2017B AMC 10 真题（P版）

1 答案 B

从可能的结果倒推。把 \(71,72,73,74,75\) 的两个数字换回去，分别得到 \(17,27,37,47,57\)。再减去 \(11\)，依次得到 \(6,16,26,36,46\)。其中只有 \(6\) 和 \(36\) 能被 \(3\) 整除，而其中只有 \(36\div3=12\) 是两位数。

所以正确答案是 B。

English

Work backward from the possible results. Reversing \(71,72,73,74,75\) gives \(17,27,37,47,57,\) respectively. Subtracting \(11\) gives \(6,16,26,36,46.\) Only \(6\) and \(36\) are divisible by \(3,\) and only \(36\div3=12\) is a two-digit number.

Thus, the correct answer is B .

2 答案 C

Sofia 总共以每秒 \(4\) 米跑了 \(5\cdot 100=500\) 米，以每秒 \(5\) 米跑了 \(5\cdot 300=1500\) 米。

因此总时间为 \[\frac{500}4 + \frac{1500}5 = 425\] 秒。

这个时间等于 \(7\) 分 \(5\) 秒。

所以正确答案是 C。

English

She ran a total of \(5\cdot 100=500\) meters at \(4\) meters per second and \(5\cdot 300=1500\) meters at \(5\) meters per second.

Therefore, her time is \[\frac{500}4 + \frac{1500}5 = 425\] seconds.

This is equal to a total of \(7\) minutes and \(5\) seconds.

Thus, the correct answer is C .

3 答案 E

因为 \(-1 < y\) 且 \(1 < z\)，所以 \(0 < y+z\)，因此选项 E 一定正确。

此外，用下面这组取值可以排除其余每个选项：\[x=0.1\text{，}\]\[y=-0.25\text{，}\]\[z=1.25\text{。}\]

所以正确答案是 E。

English

Since \(-1 < y\) and \(1 < z,\) we can add the inequalities to see that \(0 < y+z.\) This naturally proves choice E correct.

Furthermore, we can eliminate every other choice with the following values: \[x=0.1,\]\[y=-0.25,\]\[z=1.25.\]

Thus, the correct answer is E .

4 答案 D

由 \[\frac{3x+y}{x-3y}=-2\text{，}\] 两边同乘分母，得到 \[3x+y = 6y-2x\text{。}\] 整理可得 \(x=y\)。

因此 \[\frac{x+3y}{3x-y} = \frac{x+3x}{3x-x}=2\text{。}\]

所以正确答案是 D。

English

Given that \[\frac{3x+y}{x-3y}=-2,\] we can multiply by the denominator to get \[3x+y = 6y-2x.\] Solving, we can see that \(x=y.\)

Therefore, \[\frac{x+3y}{3x-y} = \frac{x+3x}{3x-x}=2.\]

Thus, the correct answer is D .

5 答案 D

设原来樱桃味有 \(c\) 颗，蓝莓味有 \(b\) 颗。

由第一句和第二句分别可得 \[b = 2c\]\[b-10=3(c-10)\] 这两个方程。

因此 \[2c-10 = 3c-30\]\[c=20\text{。}\] 这说明 \[b = 2\cdot 20=40\text{。}\]

所以正确答案是 D。

English

Let the number of cherry jelly beans be \(c\) and let the number of blueberry jelly beans be \(b.\)

Then, we know \[b = 2c\]\[b-10=3(c-10)\] from the first and second statements respectively.

Therefore, \[2c-10 = 3c-30\]\[c=20.\] This means that \[b = 2\cdot 20=40.\]

Thus, the correct answer is D .

6 答案 B

大盒子的体积是 \(3\cdot 3\cdot 2 = 18\)，每块小积木体积是 \(2\cdot 2\cdot 1=4\)。所以最多能放 \(4\) 块。

这个上界可以达到：先把 \(3\) 块小积木放入一个 \(3 \times 2 \times 2\) 的长方体，再在剩下的 \(3 \times 2 \times 1\) 空间中放入一块。

所以正确答案是 B。

English

The volume of the large solid object is \(3\cdot 3\cdot 2 = 18\) and volume of the smaller object is \(2\cdot 2\cdot 1=4.\) This means we can fit at most \(4\) of the small objects.

