2010B AMC 12 真题（P版）

1 答案 C

两场会议用了 \(45+90=135\) 分钟，而工作日共有 \(9\cdot60=540\) 分钟。

用于开会的时间占全天的比例是 \[ \frac{135}{540}=\frac14=25\%\text{。}\]

因此，正确答案是 C。

English

The two meetings lasted \(45+90=135\) minutes, and the work day is \(9\cdot60=540\) minutes.

The fraction of the day spent in meetings is \[ \frac{135}{540}=\frac14=25\%. \]

Thus, the correct answer is C.

2 答案 A

这个区域可以分成一个 \(8\times2\) 的竖直长方形和一个 \(2\times3\) 的水平底部，其中底部的宽为 \(5-2=3\)。

总面积为 \[ 8\cdot2+3\cdot2=16+6=22\text{。}\]

因此，正确答案是 A。

English

The region splits into an \(8\times2\) vertical rectangle and a \(2\times3\) horizontal foot, whose width is \(5-2=3.\)

The total area is \[ 8\cdot2+3\cdot2=16+6=22. \]

Thus, the correct answer is A.

3 答案 E

票价 \(x\) 必须同时整除两个总价，所以 \(x\) 是 \(48\) 和 \(64\) 的公因数。

因为 \(\gcd(48,64)=16\)，公因数为 \(1, 2, 4, 8\) 和 \(16\)。因此有 \(5\) 个可能的取值。

因此，正确答案是 E。

English

The price \(x\) must divide both totals, so \(x\) is a common divisor of \(48\) and \(64.\)

Since \(\gcd(48,64)=16,\) the common divisors are \(1, 2, 4, 8,\) and \(16.\) There are \(5\) possible values.

Thus, the correct answer is E.

4 答案 B

因为 \(31=4\cdot7+3\)，这个月的前三天各出现五次，其余四天各出现四次。

星期一和星期三次数相同，恰好发生在它们都属于出现五次的一组，或都属于出现四次的一组时。

如果第一天是星期一，出现五次的是星期一、星期二、星期三。若第一天是星期四或星期五，出现五次的三天都不包含星期一和星期三。其他起始日都会只包含星期一和星期三中的一个。

所以第一天可以是星期一、星期四或星期五，共 \(3\) 种可能。

因此，正确答案是 B。

English

Since \(31=4\cdot7+3,\) the first three days of the month each occur five times, and the other four days occur four times.

Mondays and Wednesdays are equal in number exactly when both fall in the five-time group or both fall in the four-time group.

If the first day is Monday, the five-time days are Mon, Tue, Wed (both appear five times). If the first day is Thursday or Friday, the five-time days miss both Monday and Wednesday (both appear four times). Every other starting day includes exactly one of Monday or Wednesday.

So the first day can be Monday, Thursday, or Friday, giving \(3\) possibilities.

Thus, the correct answer is B.

5 答案 D

正确的值是 \(a-(b-(c-(d+e)))=\) \(a-b+c-d-e\)。代入 \(a, b, c, d=1, 2, 3, 4\) 得到 \(1-2+3-4-e=-2-e\)。

Larry 去掉括号后算的是 \(1-2-3-4+e=-8+e\)。

令 \(-2-e=-8+e\)，得 \(2e=6\)，所以 \(e=3\)。

因此，正确答案是 D。

English

The correct value is \(a-(b-(c-(d+e)))=\) \(a-b+c-d-e.\) With \(a, b, c, d=1, 2, 3, 4,\) this equals \(1-2+3-4-e=-2-e.\)

Larry dropped the parentheses and computed \(1-2-3-4+e=-8+e.\)

Setting \(-2-e=-8+e\) gives \(2e=6,\) so \(e=3.\)

Thus, the correct answer is D.

6 答案 D

假设有 \(100\) 名学生。回答“喜欢”的人数从 \(50\) 增加到 \(70\)，所以至少有 \(70-50=20\) 名学生从“不喜欢”改为“喜欢”；因此 \(x\ge20\)。

因为结束时只有 \(30\) 名学生回答“不喜欢”，原来回答“喜欢”的五十人中至少有 \(50-30=20\) 人仍回答“喜欢”，所以最多有 \(80\) 名学生改变回答；因此 \(x\le80\)。

两个极端都可以实现，所以差为 \(80-20=60\)。

因此，正确答案是 D。

English

Assume \(100\) students. The number of “Yes” answers rises from \(50\) to \(70,\) so at least \(70-50=20\) students switched from “No” to “Yes”; thus \(x\ge20.\)

Since only \(30\) students answer “No” at the end, at least \(50-30=20\) of the original “Yes” students still answer “Yes,” so at most \(80\) students switched; thus \(x\le80.\)

Both extremes are achievable, so the difference is \(80-20=60.\)

Thus, the correct answer is D.

7 答案 C

设 \(t\) 为雨中骑行的分钟数。她在雨中骑了 \(20\cdot\frac{t}{60}\) 英里， 在晴天中骑了 \(30\cdot\frac{40-t}{60}\) 英里。

令总距离为 \(16\)，得到 \[ \frac{20t+30(40-t)}{60}=16\text{，} \] 所以 \(1200-10t=960\)，\(t=24\)。

因此，正确答案是 C。

English

Let \(t\) be the number of minutes driven in the rain. She covers \(20\cdot\frac{t}{60}\) miles in the rain and \(30\cdot\frac{40-t}{60}\) miles in the sun.

