2010A AMC 12 真题（P版）

1 答案 C

展开负号可得 \[ \begin{aligned} &(20-2010+201) \\ &\quad {}+(2010-201+20)=40 \end{aligned}\text{。}\]

所以正确答案是 C。

English

Distributing the negative signs gives \[ \begin{aligned} &(20-2010+201) \\ &\quad {}+(2010-201+20)=40. \end{aligned} \]

Thus, C is the correct answer.

2 答案 A

渡船共有 \(6\) 班：\(10\) 点、\(11\) 点、\(12\) 点、\(1\) 点、\(2\) 点和 \(3\) 点。

游客数分别为 \(100,99,98,97,96,95\)，所以总数为 \[ \begin{gathered} 6\cdot100-(1+2+3+4+5) \\ =600-15 \\ =585 \end{gathered}\text{。}\]

所以正确答案是 A。

English

The ferry makes \(6\) trips: at \(10,\) \(11,\) \(12,\) \(1,\) \(2,\) and \(3.\)

The numbers of tourists are \(100,99,98,97,96,95,\) so the total is \[ \begin{gathered} 6\cdot100-(1+2+3+4+5) \\ =600-15 \\ =585. \end{gathered} \]

Thus, A is the correct answer.

3 答案 E

设正方形 \(EFGH\) 的边长为 \(s\)。重叠部分宽为 \(s\)，高为 \(AD\)，所以面积为 \(s\cdot AD\)。

因为重叠部分占长方形面积的 \(50\%\)，所以 \(s\cdot AD=\tfrac12\,AB\cdot AD\)，从而 \(AB=2s\)。又因为它占正方形面积的 \(20\%\)，所以 \(s\cdot AD=\tfrac15 s^2\)，从而 \(AD=\tfrac{s}{5}\)。

因此 \[\dfrac{AB}{AD}=\dfrac{2s}{\frac{s}{5}}=10\text{。}\]

所以 E 是正确答案。

English

Let \(s\) be the side length of square \(EFGH.\) The shaded overlap has width \(s\) and height \(AD,\) so its area is \(s\cdot AD.\)

Because the overlap is \(50\%\) of the rectangle, \(s\cdot AD=\tfrac12\,AB\cdot AD,\) so \(AB=2s.\) Because it is \(20\%\) of the square, \(s\cdot AD=\tfrac15 s^2,\) so \(AD=\tfrac{s}{5}.\)

Therefore \[\dfrac{AB}{AD}=\dfrac{2s}{\frac{s}{5}}=10.\]

Thus, E is the correct answer.

4 答案 D

选项 D 为 \(-x^{-1}=-\dfrac1x\)。当 \(x\lt0\) 时，\(\dfrac1x\lt0\)，所以 \(-\dfrac1x\gt0\)。

取 \(x=-1\) 可知其他选项不一定为正：\(\dfrac{x}{|x|}=-1\)，\(-x^2=-1\)，\(-2^x=-\tfrac12\)，且 \(\sqrt[3]{x}=-1\)。

因此 D 是正确答案。

English

Choice (D) is \(-x^{-1}=-\dfrac1x.\) When \(x\lt0,\) \(\dfrac1x\lt0,\) so \(-\dfrac1x\gt0.\)

Testing \(x=-1\) shows the other choices need not be positive: \(\dfrac{x}{|x|}=-1,\) \(-x^2=-1,\) \(-2^x=-\tfrac12,\) and \(\sqrt[3]{x}=-1.\)

Thus, D is the correct answer.

5 答案 C

对手在最后 \(50\) 箭中最多可得 \(50\cdot10=500\) 分。Chelsea 已领先 \(50\) 分，所以她还需要超过 \(500-50=450\) 分。

设 Chelsea 有 \(n\) 箭靶心，得 \(10n\) 分，其余 \(50-n\) 箭至少得 \(4(50-n)\) 分。要保证获胜，需要 \[10n+4(50-n)\gt450\text{。}\] 化简得 \(6n\gt250\)，即 \(n\gt41\tfrac23\)。

因此至少需要 \(42\) 箭靶心。

所以正确答案是 C。

English

The opponent can score at most \(50\cdot10=500\) on the last \(50\) shots. Since Chelsea leads by \(50,\) she must score more than \(500-50=450\) points on her remaining shots to guarantee victory.

Her \(n\) bullseyes give \(10n\) points, and her other \(50-n\) shots give at least \(4(50-n)\) points, so \[10n+4(50-n)\gt450.\] This simplifies to \(6n\gt250,\) i.e. \(n\gt41\tfrac23.\)

Therefore Chelsea needs at least \(42\) bullseyes.

Thus, C is the correct answer.

