2013A AMC 10 真题（P版）

1 答案 C

里程费用为 \(\$0.25\cdot 5=\$1.25\)。

加上起步价，费用为 \(\$1.50+\$1.25=\$2.75\)。

所以正确答案是 C。

English

The mileage charge is \(\$0.25\cdot 5=\$1.25\).

Adding the fixed charge gives \(\$1.50+\$1.25=\$2.75\).

Thus, C is the correct answer.

2 答案 B

Alice 需要 \(2\frac12=\frac52\) 杯糖。

每次量杯装 \(\frac14\) 杯，所以需要 \(\frac52\div\frac14=\frac52\cdot4=10\) 次。

所以正确答案是 B。

English

Alice needs \(2\frac12=\frac52\) cups of sugar.

Each full measuring cup gives \(\frac14\) cup, so the number of fillings is \(\frac52\div\frac14=\frac52\cdot4=10\).

Thus, B is the correct answer.

3 答案 E

由三角形面积公式， \[ 40 = \dfrac{1}{2} \cdot 10 \cdot BE\text{。} \] 因此 \[ 40 = 5BE \]\[ BE = 8\text{。} \] 所以正确答案是 E。

English

We have by the formula for the area of a triangle that \[ 40 = \dfrac{1}{2} \cdot 10 \cdot BE. \] This gives us \[ 40 = 5BE \]\[ BE = 8. \] Thus, E is the correct answer.

4 答案 C

若本队得分是对手两倍，则本队得分必须为偶数。

因此得 \(2, 4, 6, 8,\) 和 \(10\) 分的比赛中，对手分别得 \(1, 2, 3, 4,\) 和 \(5\) 分。

这些分数的和为 \[ 1 + 2 + 3 + 4 + 5 = 15\text{。} \]

其余比赛中，对手分别得 \(2, 4, 6, 8,\) 和 \(10\) 分。

这些分数的和为 \[ 2 + 4 + 6 + 8 + 10 = 30\text{。} \]

对手总得分为 \[ 15 + 30 = 45\text{。} \] 所以正确答案是 C。

English

Note that if they scored twice as many runs as their opponents, then they scored an even number of runs.

This means in the games where they scored \(2, 4, 6, 8,\) and \(10\) runs, their opponents scored \(1, 2, 3, 4,\) and \(5\) runs respectively.

This sums to \[ 1 + 2 + 3 + 4 + 5 = 15. \]

In the other games, their opponents scored \(2, 4, 6, 8,\) and \(10\) runs.

This sums to \[ 2 + 4 + 6 + 8 + 10 = 30. \]

The total number of runs is then \[ 15 + 30 = 45. \] Thus, C is the correct answer.

5 答案 B

总费用为 \(\$105+\$125+\$175=\$405\)，所以每人应承担 \(\$405\div3=\$135\)。

Tom 少付 \(\$30\)，Dorothy 少付 \(\$10\)，因此 \(t=30\) 且 \(d=10\)。

所以 \(t-d=30-10=20\)。

所以正确答案是 B。

English

The total cost was \(\$105+\$125+\$175=\$405\), so each person’s share was \(\$405\div3=\$135\).

Tom paid \(\$30\) less than his share, and Dorothy paid \(\$10\) less than her share, so \(t=30\) and \(d=10\).

Therefore \(t-d=30-10=20\).

Thus, B is the correct answer.

6 答案 D

年龄和为 \(16\) 的配对有 \((3,13)\)、\((5,11)\)、\((7,9)\)。

\(5\) 岁的人留在家里，所以看电影的不能是 \((5,11)\)。若 \(7\) 岁和 \(9\) 岁的兄弟去看电影，则剩下小于 \(10\) 岁的只有 \(3\) 岁的兄弟，不够两人去打棒球。

因此看电影的是 \((3,13)\)，打棒球的是 \((7,9)\)，剩下的 Joey 是 \(11\) 岁。

所以正确答案是 D。

English

The age pairs that sum to \(16\) are \((3,13)\), \((5,11)\), and \((7,9)\).