Setting the total to \(16\) gives \[ \frac{20t+30(40-t)}{60}=16, \] so \(1200-10t=960\) and \(t=24.\)

Thus, the correct answer is C.

8 答案 B

若有 \(n\) 所学校，则共有 \(3n\) 名学生。Carla 排第 \(64\) 名，所以 \(3n\ge64\)，\(n\ge22\)。

分数都不同，且 Andrea 是中位数，所以 \(3n\) 是奇数，从而 \(n\) 是奇数且 \(n\ge23\)。

Andrea 的名次是 \(\dfrac{3n+1}{2}\)，且她高于 Beth（第 \(37\) 名），所以 \(\dfrac{3n+1}{2}\lt37\)，得 \(3n\lt73\)，\(n\le24\)。唯一的奇数取值是 \(n=23\)。

因此，正确答案是 B。

English

With \(n\) schools there are \(3n\) students. Carla placed \(64\)th, so \(3n\ge64\) and \(n\ge22.\)

The scores are distinct and Andrea is the median, so \(3n\) is odd, forcing \(n\) odd and \(n\ge23.\)

Andrea’s position is \(\dfrac{3n+1}{2},\) and she beat Beth (\(37\)th), so \(\dfrac{3n+1}{2}\lt37,\) giving \(3n\lt73\) and \(n\le24.\) The only odd value is \(n=23.\)

Thus, the correct answer is B.

9 答案 E

为了最小，\(n\) 只使用 \(20\) 的质因数，所以 \(n=2^a\cdot5^b\)，其中 \(a\ge2\)，\(b\ge1\)。

因为 \(n^2=2^{2a}5^{2b}\) 是完全立方数，\(a\) 和 \(b\) 都是 \(3\) 的倍数。因为 \(n^3=2^{3a}5^{3b}\) 是完全平方数，\(a\) 和 \(b\) 都是 \(2\) 的倍数。因此 \(a\) 和 \(b\) 都是 \(6\) 的倍数。

最小选择是 \(a=b=6\)，所以 \(n=2^6\cdot5^6=10^6=1{,}000{,}000\)，它有 \(7\) 位数字。

因此，正确答案是 E。

English

To be smallest, \(n\) uses only the primes of \(20,\) so \(n=2^a\cdot5^b\) with \(a\ge2\) and \(b\ge1.\)

Since \(n^2=2^{2a}5^{2b}\) is a perfect cube, \(a\) and \(b\) are multiples of \(3.\) Since \(n^3=2^{3a}5^{3b}\) is a perfect square, \(a\) and \(b\) are multiples of \(2.\) Hence \(a\) and \(b\) are multiples of \(6.\)

The smallest choice is \(a=b=6,\) so \(n=2^6\cdot5^6=10^6=1{,}000{,}000,\) which has \(7\) digits.

Thus, the correct answer is E.

10 答案 B

数 \(1\) 到 \(99\) 的和是 \(\dfrac{99\cdot100}{2}=4950\)。

由平均数的条件可得 \[ \frac{4950+x}{100}=100x\text{，} \] 所以 \(4950+x=10000x\)，\(9999x=4950\)。

因此 \(x=\dfrac{4950}{9999}=\dfrac{50}{101}\)。

因此，正确答案是 B。

English

The numbers \(1\) through \(99\) sum to \(\dfrac{99\cdot100}{2}=4950.\)

The average condition is \[ \frac{4950+x}{100}=100x, \] so \(4950+x=10000x\) and \(9999x=4950.\)

Thus \(x=\dfrac{4950}{9999}=\dfrac{50}{101}.\)

Thus, the correct answer is B.

11 答案 E

四位回文数形如 \(\overline{abba}=1001a+110b\)，其中 \(1\le a\le9\)，\(0\le b\le9\)。

因为 \(1001=7\cdot11\cdot13\) 可被 \(7\) 整除，而 \(110\) 不能，所以该数能被 \(7\) 整除当且仅当 \(b\) 是 \(7\) 的倍数，即 \(b=0\) 或 \(b=7\)。

对每个 \(a\)，这是 \(b\) 的 \(10\) 个选择中的 \(2\) 个，概率为 \(\dfrac{2}{10}=\dfrac15\)。

因此，正确答案是 E。

English

A four-digit palindrome has the form \(\overline{abba}=1001a+110b\) with \(1\le a\le9\) and \(0\le b\le9.\)

Since \(1001=7\cdot11\cdot13\) is divisible by \(7\) and \(110\) is not, the number is divisible by \(7\) exactly when \(b\) is divisible by \(7,\) that is \(b=0\) or \(b=7.\)

For each \(a,\) that is \(2\) of the \(10\) choices of \(b,\) a probability of \(\dfrac{2}{10}=\dfrac15.\)

Thus, the correct answer is E.

12 答案 D

设 \(L=\log_2 x\)。把每一项都换成以 \(2\) 为底：

\(\log_{\sqrt2}\sqrt{x}=\dfrac{\frac{L}{2}}{\frac{1}{2}}=L\)，\(\log_2 x=L\)，\(\log_4 x^2=\dfrac{2L}{2}=L\)，\(\log_8 x^3=\dfrac{3L}{3}=L\)，且 \(\log_{16}x^4=\dfrac{4L}{4}=L\)。

方程变为 \(5L=40\)，所以 \(L=8\)，\(x=2^8=256\)。

因此，正确答案是 D。

English

Let \(L=\log_2 x.\) Converting each term to base \(2:\)

\(\log_{\sqrt2}\sqrt{x}=\dfrac{\frac{L}{2}}{\frac{1}{2}}=L,\) \(\log_2 x=L,\) \(\log_4 x^2=\dfrac{2L}{2}=L,\) \(\log_8 x^3=\dfrac{3L}{3}=L,\) and \(\log_{16}x^4=\dfrac{4L}{4}=L.\)

The equation becomes \(5L=40,\) so \(L=8\) and \(x=2^8=256.\)

Thus, the correct answer is D.

