T+3}{14},\] so \(10T=3(T+7)\) \(+5(T+3)\) \(=8T+36,\) giving \(T=18.\)

Thus, the correct answer is D.

24 答案 B

这个六边形关于 \(\overline{OP}\) 对称，所以其面积是梯形 \(AOPD\) 面积的两倍。

作 \(OF\parallel AD\)，其中 \(F\) 在 \(\overline{PD}\) 上。则 \(AOFD\) 是矩形，所以 \(DF=OA=2\)，且 \(FP=PD-DF=4-2=2\)。

由于两圆外切，\(OP=2+4=6\)，所以在直角三角形 \(OFP\) 中，\(OF=\sqrt{36-4}=4\sqrt2\)。

梯形 \(AOPD\) 的平行边为 \(OA=2\) 和 \(PD=4\)，高为 \(OF=4\sqrt2\)，面积为 \(\tfrac12(2+4)(4\sqrt2)=12\sqrt2\)。六边形面积为 \(2\cdot12\sqrt2=24\sqrt2\)。

所以正确答案是 B。

English

The hexagon is symmetric about \(\overline{OP},\) so its area is twice that of trapezoid \(AOPD.\)

Draw \(OF\parallel AD\) with \(F\) on \(\overline{PD}.\) Then \(AOFD\) is a rectangle, so \(DF=OA=2\) and \(FP=PD-DF=4-2=2.\)

Since the circles are externally tangent, \(OP=2+4=6,\) so in right triangle \(OFP,\) \(OF=\sqrt{36-4}=4\sqrt2.\)

Trapezoid \(AOPD\) has parallel sides \(OA=2\) and \(PD=4\) with height \(OF=4\sqrt2,\) giving area \(\tfrac12(2+4)(4\sqrt2)=12\sqrt2.\) The hexagon area is \(2\cdot12\sqrt2=24\sqrt2.\)

Thus, the correct answer is B.

25 答案 B

因为有一个孩子 \(9\) 岁，车牌号能被 \(9\) 整除，所以它的各位数字之和 \(2(a+b)\) 是 \(9\) 的倍数，这迫使 \(a+b=9\)。

八个不同年龄是从 \(1\) 到 \(9\) 的九个整数中的八个，所以 \(4\) 和 \(8\) 中至少有一个是孩子的年龄。因此车牌号能被 \(4\) 整除。不计两个数字互换，排列形式为 \(aabb\)、\(abab\) 或 \(baab\)。把这些形式与 \(a+b=9\) 及能被 \(4\) 整除结合，剩下 \[ \begin{gathered} 1188,\ 2772,\ 3636,\ 5544,\\ 6336,\ 7272,\ 9900 \end{gathered}\text{。}\]

最后一个候选会使 Jones 先生的年龄为 \(00\)，所以不可能。其余六个候选都不能被 \(5\) 整除，因此 \(5\) 不可能是孩子的年龄。条件确实可以满足：\(5544\) 能被 \(\{1,2,3,4,6,7,8,9\}\) 中的每个年龄整除，而末两位给出 Jones 先生的年龄 \(44\)。

所以正确答案是 B。

English

Since a child is \(9,\) the number is divisible by \(9,\) so its digit sum \(2(a+b)\) is a multiple of \(9,\) which forces \(a+b=9.\)

The eight distinct ages are eight of the nine integers from \(1\) through \(9,\) so at least one of \(4\) and \(8\) is an age. Hence the plate number is divisible by \(4.\) Up to interchanging the two digits, its pattern is \(aabb,\) \(abab,\) or \(baab.\) Combining these patterns with \(a+b=9\) and divisibility by \(4\) leaves \[ \begin{gathered} 1188,\ 2772,\ 3636,\ 5544,\\ 6336,\ 7272,\ 9900. \end{gathered} \]

The last candidate would make Mr. Jones’s age \(00,\) so it is impossible. None of the other six candidates is divisible by \(5,\) so \(5\) cannot be one of the children’s ages. The conditions are attainable: \(5544\) is divisible by each age in \(\{1,2,3,4,6,7,8,9\}\) and its last two digits give Mr. Jones’s age as \(44.\)

Thus, the correct answer is B.