6 答案 E

因为 \(x\) 至多为 \(999\)，所以 \(x + 32\) 至多为 \(1031\)。

同时 \(x + 32\) 至少为 \(1000\)。

这个范围内唯一的回文数是 \(1001\)，所以 \(x + 32\) 等于这个数。

于是 \(x=1001-32=969\)，它的各位数字之和为 \(9+6+9=24\)。

所以正确答案是 E。

English

Note that \(x\) is at most \(999.\) This means that \(x + 32\) has a maximum of \(1031.\)

Similarly, we have that the minimum value of \(x + 32\) is \(1000.\)

The only palindrome in this range is \(1001,\) so this is what \(x + 32\) equals.

Thus \(x=1001-32=969,\) whose digit sum is \(9+6+9=24.\)

Thus, E is the correct answer.

7 答案 C

模型与原物的体积比为 \(\dfrac{0.1}{100{,}000}=\dfrac{1}{1{,}000{,}000}\)。长度按体积比的立方根缩放，所以模型与原物的高度比为 \(\dfrac1{100}\)。因此模型的高度应为 \(\dfrac{40}{100}=0.4\) 米。

所以正确答案是 C。

English

The model-to-original volume ratio is \(\dfrac{0.1}{100{,}000}=\dfrac{1}{1{,}000{,}000}.\) Lengths scale by the cube root, so the model-to-original height ratio is \(\dfrac1{100}.\) Therefore the model should be \(\dfrac{40}{100}=0.4\) meters tall.

Thus, C is the correct answer.

8 答案 C

设 \(\angle BAE=\angle ACD=x\)。因为 \(\triangle CFE\) 是等边三角形，所以 \(\angle CFE=60^\circ\)。射线 \(FA\) 与 \(FE\) 方向相反，因此 \(\angle AFC=120^\circ\)。

在 \(\triangle AFC\) 中，\(\angle FAC=180^\circ-120^\circ-x\) \(=60^\circ-x\)。因为 \(F\) 位于 \(AE\) 上，这个角就是 \(\angle EAC\)。所以 \(\angle BAC=x+(60^\circ-x)=60^\circ\)。

设 \(AC=s\)，则 \(AB=2s\)。由余弦定理，\[ \begin{aligned} BC^2 &= s^2+(2s)^2 \\ &\quad-2(s)(2s)\cos60^\circ \\ &=3s^2 \end{aligned}\text{。}\]因此三边之比为 \(1:\sqrt3:2\)，且 \(AB\) 为斜边，所以 \(\angle ACB=90^\circ\)。

所以正确答案是 C。

English

Let \(\angle BAE=\angle ACD=x.\) Because \(\triangle CFE\) is equilateral, \(\angle CFE=60^\circ.\) Rays \(FA\) and \(FE\) are opposite, so \(\angle AFC=120^\circ.\)

In \(\triangle AFC,\) we get \(\angle FAC=180^\circ-120^\circ-x\) \(=60^\circ-x.\) Since \(F\) lies on \(AE,\) this is \(\angle EAC.\) Hence \(\angle BAC=x+(60^\circ-x)=60^\circ.\)

Put \(AC=s,\) so \(AB=2s.\) The Law of Cosines gives \[ \begin{aligned} BC^2 &= s^2+(2s)^2 \\ &\quad-2(s)(2s)\cos60^\circ \\ &=3s^2. \end{aligned} \] Thus the side lengths are in the ratio \(1:\sqrt3:2,\) with \(AB\) the hypotenuse, so \(\angle ACB=90^\circ.\)

Thus, C is the correct answer.

9 答案 A

三个被切出的长方体都在立方体中心相交。

交集是边长为 \(2\) 的立方体。因此切除区域的体积为 \[ \begin{align*}3 \cdot 2 \cdot 2 \cdot 3 - 2 \cdot 2^3 &= 36 - 16 \\&= 20\end{align*}\text{。}\]

中心区域被全部 \(3\) 个孔包含，因此要减去重复计入的两次。

剩余体积为 \[ 3^3 - 20 = 27 - 20 = 7\text{。}\]

所以正确答案是 A。

English

Note that all the cut out solids intersect in the middle of the cube.

This region of intersection is a cube with side length \(2.\) Then the volume of the cutout region is \[ \begin{align*}3 \cdot 2 \cdot 2 \cdot 3 - 2 \cdot 2^3 &= 36 - 16 \\&= 20.\end{align*} \]

We have to subtract out the center region twice since it is included in all \(3\) regions.

The remaining volume is then \[ 3^3 - 20 = 27 - 20 = 7. \]

Thus, A is the correct answer.

10 答案 A

相邻两项之差都是公差 \(d\)，且 \(d=(3p+q)-(3p-q)=2q\)。

由前两项可得 \(9-p=d=2q\)，由第二、三项可得 \((3p-q)-9=d=2q\)。解得 \(p=5\)，\(q=2\)，\(d=4\)。

因此第 \(2010\) 项为 \[ \begin{aligned} p+2009d &= 5+2009\cdot4 \\ &= 8041 \end{aligned}\text{。}\]

所以正确答案是 A。

English

Consecutive terms differ by a common difference \(d.\) From the last two terms, \(d=(3p+q)-(3p-q)=2q.\)

From the first two terms, \(9-p=d=2q,\) and from the second and third, \((3p-q)-9=d=2q.\) Solving this system gives \(p=5,\) \(q=2,\) and \(d=4.\)

The \(2010\)th term is \[ \begin{aligned} p+2009d &= 5+2009\cdot4 \\ &= 8041. \end{aligned} \]

Thus, A is the correct answer.