The \(5\)-year-old stayed home, so the movie pair cannot be \((5,11)\). If \(7\) and \(9\) went to the movies, then the only remaining brother younger than \(10\) would be the \(3\)-year-old, not enough for baseball.

Thus the movie pair was \((3,13)\), the baseball pair was \((7,9)\), and Joey is the remaining brother, age \(11\).

Thus, D is the correct answer.

7 答案 C

英语必须选，所以还要从代数、几何、历史、美术、拉丁语这 \(5\) 门中选 \(3\) 门。

这样的选法共有 \(\binom53=10\) 种，但其中只有历史、美术、拉丁语这一种不含数学课。

因此有效选法为 \(10-1=9\) 种。

所以正确答案是 C。

English

English is required, so choose the other \(3\) courses from the \(5\) courses Algebra, Geometry, History, Art, and Latin.

There are \(\binom53=10\) such choices, but one of them, History-Art-Latin, contains no mathematics course.

Therefore the number of valid programs is \(10-1=9\).

Thus, C is the correct answer.

8 答案 C

提取 \(2^{2012}\)，\[ \dfrac{2^{2012}(2^2 + 1)}{2^{2012}(2^2 - 1)} = \dfrac{5}{3}\text{。}\]

所以正确答案是 C。

English

Factoring out a \(2^{2012},\) we get: \[ \dfrac{2^{2012}(2^2 + 1)}{2^{2012}(2^2 - 1)} = \dfrac{5}{3}. \]

Thus, C is the correct answer.

9 答案 B

设两分球出手 \(x\) 次，三分球出手 \(y\) 次。

那么 Shenille 命中 \(0.3x\) 个两分球和 \(0.2y\) 个三分球，总得分为 \[ 2 \cdot 0.3x + 3 \cdot 0.2y = 0.6x + 0.6y\text{。}\]

我们知道 \[ x + y = 30 \] \[ 0.6(x + y) = 18\text{。}\]

所以正确答案是 B。

English

Let \(x\) be the number of two-point shots and \(y\) be the number of three-point shots.

Then, Shenille makes \(0.3x\) two-point shots and \(0.2y\) three-point shots, for a total score of \[ 2 \cdot 0.3x + 3 \cdot 0.2y = 0.6x + 0.6y. \]

We know that \[ x + y = 30 \]\[ 0.6(x + y) = 18. \]

Thus, B is the correct answer.

10 答案 E

设总花数为 \(x\)。粉色花有 \(0.6x\) 朵，红色花有 \(0.4x\) 朵。

于是粉玫瑰有 \[ \dfrac{1}{3} \cdot 0.6x = 0.2x \] 朵，所以粉康乃馨有 \(0.4x\) 朵。

另外红康乃馨有 \[ \dfrac{3}{4} \cdot 0.4x = 0.3x \] 朵。因此康乃馨共有 \(0.3x + 0.4x = 0.7x\) 朵，占全部花的 \(70\%\)。

所以正确答案是 E。

English

Let the total number of flowers be \(x.\) There are \(0.6x\) pink flowers and \(0.4x\) red flowers.

Then there are \[ \dfrac{1}{3} \cdot 0.6x = 0.2x \] pink roses, which means there are \(0.4x\) pink carnations.

There are also \[ \dfrac{3}{4} \cdot 0.4x = 0.3x \] red carnations. This means there are \(0.3x + 0.4x = 0.7x\) carnations. This is \(70\%\) of the total flowers.

Thus, E is the correct answer.

11 答案 A

设学生会有 \(x\) 名成员。选二人委员会的方法数为 \[ \binom{x}{2} = \dfrac{x(x - 1)}{2}\text{。} \] 这个值等于 \(10\)，所以 \[ x^2 - x = 20 \] \[ x^2 - x - 20 = 0\text{。} \] 因式分解得 \[ (x - 5)(x + 4) = 0 \] \[ x = 5\text{，} \] 因为学生人数不能为负数。

于是选 \(3\) 人策划委员会的方法数为 \[ \binom{5}{3} = \binom{5}{2} = 10\text{。} \]

所以正确答案是 A。

English

Let \(x\) be the number of students. Then the number of ways to pick a two-person committee is \[ \binom{x}{2} = \dfrac{x(x - 1)}{2}. \] We know that this equals \(10,\) so \[ x^2 - x = 20 \]\[ x^2 - x - 20 = 0. \] Factoring yields \[ (x - 5)(x + 4) = 0 \]\[ x = 5, \] since there cannot be a negative number of students.