We can make this happen by putting \(3\) of the small objects in a \(3 \times 2 \times 2\) rectangular prism, and then we have a \(3 \times 2 \times 1\) space left where we can place one small object.

Thus, the correct answer is B .

7 答案 C

设 \(d\) 为 Samia 步行的距离。她骑车走过的距离与此相同，所以由总用时可得 \[\dfrac{d}{17}+\dfrac{d}{5}=\dfrac{44}{60}\text{。}\] 解得 \(d=\dfrac{17}{6}=2.833\ldots\)，四舍五入到十分位约为 \(2.8\) 千米。

所以正确答案是 C。

English

Let \(d\) be the distance Samia walked. She bicycled the same distance, so her total travel time gives \[\dfrac{d}{17}+\dfrac{d}{5}=\dfrac{44}{60}.\] Solving yields \(d=\dfrac{17}{6}=2.833\ldots,\) which rounds to \(2.8\) kilometers.

Thus, the correct answer is C .

8 答案 C

因为 \(AB=AC\)，所以从 \(A\) 作出的高也平分底边 \(BC\)。因此 \(D\) 是 \(BC\) 的中点。若 \(C=(x,y)\)，则 \[\frac{x+2}2 = -1\text{，}\]\[\frac{y-3}2=3\text{。}\] 于是 \[C=(x,y)=(-4,9)\text{。}\]

所以正确答案是 C。

English

Since \(AB=AC,\) the altitude from \(A\) also bisects the base \(BC.\) Therefore, \(D\) is the midpoint of \(BC.\) If \(C=(x,y),\) then we have \[\frac{x+2}2 = -1,\]\[\frac{y-3}2=3.\] As such, \[C=(x,y)=(-4,9).\]

Thus, the correct answer is C .

9 答案 D

全部 \(3\) 题答对的概率为 \[\left(\dfrac13\right)^3=\dfrac1{27}\text{。}\] 恰好答对 \(2\) 题的概率为 \[\binom32\left(\dfrac13\right)^2\left(\dfrac23\right)=\dfrac6{27}\text{。}\] 合计概率为 \[\dfrac6{27}+\dfrac1{27}=\dfrac7{27}\text{。}\]

所以正确答案是 D。

English

The probability that a contestant gets all \(3\) correct is \[\left(\dfrac13\right)^3=\dfrac1{27}.\] The probability of getting exactly \(2\) correct is \[\binom32\left(\dfrac13\right)^2\left(\dfrac23\right)=\dfrac6{27}.\] The combined probability is \[\dfrac6{27}+\dfrac1{27}=\dfrac7{27}.\]

Thus, the correct answer is D .

10 答案 E

第一个方程可改写为 \(y=\dfrac a2x-\dfrac c2\)，第二个可改写为 \(y=-\dfrac2b x-\dfrac cb\)。因为两条直线垂直，它们的斜率之积为 \(-1\)，所以 \(a=b\)。

把 \((x,y)=(1,-5)\) 代入原来的两个方程，得到 \(a+10=c\) 和 \(2-5a=-c\)。两式相加得 \(12-4a=0\)，所以 \(a=3\)，\(c=13\)。

所以正确答案是 E。

English

The first equation can be rewritten as \(y=\dfrac a2x-\dfrac c2,\) and the second as \(y=-\dfrac2b x-\dfrac cb.\) Because the lines are perpendicular, their slopes multiply to \(-1,\) so \(a=b.\)

Substituting \((x,y)=(1,-5)\) into the two original equations gives \(a+10=c\) and \(2-5a=-c.\) Adding these equations yields \(12-4a=0,\) so \(a=3\) and \(c=13.\)

Thus, the correct answer is E .

