2011A AMC 10 真题（P版）

月费为 \(\$20\)。短信费用为 \[100\cdot 5=500\] 美分，即 \(\$5\)。她还多通话 \(0.5\) 小时，即 \(30\) 分钟，超出了套餐包含的 \(30\) 小时。

因此超时通话费用为 \[30\cdot10=300\] 美分，即 \(\$3\)。

总费用为 \[ \$20 + \$5 + \$3 = \$28\text{。}\]

所以正确答案是 D。

English

Michelle has to pay \(\$20\) for the monthly fee. Her text messages cost \[100\cdot 5=500\] cents, or \(\$5.\) Finally she talked for \(0.5\) hours, \(30\) minutes, over \(30\) hours.

This means her extra charge is \[30\cdot10=300\] cents, or \(\$3.\)

Her total cost is \[ \$20 + \$5 + \$3 = \$28. \]

Thus, D is the correct answer.

所需瓶数为 \[ \dfrac{500}{35} = \dfrac{100}{7} = 14 \dfrac{2}{7}\text{。}\]

因此最少必须购买 \(15\) 个小瓶。

所以正确答案是 E。

English

The desired amount is \[ \dfrac{500}{35} = \dfrac{100}{7} = 14 \dfrac{2}{7}. \]

This means that the smallest number of small bottles she must buy is \(15.\)

Thus, E is the correct answer.

先计算得 \[ \{1 \ 1 \ 0\} = \dfrac{1 + 1 + 0}{3} = \dfrac{2}{3}\text{。}\]

另一个内层平均数为 \[ [0 \ 1] = \dfrac{1 + 0}{2} = \dfrac{1}{2}\text{。}\]

最后得 \[ \left\{\dfrac{2}{3} \ \dfrac{1}{2} \ 0\right\} = \dfrac{\dfrac{2}{3} + \dfrac{1}{2} + 0}{3} = \dfrac{7}{18}\text{。}\]

所以正确答案是 D。

English

We have that \[ \{1 \ 1 \ 0\} = \dfrac{1 + 1 + 0}{3} = \dfrac{2}{3}. \]

We also get that \[ [0 \ 1] = \dfrac{1 + 0}{2} = \dfrac{1}{2}. \]

Finally, \[ \left\{\dfrac{2}{3} \ \dfrac{1}{2} \ 0\right\} = \dfrac{\dfrac{2}{3} + \dfrac{1}{2} + 0}{3} = \dfrac{7}{18}.\]

Thus, D is the correct answer.

从 \(12\) 到 \(100\) 的所有偶数项在两个和中都出现，作差时会全部抵消。

因此 \[ Y - X = 102 - 10 = 92\text{。}\]

所以正确答案是 A。

English

Note that the terms \(12\) through \(100\) are common to both sums. When we subtract, all these terms cancel out.

This means that \[ Y - X = 102 - 10 = 92. \]

Thus, A is the correct answer.

不妨设五年级有一人，则四年级有两人，三年级有四人。

这样做可以，因为平均数不受学生人数整体倍数的影响。

总跑步时间为 \[12 \cdot 4 + 15 \cdot 2 + 10 = 88\] 分钟。总人数为 \(1 + 2 + 4 = 7\)，所以平均数为 \(\dfrac{88}{7}\)。

所以正确答案是 C。

English

WLOG, let there be one fifth grader. This then tells us that there are two fourth graders and four third graders.

We can do this, since we are only interested in the average, which is not impacted by the exact number of students.

The total number of minutes the students spend running is \[ 12 \cdot 4 + 15 \cdot 2 + 10 = 88 \] minutes. The total number of students is \(1 + 2 + 4 = 7.\) The average is then \(\dfrac{88}{7}.\)

Thus, C is the correct answer.

要使并集元素数最少，就要让两个集合重叠尽可能多。

可以让 \(B\) 完全包含于 \(A\)，此时并集就是 \(A\)，共有 \(20\) 个元素。

所以正确答案是 C。

English

To minimize the number of elements in the union, we want to maximize the overlap between the two sets.

We can then assume that \(B\) is contained completely within \(A,\) which means that the union is the same as \(A,\) which has \(20\) elements.