11 答案 C

因为 \(x=\log_b 7^7\)，所以 \(b^x=7^7\)。

因此 \[ \begin{aligned} (7b)^x=7^x\cdot b^x &= 7^x\cdot7^7 \\ &= 7^{x+7}=8^x \end{aligned}\text{。}\]

由于 \(x\gt0\)，可得 \(7b=8\)，所以 \(b=\dfrac{8}{7}\)。

所以正确答案是 C。

English

Since \(x=\log_b 7^7,\) we have \(b^x=7^7.\)

Then \[ \begin{aligned} (7b)^x=7^x\cdot b^x &= 7^x\cdot7^7 \\ &= 7^{x+7}=8^x. \end{aligned} \]

Because \(x\gt0,\) it follows that \(7b=8,\) so \(b=\dfrac{8}{7}.\)

Thus, C is the correct answer.

12 答案 D

Chris 和 LeRoy 必定属于不同物种：如果 Chris 是蟾蜍，他的话说明 LeRoy 是青蛙；如果 Chris 是青蛙，他这句假话说明 LeRoy 是蟾蜍。因此他们两个中恰好有一只是蟾蜍。

如果 Brian 是蟾蜍，他说的是真话，于是 Mike 是青蛙。如果 Brian 是青蛙，他这句假话意味着他和 Mike 是同一物种，此时 Mike 仍然是青蛙。所以 Mike 一定是青蛙。

因此 Mike 的陈述是假话，也就是说蟾蜍不足两只。而 Chris 和 LeRoy 中已经恰好有一只蟾蜍，所以 Brian 只能是青蛙。于是四只中有一只蟾蜍和 \(3\) 只青蛙。

所以正确答案是 D。

English

Chris and LeRoy are of opposite species: if Chris is a toad, his statement makes LeRoy a frog; if Chris is a frog, his false statement makes LeRoy a toad. Thus exactly one of them is a toad.

If Brian is a toad, his true statement makes Mike a frog. If Brian is a frog, his false statement says that he and Mike are the same species, again making Mike a frog. Therefore Mike is always a frog.

Mike’s statement is therefore false, so there are fewer than two toads. Chris and LeRoy already supply exactly one toad, forcing Brian to be a frog. Hence there is one toad and \(3\) frogs.

Thus, D is the correct answer.

13 答案 C

当 \(k=0\) 时，\(x^2+y^2=0\) 的图像是单点 \((0,0)\)，而 \(xy=0\) 是两条坐标轴，它们在原点相交。

当 \(k\ne0\) 时，圆的半径为 \(|k|\)，双曲线 \(xy=k\) 离原点最近的两个顶点到原点距离为 \(\sqrt{2|k|}\)。两图像相交当且仅当 \(|k|\ge\sqrt{2|k|}\)，即 \(|k|\ge2\)。

所以只有 \(|k|=1\) 时两图像不相交，也就是 \(k=1\) 和 \(k=-1\)，共有 \(2\) 个值。

因此 C 是正确答案。

English

For \(k=0,\) the graph of \(x^2+y^2=0\) is the single point \((0,0)\) and \(xy=0\) is the two axes, which meet at the origin, so the graphs intersect.

For \(k\ne0,\) the circle has radius \(|k|,\) and the hyperbola \(xy=k\) has its two vertices nearest the origin at distance \(\sqrt{2|k|}.\) The graphs meet exactly when \(|k|\ge\sqrt{2|k|},\) that is \(|k|\ge2.\)

So they fail to intersect only when \(|k|=1,\) namely \(k=1\) and \(k=-1,\) giving \(2\) values.

Thus, C is the correct answer.

14 答案 B

由角平分线定理，\[ \dfrac{AB}{3} = \dfrac{BC}{8} \] \[ AB = \dfrac{3}{8} BC\text{。}\]

为使 \(AB\) 和 \(BC\) 都是整数，\(BC\) 必须是 \(8\) 的倍数。

若为了最小化周长取 \(BC = 8\)、\(AB = 3\)，三角形会退化。

因此 \(BC\) 必须取 \(16\)，此时 \(AB = 6\)。又 \(AC = AD + DC = 11\)，周长为 \[ 16 + 6 + 11 = 33\text{。}\]

所以正确答案是 B。

English

Using the Angle Bisector Theorem, we have that \[ \dfrac{AB}{3} = \dfrac{BC}{8} \]\[ AB = \dfrac{3}{8} BC. \]

For \(AB\) and \(BC\) to be integers, we must have that \(BC\) is a multiple of \(8.\)

To minimize the perimeter, we can set \(BC = 8\) and \(AB = 3.\) This, however, makes the triangle degenerate.