11 答案 D

真正喜欢跳舞的人占 \(60\%\)，其中只有 \(80\%\) 说自己喜欢跳舞。因此全体学生中有 \(48\%\) 说自己喜欢跳舞，而真正喜欢但说不喜欢的占 \(60\%-48\% = 12\% \)。

真正不喜欢跳舞的人占 \(40\%\)，其中 \(90\%\) 说自己不喜欢，因而这部分占全体的 \(36\%\)。

所以说自己不喜欢跳舞的人共占 \(12\%+36\% = 48\%\)。

因此，说自己不喜欢、但实际上喜欢跳舞的人所占比例为 \[\frac{12}{48} = \frac 14 = 25\%\text{。}\]

所以正确答案是 D。

English

Observe that of the \(60\%\) of people that actually like dancing, only \(80\%\) say they like dancing. This suggests that \(48\%\) of the students say that they like dancing, and as such, \(60\%-48\% = 12\% \) of the students who like dancing say they don’t like it.

Then, we know that \(90\%\) of the \(40\%\) of people who don’t like dancing say they don’t like it, which is \(36\%\) of the total student population.

This means the total amount of people who say they don’t like dancing is \(12\%+36\% = 48\%.\)

We know then that the fraction of people who say they dislike dancing but actually like it is equal to: \[\frac{12}{48} = \frac 14 = 25\% .\]

Thus, the correct answer is D .

12 答案 A

设旧车的燃油效率为每升 \(M\) 千米，汽油价格为每升 \(C\) 美元。于是旧车每千米的燃油费用为 \(\dfrac CM\) 美元。

新车每升可行驶 \(1.5M\) 千米，而它所用燃料的价格为每升 \(1.2C\) 美元，所以新车每千米的燃油费用为 \[\dfrac{1.2C}{1.5M}=0.8\dfrac CM\text{。}\]

新车的燃油费用是旧车的 \(80\%\)，所以 Elmer 节省了 \(20\%\)。

所以正确答案是 A。

English

Let the old car’s fuel efficiency be \(M\) kilometers per liter and let gasoline cost \(C\) dollars per liter. The old car therefore costs \(\dfrac CM\) dollars per kilometer to fuel.

The new car gets \(1.5M\) kilometers per liter and its fuel costs \(1.2C\) dollars per liter, so its fuel cost per kilometer is \[\dfrac{1.2C}{1.5M}=0.8\dfrac CM.\]

The new fuel cost is \(80\%\) of the old one, so Elmer saves \(20\%.\)

Thus, the correct answer is A .

13 答案 C

总选课人次为 \(10+13+9=32\)。

设 \(x\) 表示选 \(1\) 门课的人数，\(y\) 表示选 \(2\) 门课的人数，\(z\) 表示选 \(3\) 门课的人数。

于是 \(x+2y+3z = 32\)。

因此总人数为 \(20\)，所以 \(x+y+z = 20\)，从而 \(y+2z=12\)。

至少选两门课的人数为 \(9\)，所以 \(y+z = 9\)。

因此 \(z=3\)，这就是答案。

所以正确答案是 C。

English

The number of classes taken total is \(10+13+9=32.\)

Let \(x\) represent the number of people who take \(1,\) let \(y\) represent the number of people who take \(2\) classes, and let \(z\) represent the number of people who take \(3\) classes.

Then, we know \(x+2y+3z = 32.\)

As such, the total number of people is \(20,\) so \(x+y+z = 20.\) This makes \(y+2z=12.\)

The number of people who take at least two classes is \(9,\) so \(y+z = 9.\)

Therefore, \(z=3,\) making that the answer.

Thus, the correct answer is C .

14 答案 D

根据费马小定理，只要 \(N\) 不能被 \(5\) 整除，就有 \(N^4\equiv1\pmod5\)。因此 \[N^{16}\equiv(N^4)^4\equiv1\pmod5\text{。}\]

\(5\) 的倍数共有 \(\frac{2020}{5}=404\) 个，所以 \(N\) 的可取值共有 \(2020-404=1616\) 个。

\(5\) 的倍数满足 \(N^{16}\equiv0\pmod5\)，所以其余的值都不符合要求。因此所求概率为 \[\dfrac{1616}{2020}=\dfrac45\text{。}\]

所以正确答案是 D。

English

By Fermat’s Little Theorem, \(N^4\equiv1\pmod5\) whenever \(N\) is not divisible by \(5.\) Therefore, \[N^{16}\equiv(N^4)^4\equiv1\pmod5.\]

There are \(\frac{2020}{5}=404\) multiples of \(5,\) so there are \(2020-404=1616\) allowable values of \(N.\)

A multiple of \(5\) has \(N^{16}\equiv0\pmod5,\) so no other values work. Thus the probability is \[\dfrac{1616}{2020}=\dfrac45.\]

Thus, the correct answer is D .