13 答案 C

一个余弦值加一个正弦值等于 \(2\)，只能是两者都等于 \(1\)。因此 \(\cos(2A-B)=1\) 且 \(\sin(A+B)=1\)，得 \(2A-B=0^\circ\) 和 \(A+B=90^\circ\)。

解得 \(A=30^\circ\)，\(B=60^\circ\)，所以 \(\triangle ABC\) 是直角在 \(C\) 的 \(30\text{-}60\text{-}90\) 直角三角形。

斜边 \(AB=4\) 而 \(BC\) 是 \(30^\circ\) 角所对的边，等于斜边的一半，所以 \(BC=2\)。

因此，正确答案是 C。

English

A cosine plus a sine equals \(2\) only when each equals \(1.\) So \(\cos(2A-B)=1\) and \(\sin(A+B)=1,\) giving \(2A-B=0^\circ\) and \(A+B=90^\circ.\)

Solving, \(A=30^\circ\) and \(B=60^\circ,\) so \(\triangle ABC\) is a \(30\text{-}60\text{-}90\) right triangle with the right angle at \(C.\)

With hypotenuse \(AB=4,\) the side \(BC\) opposite the \(30^\circ\) angle is half the hypotenuse, so \(BC=2.\)

Thus, the correct answer is C.

14 答案 B

\(a+b\)，\(d+e\)，和 \(c\) 都至多为 \(M\)（注意 \(c\le c+d\le M\)）。相加得 \(2010=(a+b)+c+(d+e)\) \(\le3M\)，所以 \(M\ge670\)。

如果 \(M=670\) 则 \(c=670\)，但此时 \(b+c\ge671\gt M\)，矛盾。因此 \(M\ge671\)。

取 \((a,b,c,d,e)=\) \((669,1,670,1,669)\)，可达到 \(671\)，其相邻两项和为 \(670,671,671,670\)。

因此，正确答案是 B。

English

Each of \(a+b,\) \(d+e,\) and \(c\) is at most \(M\) (note \(c\le c+d\le M\)). Adding, \(2010=(a+b)+c+(d+e)\) \(\le3M,\) so \(M\ge670.\)

If \(M=670,\) then \(c=670,\) but then \(b+c\ge671\gt M,\) a contradiction. Hence \(M\ge671.\)

The value \(671\) is reached by \((a,b,c,d,e)=\) \((669,1,670,1,669),\) whose consecutive-pair sums are \(670,671,671,670.\)

Thus, the correct answer is B.

15 答案 D

我们需要 \(i^x\)，\((1+i)^y\)，\(z\) 中恰好有两个相等，第三个不同。三种情况对应三种可能的相等配对。

情况 \(i^x=(1+i)^y\)： 因为 \(|i^x|=1\)，而当 \(y\ge1\) 时 \(|(1+i)^y|=2^{\frac{y}{2}}\gt1\)，所以必须有 \(y=0\)。此时 \((1+i)^0=1\)，且 \(i^x=1\)，即 \(x\in\{0,4,8,12,16\}\)。此时 \(z\) 可以取除 \(1\) 以外的任意一个值，共有 \(19\) 种选择，得到 \(5\cdot19=95\) 个三元组。

情况 \(i^x=z\)： \(i^x\) 中唯一的非负整数值是 \(1\)（当 \(x\) 是 \(4\) 的倍数时），所以 \(z=1\)，且 \((1+i)^y\ne1\)，即 \(y\ge1\)。这给出 \(5\cdot19=95\) 个三元组。

情况 \((1+i)^y=z\)： 因为 \((1+i)^2=2i\)，所以 \((1+i)^y\) 只有在 \(y=0\)（值为 \(1\)）或 \(y=8\)（值为 \(16\)）时才是小于 \(20\) 的非负整数。如果 \(y=0,\ z=1\)，则需要 \(i^x\ne1\)，所以 \(x\) 不是 \(4\) 的倍数（\(15\) 个值）。如果 \(y=8,\ z=16\)，则 \(i^x\) 不可能等于 \(16\)，所以 \(x\) 可任取（\(20\) 个值）。这一情况给出 \(15+20=35\) 个三元组。

总数为 \(95+95+35=225\)。

因此，正确答案是 D。

English

We need exactly two of \(i^x,\) \((1+i)^y,\) \(z\) equal, with the third different. The three cases are the three possible equal pairs.

Case \(i^x=(1+i)^y:\) since \(|i^x|=1\) but \(|(1+i)^y|=2^{\frac{y}{2}}\gt1\) for \(y\ge1,\) we need \(y=0,\) so \((1+i)^0=1\) and \(i^x=1,\) i.e. \(x\in\{0,4,8,12,16\}.\) Then \(z\) is any of the \(19\) values other than \(1.\) This gives \(5\cdot19=95\) triples.

