2009A AMC 10 真题（P版）

1 答案 E

因为 \(\dfrac{128}{12} = 10\dfrac{2}{3}\)，十罐只能装 \(120\) 盎司，不够。

因此需要 \(11\) 罐。

所以正确答案是 E。

English

Since \(\dfrac{128}{12} = 10\dfrac{2}{3},\) ten cans hold only \(120\) ounces, which is not enough.

Therefore \(11\) cans are needed.

Thus, the correct answer is E.

2 答案 A

要得到 \(5\) 的倍数美分，一分币的数量必须是 \(5\) 的倍数。只有四枚硬币时，这意味着不能有一分币；但此时每枚硬币至少值 \(5\) 美分，总值至少为 \(20\) 美分。

所以 \(15\) 美分无法得到。其他金额都可以组成：\(25 = 10 + 3(5)\)，\(35 = 3(10) + 5\)，\(45 = 25 + 10 + 2(5)\)，以及 \(55 = 25 + 3(10)\)。

所以正确答案是 A。

English

To get a multiple of \(5\) cents, the number of pennies must be a multiple of \(5.\) With only four coins, that means using no pennies, but then the four coins are each worth at least \(5\) cents, for a total of at least \(20\) cents.

So \(15\) cents cannot be made. The others can: \(25 = 10 + 3(5),\) \(35 = 3(10) + 5,\) \(45 = 25 + 10 + 2(5),\) and \(55 = 25 + 3(10).\)

Thus, the correct answer is A.

3 答案 C

从内向外计算，\[ \begin{aligned} &1 + \cfrac{1}{1 + \cfrac{1}{1 + 1}} \\ &= 1 + \cfrac{1}{1 + \dfrac{1}{2}} \\ &= 1 + \dfrac{1}{\frac{3}{2}} \\ &= 1 + \dfrac{2}{3} = \dfrac{5}{3} \end{aligned}\text{。}\]

所以正确答案是 C。

English

Working outward, \[ \begin{aligned} &1 + \cfrac{1}{1 + \cfrac{1}{1 + 1}} \\ &= 1 + \cfrac{1}{1 + \dfrac{1}{2}} \\ &= 1 + \dfrac{1}{\frac{3}{2}} \\ &= 1 + \dfrac{2}{3} = \dfrac{5}{3}. \end{aligned} \]

Thus, the correct answer is C.

4 答案 A

游泳用时 \(\dfrac{\frac{1}{4}}{2} = \dfrac18\) 小时，跑步用时 \(\dfrac{3}{6} = \dfrac12\) 小时。这给自行车赛段留下 \[2 - \dfrac18 - \dfrac12 = \dfrac{11}{8}\] 小时。

所需平均速度为 \[\dfrac{15}{\frac{11}{8}} = \dfrac{120}{11}\] 英里/小时。

所以正确答案是 A。

English

The swim takes \(\dfrac{\frac{1}{4}}{2} = \dfrac18\) hour and the run takes \(\dfrac{3}{6} = \dfrac12\) hour. This leaves \[2 - \dfrac18 - \dfrac12 = \dfrac{11}{8}\] hours for the bicycle ride.

His average speed must be \[\dfrac{15}{\frac{11}{8}} = \dfrac{120}{11}\] miles per hour.

Thus, the correct answer is A.

5 答案 E

这个各位都是一的九位数的平方是回文数 \[ \begin{aligned} &111{,}111{,}111^2 \\ &= 12{,}345{,}678{,}987{,}654{,}321 \end{aligned}\text{。}\]

它的各位数字依次为 \(1, 2, \ldots, 9, 8, \ldots, 1\)，所以数字和为 \[ \begin{aligned} &2(1 + 2 + \cdots + 8) + 9 \\ &= 2 \cdot 36 + 9 = 81 \end{aligned}\text{。}\]

所以正确答案是 E。

English

The square of the nine-digit repunit is the palindrome \[ \begin{aligned} &111{,}111{,}111^2 \\ &= 12{,}345{,}678{,}987{,}654{,}321. \end{aligned} \]

Its digits are \(1, 2, \ldots, 9, 8, \ldots, 1,\) so the sum is \[ \begin{aligned} &2(1 + 2 + \cdots + 8) + 9 \\ &= 2 \cdot 36 + 9 = 81. \end{aligned} \]

Thus, the correct answer is E.