15 答案 E

\(\triangle CDA\) 的面积是 \(\dfrac{3\cdot4}{2}=6\)。因为 \(E\) 在 \(AC\) 上，三角形 \(EAD\) 和 \(CDA\) 共用从 \(D\) 作出的同一条高，所以 \([EAD]=[CDA]\cdot \dfrac{AE}{AC}\)。

由勾股定理，\(AC=5\)。又由 \(\triangle ABE\sim \triangle ACB\)，\(\dfrac{AE}{AB}=\dfrac{AB}{AC}=\dfrac35\)，所以 \(AE=\dfrac95\)，\(\dfrac{AE}{AC}=\dfrac{9}{25}\)。

因此 \([EAD]=6\cdot \dfrac{9}{25}=\dfrac{54}{25}\)。所以正确答案是 E。

English

The area of \(\triangle CDA\) is \(\dfrac{3\cdot4}{2}=6\). Since \(E\) lies on \(AC\), triangles \(EAD\) and \(CDA\) share the same altitude from \(D\), so \([EAD]=[CDA]\cdot \dfrac{AE}{AC}\).

By the Pythagorean Theorem, \(AC=5\). Also \(\triangle ABE\sim \triangle ACB\), so \(\dfrac{AE}{AB}=\dfrac{AB}{AC}=\dfrac35\), giving \(AE=\dfrac95\). Thus \(\dfrac{AE}{AC}=\dfrac{9}{25}\).

Therefore \([EAD]=6\cdot \dfrac{9}{25}=\dfrac{54}{25}\). Thus, E is the correct answer.

16 答案 A

小于 \(100\) 的数中，只有 \(10\) 的倍数才含有数字 \(0\)，这样的数有 \(9\) 个。

对 \(100\) 到 \(999\)（含端点）的数用补集计数。这个范围共有 \(900\) 个数。没有 \(0\) 的数有 \(9\cdot 9\cdot 9=729\) 个，因为每一位都有 \(9\) 种选择使它不是 \(0\)。因此这个范围内共有 \(171\) 个。

对 \(1000\) 到 \(1999\)（含端点）的数再次用补集计数。这个范围共有 \(1000\) 个数。没有 \(0\) 的数有 \(1\cdot 9\cdot 9\cdot 9=729\) 个，因为后 \(3\) 位各有 \(9\) 种选择使它不是 \(0\)，首位必须是 \(1\)。因此这个范围内共有 \(271\) 个。

从 \(2000\) 到 \(2017\)（含端点）共有 \(18\) 个数，每个数从左数第二位都是 \(0\)。

总数为 \[9+171+271+18=469\text{。}\]

所以正确答案是 A。

English

For numbers less than \(100,\) we only have a \(0\) if it is a multiple of \(10,\) of which there are \(9.\)

For numbers between \(100\) and \(999\) inclusive, we will use complementary counting. There are \(900\) total numbers in this range. Also, there are \(9\cdot 9\cdot 9=729\) numbers in this range with no \(0\) since there are \(9\) ways to choose each digit to not be \(0.\) Thus, the total in this range is \(171.\)

For numbers between \(1000\) and \(1999\) inclusive, we will use complementary counting again. There are \(1000\) total numbers in this range. Also, there are \(1\cdot 9\cdot 9\cdot 9=729\) numbers in this range with no \(0\) since there are \(9\) ways to choose each of the last \(3\) digits to not be \(0\) and the first digit must be \(1.\) Thus, the total in this range is \(271.\)

There are \(18\) numbers between \(2000\) and \(2017\) inclusive, each with a \(0\) in the second digit from the left.

This makes the total \[9+171+271+18=469.\]

Thus, the correct answer is A .