Case \(i^x=z:\) the only nonnegative-integer value of \(i^x\) is \(1\) (with \(x\) a multiple of \(4\)), so \(z=1\) and \((1+i)^y\ne1,\) meaning \(y\ge1.\) This gives \(5\cdot19=95\) triples.

Case \((1+i)^y=z:\) since \((1+i)^2=2i,\) the power \((1+i)^y\) is a nonnegative integer below \(20\) only for \(y=0\) (value \(1\)) or \(y=8\) (value \(16\)). If \(y=0,\ z=1,\) we need \(i^x\ne1,\) so \(x\) is not a multiple of \(4\) (\(15\) values). If \(y=8,\ z=16,\) then \(i^x\) is never \(16,\) so \(x\) is free (\(20\) values). This gives \(15+20=35\) triples.

Altogether \(95+95+35=225.\)

Thus, the correct answer is D.

16 答案 E

分解 \(abc+ab+a=a(bc+b+1)\)。因为 \(2010\) 是 \(3\) 的倍数，\(a, b, c\) 在模 \(3\) 下各个余数等可能。

如果 \(a\) 是 \(3\) 的倍数（概率为 \(\tfrac13\)），乘积能被 \(3\) 整除。

如果 \(a\) 不是 \(3\) 的倍数（概率为 \(\tfrac23\)），需要 \(bc+b+1\) 是 \(3\) 的倍数。检查余数可知，这恰好在 \((b,c)\equiv(1,1)\) 或 \((2,0)\pmod3\) 时成立，概率为 \(\tfrac13\cdot\tfrac13+\tfrac13\cdot\tfrac13=\tfrac29\)。

总概率是 \[ \frac13+\frac23\cdot\frac29=\frac13+\frac{4}{27}=\frac{13}{27}\text{。}\]

因此，正确答案是 E。

English

Factor \(abc+ab+a=a(bc+b+1).\) Since \(2010\) is a multiple of \(3,\) each of \(a, b, c\) is uniform modulo \(3.\)

If \(a\) is divisible by \(3\) (probability \(\tfrac13\)), the product is divisible by \(3.\)

If \(a\) is not divisible by \(3\) (probability \(\tfrac23\)), we need \(bc+b+1\) to be divisible by \(3.\) Checking residues, this holds exactly when \((b,c)\equiv(1,1)\) or \((2,0)\pmod3,\) a probability of \(\tfrac13\cdot\tfrac13+\tfrac13\cdot\tfrac13=\tfrac29.\)

The total probability is \[ \frac13+\frac23\cdot\frac29=\frac13+\frac{4}{27}=\frac{13}{27}. \]

Thus, the correct answer is E.

17 答案 D

记第 \(i\) 行第 \(j\) 列的项为 \(a_{ij}\)。题设条件迫使 \(a_{11}=1\)、\(a_{33}=9\)，且 \(a_{22}\in\{4,5,6\}\)。

如果 \(a_{22}=4\) 则 \(\{a_{12},a_{21}\}=\{2,3\}\) 而 \(\{5,6,7,8\}\) 被分成互补的两对，填入最后一行和最后一列剩下的位置：分法有 \(\binom42=6\) 种，再乘 \(\{2,3\}\) 的 \(2\) 种顺序，得到 \(12\) 个数组。由对称性，\(a_{22}=6\) 也给出 \(12\) 个。

如果 \(a_{22}=5\) 则 \(\{a_{12},a_{13},a_{23}\}\) 和 \(\{a_{21},a_{31},a_{32}\}\) 是 \(\{2,3,4,6,7,8\}\) 的互补子集，并受递增顺序限制；第一组可以是任意三元子集，只要它不是 \(\{2,3,4\}\) 或 \(\{6,7,8\}\)，因此有 \(\binom63-2=18\) 个数组。

总共有 \(12+12+18=42\)。

因此，正确答案是 D。

English

Write \(a_{ij}\) for the entry in row \(i,\) column \(j.\) The conditions force \(a_{11}=1,\) \(a_{33}=9,\) and \(a_{22}\in\{4,5,6\}.\)

If \(a_{22}=4,\) then \(\{a_{12},a_{21}\}=\{2,3\}\) and \(\{5,6,7,8\}\) split as complementary pairs filling the rest of the last row and column: \(\binom42=6\) splits times \(2\) orders for \(\{2,3\}\) gives \(12\) arrays. By symmetry \(a_{22}=6\) also gives \(12.\)

If \(a_{22}=5,\) then \(\{a_{12},a_{13},a_{23}\}\) and \(\{a_{21},a_{31},a_{32}\}\) are complementary subsets of \(\{2,3,4,6,7,8\}\) subject to the ordering constraints. The first set can be any three-element subset except \(\{2,3,4\}\) or \(\{6,7,8\},\) giving \(\binom63-2=18\) arrays.

Altogether \(12+12+18=42.\)

Thus, the correct answer is D.