6 答案 A

内切圆落在直径上并与半圆弧相切，所以半圆的半径是 \(4\)。半圆面积为 \[\dfrac12 \pi (4)^2 = 8\pi\text{。}\]

小圆面积为 \(\pi(2)^2 = 4\pi\)，所以阴影面积为 \(8\pi - 4\pi = 4\pi\)。

阴影部分占半圆的比例为 \(\dfrac{4\pi}{8\pi} = \dfrac12\)。

所以正确答案是 A。

English

The inscribed circle rests on the diameter and is tangent to the arc, so the semicircle has radius \(4.\) Its area is \[\dfrac12 \pi (4)^2 = 8\pi.\]

The circle’s area is \(\pi(2)^2 = 4\pi,\) so the shaded area is \(8\pi - 4\pi = 4\pi.\)

The shaded fraction is \(\dfrac{4\pi}{8\pi} = \dfrac12.\)

Thus, the correct answer is A.

7 答案 C

设全脂牛奶含脂肪 \(x\%\)。因为 \(2\) 比 \(x\) 少 \(40\%\)，所以 \[0.6x = 2\text{，}\]从而 \[x = \dfrac{2}{0.6} = \dfrac{10}{3}\text{。}\]

所以正确答案是 C。

English

Let whole milk be \(x\%\) fat. Since \(2\) is \(40\%\) less than \(x,\) we have \[0.6x = 2,\] so \[x = \dfrac{2}{0.6} = \dfrac{10}{3}.\]

Thus, the correct answer is C.

8 答案 B

老人票价为 \(\$6\)，这是全价的 \(\dfrac34\)，所以全价票为 \(\dfrac43 \cdot 6 = \$8\)，儿童票为 \(\dfrac12 \cdot 8 = \$4\)。

总价为 \[2(6 + 8 + 4) = \$36\text{。}\]

所以正确答案是 B。

English

The senior ticket costs \(\$6,\) which is \(\dfrac34\) of the full price, so a full ticket costs \(\dfrac43 \cdot 6 = \$8,\) and a child ticket costs \(\dfrac12 \cdot 8 = \$4.\)

The total is \[2(6 + 8 + 4) = \$36.\]

Thus, the correct answer is B.

9 答案 B

设公比为 \(r\)。则 \(a r^2 = 2009 = 7^2 \cdot 41\)。

因为 \(r\) 必须是大于 \(1\) 的整数，唯一可能是 \(r = 7\)，从而 \(a = 41\)，数列为 \(41, 287, 2009\)。

所以正确答案是 B。

English

Let the common ratio be \(r.\) Then \(a r^2 = 2009 = 7^2 \cdot 41.\)

Since \(r\) must be an integer greater than \(1,\) the only possibility is \(r = 7,\) giving \(a = 41\) and the sequence \(41, 287, 2009.\)

Thus, the correct answer is B.

10 答案 B

对直角三角形中从直角顶点到斜边的高，有 \[BD^2 = AD \cdot DC = 3 \cdot 4 = 12\text{，}\]所以 \(BD = 2\sqrt{3}\)。

斜边满足 \(AC = 3 + 4 = 7\)，所以面积为 \[\dfrac12 \cdot 7 \cdot 2\sqrt3 = 7\sqrt3\text{。}\]

所以正确答案是 B。

English

For the altitude from the right angle to the hypotenuse, \[BD^2 = AD \cdot DC = 3 \cdot 4 = 12,\] so \(BD = 2\sqrt{3}.\)

The hypotenuse is \(AC = 3 + 4 = 7,\) so the area is \[\dfrac12 \cdot 7 \cdot 2\sqrt3 = 7\sqrt3.\]

Thus, the correct answer is B.