Then, the number of ways to pick a \(3\)-person committee is \[ \binom{5}{3} = \binom{5}{2} = 10. \]

Thus, A is the correct answer.

12 答案 C

由平行线可知 \(\triangle DBE \sim \triangle ABC\)，且 \(\triangle FEC \sim \triangle ABC\)。

因为原三角形两腰相等，所以在这些相似三角形中 \(DB = DE\)，\(FE = FC\)。平行四边形 \(ADEF\) 的周长为 \[ AD + DE + EF + AF \] \[= AD + DB + FC + AF \] \[ = AB + AC \] \[= 56\text{。}\]

所以正确答案是 C。

English

Note that \(\triangle DBE \sim \triangle ABC\) and \(\triangle FEC \sim \triangle ABC\) due to the parallel lines.

This tells us that \(DB = DE\) and \(FE = FC.\) We have that the perimeter of \(ADEF\) is \[ AD + DE + EF + AF \]\[= AD + DB + FC + AF \] \[ = AB + AC \]\[= 56. \]

Thus, C is the correct answer.

13 答案 B

要使这个数不能被 \(5\) 整除，个位数字不能是 \(0\) 或 \(5\)。

设百位和个位数字为 \(x\)，十位数字为 \(y\)。需要 \[ 2x + y \lt 20\text{。}\] 按 \(x\) 的 \(8\) 种可能取值分类：

若 \(x\) 为 \(1, 2, 3\) 或 \(4\)，则任意数字 \(y\) 都满足，因为 \(y \lt 10\)。

若 \(x = 6\)，则 \(y \lt 8\)，有 \(8\) 种选择。

若 \(x = 7\)，则 \(y \lt 6\)，有 \(6\) 种选择。

若 \(x = 8\)，则 \(y \lt 4\)，有 \(4\) 种选择。

若 \(x = 9\)，则 \(y \lt 2\)，有 \(2\) 种选择。

这样总共有 \[ 4 \cdot 10 + 8 + 6 + 4 + 2 = 60\] 组解。

所以正确答案是 B。

English

Note that for the number to not be divisible by \(5,\) the units digits cannot be either \(0\) or \(5.\)

Let \(x\) be the hundreds and units digit and \(y\) be the tens digit. Then we want \[ 2x + y \lt 20. \] Casing on the \(8\) possible values of \(x,\) we get:

If \(x\) is \(1, 2, 3,\) or \(4,\) then \(y\) can be anything since \(y \lt 10.\)

If \(x = 6,\) then \(y \lt 8,\) which gives us \(8\) solutions.

If \(x = 7,\) then \(y \lt 6,\) which gives us \(6\) solutions.

If \(x = 8,\) then \(y \lt 4,\) which gives us \(4\) solutions.

If \(x = 9,\) then \(y \lt 2,\) which gives us \(2\) solutions.

This gives us a total of \[ 4 \cdot 10 + 8 + 6 + 4 + 2 = 60\] solutions.

Thus, B is the correct answer.

14 答案 D

去掉角上的小立方体不会消除原来大立方体的十二条外边界棱，只会增加新的棱。

每去掉一个角，会新增 \(9\) 条棱。

共有 \(8\) 个角，所以新增 \(8 \cdot 9 = 72\) 条棱。再加上原来的 \(12\) 条棱，总数为 \(72 + 12 = 84\)。

所以正确答案是 D。

English

Removing the cubes does not remove any edges from the original cube. It only adds edges.

After removing each cube, we can see that \(9\) extra edges are added to the solid.

