2021D AMC 10 真题（P版）

逐位相加，得到 \(11,110\)。

也可以注意到每个数位上的数字和均为 \(10\)，所以总和为 \(10\cdot 1111 = 11,110\)。

所以答案是 E。

English

We can add each individual digit, yielding \(11,110.\)

We can also get the sum by noticing that each digit has a sum of \(10,\) so the sum is equal to \(10\cdot 1111 = 11,110.\)

Thus, the answer is E .

阴影部分的面积是一个三角形的面积 \[\dfrac {4\cdot 5}2 - \dfrac{2\cdot 4}2 = 10-4 \]\[= 6\] 因为这是用较大三角形的面积减去较小三角形的面积得到的。

所以答案是 B。

English

The area is a triangle of area \[\dfrac {4\cdot 5}2 - \dfrac{2\cdot 4}2 = 10-4 \]\[= 6\] since we subtract the area of a smaller triangle from a larger triangle.

Thus, the answer is B .

通分得 \[\dfrac{2021\cdot 2021}{2020\cdot 2021} -\dfrac{2020\cdot 2020}{2020\cdot 2021}\] \[= \dfrac{2021^2-2020^2}{2020\cdot 2021}\text{。}\]

利用平方差公式，可进一步化为 \[\dfrac{(2021-2020)(2021+2020)}{2020\cdot 2021} \] \[=\dfrac{4041}{2020\cdot 2021}\text{。}\]

因为 \(4041\) 与 \(2020\)、\(2021\) 互质，所以分子为 \(4041\)。

所以答案是 E。

English

We can rewrite this as \[\dfrac{2021\cdot 2021}{2020\cdot 2021} -\dfrac{2020\cdot 2020}{2020\cdot 2021}\] \[= \dfrac{2021^2-2020^2}{2020\cdot 2021}.\]

This can be simplified to \[\dfrac{(2021-2020)(2021+2020)}{2020\cdot 2021} \] \[=\dfrac{4041}{2020\cdot 2021}.\]

Since \(4041\) is coprime with both \(2020\) and \(2021,\) we know \(4041\) is the numerator.

Thus, the answer is E .

设中午两城气温差为 \(N=m-s\)。到 \(4{:}00\) 时，新温差为 \((m-5)-(s+3)=N-8\)。

此时两城气温相差 \(2\) 度，所以 \(|N-8|=2\)。因此 \(N=6\) 或 \(N=10\)，所有可能值的乘积为 \(6\cdot10=60\)。

所以正确答案是 C。

English

Let \(N=m-s\) be the noon temperature difference. At \(4{:}00,\) the new difference is \((m-5)-(s+3)=N-8.\)

The temperatures then differ by \(2\) degrees, so \(|N-8|=2.\) Hence \(N=6\) or \(N=10,\) and the product of all possible values is \(6\cdot10=60.\)

Thus, the answer is C .

因为 \(8=2^3\)，所以 \[n=8^{2022}=2^{6066}=4^{3033}\text{。}\]

因此 \[\frac n4=\frac{4^{3033}}4=4^{3032}\text{。}\]

所以答案是 E。

English

Since \(8=2^3,\) we have \[n=8^{2022}=2^{6066}=4^{3033}.\]

Therefore \[\frac n4=\frac{4^{3033}}4=4^{3032}.\]

Thus, the answer is E .

开始之前先注意：如果整数 \(z\) 的质因数分解可以写成 \[z=p_1^{e_1} p_2^{e_2} \cdots\text{，}\] 那么它共有 \((e_1+1)(e_2+1) \cdots\) 个不同的正因数。

如果所求的数有 \(2021\) 个因数，由上面的道理可知 \[2021=(e_1+1)(e_2+1) \cdots\text{，}\] 而 \(2021 = 43\cdot 47\)，所以这个数必定形如 \(p_1^{46}p_2^{42}\) 或 \(p^{2020}\)。

在这两种形式中，能得到的最小的数是在第一种形式中取 \(p_1 = 2,p_2=3\)，即 \[2^{46}3^{42} = 16\cdot 6^{42}\text{。}\]

因此 \[m=16\]\[k=42\text{，}\] 所以 \[m+k = 42+16 = 58\text{。}\]

所以答案是 B。

English

Before starting, note that if we can represent the prime factorization of an integer \(z\) as \[z=p_1^{e_1} p_2^{e_2} \cdots,\] then there are \((e_1+1)(e_2+1) \cdots\) distinct positive factors.