17 答案 B

严格递增的正整数对应于 \(\{1,2,3,4,5,6,7,8,9\}\) 的非空子集，把选出的数字按递增顺序写出。因此有 \(2^9-1=511\) 个。

严格递减的正整数对应于 \(\{0,1,2,3,4,5,6,7,8,9\}\) 的子集，把选出的数字按递减顺序写出，但要排除空集和 \(\{0\}\)。因此有 \(2^{10}-2=1022\) 个。

一位数 \(1\) 到 \(9\) 被两边都算了一次，所以总数为 \(511+1022-9=1524\)。所以正确答案是 B。

English

The strictly increasing positive integers correspond to the nonempty subsets of \(\{1,2,3,4,5,6,7,8,9\}\), written in increasing order. There are \(2^9-1=511\) of these.

The strictly decreasing positive integers correspond to subsets of \(\{0,1,2,3,4,5,6,7,8,9\}\), written in decreasing order, except for the empty set and \(\{0\}\). There are \(2^{10}-2=1022\) of these.

The one-digit numbers \(1\) through \(9\) were counted in both groups, so the total is \(511+1022-9=1524\). Thus, B is the correct answer.

18 答案 D

由对称性，绿色圆盘的位置只有两类：角上的位置或边中点的位置。每类各固定一个代表位置。此时选出两个红色圆盘的方法有 \(\binom52=10\) 种。

保持绿色位置不动的那个反射会固定其余圆盘中的一个，并把另外四个圆盘两两交换。恰好有 \(2\) 种红色圆盘的选法在这个反射下不变，即选中被交换的某一对。其余 \(8\) 种选法两两互为镜像，构成 \(4\) 对。因此绿色圆盘的每一类位置都给出 \(2+4=6\) 种涂色方案。

于是两类位置合起来给出 \(6+6=12\) 种涂色方案。

所以正确答案是 D。

English

By symmetry, the green disk has two possible types of position: a corner or a side midpoint. Fix one representative of either type. There are \(\binom52=10\) ways to choose the two red disks.

The reflection that fixes the green position fixes one of the other disks and exchanges the other four disks in two pairs. Exactly \(2\) red-disk choices are unchanged by this reflection: choosing either exchanged pair. The other \(8\) choices form \(4\) mirror-image pairs. Hence there are \(2+4=6\) paintings for each type of green position.

The two types therefore give \(6+6=12\) paintings.

Thus, the correct answer is D .

19 答案 E

设 \(\triangle ABC\) 的面积为 \(X\)。\(\triangle BB'C\)、\(\triangle CC'A\) 和 \(\triangle AA'B\) 的底边都是 \(\triangle ABC\) 对应边的三倍，而对应的高相同。因此它们的面积都是 \(3X\)。

接着，\(\triangle AA'C'\) 的底是 \(\triangle ACC'\) 的三倍，高相同；后者面积为 \(3X\)。所以 \(\triangle AA'C'\) 的面积为 \(9X\)。同理，\(\triangle CC'B'\) 和 \(\triangle BB'A'\) 的面积也都是 \(9X\)。

这七个区域恰好分割大三角形，所以 \[\begin{aligned}[A'B'C']&=X+3(3X)+3(9X)\\&=37X\end{aligned}\text{。}\]所求比为 \(37:1\)。

所以正确答案是 E。

English

Let \(X\) be the area of \(\triangle ABC.\) Each of \(\triangle BB'C,\) \(\triangle CC'A,\) and \(\triangle AA'B\) has a base three times as long as a side of \(\triangle ABC\) and the same corresponding altitude. Each therefore has area \(3X.\)

Next, \(\triangle AA'C'\) has three times the base and the same altitude as \(\triangle ACC',\) whose area is \(3X.\) Thus \(\triangle AA'C'\) has area \(9X.\) Similarly, \(\triangle CC'B'\) and \(\triangle BB'A'\) each have area \(9X.\)

These seven regions partition the large triangle, so \[\begin{aligned}[A'B'C']&=X+3(3X)+3(9X)\\&=37X.\end{aligned}\] The requested ratio is \(37:1.\)

Thus, the correct answer is E .