11 答案 D

设立方体边长为 \(x\)。新长方体的体积为 \[x(x+1)(x-1) = x^3 - x\text{。}\]

新长方体的体积为 \(x^3 - 5\)，所以 \(x^3 - x = x^3 - 5\)，得到 \(x = 5\)。

原立方体体积为 \(5^3 = 125\)。

所以正确答案是 D。

English

Let the cube have side length \(x.\) The new solid has volume \[x(x+1)(x-1) = x^3 - x.\]

Setting this equal to \(x^3 - 5\) gives \(x^3 - x = x^3 - 5,\) so \(x = 5.\)

The cube’s volume is \(5^3 = 125.\)

Thus, the correct answer is D.

12 答案 C

在 \(\triangle BCD\) 中，三角形不等式给出 \(5 + BD \gt 17\)，所以 \(BD \gt 12\)。

在 \(\triangle ABD\) 中，三角形不等式给出 \(5 + 9 \gt BD\)，所以 \(BD \lt 14\)。

唯一满足 \(12 \lt BD \lt 14\) 的整数是 \(BD = 13\)。

所以正确答案是 C。

English

In \(\triangle BCD,\) the triangle inequality gives \(5 + BD \gt 17,\) so \(BD \gt 12.\)

In \(\triangle ABD,\) it gives \(5 + 9 \gt BD,\) so \(BD \lt 14.\)

The only integer with \(12 \lt BD \lt 14\) is \(BD = 13.\)

Thus, the correct answer is C.

13 答案 E

因为 \(12 = 2^2 \cdot 3\)，\[ \begin{aligned} 12^{mn} &= 2^{2mn} \cdot 3^{mn} \\ &= (2^m)^{2n} \cdot (3^n)^m \\ &= P^{2n} Q^m \end{aligned}\text{。}\]

所以正确答案是 E。

English

Since \(12 = 2^2 \cdot 3,\) \[ \begin{aligned} 12^{mn} &= 2^{2mn} \cdot 3^{mn} \\ &= (2^m)^{2n} \cdot (3^n)^m \\ &= P^{2n} Q^m. \end{aligned} \]

Thus, the correct answer is E.

14 答案 A

设每个长方形短边为 \(x\)、长边为 \(y\)。外部正方形边长为 \(y + x\)，内部正方形边长为 \(y - x\)。

因为面积比为 \(4\)，所以边长比为 \(2\)，于是 \[y + x = 2(y - x)\text{，}\]得到 \(y = 3x\)。

长边与短边的比为 \(\dfrac{y}{x} = 3\)。

所以正确答案是 A。

English

Let each rectangle have shorter side \(x\) and longer side \(y.\) The outer square has side length \(y + x\) and the inner square has side length \(y - x.\)

Since the area ratio is \(4,\) the side ratio is \(2,\) so \[y + x = 2(y - x),\] which gives \(y = 3x.\)

The ratio of longer to shorter side is \(\dfrac{y}{x} = 3.\)

Thus, the correct answer is A.

15 答案 E

从 \(F_{n-1}\) 到 \(F_n\)，新的外部正方形带来 \(4(n-1)\) 个菱形。从 \(F_1\) 的单个菱形开始，\[ \begin{aligned} F_n &= 1 \\ &\quad {}+ 4\big(1 + 2 + \cdots + (n-1)\big) \\ &= 1 + 4 \cdot \dfrac{(n-1)n}{2} \\ &= 1 + 2n(n-1) \end{aligned}\text{。}\]

因此 \[F_{20} = 1 + 2 \cdot 20 \cdot 19 = 761\text{。}\]

所以正确答案是 E。

English

Going from \(F_{n-1}\) to \(F_n,\) the new outside square carries \(4(n-1)\) diamonds. Starting from the single diamond of \(F_1,\) \[ \begin{aligned} F_n &= 1 \\ &\quad {}+ 4\big(1 + 2 + \cdots + (n-1)\big) \\ &= 1 + 4 \cdot \dfrac{(n-1)n}{2} \\ &= 1 + 2n(n-1). \end{aligned} \]

Therefore \[F_{20} = 1 + 2 \cdot 20 \cdot 19 = 761.\]

Thus, the correct answer is E.