18 答案 C

这是一个连续的几何概率问题。把第二跳固定为从 \(P=(0,0)\) 到 \(Q=(1,0)\)，令 \(\alpha,\beta\) 为第一跳和第三跳的方向，则起点为 \(A=(\cos\alpha,\sin\alpha)\)，终点为 \(B=(1+\cos\beta,\sin\beta)\)。

取 \(0\le\alpha\le\pi\) 和 \(0\le\beta\le2\pi\)，条件 \(AB\le1\) 恰好等价于 \(\alpha\le\beta\le\pi\)。

在面积为 \(2\pi^2\) 的 \(\alpha\beta\)-矩形中，有利区域是面积 \(\tfrac{\pi^2}{2}\) 的三角形，所以概率为 \(\dfrac{\frac{\pi^2}{2}}{2\pi^2}=\dfrac14\)。

因此，正确答案是 C。

English

This is a continuous (geometric) probability. Anchor the second jump from \(P=(0,0)\) to \(Q=(1,0),\) and let \(\alpha,\beta\) be the directions of the first and third jumps, so the start is \(A=(\cos\alpha,\sin\alpha)\) and the end is \(B=(1+\cos\beta,\sin\beta).\)

Taking \(0\le\alpha\le\pi\) and \(0\le\beta\le2\pi,\) the requirement \(AB\le1\) holds exactly when \(\alpha\le\beta\le\pi.\)

In the \(\alpha\beta\)-rectangle of area \(2\pi^2,\) the favorable region is a triangle of area \(\tfrac{\pi^2}{2},\) so the probability is \(\dfrac{\frac{\pi^2}{2}}{2\pi^2}=\dfrac14.\)

Thus, the correct answer is C.

19 答案 E

设 Raiders 四节的得分依次为 \(a, ar, ar^2, ar^3\)（递增等比数列，\(r\gt1\)），Wildcats 的得分依次为 \(a, a+d, a+2d, a+3d\)（递增等差数列），两队第一节都得 \(a\)。

将 \(r=\frac{m}{n}\) 写成最简分数，其中 \(m\gt n\)。因为 \(ar^3\) 是整数，所以 \(a\) 是 \(n^3\) 的倍数；令 \(a=A n^3\)。Raiders 的总分为 \[ R=A(n^3+mn^2+m^2n+m^3)\text{。}\]由于 \(m^3\lt R\le100\)，可知 \(m\le4\)。数对 \((4,3)\) 已使括号内的和达到 \(175\)，所以互质数对 \((m,n)\) 只可能是 \((2,1),(3,1),(3,2),(4,1)\)。

相应的 \(\frac{R}{A}\) 分别为 \(15,40,65,85\)。对于 \((4,1)\) 和 \((3,2)\)，上界迫使 \(A=1\)，但由 \(R-1=4a+6d\) 得到的 \(d\) 不是整数。对于 \((3,1)\)，会得到 \(36A-1=6d\)，这在模 \(6\) 意义下不可能。

当 \((m,n)=(2,1)\) 时，\(R=15A\) 且 \(a=A\)，所以 \(11A-1=6d\)。因而 \(A\equiv5\pmod6\)，再由 \(R\le100\) 得 \(A=5\)。此时 \(d=9\)，Raiders 的得分是 \(5,10,20,40\)，Wildcats 的得分是 \(5,14,23,32\)。Raiders 以 \(75\) 比 \(74\) 获胜。

上半场总分为 \((5+10)+(5+14)=34\)。

因此，正确答案是 E。

English

Let the Raiders score \(a, ar, ar^2, ar^3\) (increasing geometric, \(r\gt1\)) and the Wildcats \(a, a+d, a+2d, a+3d\) (increasing arithmetic), tied in the first quarter at \(a.\)

Write \(r=\frac{m}{n}\) in lowest terms, with \(m\gt n.\) Since \(ar^3\) is an integer, \(a\) is divisible by \(n^3;\) put \(a=A n^3.\) The Raiders’ total is \[ R=A(n^3+mn^2+m^2n+m^3). \] Since \(m^3\lt R\le100,\) we have \(m\le4.\) The pair \((4,3)\) already makes the parenthesized sum \(175,\) so the only possible coprime pairs \((m,n)\) are \((2,1),(3,1),(3,2),(4,1).\)

The corresponding base values of \(\frac{R}{A}\) are \(15,40,65,85.\) For \((4,1)\) and \((3,2),\) the bound forces \(A=1,\) but \(R-1=4a+6d\) gives a nonintegral \(d.\) For \((3,1),\) it would give \(36A-1=6d,\) which is impossible modulo \(6.\)

For \((m,n)=(2,1),\) we have \(R=15A\) and \(a=A,\) so \(11A-1=6d.\) Thus \(A\equiv5\pmod6,\) and \(R\le100\) forces \(A=5.\) Then \(d=9,\) giving Raiders scores \(5,10,20,40\) and Wildcats scores \(5,14,23,32.\) The Raiders win \(75\) to \(74.\)

The first-half total is \((5+10)+(5+14)=34.\)

Thus, the correct answer is E.

20 答案 E

公比为 \(r=\dfrac{a_2}{a_1}=\cot x\)。因此 \(a_4=a_3\cdot r=\tan x\cot x=1\)。

由 \(a_3=a_1r^2\)，得 \(\tan x=\sin x\cot^2 x=\dfrac{\cos^2 x}{\sin x}\)，所以 \(\sin^2 x=\cos^3 x\)，即 \((\cos^2 x)(1+\cos x)=1\)。

因此 \(1+\cos x=\dfrac{1}{\cos^2 x}\)。又 \(r^2=\dfrac{\cos^2 x}{\sin^2 x}=\dfrac{\cos^2 x}{\cos^3 x}=\dfrac{1}{\cos x}\)，所以 \(r^4=\dfrac{1}{\cos^2 x}=1+\cos x\)。

所以 \(1+\cos x=a_4\cdot r^4=a_8\)，因此 \(n=8\)。

因此，正确答案是 E。

English

The common ratio is \(r=\dfrac{a_2}{a_1}=\cot x.\) Then \(a_4=a_3\cdot r=\tan x\cot x=1.\)