20 答案 B

\(21!\) 中 \(2\) 的指数为 \[\begin{aligned} &\left\lfloor\dfrac{21}{2}\right\rfloor+ \left\lfloor\dfrac{21}{4}\right\rfloor\\ &\quad{}+\left\lfloor\dfrac{21}{8}\right\rfloor+ \left\lfloor\dfrac{21}{16}\right\rfloor\\ &=10+5+2+1=18 \end{aligned}\text{。}\] 因此 \(21!=2^{18}d\)，其中 \(d\) 是某个奇整数。

对 \(d\) 的每个因子 \(x\)，\(21!\) 中奇数部分为 \(x\) 的因子恰好是 \(x,2x,\ldots,2^{18}x\)。这 \(19\) 个因子中只有一个是奇数，所以所求概率为 \(\dfrac1{19}\)。

所以正确答案是 B。

English

The exponent of \(2\) in \(21!\) is \[\begin{aligned} &\left\lfloor\dfrac{21}{2}\right\rfloor+ \left\lfloor\dfrac{21}{4}\right\rfloor\\ &\quad{}+\left\lfloor\dfrac{21}{8}\right\rfloor+ \left\lfloor\dfrac{21}{16}\right\rfloor\\ &=10+5+2+1=18. \end{aligned}\] Thus \(21!=2^{18}d\) for some odd integer \(d.\)

For every divisor \(x\) of \(d,\) the divisors of \(21!\) with odd part \(x\) are \(x,2x,\ldots,2^{18}x.\) Exactly one of these \(19\) divisors is odd, so the probability is \(\dfrac1{19}.\)

Thus, the correct answer is B .

21 答案 D

三角形 \(ABC\) 是在 \(A\) 处为直角的直角三角形。由于 \(D\) 是斜边中点，所以它是这个三角形的外心。

因此 \[AD = BD = DC = 5\text{。}\]同时，\(ABC\) 的面积为 \[\dfrac{6\cdot 8}2 = 24\text{。}\]

底边 \(BD\) 和 \(DC\) 相等，两个三角形从 \(A\) 作出的高也相同。因此，\(\triangle ABD\) 和 \(\triangle ACD\) 的面积都为 \(12\)。

对每个三角形，都有 \(A = rs\)，其中 \(A\) 是面积，\(r\) 是内切圆半径，\(s\) 是半周长。等价地，\(12=\dfrac12rP\)，其中 \(P\) 是周长，所以 \(r=\dfrac{24}{P}\)。对 \(\triangle ABD\)，内切圆半径为 \(\dfrac{24}{5+5+6}=\dfrac32\)。对 \(\triangle ACD\)，内切圆半径为 \(\dfrac{24}{5+5+8}=\dfrac43\)。

两者之和为 \(\dfrac 32 + \dfrac 43 = \dfrac{17}6\)。

所以正确答案是 D。

English

The triangle \(ABC\) is a right triangle with a right angle at \(A.\) This makes \(D\) the circumcenter of the triangle since it is the midpoint of the hypotenuse.

Therefore, \[AD = BD = DC = 5.\] Also, the area of \(ABC\) is \[\dfrac{6\cdot 8}2 = 24.\]

The bases \(BD\) and \(DC\) are equal, and the two triangles share the same altitude from \(A.\) Therefore, \(\triangle ABD\) and \(\triangle ACD\) each have area \(12.\)

Then, for each triangle, we have \(A = rs\) where \(A\) is the area, \(r\) is the inradius, and \(s\) is the semiperimeter. Equivalently, \(12=\dfrac12rP,\) where \(P\) is the perimeter, so \(r=\dfrac{24}{P}.\) For \(\triangle ABD,\) the inradius is \(\dfrac{24}{5+5+6}=\dfrac32.\) For \(\triangle ACD,\) it is \(\dfrac{24}{5+5+8}=\dfrac43.\)

Their sum is \(\dfrac 32 + \dfrac 43 = \dfrac{17}6 .\)

Thus, the correct answer is D .