16 答案 D

因为 \(a - d\) \(= (a-b) + (b-c) + (c-d)\) \(= \pm 2 \pm 3 \pm 4\)，所以可能的绝对值为 \[ \begin{aligned} 2+3+4 &= 9, \\ 2+3-4 &= 1, \\ 2-3+4 &= 3, \\ -2+3+4 &= 5 \end{aligned}\text{。}\]

它们的和为 \(9 + 1 + 3 + 5 = 18\)。

所以正确答案是 D。

English

Since \(a - d\) \(= (a-b) + (b-c) + (c-d)\) \(= \pm 2 \pm 3 \pm 4,\) the possible absolute values are \[ \begin{aligned} 2+3+4 &= 9, \\ 2+3-4 &= 1, \\ 2-3+4 &= 3, \\ -2+3+4 &= 5. \end{aligned} \]

Their sum is \(9 + 1 + 3 + 5 = 18.\)

Thus, the correct answer is D.

17 答案 C

对角线 \(DB = \sqrt{4^2 + 3^2} = 5\)。

直角三角形 \(EBA\)、\(DBC\) 和 \(BFC\) 都与 \(\triangle DBA\) 相似。由 \(\triangle EBA\)，\[ \begin{aligned} \dfrac{EB}{AB} &= \dfrac{DB}{BC} \\ &\implies \dfrac{EB}{4} = \dfrac{5}{3} \\ &\implies EB = \dfrac{20}{3} \end{aligned}\text{。}\]

由 \(\triangle BFC\)，\[ \begin{aligned} \dfrac{BF}{BC} &= \dfrac{DB}{AB} \\ &\implies \dfrac{BF}{3} = \dfrac{5}{4} \\ &\implies BF = \dfrac{15}{4} \end{aligned}\text{。}\]

因此 \[ \begin{aligned} EF &= EB + BF \\ &= \dfrac{20}{3} + \dfrac{15}{4} \\ &= \dfrac{125}{12} \end{aligned}\text{。}\]

所以正确答案是 C。

English

The diagonal is \(DB = \sqrt{4^2 + 3^2} = 5.\)

Right triangles \(EBA,\) \(DBC,\) and \(BFC\) are all similar to \(\triangle DBA.\) From \(\triangle EBA,\) \[ \begin{aligned} \dfrac{EB}{AB} &= \dfrac{DB}{BC} \\ &\implies \dfrac{EB}{4} = \dfrac{5}{3} \\ &\implies EB = \dfrac{20}{3}. \end{aligned} \]

From \(\triangle BFC,\) \[ \begin{aligned} \dfrac{BF}{BC} &= \dfrac{DB}{AB} \\ &\implies \dfrac{BF}{3} = \dfrac{5}{4} \\ &\implies BF = \dfrac{15}{4}. \end{aligned} \]

Therefore \[ \begin{aligned} EF &= EB + BF \\ &= \dfrac{20}{3} + \dfrac{15}{4} \\ &= \dfrac{125}{12}. \end{aligned} \]

Thus, the correct answer is C.

18 答案 D

假设有 \(100\) 个孩子：\(60\) 人踢足球，其中 \(40\%\)，也就是 \(24\) 人，同时游泳。所以 \(60 - 24 = 36\) 名足球队员不游泳。

共有 \(30\) 名游泳者和 \(70\) 名不游泳者，因此不游泳者中踢足球的比例为 \[\dfrac{36}{70} \approx 0.514 \approx 51\%\text{。}\]

所以正确答案是 D。

English

Take \(100\) children: \(60\) play soccer, and \(40\%\) of them, or \(24,\) also swim. So \(60 - 24 = 36\) soccer players do not swim.

There are \(30\) swimmers and \(70\) non-swimmers, so the fraction of non-swimmers who play soccer is \[\dfrac{36}{70} \approx 0.514 \approx 51\%.\]

Thus, the correct answer is D.