From \(a_3=a_1r^2,\) we get \(\tan x=\sin x\cot^2 x=\dfrac{\cos^2 x}{\sin x},\) so \(\sin^2 x=\cos^3 x,\) i.e. \((\cos^2 x)(1+\cos x)=1.\)

Hence \(1+\cos x=\dfrac{1}{\cos^2 x}.\) Also \(r^2=\dfrac{\cos^2 x}{\sin^2 x}=\dfrac{\cos^2 x}{\cos^3 x}=\dfrac{1}{\cos x},\) so \(r^4=\dfrac{1}{\cos^2 x}=1+\cos x.\)

Therefore \(1+\cos x=a_4\cdot r^4=a_8,\) so \(n=8.\)

Thus, the correct answer is E.

21 答案 B

因为 \(1, 3, 5, 7\) 是 \(P(x)-a\) 的根，可写作 \(P(x)-a=(x-1)(x-3)\) \(\cdot(x-5)(x-7)Q(x)\)，其中 \(Q\) 的系数都是整数。

在 \(x=2, 4, 6, 8\) 处代入（此时 \(P=-a\)）得到 \[ \begin{aligned} -2a &=-15\,Q(2) \\ &=9\,Q(4) \\ &=-15\,Q(6) \\ &=105\,Q(8) \end{aligned}\text{。}\]

因此 \(15, 9\) 和 \(105\) 都整除 \(2a\)，所以 \(\operatorname{lcm}(15,9,105)=315\) 整除 \(2a\)。因为 \(315\) 是奇数，\(a\) 是 \(315\) 的倍数，所以 \(a\ge315\)。

为了达到这个下界，取 \[ \begin{aligned} F(x)&=(x-1)(x-3) \\ &\quad\cdot(x-5)(x-7), \\ Q(x)&=42+(x-2)(x-6) \\ &\quad\cdot(60-8x), \\ P(x)&=315+F(x)Q(x) \end{aligned}\text{。}\] 这个整系数多项式在 \(1,3,5,7\) 处的取值都是 \(315\)。在 \(2,4\) 处，数对 \((F(x),Q(x))\) 分别为 \((-15,42),(9,-70)\)；在 \(6,8\) 处则分别为 \((-15,42),(105,-6)\)。因此每次都有 \(F(x)Q(x)=-630\)，从而 \(P(x)=-315\)。所以这个下界确实可以达到。

因此，正确答案是 B。

English

Since \(1, 3, 5, 7\) are roots of \(P(x)-a,\) write \(P(x)-a=(x-1)(x-3)\) \(\cdot(x-5)(x-7)Q(x)\) with \(Q\) having integer coefficients.

Evaluating at \(x=2, 4, 6, 8\) (where \(P=-a\)) gives \[ \begin{aligned} -2a &=-15\,Q(2) \\ &=9\,Q(4) \\ &=-15\,Q(6) \\ &=105\,Q(8). \end{aligned} \]

So \(15, 9,\) and \(105\) all divide \(2a,\) hence \(\operatorname{lcm}(15,9,105)=315\) divides \(2a.\) Since \(315\) is odd, \(a\) is divisible by \(315,\) so \(a\ge315.\)

To attain the bound, let \[ \begin{aligned} F(x)&=(x-1)(x-3) \\ &\quad\cdot(x-5)(x-7), \\ Q(x)&=42+(x-2)(x-6) \\ &\quad\cdot(60-8x), \\ P(x)&=315+F(x)Q(x). \end{aligned} \] This integer polynomial equals \(315\) at \(1,3,5,7.\) At \(2,4,\) the pairs \((F(x),Q(x))\) are \((-15,42),(9,-70),\) and at \(6,8\) they are \((-15,42),(105,-6).\) Thus \(F(x)Q(x)=-630\) each time and \(P(x)=-315.\) Hence the bound is attainable.

Thus, the correct answer is B.

22 答案 D

令 \(a=AB\)，\(b=BC\)，\(c=CD\)，\(d=DA\)，并令 \(k=bc=ad\)。用外接圆半径表示各三角形的面积，再利用 \([ABC]+[CDA]=\) \([BCD]+[ABD]\)，可得 \((ab+cd)\cdot AC=2k\cdot BD\)。

托勒密定理给出 \(AC\cdot BD=ac+bd\)。消去 \(AC\)，得 \[ \begin{aligned} BD^2 &=\frac{(ac+bd)(ab+cd)}{2k} \\ &=\frac12\left(a^2+b^2+c^2+d^2\right) \end{aligned}\text{。}\]

四条边是小于 \(15\) 的互异整数，并满足 \(bc=ad\)，所以 \(11\) 和 \(13\) 都不能出现（它们都是质数，需要等式另一侧有相应因子）。

如果最大边至多为 \(12\)，因为不能使用 \(11\)，四边平方和至多为 \(12^2+10^2+9^2+8^2=389\)。

现在设最大边为 \(14\)，其余三边记为 \(s_1\gt s_2\gt s_3\)。乘积条件必须把 \(14\) 与 \(s_3\) 配对，所以 \(14s_3=s_1s_2\)。因而 \(s_1,s_2\) 中一个等于 \(7\)。若 \(s_1=7\)，平方和小于 \(14^2+7^2+6^2+5^2=306\)。若 \(s_2=7\)，则 \(s_1=2s_3\)，最大的可能是 \((s_1,s_2,s_3)=(12,7,6)\)。因此 \[ \begin{aligned} 2BD^2 &=14^2+12^2+7^2+6^2 \\ &=425 \end{aligned}\text{，}\]所以 \(BD\le\sqrt{\dfrac{425}{2}}\)。边依次为 \((a,b,c,d)=(14,12,7,6)\) 的圆内接四边形达到等号，此时 \(bc=ad=84\)。