\(8\) cubes are removed, which means \(8 \cdot 9 = 72\) edges are added to the original \(12\) edges, for a total of \(72 + 12 = 84\) edges.

Thus, D is the correct answer.

15 答案 D

设边长 \(10\) 的边对应的高为 \(h_1\)，另一条已知边对应的高为 \(h_2\)。

同一三角形面积相同，所以 \[ 10h_1 = 15h_2 \] \[ h_1 = \dfrac{3}{2}h_2\text{。}\]

第三边对应的高是另外两个高的平均数，因此为 \[ \dfrac{h_2 + \frac{3}{2}h_2}{2} = \dfrac{5}{4}h_2\text{。}\]

设第三边长为 \(x\)。由面积相等，\[ \dfrac{5}{4}h_2x = 15h_2 \] \[ x = 12\text{。}\]

所以正确答案是 D。

English

Let \(h_1\) be the length of the altitude to the side of length \(10\) and similarly define \(h_2\) for the other given side.

We have that \[ 10h_1 = 15h_2 \]\[ h_1 = \dfrac{3}{2}h_2. \]

The third altitude is the average of the other two, which makes its length \[ \dfrac{h_2 + \frac{3}{2}h_2}{2} = \dfrac{5}{4}h_2. \]

Let the third side have length \(x.\) Then \[ \dfrac{5}{4}h_2x = 15h_2 \]\[ x = 12. \]

Thus, D is the correct answer.

16 答案 E

反射后三角形顶点为 \((7,1)\)、\((8,-3)\)、\((10,5)\)。

直线经过 \((6,5)\) 和 \((9,1)\)，方程为 \(y=-\frac43x+13\)，它与 \(x=8\) 相交于 \(y=\frac73\)。由对称性，并集可看作两个全等三角形，每个三角形的竖直底边从 \((8,-3)\) 到 \((8,\frac73)\)。

这条底边长度为 \(\frac73+3=\frac{16}{3}\)，每个三角形的水平高为 \(2\)。因此并集面积为 \(2\left(\frac12\cdot\frac{16}{3}\cdot2\right)=\frac{32}{3}\)。

所以正确答案是 E。

English

The reflected triangle has vertices \((7,1)\), \((8,-3)\), and \((10,5)\).

The line through \((6,5)\) and \((9,1)\) is \(y=-\frac43x+13\), so it meets \(x=8\) at \(y=\frac73\). By symmetry, the union is two congruent triangles with vertical base from \((8,-3)\) to \((8,\frac73)\).

That base has length \(\frac73+3=\frac{16}{3}\), and each triangle has horizontal height \(2\). Hence the union area is \(2\left(\frac12\cdot\frac{16}{3}\cdot2\right)=\frac{32}{3}\).

Thus, E is the correct answer.

17 答案 B

两两同时来访的周期分别为 \(\operatorname{lcm}(3,4)=12\)、\(\operatorname{lcm}(3,5)=15\)、\(\operatorname{lcm}(4,5)=20\) 天。

在接下来 \(365\) 天中，这三种两两同时来访的天数分别为 \(\left\lfloor\frac{365}{12}\right\rfloor=30\)、\(\left\lfloor\frac{365}{15}\right\rfloor=24\)、\(\left\lfloor\frac{365}{20}\right\rfloor=18\)。

三人都来访的周期为 \(60\) 天，共 \(\left\lfloor\frac{365}{60}\right\rfloor=6\) 天。恰好两人来访的天数为 \((30-6)+(24-6)\) \(+(18-6)=54\)。

所以正确答案是 B。

English

Pairwise visits occur every \(\operatorname{lcm}(3,4)=12\), \(\operatorname{lcm}(3,5)=15\), and \(\operatorname{lcm}(4,5)=20\) days.

In the next \(365\) days, these pair counts are \(\left\lfloor\frac{365}{12}\right\rfloor=30\), \(\left\lfloor\frac{365}{15}\right\rfloor=24\), and \(\left\lfloor\frac{365}{20}\right\rfloor=18\).