19 答案 B

两个圆的周长分别为 \(200\pi\) 和 \(2\pi r\)，所以初始切点在 \[\dfrac{200\pi}{2\pi r} = \dfrac{100}{r}\] 圈后回到原处。

为使它是大于 \(1\) 的整数，\(r\) 必须是小于 \(100\) 的 \(100\) 的因数：\(1, 2, 4, 5, 10, 20, 25\)，和 \(50\)，共有 \(8\) 个值。

所以正确答案是 B。

English

The circumferences are \(200\pi\) and \(2\pi r,\) so the initial point of tangency returns after \[\dfrac{200\pi}{2\pi r} = \dfrac{100}{r}\] rolls.

For this to be an integer greater than \(1,\) \(r\) must be a divisor of \(100\) less than \(100:\) namely \(1, 2, 4, 5, 10, 20, 25,\) and \(50.\) That is \(8\) values.

Thus, the correct answer is B.

20 答案 D

设 Lauren 的速度为 \(r\) 千米/分钟。则 \(r + 3r = 1\)，所以 \(r = \dfrac14\)。

前 \(5\) 分钟间距缩短 \(5\) 千米，还剩 \(15\) 千米。Lauren 独自以 \(\dfrac14\) 千米/分钟走完，需 \[\dfrac{15}{\frac{1}{4}} = 60\] 分钟。

总时间为 \(5 + 60 = 65\) 分钟。

所以正确答案是 D。

English

Let Lauren’s rate be \(r\) km/min. Then \(r + 3r = 1,\) so \(r = \dfrac14.\)

In the first \(5\) minutes the gap shrinks by \(5\) km, leaving \(15\) km. Lauren covers this alone at \(\dfrac14\) km/min, taking \[\dfrac{15}{\frac{1}{4}} = 60\] minutes.

The total time is \(5 + 60 = 65\) minutes.

Thus, the correct answer is D.

21 答案 C

设每个小圆半径为 \(1\)。它们的圆心构成边长为 \(2\) 的正方形，该正方形对角线为 \(2\sqrt2\)。

大圆的直径为 \(2 + 2\sqrt2\)，所以半径为 \(1 + \sqrt2\)。

所求比值为 \[ \begin{aligned} \dfrac{4 \cdot \pi (1)^2}{\pi (1 + \sqrt2)^2} &= \dfrac{4}{3 + 2\sqrt2} \\ &= 4(3 - 2\sqrt2) \end{aligned}\text{。}\]

所以正确答案是 C。

English

Let each small circle have radius \(1.\) Their centers form a square of side \(2,\) whose diagonal is \(2\sqrt2.\)

The large circle’s diameter is \(2 + 2\sqrt2,\) so its radius is \(1 + \sqrt2.\)

The desired ratio is \[ \begin{aligned} \dfrac{4 \cdot \pi (1)^2}{\pi (1 + \sqrt2)^2} &= \dfrac{4}{3 + 2\sqrt2} \\ &= 4(3 - 2\sqrt2). \end{aligned} \]

Thus, the correct answer is C.

22 答案 D

随机贴数字再掷骰子，等价于从这十二个数字中随机选两个并相加。

假设第一个顶面显示 \(N\)。若和为 \(7\)，第二个必须是 \(7 - N\)；符合要求的牌恰有 \(2\) 张，都等于 \(7 - N\)，而此时总共剩余 \(11\) 张牌。

所以概率为 \(\dfrac{2}{11}\)。

所以正确答案是 D。

English

Randomly attaching the tiles and then rolling is equivalent to choosing two of the twelve numbers at random and adding them.

Suppose the first top face shows \(N.\) For a sum of \(7,\) the second must be \(7 - N,\) and there are exactly \(2\) tiles equal to \(7 - N\) among the remaining \(11.\)

So the probability is \(\dfrac{2}{11}.\)

Thus, the correct answer is D.