\(BC\) must then be \(16\) and \(AB = 6.\) Since \(AC = AD + DC = 11,\) the perimeter is \[ 16 + 6 + 11 = 33. \]

Thus, B is the correct answer.

15 答案 D

设正面朝上的概率为 \(p\)。抛掷四次恰好出现两次正面和两次反面的概率为 \[ \begin{aligned} \binom{4}{2}p^2(1-p)^2 &= 6p^2(1-p)^2 \\ &= \frac16 \end{aligned}\text{。}\]

因此 \(p^2(1-p)^2=\dfrac{1}{36}\)，所以 \(p(1-p)=\dfrac16\)。

这给出 \(6p^2-6p+1=0\)，故 \(p=\dfrac{3\pm\sqrt3}{6}\)。由于 \(p\lt\dfrac12\)，取 \(p=\dfrac{3-\sqrt3}{6}\)。

因此 D 是正确答案。

English

Let \(p\) be the probability of heads. The chance of two heads and two tails in four flips is \[ \begin{aligned} \binom{4}{2}p^2(1-p)^2 &= 6p^2(1-p)^2 \\ &= \frac16. \end{aligned} \]

Thus \(p^2(1-p)^2=\dfrac{1}{36},\) so \(p(1-p)=\dfrac16.\)

This gives \(6p^2-6p+1=0,\) so \(p=\dfrac{3\pm\sqrt3}{6}.\) Since \(p\lt\dfrac12,\) we take \(p=\dfrac{3-\sqrt3}{6}.\)

Thus, D is the correct answer.

16 答案 B

分两种情况：Bernardo 选到 \(9\)，或者没有选到。

情况 \(1\)： Bernardo 选到 \(9\)。

固定一个数后，另外两个数有 \(\binom{8}{2} = 28\) 种选法。

总选法为 \(\binom{9}{3} = 84\)，所以此情况的概率是 \[ \dfrac{28}{84} = \dfrac{1}{3}\text{。}\]

注意，如果 Bernardo 选到 \(9\)，他一定比 Silvia 的数大。

因此本情况中 Bernardo 一定获胜。

情况 \(2\)： Bernardo 没有选到 \(9\)。

这种情况发生的概率为 \(1 - \frac{1}{3} = \frac{2}{3}\)。此时两人从同一组数字中选择，获胜机会对称。

我们还需要求出两人选到相同数字的概率。Silvia 与 Bernardo 选到同一组数字的概率为 \[ \dfrac{1}{\binom{8}{3}} = \dfrac{1}{56} \] 于是此时 Bernardo 的数更大的概率为 \[ \dfrac{1 - \frac{1}{56}}{2} = \dfrac{55}{112}\text{。}\]

因此 Bernardo 得到较大数的总概率为 \[ \dfrac{1}{3} + \dfrac{2}{3} \cdot \dfrac{55}{112} = \dfrac{37}{56}\text{。}\]

所以正确答案是 B。

English

There are two cases: Bernardo picks a \(9\) or he doesn’t.

Case \(1:\) Bernardo picks a \(9\)

Since a number is fixed, there are \(\binom{8}{2} = 28\) ways to choose the other two numbers.

There are a total of \(\binom{9}{3} = 84\) ways to pick all three numbers. The probability is then \[ \dfrac{28}{84} = \dfrac{1}{3}. \]

Note that if Bernardo picks a \(9,\) he automatically has a greater number than Silvia.

This means that Bernardo always wins in this case.

Case \(2:\) Bernardo doesn’t pick a \(9\)

There is a \(1 - \frac{1}{3} = \frac{2}{3}\) chance of this happening. Since both people are choosing from the same numbers, they have an equal chance of winning.

We still need to find the probability that the numbers are the same. There is a \[ \dfrac{1}{\binom{8}{3}} = \dfrac{1}{56} \] chance that Silvia chooses the same numbers as Bernardo. The probability that Bernardo gets a higher number is then \[ \dfrac{1 - \frac{1}{56}}{2} = \dfrac{55}{112}. \]

The total probability of Bernardo getting a higher number is then \[ \dfrac{1}{3} + \dfrac{2}{3} \cdot \dfrac{55}{112} = \dfrac{37}{56}. \]

Thus, B is the correct answer.

17 答案 E

注意 \(\triangle ACE\) 是等边三角形。在 \(\triangle ABC\) 中用余弦定理，得 \(AC^2=r^2+1^2-2r\cos120^\circ\) \(=r^2+r+1\)。

因此 \(\triangle ACE\) 的面积为 \[ \dfrac{\sqrt3}{4} (r^2 + r + 1)\text{。} \]

三个角上的三角形 \(\triangle ABC\)、\(\triangle CDE\) 和 \(\triangle EFA\) 各自面积为 \(\frac12\cdot1\cdot r\cdot\sin120^\circ=\frac{r\sqrt3}{4}\)。

所以六边形面积为 \(\dfrac{\sqrt3}{4}(r^2+r+1)\) \(+3\cdot\dfrac{r\sqrt3}{4}\) \(=\dfrac{\sqrt3}{4}(r^2+4r+1)\)。