All three visit every \(60\) days, which happens \(\left\lfloor\frac{365}{60}\right\rfloor=6\) times. Subtracting those days from each pair count gives \((30-6)+(24-6)\) \(+(18-6)=54\).

Thus, B is the correct answer.

18 答案 B

设分割线与 \(\overline{CD}\) 交于 \(G\)。如图，从 \(B\)、\(C\) 和 \(G\) 向 \(x\)-轴作垂线。

由图中分割可得 \(\triangle ABF\)、梯形 \(BCEF\)、\(\triangle CDE\) 的面积分别为 \(1\)、\(5\) 和 \(\frac32\)，所以 \([ABCD]=\frac{15}{2}\)。

因此 \(\triangle ADG\) 面积为 \(\frac{15}{4}\)。因为 \(AD=4\)，点 \(G\) 的高度为 \(\frac{15}{8}\)。

直线 \(CD\) 的方程为 \(y=-3x+12\)，所以 \(\frac{15}{8}=-3x+12\)，得到 \(x=\frac{27}{8}\)。因此 \(p+q+r+s\) \(=27+8+15+8\) \(=58\)。

所以正确答案是 B。

English

Let the cutting line meet \(\overline{CD}\) at \(G\). Drop perpendiculars from \(B\), \(C\), and \(G\) to the \(x\)-axis as in the diagram.

The areas of \(\triangle ABF\), trapezoid \(BCEF\), and \(\triangle CDE\) are \(1\), \(5\), and \(\frac32\), respectively, so \([ABCD]=\frac{15}{2}\).

Thus \(\triangle ADG\) has area \(\frac{15}{4}\). Since \(AD=4\), the height of \(G\) is \(\frac{15}{8}\).

The line \(CD\) has equation \(y=-3x+12\), so \(\frac{15}{8}=-3x+12\), giving \(x=\frac{27}{8}\). Therefore \(p+q+r+s\) \(=27+8+15+8\) \(=58\).

Thus, B is the correct answer.

19 答案 C

进制表示的末位数字就是除以底数后的余数。

因此问题等价于寻找所有正整数 \(b\)，使 \(2013\) 除以 \(b\) 余 \(3\)。

这意味着 \(b\) 必须整除 \(2010\)。此外 \(b \geq 4\)，否则余数不可能是 \(3\)。

\(2010\) 的质因数分解为 \[ 2010 = 2 \cdot 3 \cdot 5 \cdot 67\text{。} \] 因此 \(2010\) 有 \[ (1 + 1)^4 = 2^4 = 16 \] 个正因数。它有 \(3\) 个小于 \(4\) 的因数，即 \(1, 2,\) 和 \(3\)，都不能作为底数。因此有效的 \(b\) 有 \(16 - 3 = 13\) 个。

所以正确答案是 C。

English

Note that the units digit represents the remainder when the number is divided by the base.

The question then boils down to finding all numbers, \(b,\) such that \(2013\) leaves a remainder of \(3\) when divided by \(b.\)

This means that \(b\) must divide \(2010.\) Also note that \(b \geq 4,\) since otherwise the remainder cannot be \(3.\)

The prime factorization of \(2010\) is \[ 2010 = 2 \cdot 3 \cdot 5 \cdot 67. \] Then, \(2010\) has \[ (1 + 1)^4 = 2^4 = 16 \] factors. It has \(3\) factors less than \(4,\) namely \(1, 2,\) and \(3.\) This means there are \(16 - 3 = 13\) valid values for \(b.\)

Thus, C is the correct answer.

20 答案 C

考虑扫过区域的四分之一，最后乘以 \(4\)。

其中扇形圆心角为 \(45^\circ\)，半径为 \(\frac{\sqrt2}{2}\)，面积为 \(\frac18\pi\left(\frac{\sqrt2}{2}\right)^2=\frac{\pi}{16}\)。

该四分之一区域中的两个直角三角形面积分别为 \(\frac12\cdot\frac12\cdot\frac12=\frac18\) 和 \(\frac12\left(\frac{\sqrt2-1}{2}\right)^2\)。

把四分之一区域的面积和乘以 \(4\)，得到 \(2-\sqrt2+\frac{\pi}{4}\)。

所以正确答案是 C。

English

Consider one quarter of the swept region and multiply its area by \(4\).