Thus, C is the correct answer.

选项 A 化为 \[x + 7 = 0\] \[x = -7\text{，}\]因此有解。

选项 B 化为 \[|-3x| = -5\text{，}\]绝对值不可能为负数，因此无解。

其他选项都有解，如下面计算所示。

选项 C 化为 \[\sqrt{-x} = 2\] \[-x = 4\] \[x = -4\text{，}\]所以有解。

选项 D 化为 \[\sqrt{x} = 8\] \[x = 64\text{，}\]所以也有解。

最后，选项 E 化为 \[|-3x| = 4\] \[-3x = \pm 4\] \[x = \pm \dfrac{4}{3}\text{，}\]同样有解。

所以正确答案是 B。

English

A simplifies to \[ x + 7 = 0 \]\[ x = -7, \] so it has a solution.

B simplifies to \[ |-3x| = -5, \] which has no solution since absolute value makes everything positive.

Let us make sure that all the other choices have solutions.

C simplifies to \[ \sqrt{-x} = 2 \]\[ -x = 4 \]\[ x = -4, \] which is fine.

D simplifies to \[ \sqrt{x} = 8 \]\[ x = 64, \] which works.

Finally, E simplifies to \[ |-3x| = 4 \]\[ -3x = \pm 4 \] \[ x = \pm \dfrac{4}{3}, \] which has a solution as well.

Thus, B is the correct answer.

不妨设共有 \(100\) 只鸟，则不是天鹅的有 \(75\) 只。所求百分比为 \[ \dfrac{30}{75} \cdot 100 = 40 \%\text{。}\]

所以正确答案是 C。

English

WLOG, let there be \(100\) birds. Then \(75\) birds are not swans. The desired percentage is then \[ \dfrac{30}{75} \cdot 100 = 40 \%. \]

Thus, C is the correct answer.

该区域是长方形，两条边长分别为 \[a - (-b) = a + b\] 和 \[d - (-c) = c + d\text{。}\]因此面积为 \[(a + b)(c + d) =\] \[ac + ad + bc + bd\text{。}\]

所以正确答案是 A。

English

Note that the region is a rectangle with side lengths \[ a - (-b) = a + b \] and \[ d - (-c) = c + d. \] The area is then \[ (a + b)(c + d) =\]\[ ac + ad + bc + bd. \]

Thus, A is the correct answer.

设每人买 \(p\) 支铅笔，买铅笔的学生有 \(s\) 人，每支铅笔 \(c\) 美分。

则 \[ psc = 1771 = 7 \cdot 11 \cdot 23\text{。}\]

还必须满足 \[ 30 \geq s \gt 15, p \gt 1, c \gt p\text{。}\]

由质因数分解，唯一满足学生人数条件的是 \(s = 23\)。

剩下的因数是七和十一；结合价格大于购买数量的条件，得到 \(p = 7\)、\(c = 11\)。

所以正确答案是 B。

English

Let \(p\) be the number of pencils that each student bought, \(s\) be the number of students that bought pencils, and \(c\) be the cost of a pencil.

We have that \[ psc = 1771 = 7 \cdot 11 \cdot 23. \]

We also have the following restrictions: \[ 30 \geq s \gt 15, p \gt 1, c \gt p. \]

From the above prime factorization, we have that \(s = 23\) is the only value that satisfies the conditions.

Finally, we get that \(p = 7\) and \(c = 11\) are the only remaining values that satisfy the other conditions.

Thus, B is the correct answer.

设 \(x = EB\)，则 \(AB = 8x\)。对 \(EFGH\) 的一条边所在的直角三角形使用勾股定理，得该边长为 \[ \sqrt{(7x)^2 + x^2} = \sqrt{50x^2} \]

因此所求面积比为 \[ \dfrac{\sqrt{50x^2}^2}{(8x)^2} = \dfrac{50}{64} = \dfrac{25}{32}\text{。}\]

所以正确答案是 B。

English

Let \(x = EB.\) Then \(AB = 8x.\) Applying the Pythagorean Theorem to a side of \(EFGH,\) we get \[ \sqrt{(7x)^2 + x^2} = \sqrt{50x^2} \]

The desired ratio is then \[ \dfrac{\sqrt{50x^2}^2}{(8x)^2} = \dfrac{50}{64} = \dfrac{25}{32}. \]

Thus, B is the correct answer.