条件 \([ACE]=70\%\cdot[ABCDEF]\) 给出 \[r^2+r+1=\dfrac{7}{10}(r^2+4r+1)\text{，}\] 所以 \(r^2-6r+1=0\)。

由韦达定理，\(r\) 的所有可能值之和为 \(6\)。

因此 E 是正确答案。

English

Note that \(\triangle ACE\) is equilateral. Using the Law of Cosines in \(\triangle ABC,\) we get \(AC^2=r^2+1^2-2r\cos120^\circ\) \(=r^2+r+1.\)

The area of \(\triangle ACE\) is then \[ \dfrac{\sqrt3}{4} (r^2 + r + 1). \]

The three corner triangles \(\triangle ABC,\) \(\triangle CDE,\) and \(\triangle EFA\) each have area \(\frac12\cdot1\cdot r\cdot\sin120^\circ=\frac{r\sqrt3}{4}.\)

Thus the hexagon has area \(\dfrac{\sqrt3}{4}(r^2+r+1)\) \(+3\cdot\dfrac{r\sqrt3}{4}\) \(=\dfrac{\sqrt3}{4}(r^2+4r+1).\)

The condition \([ACE]=70\%\cdot[ABCDEF]\) gives \[r^2+r+1=\dfrac{7}{10}(r^2+4r+1),\] so \(r^2-6r+1=0.\)

By Vieta’s formulas, the sum of the possible values of \(r\) is \(6.\)

Thus, E is the correct answer.

18 答案 D

每一步都使 \(x+y\) 增加 \(1\)，而坐标和从 \(-8\) 变到 \(8\)，所以每条路径恰好经过一个满足 \(x+y=0\) 的格点。

为了不进入开正方形内部，该点 \((t,-t)\) 必须满足 \(|t|\ge2\)，所以它是 \((\pm2,\mp2),(\pm3,\mp3),(\pm4,\mp4)\) 之一。

由对称性，只考虑三点 \((-4,4),(-3,3),(-2,2)\)，然后将结果加倍。从 \((-4,-4)\) 到 \((-(4-j),4-j)\) 的路径数为 \(\binom{8}{j}\)，继续到 \((4,4)\) 的路径数也是 \(\binom{8}{j}\)。

因此总数为 \[ \begin{gathered} 2\left(\binom80^2+\binom81^2+\binom82^2\right) \\ =2(1+64+784) \\ =1698 \end{gathered}\text{。}\]

所以正确答案是 D。

English

Every step increases \(x+y\) by \(1,\) which runs from \(-8\) to \(8,\) so each path passes through exactly one lattice point with \(x+y=0.\)

To stay out of the open square, that point \((t,-t)\) must have \(|t|\ge2,\) so it is one of \((\pm2,\mp2),(\pm3,\mp3),(\pm4,\mp4).\)

By symmetry consider the three points \((-4,4),(-3,3),(-2,2)\) and double. The number of paths from \((-4,-4)\) to \((-(4-j),4-j)\) is \(\binom{8}{j},\) and the number continuing on to \((4,4)\) is also \(\binom{8}{j}.\)

Therefore the total is \[ \begin{gathered} 2\left(\binom80^2+\binom81^2+\binom82^2\right) \\ =2(1+64+784) \\ =1698. \end{gathered} \]

Thus, D is the correct answer.

19 答案 A

第 \(k\) 个盒子中有 \(k + 1\) 颗弹珠，所以 Isabella 从中抽到白弹珠的概率为 \(\dfrac{k}{k + 1}\)。

抽到红弹珠的概率为 \(\dfrac{1}{k + 1}\)。要在抽到第 \(n\) 颗弹珠后停止，前 \(n - 1\) 颗必须都是白色。

其概率为 \[ \dfrac{1}{2} \cdot \dfrac{2}{3} \cdot \ldots \cdot \dfrac{n - 1}{n} \cdot \dfrac{1}{n + 1}\text{。}\]

所有分子都与相邻分母相消，所以这个表达式化为 \(\dfrac{1}{n(n + 1)}\)。

需要寻找最小的 \(n\)，使得 \[ \dfrac{1}{n(n + 1)} \lt \dfrac{1}{2010} \]\[ n(n + 1) \gt 2010\text{。}\]

因为 \(44\cdot45=1980\lt2010\)，但 \(45\cdot46=2070\gt2010\)，所以最小的 \(n\) 是 \(45\)。

所以正确答案是 A。

English

Since there are \(k + 1\) marbles in the \(k\) th box, there is a \(\dfrac{k}{k + 1}\) chance Isabella draws a white marble from it.

The probability of drawing a red marble is then \(\dfrac{1}{k + 1}.\) To stop after drawing the \(n\) th marble, the first \(n - 1\) marbles must have been white.