因此，正确答案是 D。

English

Let \(a=AB,\) \(b=BC,\) \(c=CD,\) \(d=DA\) and \(k=bc=ad.\) Writing each triangle’s area in terms of the circumradius and using \([ABC]+[CDA]=\) \([BCD]+[ABD]\) gives \((ab+cd)\cdot AC=2k\cdot BD.\)

Ptolemy’s theorem gives \(AC\cdot BD=ac+bd.\) Eliminating \(AC,\) \[ \begin{aligned} BD^2 &=\frac{(ac+bd)(ab+cd)}{2k} \\ &=\frac12\left(a^2+b^2+c^2+d^2\right). \end{aligned} \]

The sides are distinct integers below \(15\) with \(bc=ad,\) so neither \(11\) nor \(13\) can appear (each is prime and would need a matching factor on the other side).

If the largest side is at most \(12,\) the four squares sum to at most \(12^2+10^2+9^2+8^2=389,\) because \(11\) is unavailable.

Now suppose the largest side is \(14,\) and write the others as \(s_1\gt s_2\gt s_3.\) The product condition must pair \(14\) with \(s_3,\) so \(14s_3=s_1s_2.\) Hence one of \(s_1,s_2\) equals \(7.\) If \(s_1=7,\) the sum of squares is less than \(14^2+7^2+6^2+5^2=306.\) If \(s_2=7,\) then \(s_1=2s_3,\) so the largest possibility is \((s_1,s_2,s_3)=(12,7,6).\) Thus \[ \begin{aligned} 2BD^2 &=14^2+12^2+7^2+6^2 \\ &=425, \end{aligned} \] so \(BD\le\sqrt{\dfrac{425}{2}}.\) Equality is attained by the cyclic quadrilateral with side order \((a,b,c,d)=(14,12,7,6),\) for which \(bc=ad=84.\)

Thus, the correct answer is D.

23 答案 A

如果 \(P\) 只有一个实根，那么 \(P(Q(x))=0\) 至多有两个实数解，而不是四个。因此 \(P\) 有两个不同的实根；同理，\(Q\) 也有两个不同的实根。写成 \(P(x)=(x-h_1)^2-k_1^2\) 和 \(Q(x)=(x-h_2)^2-k_2^2\)，其中 \(k_1,k_2\gt0\)，最小值分别为 \(-k_1^2\) 和 \(-k_2^2\)。

方程 \(P(Q(x))\) 的零点满足 \(Q(x)=h_1\pm k_1\)；四个解关于 \(h_2\) 对称，所以 \(h_2\) 是平均数 \(\tfrac{-23-21-17-15}{4}=-19\)。于是 \(Q(-15)-Q(-17)\) \(=(16-k_2^2)-(4-k_2^2)\) \(=12\)，而这个差等于 \(2k_1\)，所以 \(k_1=6\)。

对称地，\(h_1=\tfrac{-59-57-51-49}{4}=-54\)，且 \(P(-49)-P(-51)\) \(=(25-k_1^2)-(9-k_1^2)\) \(=16=2k_2\)，所以 \(k_2=8\)。

两个最小值之和为 \(-k_1^2-k_2^2=-36-64=-100\)。

因此，正确答案是 A。

English

If \(P\) had only one real root, then \(P(Q(x))=0\) would have at most two real solutions, not four. Thus \(P\) has two distinct real roots, and the same argument applies to \(Q.\) Write \(P(x)=(x-h_1)^2-k_1^2\) and \(Q(x)=(x-h_2)^2-k_2^2,\) with \(k_1,k_2\gt0\) and minimum values \(-k_1^2\) and \(-k_2^2.\)

The zeros of \(P(Q(x))\) occur where \(Q(x)=h_1\pm k_1;\) their four solutions are symmetric about \(h_2,\) so \(h_2\) is the average \(\tfrac{-23-21-17-15}{4}=-19.\) Then \(Q(-15)-Q(-17)\) \(=(16-k_2^2)-(4-k_2^2)\) \(=12,\) and this difference equals \(2k_1,\) so \(k_1=6.\)

Symmetrically, \(h_1=\tfrac{-59-57-51-49}{4}=-54,\) and \(P(-49)-P(-51)\) \(=(25-k_1^2)-(9-k_1^2)\) \(=16=2k_2,\) so \(k_2=8.\)

The sum of the minimum values is \(-k_1^2-k_2^2=-36-64=-100.\)

Thus, the correct answer is A.