设投进的两分球数为 \(x\)，则 \[2x + 2x + (x + 1) = 61\text{，}\]化简得 \[5x + 1 = 61\] \[x = 12\text{。}\]

因此罚球数为 \(12 + 1 = 13\)。

所以正确答案是 A。

English

Let \(x\) be the number of successful two-point shots. Then we have that \[ 2x + 2x + (x + 1) = 61, \] which simplifies to \[ 5x + 1 = 61 \]\[ x = 12. \]

The number of successful free throws is then \(12 + 1 = 13.\)

Thus, A is the correct answer.

百位只能是 \(2\) 或 \(5\)，因此按这个数值分情况讨论。

情况 \(1\)：百位是 \(2\)。

此时个位只能是 \(8\)，十位还有 \(4\) 种选择。

这一情况得到 \(1 \cdot 4 = 4\) 个数。

情况 \(2\)：百位是 \(5\)。

同理，个位只能选 \(2\) 或 \(8\)，十位还有 \(4\) 种选择。

这一情况给出 \(2 \cdot 4 = 8\) 个数。

因此整数总数为 \(4 + 8 = 12\)。

所以正确答案是 A。

English

Since the hundreds digit can only be a \(2\) or \(5,\) we can case on this value.

Case \(1:\) hundreds digit is \(2\)

The only option for the units digit is \(8,\) since the number must be even. This leaves \(4\) options for the tens digit.

This gives us \(1 \cdot 4 = 4\) numbers for this case.

Case \(2:\) hundreds digit is \(5\)

Similarly to above, \(2\) and \(8\) are the only options for the units digit, leaving \(4\) options for the tens digit.

This gives us \(2 \cdot 4 = 8\) numbers for this case.

The total number of integers is then \(4 + 8 = 12.\)

Thus, A is the correct answer.

要使面积小于周长，必须有 \[ \pi r^2 \lt 2\pi r \] \[ r \lt 2\text{。}\]

因此直径必须小于 \(4\)。满足条件的掷法有三种：\[ (1, 1), (1, 2), (2, 1)\text{。}\]

因此概率为 \[\dfrac{3}{36} = \dfrac{1}{12}\text{。}\]

所以正确答案是 B。

English

For the area to be less than the circumference, we must have \[ \pi r^2 \lt 2\pi r \]\[ r \lt 2. \]

This means the diameter must be less than \(4.\) There are three possible rolls that satisfy this: \[ (1, 1), (1, 2), (2, 1). \]

The probability is then \[\dfrac{3}{36} = \dfrac{1}{12}.\]

Thus, B is the correct answer.

设汽车只用汽油行驶的距离为 \(x\)，则 \[\dfrac{40 + x}{0.02x} = 55\text{。}\]交叉相乘并化简得 \[40 = 0.1x\] \[x = 400\text{。}\]因此旅程总长为 \(400 + 40 = 440\)。

所以正确答案是 C。

English

Let \(x\) be the distance the car drove solely on gasoline. We have that \[ \dfrac{40 + x}{0.02x} = 55. \] Cross-multiplying and simplifying gives \[ 40 = 0.1x \]\[ x = 400. \] The total length of the trip is then \(400 + 40 = 440.\)

Thus, C is the correct answer.

由于式子中含有平方根，可尝试把每个根号内的式子写成完全平方。

注意原式可改写为 \[ \begin{aligned} & \sqrt{6 - 6\sqrt2 + 3} \\ &{}+ \sqrt{6 + 6\sqrt2 + 3} \end{aligned}\text{。}\]

因式分解并化简得 \[ \sqrt{(\sqrt6 - \sqrt3)^2} + \sqrt{(\sqrt6 + \sqrt3)^2} \] \[ = \sqrt6 - \sqrt3 + \sqrt6 + \sqrt3 = 2\sqrt6\text{。}\]

所以正确答案是 B。

English

Since we have square roots, we can try to change the inside of each radical to be a perfect square.