This happens with a probability of \[ \dfrac{1}{2} \cdot \dfrac{2}{3} \cdot \ldots \cdot \dfrac{n - 1}{n} \cdot \dfrac{1}{n + 1}. \]

Note that all the numerators cancel with the adjacent denominator, which means that this expression reduces to \(\dfrac{1}{n(n + 1)}.\)

We have to find the smallest \(n\) such that \[ \dfrac{1}{n(n + 1)} \lt \dfrac{1}{2010} \] \[ n(n + 1) \gt 2010. \]

Since \(44\cdot45=1980\lt2010\) but \(45\cdot46=2070\gt2010,\) the smallest such \(n\) is \(45.\)

Thus, A is the correct answer.

20 答案 C

因为 \(a_n=1+(n-1)d_1\)、\(b_n=1+(n-1)d_2\)，其中 \(d_1,d_2\) 为整数，所以 \(n-1\) 同时整除 \(a_n-1\) 和 \(b_n-1\)，从而整除 \(\gcd(a_n-1,b_n-1)\)。

满足 \(2\le a_n\le b_n\) 的 \(2010\) 因数对为 \((2,1005)\)、\((3,670)\)、\((5,402)\)、\((6,335)\)、\((10,201)\)、\((15,134)\) 和 \((30,67)\)。

除了 \((15,134)\) 外，每对的 \(a_n-1\) 和 \(b_n-1\) 都互质，迫使 \(n=2\)。对于 \((15,134)\)，\(\gcd(14,133)=7\)，所以 \(n-1\) 可以等于 \(7\)，得到 \(n=8\)。

数列 \(a_n=2n-1\) 和 \(b_n=19n-18\) 可以达到这个值，因此最大值为 \(8\)。

所以正确答案是 C。

English

Since \(a_n=1+(n-1)d_1\) and \(b_n=1+(n-1)d_2\) for integers \(d_1,d_2,\) the value \(n-1\) divides both \(a_n-1\) and \(b_n-1,\) hence divides \(\gcd(a_n-1,b_n-1).\)

The factor pairs of \(2010\) with \(2\le a_n\le b_n\) are \((2,1005),\) \((3,670),\) \((5,402),\) \((6,335),\) \((10,201),\) \((15,134),\) and \((30,67).\)

For every pair except \((15,134),\) the numbers \(a_n-1\) and \(b_n-1\) are relatively prime, forcing \(n=2.\) For \((15,134),\) \(\gcd(14,133)=7,\) so \(n-1\) can equal \(7,\) giving \(n=8.\)

The sequences \(a_n=2n-1\) and \(b_n=19n-18\) realize this, so the largest value is \(8.\)

Thus, C is the correct answer.

21 答案 A

设 \(f(x)\) 为该图像减去直线所得的函数。它非负，并在三个点处为零，每个零点都是二重根，所以 \[f(x)=\big(x^3-Ax^2+Bx-C\big)^2\text{。}\]

比较系数可得 \(-2A=-10\Rightarrow A=5\)，接着 \(A^2+2B=29\Rightarrow B=2\)，再由 \(-2C-2AB=-4\Rightarrow C=-8\)。

因此三次式为 \(x^3-5x^2+2x+8\) \(=(x+1)(x-2)(x-4)\)，根为 \(-1,2\) 和 \(4\)。最大值为 \(4\)。

所以正确答案是 A。

English

Let \(f(x)\) be the graph minus the line. It is nonnegative and vanishes at three points, each a double root, so \[f(x)=\big(x^3-Ax^2+Bx-C\big)^2.\]

Matching coefficients gives \(-2A=-10\Rightarrow A=5,\) then \(A^2+2B=29\Rightarrow B=2,\) then \(-2C-2AB=-4\Rightarrow C=-8.\)

Thus the cubic is \(x^3-5x^2+2x+8\) \(=(x+1)(x-2)(x-4),\) with roots \(-1,2,\) and \(4.\) The largest is \(4.\)

Thus, A is the correct answer.

22 答案 A

函数 \(f\) 是分段线性的，断点在 \(x=\tfrac1k\)。在区间 \(\left[\tfrac1m,\tfrac1{m-1}\right]\) 上，它的斜率为 \[ \begin{aligned} &\sum_{k=m}^{119}k \\ &\quad {}-\sum_{k=1}^{m-1}k=7140-(m-1)m\text{，} \end{aligned} \] 其中 \(7140=\tfrac{119\cdot120}{2}\)。

当 \((m-1)m=7140\) 时斜率为零，即 \(m=85\)，所以最小值出现在右端点 \(x=\tfrac1{84}\)。

在那里，\(k\le84\) 的项贡献 \(\tfrac{84-k}{84}\)，\(k\ge85\) 的项贡献 \(\tfrac{k-84}{84}\)，所以 \[ \begin{aligned} f\left(\tfrac1{84}\right) &= \frac{3486}{84}+\frac{630}{84} \\ &= 41.5+7.5=49\text{。} \end{aligned} \]