The sector has angle \(45^\circ\) and radius \(\frac{\sqrt2}{2}\), so its area is \(\frac18\pi\left(\frac{\sqrt2}{2}\right)^2=\frac{\pi}{16}\).

The two right-triangle pieces in that quarter have areas \(\frac12\cdot\frac12\cdot\frac12=\frac18\) and \(\frac12\left(\frac{\sqrt2-1}{2}\right)^2\).

Multiplying the quarter-area sum by \(4\) gives \(2-\sqrt2+\frac{\pi}{4}\).

Thus, C is the correct answer.

21 答案 D

反向计算。若第 \(12\) 个海盗分到时还剩 \(n\) 枚金币，那么在第 \(k\) 个海盗拿走份额之前，箱中的金币数是他拿完之后的 \(\frac{12}{12-k}\) 倍。

因此最初的金币数为 \(n\cdot\frac{12^{11}}{11!}\)。

由于 \(\frac{12^{11}}{11!}=\frac{2^{14}3^7}{5\cdot7\cdot11}\)，使最初金币数为整数的最小 \(n\) 是 \(5^2\cdot7\cdot11=1925\)。

这个值确实可以达到。在第 \(k\) 个海盗拿走份额之前，金币数为 \[ 2^{14}3^7\cdot \frac{11!}{(12-k)!\,12^{k-1}}\text{。}\]对 \(k=1,2,\ldots,12\)，这些数全是整数。每个海盗拿到的份额是相邻两个剩余量之差，所以每份也都是整数。

因此第 \(12\) 个海盗得到 \(1925\) 枚金币，正确答案是 D。

English

Work backward. If \(n\) coins remain for the \(12\)th pirate, then before pirate \(k\) took a share, the chest had \(\frac{12}{12-k}\) times as many coins as it had afterward.

Therefore the initial number of coins is \(n\cdot\frac{12^{11}}{11!}\).

Since \(\frac{12^{11}}{11!}=\frac{2^{14}3^7}{5\cdot7\cdot11}\), the smallest \(n\) that makes the initial number an integer is \(5^2\cdot7\cdot11=1925\).

This value is attainable. The number of coins present just before pirate \(k\) takes a share is \[ 2^{14}3^7\cdot \frac{11!}{(12-k)!\,12^{k-1}}. \] For \(k=1,2,\ldots,12\), all these amounts are integers. Each pirate’s share is the difference between two consecutive remaining amounts, so every share is an integer as well.

Thus, the \(12\)th pirate receives \(1925\) coins, and D is the correct answer.

22 答案 B

六个半径为 \(1\) 的小球球心构成边长为 \(2\) 的正六边形，所以每个小球球心到大球球心的距离为 \(2\)。因此大球半径为 \(3\)。

设第八个球半径为 \(r\)，球心到大球球心的距离为 \(x\)。内切给出 \(x+r=3\)，所以 \(x=3-r\)。

由图中的直角三角形，\((r+1)^2=2^2+(3-r)^2\)。化简得 \(2r+1=13-6r\)，所以 \(r=\frac32\)。

所以正确答案是 B。

English

The centers of the six radius-\(1\) spheres form a regular hexagon of side length \(2\), so each is \(2\) units from the large sphere’s center. Hence the large sphere has radius \(3\).

Let the eighth sphere have radius \(r\), and let its center be distance \(x\) from the large sphere’s center. Internal tangency gives \(x+r=3\), so \(x=3-r\).

Using the right triangle between the large center, a small-sphere center, and the eighth-sphere center, \((r+1)^2=2^2+(3-r)^2\). Thus \(2r+1=13-6r\), so \(r=\frac32\).

Thus, B is the correct answer.