Note that we can rewrite the expression as \[ \begin{aligned} & \sqrt{6 - 6\sqrt2 + 3} \\ &{}+ \sqrt{6 + 6\sqrt2 + 3}. \end{aligned} \]

Factoring and simplifying gives us \[ \sqrt{(\sqrt6 - \sqrt3)^2} + \sqrt{(\sqrt6 + \sqrt3)^2} \] \[ = \sqrt6 - \sqrt3 + \sqrt6 + \sqrt3 = 2\sqrt6. \]

Thus, B is the correct answer.

由题设得到 \[A + B + C = 30\] \[B = 25 - A\text{。}\]同理，\[B + C + D = 30\] \[D = A\text{。}\]

继续对每三个连续项应用同一条件，得到 \[E = 25 - A, F = 5, G = A\text{，}\]最后 \(H = 25 - A\)。

因此所求和为 \[ A + 25 - A = 25\text{。}\]

所以正确答案是 C。

English

From the condition about the sequence, we get that \[ A + B + C = 30 \]\[ B = 25 - A. \] Similarly, we get \[ B + C + D = 30 \]\[ D = A. \]

Propagating these values through the sequence and repeating the condition for every consecutive triple, we get that \[ E = 25 - A, F = 5, G = A, \] and finally, \(H = 25 - A.\)

The desired sum is then \[ A + 25 - A = 25. \]

Thus, C is the correct answer.

所求区域面积等于圆 \(C\) 的面积减去它与圆 \(A\) 和圆 \(B\) 的重叠区域面积。

由图可知，一个重叠区域的一半可由一个四分之一圆扇形减去一个直角三角形得到。

这个面积为 \[ \dfrac{1}{4} \pi \cdot 1^2 - \dfrac{1}{2} \cdot 1 \cdot 1 = \dfrac{\pi}{4} - \dfrac{1}{2}\text{。}\]

共需减去四块这样的区域，所以最后的面积为 \[ \pi \cdot 1^2 - 4(\dfrac{\pi}{4} - \dfrac{1}{2}) = 2\text{。}\]

所以正确答案是 C。

English

The area of this region is the area of circle \(C\) minus the area of the overlapping regions with \(A\) and \(B.\)

From the diagram, we can find the area of half of one of the overlapping regions by finding the area of the sector and subtracting the area of the triangle.

This area is then \[ \dfrac{1}{4} \pi \cdot 1^2 - \dfrac{1}{2} \cdot 1 \cdot 1 = \dfrac{\pi}{4} - \dfrac{1}{2}. \]

There are four of these that we must subtract, which leaves us with a final answer of \[ \pi \cdot 1^2 - 4(\dfrac{\pi}{4} - \dfrac{1}{2}) = 2. \]

Thus, C is the correct answer.

设 \(1991\) 年人口为 \(p^2\)。\(2001\) 年人口为 \(q^2 + 9\)。

于是 \[ p^2 + 150 = q^2 + 9 \] \[ q^2 - p^2 = 141\text{。}\]

因式分解得 \[ (q - p)(q + p) = 141\text{。}\]

因为 \(p\) 和 \(q\) 都是整数，所以 \(q - p\) 与 \(q + p\) 的取值只可能是 \((1, 141)\) 和 \((3, 47)\)。

检验第一对，得到 \[q-p=1\] 和 \[q+p=141\text{，}\]两式相加后除以二，得到 \(q = 71\)、\(p = 70\)。

但此时 \(p^2 + 300\) 不是完全平方数，所以这一对不合要求。

检验另一对，得到 \(q = 25\) 和 \(p = 22\)。

此时 \(p^2 + 300 = 784\) 是完全平方数，人口增长率为 \[ \dfrac{300}{22^2} \cdot 100 \% \approx 62 \%\text{。}\]

所以正确答案是 E。

English

Let the population in \(1991\) be \(p^2.\) Then let the population in \(2001\) be \(q^2 + 9.\)

Using these values, we have \[ p^2 + 150 = q^2 + 9 \] \[ q^2 - p^2 = 141. \]

Factoring, we get \[ (q - p)(q + p) = 141. \]

As \(p\) and \(q\) are integers, we have that the only possible values for \(q - p\) and \(q + p\) are \((1, 141)\) and \((3, 47).\)

Trying the first pair, we have \[q-p=1\] and \[q+p=141,\] which adding together and dividing gives us \(q = 71\) and \(p = 70.\)

We have that \(p^2 + 300\) is not a square number, which means that this pair is the wrong one.