所以正确答案是 A。

English

The function \(f\) is piecewise linear with breakpoints at \(x=\tfrac1k.\) On the interval \(\left[\tfrac1m,\tfrac1{m-1}\right]\) its slope is \[ \begin{aligned} &\sum_{k=m}^{119}k \\ &\quad {}-\sum_{k=1}^{m-1}k=7140-(m-1)m, \end{aligned} \] where \(7140=\tfrac{119\cdot120}{2}.\)

This slope is zero when \((m-1)m=7140,\) i.e. \(m=85,\) so the minimum occurs at the right endpoint \(x=\tfrac1{84}.\)

There, terms with \(k\le84\) contribute \(\tfrac{84-k}{84}\) and terms with \(k\ge85\) contribute \(\tfrac{k-84}{84},\) so \[ \begin{aligned} f\left(\tfrac1{84}\right) &= \frac{3486}{84}+\frac{630}{84} \\ &= 41.5+7.5=49. \end{aligned} \]

Thus, A is the correct answer.

23 答案 A

\(90!\) 末尾零的个数为 \(\left\lfloor\dfrac{90}{5}\right\rfloor+\left\lfloor\dfrac{90}{25}\right\rfloor=21\)。令 \(N=\dfrac{90!}{10^{21}}\)。

去掉 \(10^{21}\) 后仍剩下超过两个因数 \(2\)，所以 \(N\equiv0 \pmod4\)。

设 \(A\) 为 \(90!\) 中不能被 \(5\) 整除的那些因数之积，\(B\) 为能被 \(5\) 整除的那些因数之积。每一块 \((5j+1)(5j+2)(5j+3)(5j+4)\) 都满足 \(24\equiv-1\pmod{25}\)，一共有 \(18\) 块，所以 \(A\equiv1\pmod{25}\)。

从 \(B\) 中去掉 \(21\) 个因数 \(5\) 后，剩余因数可分组为 \[ \begin{aligned} \dfrac{B}{5^{21}}={}&(1\cdot2\cdot3\cdot4) \\ &\cdot(6\cdot7\cdot8\cdot9) \\ &\cdot(11\cdot12\cdot13\cdot14) \\ &\cdot(16\cdot17\cdot18)(1\cdot2\cdot3) \\ &\equiv-1\pmod{25} \end{aligned}\text{。}\]

因此 \(\dfrac{90!}{5^{21}}\equiv-1\pmod{25}\)。因为 \(2^{21}\equiv2\pmod{25}\)，而 \(2\) 模 \(25\) 的逆元为 \(13\)，所以 \(N\equiv-13\equiv12\pmod{25}\)。

同时满足 \(0\pmod4\) 和 \(12\pmod{25}\) 的数为 \(12\pmod{100}\)，所以最后两个非零数字组成 \(12\)。

所以正确答案是 A。

English

The number of trailing zeroes in \(90!\) is \(\left\lfloor\dfrac{90}{5}\right\rfloor+\left\lfloor\dfrac{90}{25}\right\rfloor=21.\) Let \(N=\dfrac{90!}{10^{21}}.\)

There are still more than two factors of \(2\) left after removing \(10^{21},\) so \(N\equiv0 \pmod4.\)

Let \(A\) be the product of factors of \(90!\) not divisible by \(5,\) and let \(B\) be the product of the factors divisible by \(5.\) Each block \((5j+1)(5j+2)(5j+3)(5j+4)\) is \(24\equiv-1\pmod{25},\) and there are \(18\) blocks, so \(A\equiv1\pmod{25}.\)

After removing the \(21\) factors of \(5\) from \(B,\) the remaining factors can be grouped as \[ \begin{aligned} \dfrac{B}{5^{21}}={}&(1\cdot2\cdot3\cdot4) \\ &\cdot(6\cdot7\cdot8\cdot9) \\ &\cdot(11\cdot12\cdot13\cdot14) \\ &\cdot(16\cdot17\cdot18)(1\cdot2\cdot3) \\ &\equiv-1\pmod{25}. \end{aligned} \]

Therefore \(\dfrac{90!}{5^{21}}\equiv-1\pmod{25}.\) Since \(2^{21}\equiv2\pmod{25}\) and the inverse of \(2\) modulo \(25\) is \(13,\) we get \(N\equiv-13\equiv12\pmod{25}.\)

The number congruent to \(0\pmod4\) and \(12\pmod{25}\) is \(12\pmod{100},\) so the last two nonzero digits form \(12.\)

Thus, A is the correct answer.