23 答案 D

由点 \(C\) 的圆幂，\(CB\cdot CX\) \(=AC^2-AB^2\) \(=97^2-86^2\)。

右边等于 \((97-86)(97+86)\) \(=11\cdot183\) \(=2013\) \(=3\cdot11\cdot61\)。

\(CX\) 和 \(BX\) 都是整数，所以 \(BC=BX+CX\) 是与 \(CX\) 配对的整数因数。还需满足 \(CX<BC<86+97=183\)，唯一可行因数对为 \(CX=33\)、\(BC=61\)。

所以正确答案是 D。

English

By power of a point from \(C\), \(CB\cdot CX\) \(=AC^2-AB^2\) \(=97^2-86^2\).

This equals \((97-86)(97+86)\) \(=11\cdot183\) \(=2013\) \(=3\cdot11\cdot61\).

Both \(CX\) and \(BX\) are integers, so \(BC=BX+CX\) is an integer factor paired with \(CX\). Also \(CX<BC<86+97=183\), so the only possible pair is \(CX=33\), \(BC=61\).

Thus, D is the correct answer.

24 答案 E

记 Central 的选手为 \(A,B,C\)，Northern 的选手为 \(X,Y,Z\)。

选手 \(A\) 的六轮对手字符串中 \(X,Y,Z\) 各出现两次，所以有 \(\frac{6!}{2!2!2!}=90\) 种。

固定 \(A\) 的字符串，例如 \(XXYYZZ\)。\(B\) 的字符串也必须含两个 \(X,Y,Z\)，且每一轮不能与 \(A\) 的对手相同。

若 \(B\) 的前两位为 \(Y,Z\) 的某种顺序，中间两位必须为 \(X,Z\) 的某种顺序，后两位必须为 \(X,Y\) 的某种顺序，共 \(2^3=8\) 种。另有 \(YYZZXX\) 和 \(ZZXXYY\) 两种，共 \(10\) 种 \(B\) 的字符串。

一旦 \(A\) 和 \(B\) 的赛程确定，\(C\) 的赛程被迫确定。因此总数为 \(90\cdot10=900\)。

所以正确答案是 E。

English

Label Central’s players \(A,B,C\) and Northern’s players \(X,Y,Z\).

Player \(A\)’s six-round opponent string contains two each of \(X,Y,Z\), so it can be chosen in \(\frac{6!}{2!2!2!}=90\) ways.

For a fixed \(A\)-string, say \(XXYYZZ\), player \(B\)’s string must also contain two each of \(X,Y,Z\) and cannot match \(A\)’s opponent in any position.

If \(B\)’s first two entries are \(Y,Z\) in either order, then the middle two entries must be \(X,Z\) in either order and the last two must be \(X,Y\) in either order, giving \(2^3=8\) strings. The two remaining possibilities are \(YYZZXX\) and \(ZZXXYY\), for \(10\) total \(B\)-strings.

Once \(A\) and \(B\) are scheduled, \(C\)’s schedule is forced. Hence there are \(90\cdot10=900\) schedules.

Thus, E is the correct answer.

25 答案 A

若没有三条对角线共点，每选 \(4\) 个顶点会确定一个内部交点，共 \(\binom84=70\) 个。

通过对顶点的 \(4\) 条长对角线都在中心相交。中心原本被数了 \(\binom42=6\) 次，但应只数一次，所以减去 \(5\)。

另外还有 \(8\) 个对称位置各有 \(3\) 条对角线共点。每个位置原本被数 \(\binom32=3\) 次，但应只数一次，所以再减去 \(8(3-1)=16\)。

不同内部交点数为 \(70-5-16=49\)。

所以正确答案是 A。

English

If no three diagonals were concurrent, each choice of \(4\) vertices would determine one interior intersection, giving \(\binom84=70\).

The \(4\) long diagonals through opposite vertices all meet at the center, so the center was counted \(\binom42=6\) times and should be counted once. Subtract \(5\).

There are also \(8\) symmetric points where \(3\) diagonals meet. Each was counted \(\binom32=3\) times and should be counted once, so subtract \(8(3-1)=16\).

The number of distinct interior intersection points is \(70-5-16=49\).

Thus, A is the correct answer.