Trying the other pair and using the same strategy gives us \(q = 25\) and \(p = 22.\)

Now, \(p^2 + 300 = 784,\) which is a perfect square. The percent increase in population is then \[ \dfrac{300}{22^2} \cdot 100 \% \approx 62 \%. \]

Thus, E is the correct answer.

半径为 \(r\) 的圆中，长度为 \(r\) 的弦所对的圆心角为 \(60^\circ\)。固定第一条弦，使其端点的角度为 \(0^\circ\) 和 \(60^\circ\)。若第二条弦从角度 \(\theta\) 开始，则另一端点在其顺时针方向 \(60^\circ\) 处。

两条弦的端点恰好交替出现，当且仅当 \(\theta\) 位于紧邻固定弦两个端点的两段 \(60^\circ\) 圆弧之一。因此有利的起点共占圆周的 \(120^\circ\)。

所求概率为 \[ \dfrac{2}{6} = \dfrac{1}{3}\text{。}\]

所以正确答案是 D。

English

A chord of length \(r\) in a circle of radius \(r\) subtends a \(60^\circ\) arc. Fix the first chord, with endpoints at angles \(0^\circ\) and \(60^\circ.\) If the second chord starts at angle \(\theta,\) its other endpoint is \(60^\circ\) clockwise from there.

The endpoints of the two chords alternate exactly when \(\theta\) lies in either of the two \(60^\circ\) arcs immediately adjacent to the fixed chord’s endpoints. Thus the favorable starting positions occupy \(120^\circ\) of the circle.

The desired probability is then \[ \dfrac{2}{6} = \dfrac{1}{3}. \]

Thus, D is the correct answer.

可能情况只有两类：两对都全是真币，或每一对都含一枚假币。

全是真币时，选第一对有 \(\binom{8}{2} = 28\) 种，第二对有 \(\binom{6}{2} = 15\) 种。

两对可交换，所以还要除以 \(2\)，得到 \[28 \cdot 15 \div 2 = 210\] 种配置。

每对各有一枚假币时，选择两枚真币有 \(\binom{8}{2} = 28\) 种，而假币的选择只有一种。

把两枚假币分别与两枚真币配对有两种方法。

因此这一类共有 \(28 \cdot 2 = 56\) 种。

因此所求概率为 \[ \dfrac{210}{210 + 56} = \dfrac{210}{266} = \dfrac{15}{19}\text{。}\]

所以正确答案是 D。

English

There are two cases: either both selected pairs contain only genuine coins or each selected pair has one counterfeit coin.

For the first case, there are \(\binom{8}{2} = 28\) ways to choose the coins for the first pair and \(\binom{6}{2} = 15\) choices for the second pair.

We also have to divide by \(2\) since we can swap the pairs. This gives us \[ 28 \cdot 15 \div 2 = 210 \] configurations for this case.

For the second case, there are \(\binom{8}{2} = 28\) ways to choose the non-counterfeit coins. There is only one choice for the counterfeit coins.

There are two ways to create the two pairs, two choices for which counterfeit coin goes with a genuine coin.

This means that there are \(28 \cdot 2 = 56\) configurations for this case.

The desired probability is then \[ \dfrac{210}{210 + 56} = \dfrac{210}{266} = \dfrac{15}{19}. \]

Thus, D is the correct answer.