24 答案 B

令 \(g(x)=\prod_{k=1}^8\sin(k\pi x)\)；则 \(f\) 的定义域为 \(g(x)\gt0\) 的地方。由于 \(\sin(k\pi(1-x))\) \(=(-1)^{k+1}\sin(k\pi x)\)，且 \(\sum_{k=1}^8(k+1)\) 为偶数，所以 \(g(1-x)=g(x)\)，只需研究 \(\big(0,\tfrac12\big)\) 后再加倍。

在 \(\big(0,\tfrac12\big)\) 中，\(g\) 的零点是满足 \(\tfrac{k}{n}\)、\(2\le n\le8\)、\(1\le k\lt\tfrac n2\)、\(\gcd(k,n)=1\) 的分数。对 \(n=2,\ldots,8\) 数量分别为 \(0,1,1,2,1,3,2\)，合计 \(10\) 个。

这 \(10\) 个零点把 \(\big(0,\tfrac12\big)\) 分成 \(11\) 个小区间，在每个小区间上 \(g\) 的符号恒定。靠近 \(0\) 时每个因子为正，所以 \(g\gt0\)。每经过一个零点符号通常改变，但在 \(\tfrac14=\tfrac28\) 和 \(\tfrac13=\tfrac26\) 处有偶数个因子同时为零，符号不改变。

跟踪符号可知，这 \(11\) 个小区间中正好有 \(6\) 个满足 \(g\gt0\)。由对称性，在 \(\big(\tfrac12,1\big)\) 中还有 \(6\) 个，所以 \(n=12\)。

因此，正确答案是 B。

English

Let \(g(x)=\prod_{k=1}^8\sin(k\pi x);\) the domain of \(f\) is where \(g(x)\gt0.\) Since \(\sin(k\pi(1-x))\) \(=(-1)^{k+1}\sin(k\pi x)\) and \(\sum_{k=1}^8(k+1)\) is even, \(g(1-x)=g(x),\) so it suffices to study \(\big(0,\tfrac12\big)\) and double.

In \(\big(0,\tfrac12\big)\) the zeros of \(g\) are the fractions \(\tfrac{k}{n}\) with \(2\le n\le8,\) \(1\le k\lt\tfrac n2,\) and \(\gcd(k,n)=1.\) For \(n=2,\ldots,8\) there are \(0,1,1,2,1,3,2\) of them, totaling \(10.\)

These \(10\) zeros split \(\big(0,\tfrac12\big)\) into \(11\) subintervals on which \(g\) has constant sign. Near \(0\) every factor is positive, so \(g\gt0\) there, and the sign flips at each zero except \(\tfrac14=\tfrac28\) and \(\tfrac13=\tfrac26,\) where an even number of factors vanish.

Tracking the signs, exactly \(6\) of the \(11\) subintervals have \(g\gt0.\) By symmetry there are \(6\) more in \(\big(\tfrac12,1\big),\) so \(n=12.\)

Thus, B is the correct answer.

25 答案 C

一个凸圆内接四边形由其按圆周顺序排列的边长唯一确定，且存在的条件是最大边长小于其他三边之和。周长为 \(32\) 时，每条边至多为 \(15\)。

先数正整数有序四元组 \((a,b,c,d)\)，满足 \(a+b+c+d=32\) 且每项至多为 \(15\)。若没有上界限制，有 \(\binom{31}{3}=4495\) 个；去掉某一项至少为 \(16\) 的情况，需减去 \(4\binom{16}{3}=2240\)，剩下 \(2255\) 个。

四边形的旋转对应于 \((a,b,c,d)\) 的循环置换。由 Burnside 引理，不同四边形的数量为 \[\frac{1}{4}\big(2255+f_1+f_2+f_3\big)\text{，}\] 其中 \(f_i\) 表示旋转 \(i\) 个位置后不变的四元组数。

旋转一步或三步只固定 \((8,8,8,8)\)，所以 \(f_1=f_3=1\)。旋转两步固定形如 \((a,b,a,b)\) 且 \(a+b=16\)、\(1\le a,b\le15\) 的四元组，所以 \(f_2=15\)。

因此数量为 \[ \begin{aligned} \frac{1}{4}(2255+1+15+1) &= \frac{2272}{4} \\ &= 568\text{。} \end{aligned} \]

所以正确答案是 C。

English

A convex cyclic quadrilateral is determined up to rotation and translation by its cyclic sequence of side lengths, and it exists exactly when the largest side is less than the sum of the others. With perimeter \(32,\) this means each side is at most \(15.\)

First count ordered quadruples \((a,b,c,d)\) of positive integers with \(a+b+c+d=32\) and each entry at most \(15.\) Without the upper bound there are \(\binom{31}{3}=4495;\) removing those with some entry at least \(16\) subtracts \(4\binom{16}{3}=2240,\) leaving \(2255.\)

Rotations of the quadrilateral correspond to cyclic permutations of \((a,b,c,d).\) By Burnside’s lemma the number of distinct quadrilaterals is \[\frac{1}{4}\big(2255+f_1+f_2+f_3\big),\] where \(f_i\) counts quadruples fixed by rotating \(i\) positions.

A one- or three-step rotation fixes only \((8,8,8,8),\) so \(f_1=f_3=1.\) A two-step rotation fixes \((a,b,a,b)\) with \(a+b=16\) and \(1\le a,b\le15,\) giving \(f_2=15.\)

Hence the count is \[ \begin{aligned} \frac{1}{4}(2255+1+15+1) &= \frac{2272}{4} \\ &= 568. \end{aligned} \]

Thus, C is the correct answer.