22 答案 D

因为半径为 \(2\)，且 \(BD =3\)，所以 \(AD = 7\)。由于 \(ED = 5\)，且 \(D\) 处角为直角，\(ADE\) 的面积为 \(\dfrac{5\cdot 7}{2} = \dfrac{35}{2} \)。

由勾股定理，\(AE=\sqrt{5^2+7^2}=\sqrt{74}\)。又因为 \(AB\) 是直径，所以 \(\angle ACB\) 是直角。两个三角形共用 \(A\) 处的角，因此由角角相似可知 \(\triangle ABC\sim\triangle AED\)。

它们对应的斜边分别是 \(AB=4\) 和 \(AE=\sqrt{74}\)，所以面积之比为 \[\dfrac{[ABC]}{[AED]}=\left(\dfrac4{\sqrt{74}}\right)^2=\dfrac8{37}\text{。}\] 因此 \[[ABC]=\dfrac8{37}\cdot\dfrac{35}{2}=\dfrac{140}{37}\text{。}\]

所以正确答案是 D。

English

Since the radius is \(2\) and \(BD =3,\) we have \(AD = 7.\) Since \(ED = 5\) and the angle at \(D\) is a right angle, the area of \(ADE\) is \(\dfrac{5\cdot 7}{2} = \dfrac{35}{2} .\)

By the Pythagorean Theorem, \(AE=\sqrt{5^2+7^2}=\sqrt{74}.\) Also, \(\angle ACB\) is a right angle because \(AB\) is a diameter. The triangles share the angle at \(A,\) so \(\triangle ABC\sim\triangle AED\) by angle-angle similarity.

Their corresponding hypotenuses are \(AB=4\) and \(AE=\sqrt{74},\) so their area ratio is \[\dfrac{[ABC]}{[AED]}=\left(\dfrac4{\sqrt{74}}\right)^2=\dfrac8{37}.\] Therefore, \[[ABC]=\dfrac8{37}\cdot\dfrac{35}{2}=\dfrac{140}{37}.\]

Thus, the correct answer is D .

23 答案 C

要找除以 \(45\) 的余数，必须分别找除以 \(5\) 和 \(9\) 的余数。除以 \(5\) 的余数就是个位数字除以 \(5\) 的余数，所以是 \(4\)。

要找除以 \(9\) 的余数，通常找数字和。不过每个两位数除以 \(9\) 的余数与它的数字和相同，所以可以直接把 \(1\) 到 \(44\) 的整数相加，因为每个整数与它自己的数字和同余。这个和为 \(\dfrac{44\cdot45}{2}=990\)，它是 \(9\) 的倍数。因此 \(N\) 是 \(9\) 的倍数。

这个数是 \(9\) 的倍数，并且除以 \(5\) 的余数为 \(4\)，所以除以 \(45\) 的余数为 \(9\)。

所以正确答案是 C。

English

To find the remainder when divided by \(45,\) we must find the remainder when divided by \(5\) and \(9.\) The remainder when divided by \(5\) is the remainder when the units digit is divided by \(5,\) making it \(4.\)

To find the remainder when divided by \(9,\) we usually find the sum of the digits. However, each double digit number has the same remainder when divided by \(9\) as its digit sum, so we can just sum the integers from \(1\) to \(44,\) because each integer is congruent to its own digit sum. This sum is \(\dfrac{44\cdot45}{2}=990,\) which is a multiple of \(9.\) Thus, \(N\) is a multiple of \(9.\)

Since it is a multiple of \(9\) and has a remainder of \(4\) when divided by \(5,\) the remainder when divided by \(45\) is \(9.\)

Thus, the correct answer is C .

24 答案 C

由对称性，不妨设重心是双曲线顶点 \(G=(1,1)\)。三角形至少有两个顶点位于双曲线的同一支上。它们不能都位于负支：如果其中两个顶点的 \(x\) 坐标为负，第三个顶点的横坐标就会大于 \(3\)，而三个 \(y\) 坐标之和会小于 \(\frac{1}{3}\)，这与重心是 \((1,1)\) 矛盾。因此两个顶点位于正支。

把这两个顶点写成 \(P=(a,\frac{1}{a})\) 和 \(Q=(b,\frac{1}{b})\)，其中 \(a,b>0\)。等边三角形的重心也是外心，所以 \(P\) 和 \(Q\) 到 \(G\) 的距离相等。对 \(t>0\)，点 \((t,\frac{1}{t})\) 到 \(G\) 的距离平方为 \[\left(t+\dfrac1t\right)\left(t+\dfrac1t-2\right)\text{。}\]当 \(t+\frac{1}{t}\) 从 \(2\) 开始增大时，此式严格递增，所以两个不同的点必须满足 \(b=\frac{1}{a}\)。因此 \(P\) 和 \(Q\) 关于 \(y=x\) 对称。