只有 \(3\) 种情况：所有顶点颜色都不同；恰有一对相邻顶点同色；或有 \(2\) 对相邻顶点同色，且两对使用不同颜色。

情况 \(1\)：所有顶点颜色都不同。

此时有 \(6! = 720\) 种着色方法。

情况 \(2\)：恰有一对相邻顶点同色。

选择各顶点颜色的方式有 \[\dfrac{6!}{2} = 360\] 种，而同色的相邻顶点对有 \(5\) 种选择。

因此这种情况共有 \[360 \cdot 5 = 1800\] 种着色方法。

情况 \(3\)：有两对相邻顶点分别同色。

不在任何同色对中的顶点有 \(5\) 种选择。再为两对和该单独顶点选择三种不同颜色，有 \[6 \cdot 5 \cdot 4 = 120\] 种，所以这种情况共有 \[120 \cdot 5 = 600\] 种着色方法。

所有情况合计 \[720 + 1800 + 600 = 3120\] 种。

所以正确答案是 C。

English

Note that there are only \(3\) cases: all the vertices are different, there is one pair of adjacent vertices with the same colors, or there are \(2\) pairs (each pair has a different color).

Case \(1:\) all vertices have different colors

This case just gives us \(6! = 720\) different colorings.

Case \(2:\) one pair of adjacent vertices has the same color

There are \[ \dfrac{6!}{2} = 360 \] ways to choose the colors for this case. There are then \(5\) options for the pair of vertices.

This gives us a total of \[ 360 \cdot 5 = 1800 \] colorings for this case.

Case \(3:\) two pairs of adjacent vertices have the same color

There are \(5\) choices for the vertex that is not in a pair. There are then \[ 6 \cdot 5 \cdot 4 = 120 \] choices for the colors. There are then a total of \[ 120 \cdot 5 = 600 \] colorings for this case.

There are a total of \[ 720 + 1800 + 600 = 3120 \] colorings for all the cases.

Thus, C is the correct answer.

逐步追踪这些数列即可找出最后剩下的数。

Alice 没说的数为 \[ 2, 5, 8, 11, 14, 17, \ldots, 998\text{。}\]

Barbara 说完后剩下 \[ 5, 14, 23, 32, 41, \ldots, 995\text{。}\]

这些都是等差数列，而公差每次变为原来的 \(3\) 倍。

Candice 说完后剩下 \[ 14, 41, 68, 95, \ldots, 986\text{。}\]

Debbie 说完后剩下 \[41, 122, 203, \ldots, 932\text{，}\]Eliza 说完后剩下 \[122, 365, 608, 851\text{。}\]

最后，Fatima 说完后唯一剩下的数是 \(365\)，所以 George 说的就是这个数。

所以正确答案是 C。

English

We can walk through all the iterations to find what is left.

Alice does not say the numbers \[ 2, 5, 8, 11, 14, 17, \ldots, 998. \]

After Barbara says her numbers, the remaining ones are \[ 5, 14, 23, 32, 41, \ldots, 995. \]

Note that both of these are arithmetic sequences where the common difference is increased by a multiple of \(3.\)

This pattern continues as the numbers remaining after Candice says hers are \[ 14, 41, 68, 95, \ldots, 986. \]

Then after Debbie, they are \[ 41, 122, 203, \ldots, 932 \] and after Eliza, they are \[ 122, 365, 608, 851. \]

Finally, the only number left after Fatima goes is \(365,\) which is the number that George will have to say.

Thus, C is the correct answer.

两个正四面体使用立方体两组交替的四个顶点。每个四面体的边长都是 \(\sqrt2\)，也就是立方体的一条面对角线。

边长为 \(s\) 的正四面体体积为 \(\dfrac{\sqrt2}{12}s^3\)，所以一个大四面体体积为 \(\dfrac{\sqrt2}{12}(\sqrt2)^3=\dfrac13\)。

一个四面体的每个面都会从另一个四面体切去一个相似比为 \(\dfrac12\) 的角四面体，每个切去部分体积为原四面体的 \(\dfrac18\)。

共有四个这样的角部分被切去，所以交集体积是一个大四面体体积的 \(1-4\cdot\dfrac18=\dfrac12\)。故交集体积为 \(\dfrac12\cdot\dfrac13=\dfrac16\)。

所以正确答案是 D。

English

The two regular tetrahedra use the two alternating sets of four vertices of the cube. Each has edge length \(\sqrt2\), a face diagonal of the cube.

The volume of a regular tetrahedron with edge length \(s\) is \(\dfrac{\sqrt2}{12}s^3\). Thus one large tetrahedron has volume \(\dfrac{\sqrt2}{12}(\sqrt2)^3=\dfrac13\).