23 答案 E

因为 \([AED] = [BEC]\)，两边都加上 \([CED]\)，得到 \([ACD] = [BCD]\)。它们共用底边 \(CD\)，所以 \(A\) 和 \(B\) 到直线 \(CD\) 的距离相同，即 \(AB \parallel CD\)。

于是 \(\triangle ABE \sim \triangle CDE\)，相似比为 \(\dfrac{AB}{CD} = \dfrac{9}{12} = \dfrac34\)，所以 \(\dfrac{AE}{EC} = \dfrac34\)。

又 \(AE + EC = AC = 14\)，得到 \(AE = \dfrac{3}{7} \cdot 14 = 6\)。

所以正确答案是 E。

English

Since \([AED] = [BEC],\) adding \([CED]\) to both gives \([ACD] = [BCD].\) These share base \(CD,\) so \(A\) and \(B\) are equidistant from line \(CD,\) meaning \(AB \parallel CD.\)

Then \(\triangle ABE \sim \triangle CDE\) with ratio \(\dfrac{AB}{CD} = \dfrac{9}{12} = \dfrac34,\) so \(\dfrac{AE}{EC} = \dfrac34.\)

With \(AE + EC = AC = 14,\) we get \(AE = \dfrac{3}{7} \cdot 14 = 6.\)

Thus, the correct answer is E.

24 答案 C

三个顶点确定的平面会穿过内部，除非这三个顶点都在同一个面上。

\(6\) 个面中每个面给出 \(\binom{4}{3} = 4\) 组三顶点，所以同一面上的共有 \(6 \cdot 4 = 24\) 组；总数为 \(\binom{8}{3} = 56\)。

穿过内部的概率为 \[1 - \dfrac{24}{56} = \dfrac{4}{7}\text{。}\]

所以正确答案是 C。

English

Three vertices determine a plane that cuts through the interior unless all three lie on a single face.

Each of the \(6\) faces gives \(\binom{4}{3} = 4\) triples, so \(6 \cdot 4 = 24\) triples lie on a face out of \(\binom{8}{3} = 56\) total.

The probability of hitting the interior is \[1 - \dfrac{24}{56} = \dfrac{4}{7}.\]

Thus, the correct answer is C.

25 答案 B

写成 \[ \begin{aligned} I_k &= 10^{k+2} + 64 \\ &= 2^{k+2} \cdot 5^{k+2} + 2^6 \end{aligned}\text{。}\]

如果 \(k \lt 4\)，第一项含有少于 \(6\) 个因数 \(2\)，所以 \(N(k) = k + 2 \lt 6\)。

如果 \(k \gt 4\)，第一项至少含有 \(7\) 个因数 \(2\)，而第二项恰好含有 \(6\) 个，所以它们的和恰好含有 \(6\) 个这样的因数，即 \(N(k) = 6\)。

如果 \(k = 4\)，则 \(I_4 = 2^6(5^6 + 1)\)。因为 \(5^6 + 1\) \(= (5^2 + 1)\big((5^2)^2 - 5^2 + 1\big)\) \(= 26 \cdot 601\)，它恰好多贡献一个因数 \(2\)，所以 \(N(4) = 7\)。

最大值为 \(N(4) = 7\)。

所以正确答案是 B。

English

Write \[ \begin{aligned} I_k &= 10^{k+2} + 64 \\ &= 2^{k+2} \cdot 5^{k+2} + 2^6. \end{aligned} \]

If \(k \lt 4,\) the first term has fewer than \(6\) factors of \(2,\) so \(N(k) = k + 2 \lt 6.\)

If \(k \gt 4,\) the first term has at least \(7\) factors of \(2\) while the second has exactly \(6,\) so their sum has exactly \(6:\) \(N(k) = 6.\)

If \(k = 4,\) then \(I_4 = 2^6(5^6 + 1).\) Since \(5^6 + 1\) \(= (5^2 + 1)\big((5^2)^2 - 5^2 + 1\big)\) \(= 26 \cdot 601,\) it contributes exactly one more factor of \(2.\) Thus \(N(4) = 7.\)

The maximum value is \(N(4) = 7.\)

Thus, the correct answer is B.