第三个顶点位于垂直平分线 \(y=x\) 上。它的坐标还满足 \(xy=1\)，所以它是 \((1,1)\) 或 \((-1,-1)\)。它不能与重心重合，所以是 \((-1,-1)\)。因此外接圆半径就是从 \((1,1)\) 到 \((-1,-1)\) 的距离，即 \(2\sqrt2\)。

从中心把等边三角形分成三个三角形，其面积为 \[3\cdot\dfrac12(2\sqrt2)^2\sin120^\circ=6\sqrt3\text{。}\]面积的平方为 \((6\sqrt3)^2=108\)。

所以正确答案是 C。

English

By symmetry, assume that the centroid is the hyperbola vertex \(G=(1,1).\) At least two triangle vertices lie on the same branch of the hyperbola. They cannot both lie on the negative branch: if two of their \(x\)-coordinates were negative, the third would exceed \(3,\) while the sum of the three \(y\)-coordinates would be less than \(\frac{1}{3},\) contradicting that their centroid is \((1,1).\) Thus two vertices lie on the positive branch.

Write these vertices as \(P=(a,\frac{1}{a})\) and \(Q=(b,\frac{1}{b}),\) where \(a,b>0.\) The centroid of an equilateral triangle is also its circumcenter, so \(P\) and \(Q\) are equidistant from \(G.\) For \(t>0,\) the squared distance from \((t,\frac{1}{t})\) to \(G\) is \[\left(t+\dfrac1t\right)\left(t+\dfrac1t-2\right).\] This is strictly increasing as \(t+\frac{1}{t}\) increases from \(2,\) so the distinct points must satisfy \(b=\frac{1}{a}.\) Hence \(P\) and \(Q\) are reflections across \(y=x.\)

The third vertex lies on the perpendicular bisector \(y=x.\) Its coordinates also satisfy \(xy=1,\) so it is either \((1,1)\) or \((-1,-1).\) It cannot equal the centroid, so it is \((-1,-1).\) Therefore, the circumradius is the distance from \((1,1)\) to \((-1,-1),\) namely \(2\sqrt2.\)

Dividing the equilateral triangle into three triangles at its center gives its area as \[3\cdot\dfrac12(2\sqrt2)^2\sin120^\circ=6\sqrt3.\] The square of the area is \((6\sqrt3)^2=108.\)

Thus, the correct answer is C .

25 答案 E

设七次总分为 \(S\)。因为七次后的平均数是整数，\(S\) 可被 \(7\) 整除。七个不同分数都在 \(91\) 到 \(100\) 之间，所以 \(91+92+\cdots+97\) \(\le S\le\) \(94+95+\cdots+100\)。

因此 \(658\le S\le 679\)，可能的 \(7\) 的倍数为 \(658,665,672,679\)。由于第七次分数是 \(95\)，前六次总分为 \(S-95\)，它必须能被 \(6\) 整除。这迫使 \(S=665\)。

因此前六次总分为 \(570\)。前五次平均数也是整数，所以前五次总分能被 \(5\) 整除。因此第六次分数也必须能被 \(5\) 整除。由于第七次已经是 \(95\)，且所有分数不同，第六次分数只能是 \(100\)。所以正确答案是 E。

English

Let \(S\) be the sum of all seven scores. Since all seven averages were integers, \(S\) is divisible by \(7\). Also the seven distinct scores are between \(91\) and \(100\), so \(91+92+\cdots+97\) \(\le S\le\) \(94+95+\cdots+100\).

Thus \(658\le S\le 679\), and the possible multiples of \(7\) are \(658,665,672,679\). Since the seventh score is \(95\), the first six scores sum to \(S-95\), which must be divisible by \(6\). This forces \(S=665\).

The first six scores sum to \(570\). The first five-score average was also an integer, so the sum of the first five scores is divisible by \(5\). Therefore the sixth score is divisible by \(5\). Since the seventh score is already \(95\) and all scores are distinct, the sixth score is \(100\). Thus, E is the correct answer.