Intersect one tetrahedron with the other. Each face of the first cuts from the second a corner tetrahedron similar to the original with scale factor \(\dfrac12\), so each cut-off piece has \(\dfrac18\) of the large tetrahedron’s volume.

There are four such corner pieces, so the intersection has \(1-4\cdot\dfrac18=\dfrac12\) of the volume of one large tetrahedron. Hence the intersection volume is \(\dfrac12\cdot\dfrac13=\dfrac16\).

Thus, D is the correct answer.

把正方形缩放为边长 \(1\)，并写成 \(X=(u,v)\)，其中 \(u\) 和 \(v\) 分别是它到左边和下边的距离。每个顶点都必须与 \(X\) 相连；否则包含该顶点的某个区域就不是三角形。

这 \(n\) 个三角形的面积都是 \(\frac{1}{n}\)。以正方形下边为底的三角形高为 \(v\)，所以底长为 \(\frac{2}{nv}\)。因此下边上的三角形数为 \(\frac{nv}{2}\)，它必须是正整数。对四条边作同样分析可知，\[\begin{gathered} \dfrac{nu}{2},\quad\dfrac{n(1-u)}{2},\\ \dfrac{nv}{2},\quad\dfrac{n(1-v)}{2} \end{gathered}\] 都是正整数。反过来，只要这四个数都是整数，把各边分成相应数量的等长底边，再将分点与 \(X\) 相连，就能得到所需的三角形。

当 \(n=100\) 时，这说明 \(u=\frac{i}{50}\)，且 \(v=\frac{j}{50}\)，其中 \(i,j\in\{1,2,\ldots,49\}\)。因而所有 \(100\) 射线等分点构成 \(49\times49\) 网格。同理，所有 \(60\) 射线等分点满足 \(u=\frac{i}{30}\)、\(v=\frac{j}{30}\)，其中 \(i,j\in\{1,2,\ldots,29\}\)。

一个坐标同时属于两个网格，当且仅当 \(\frac{i}{50}=\frac{j}{30}\)，即 \(3i=5j\)。因此公共坐标为 \(\frac{1}{10},\frac{2}{10},\ldots,\frac{9}{10}\)，形成一个 \(9\times9\) 的重叠网格。所求点数为 \[49^2-9^2=2401-81=2320\text{。}\]

所以正确答案是 C。

English

Scale the square to have side length \(1,\) and write \(X=(u,v),\) where \(u\) and \(v\) are its distances from the left and bottom sides. Every corner must be joined to \(X\); otherwise one of the regions containing that corner would not be a triangle.

Each of the \(n\) triangles has area \(\frac{1}{n}.\) A triangle whose base lies on the bottom side has height \(v,\) so its base has length \(\frac{2}{nv}.\) Therefore the number of triangles along the bottom side is \(\frac{nv}{2},\) which must be a positive integer. Applying the same argument to all four sides shows that \[\begin{gathered} \dfrac{nu}{2},\quad\dfrac{n(1-u)}{2},\\ \dfrac{nv}{2},\quad\dfrac{n(1-v)}{2} \end{gathered}\] are positive integers. Conversely, whenever these four numbers are integers, subdividing each side into the indicated number of equal bases and joining the division points to \(X\) produces the required triangles.

For \(n=100,\) this says \(u=\frac{i}{50}\) and \(v=\frac{j}{50}\) for \(i,j\in\{1,2,\ldots,49\}.\) Hence the \(100\)-ray points form a \(49\times49\) grid. Similarly, the \(60\)-ray points have coordinates \(u=\frac{i}{30}\) and \(v=\frac{j}{30}\) with \(i,j\in\{1,2,\ldots,29\}.\)

A coordinate belongs to both grids exactly when \(\frac{i}{50}=\frac{j}{30},\) or \(3i=5j.\) Thus the common coordinates are \(\frac{1}{10},\frac{2}{10},\ldots,\frac{9}{10},\) giving a \(9\times9\) overlap. The requested number is \[49^2-9^2=2401-81=2320.\]

Thus, C is the correct answer.