24 答案 C

设 \(f(x)\) 为不等式的左边。在相邻的竖直渐近线 \(2009, 2010, 2011\) 之间的每个区间上，函数 \(f\) 都递减；而且对所有 \(x\lt2009\)，都有 \(f\lt1\)。

在 \((2009,2010)\)、\((2010,2011)\) 和 \((2011,\infty)\) 上，解都是从左侧的渐近线开始，直到某个满足 \(f(x_i)=1\) 的 \(x_i\) 为止。因此解集由三个区间组成，左端点为 \(2009, 2010, 2011\)，右端点为 \(x_1, x_2, x_3\)。

总长度是 \((x_1-2009)\) \(+(x_2-2010)\) \(+(x_3-2011)\) \(=x_1+x_2+x_3-6030\)。

在 \(f(x)=1\) 中清除分母，得到 \[ \begin{aligned} &x^3 \\ &\quad \small{}-(2009+2010+2011+3)x^2 \\ &\quad {}+\cdots=0\text{，} \end{aligned} \] 它的三个根是 \(x_1, x_2, x_3\)。由韦达定理，\(x_1+x_2+x_3=6033\)，所以长度之和为 \(6033-6030=3\)。

因此，正确答案是 C。

English

Let \(f(x)\) be the left-hand side. On each interval between consecutive asymptotes \(2009, 2010, 2011,\) the function \(f\) is decreasing, and \(f\lt1\) for all \(x\lt2009.\)

On each of \((2009,2010),\) \((2010,2011),\) and \((2011,\infty),\) the solution is the part from the left asymptote up to a value \(x_i\) where \(f(x_i)=1.\) So the solution set consists of three intervals with left endpoints \(2009, 2010, 2011\) and right endpoints \(x_1, x_2, x_3.\)

The total length is \((x_1-2009)\) \(+(x_2-2010)\) \(+(x_3-2011)\) \(=x_1+x_2+x_3-6030.\)

Clearing denominators in \(f(x)=1\) gives \[ \begin{aligned} &x^3 \\ &\quad \small{}-(2009+2010+2011+3)x^2 \\ &\quad {}+\cdots=0, \end{aligned} \] whose roots are \(x_1, x_2, x_3.\) By Vieta, \(x_1+x_2+x_3=6033,\) so the sum of lengths is \(6033-6030=3.\)

Thus, the correct answer is C.

25 答案 D

因为 \(2010=2\cdot3\cdot5\cdot67\)，将乘积写成 \(2^A3^B5^C67^D\) 乘以一个与这四个质数都互质的因子；于是 \(m=\min(A,B,C,D)\)。

质数 \(2\)： 只有当 \(n=2^k\) 时，\(\operatorname{pow}(n)\) 才是 \(2\) 的幂。因为 \(2^{12}=4096\lt5300\lt2^{13}\)，所以 \(k=1,\ldots,12\) 贡献 \(A=1+2+\cdots+12=78\)。

质数 \(67\)： 当 \(67\) 是最大质因数时，\(\operatorname{pow}(n)=67\)，即 \(n=67j\)，其中 \(1\le j\le79\)，且 \(j\) 的每个质因数都至多为 \(67\)；排除 \(j=67, 71, 73, 79\) 后剩下 \(75\) 个值。唯一满足 \(\operatorname{pow}(n)=67^2\) 的 \(n\) 是 \(n=67^2\lt5300\)，再贡献 \(2\)。所以 \(D=75+2=77\)。

质数 \(3\)： 对于 \(n=2^a3^b\)，当 \(b=1,2,3,4\) 时，允许的指数 \(a\) 的个数分别为 \(11,10,8,7\)，仅这些项就贡献 \[ \begin{aligned} B&\ge11+2\cdot10 \\ &\quad+3\cdot8+4\cdot7=83 \end{aligned}\text{。}\]

质数 \(5\)： 写成 \(n=2^a3^c5^b\)。当 \(b=1\) 时，对 \(c=0,\ldots,6\) 的计数总和为 \(11+9+7+6+4+3+1=41\)。当 \(b=2\) 时，计数总和为 \(8+7+5+3+2=25\)，每个贡献两个因数 \(5\)。因此 \(C\ge41+2\cdot25=91\)。

所以 \(m=\min(78,B,C,77)=77\)。

因此，正确答案是 D。

English

Since \(2010=2\cdot3\cdot5\cdot67,\) write the product as \(2^A3^B5^C67^D\) times a factor coprime to all four primes; then \(m=\min(A,B,C,D).\)

Prime \(2:\) \(\operatorname{pow}(n)\) is a power of \(2\) only when \(n=2^k.\) Since \(2^{12}=4096\lt5300\lt2^{13},\) the values \(k=1,\ldots,12\) contribute \(A=1+2+\cdots+12=78.\)

Prime \(67:\) \(\operatorname{pow}(n)=67\) when \(67\) is the largest prime factor, i.e. \(n=67j\) with \(1\le j\le79\) and every prime factor of \(j\) at most \(67;\) excluding \(j=67, 71, 73, 79\) leaves \(75\) values. The one \(n\) with \(\operatorname{pow}(n)=67^2\) is \(n=67^2\lt5300,\) adding \(2.\) So \(D=75+2=77.\)

Prime \(3:\) For \(n=2^a3^b\) with \(b=1,2,3,4,\) the numbers of allowable exponents \(a\) are \(11,10,8,7,\) respectively. These terms alone contribute \[ \begin{aligned} B&\ge11+2\cdot10 \\ &\quad+3\cdot8+4\cdot7=83. \end{aligned} \]

Prime \(5:\) Write \(n=2^a3^c5^b.\) For \(b=1,\) the counts over \(c=0,\ldots,6\) total \(11+9+7+6+4+3+1=41.\) For \(b=2,\) the counts total \(8+7+5+3+2=25,\) each contributing two factors of \(5.\) Hence \(C\ge41+2\cdot25=91.\)

Therefore \(m=\min(78,B,C,77)=77.\)

Thus, the correct answer is D.