If the number in question has \(2021\) factors, by the previous logic, \[2021=(e_1+1)(e_2+1) \cdots,\] and as the prime factorization of \(2021 = 43\cdot 47,\) then our number must be \(p_1^{46}p_2^{42} \) or \(p^{2020}.\)

The smallest number we can make in either of these is making \(p_1 = 2,p_2=3\) in the first configuration, yielding \[2^{46}3^{42} = 16\cdot 6^{42} .\]

Therefore, \[m=16\]\[k=42,\] so \[m+k = 42+16 = 58.\]

Thus, the answer is B .

分母为 \(b\) 的特殊分数等于 \(\frac{15-b}{b}=\frac{15}{b}-1\)，其中 \(1\le b\le14\)。所求整数可写成 \(\frac{15}{x}+\frac{15}{y}-2\)。

设 \(x\le y\)，逐一检查十四个可能的分母，得到下列能产生整数的数对：\[ \begin{gathered} (1,1),(1,3),(1,5),(2,2),(2,6),\\ (2,10),(3,3),(3,5),(4,12),\\ (5,5),(6,6),(6,10),(10,10) \end{gathered}\text{。}\] 它们给出的不同的和为 \(1,2,3,4,6,7,8,13,16,18,28\)。

这样的整数共有 \(11\) 个。

所以答案是 C。

English

A special fraction with denominator \(b\) equals \(\frac{15-b}{b}=\frac{15}{b}-1,\) where \(1\le b\le14.\) We need integer values of \(\frac{15}{x}+\frac{15}{y}-2.\)

Taking \(x\le y,\) a check of the fourteen possible denominators gives the following pairs that produce integers: \[ \begin{gathered} (1,1),(1,3),(1,5),(2,2),(2,6),\\ (2,10),(3,3),(3,5),(4,12),\\ (5,5),(6,6),(6,10),(10,10). \end{gathered} \] Their distinct sums are \(1,2,3,4,6,7,8,13,16,18,28.\)

There are \(11\) such integers.

Thus, the answer is C .

有 \[\begin{aligned}16,383 &= 16384-1 \\&= 2^{14}-1 \\&= (2^7-1)(2^7+1) \\&= 127\cdot 129\end{aligned}\text{。}\] 因为 \(129=3\cdot 43\)，所以 \[16383 = 3\cdot 43\cdot 127\text{。}\] 因此 \(127\) 是最大质因数，其各位数字之和为 \(10\)。

所以答案是 C。

English

We know \[\begin{aligned}16,383 &= 16384-1 \\&= 2^{14}-1 \\&= (2^7-1)(2^7+1) \\&= 127\cdot 129.\end{aligned}\] Since \(129=3\cdot 43,\) we get \[16383 = 3\cdot 43\cdot 127.\] Therefore, \(127\) is the largest prime factor, and the sum of its digits is \(10.\)

Thus, the answer is C .

设蓝骑士中有魔法的比例为 \(x\)，则红骑士中有魔法的比例为 \(2x\)。

总魔法比例为按红蓝人数加权的平均：\[\frac27(2x)+\frac57(x)=\frac16\text{。}\]

因此由 \(\frac97x=\frac16\) 得 \(x=\frac7{54}\)。红骑士中有魔法的比例为 \(2x=\frac7{27}\)。

所以正确答案是 C。

English

Let \(x\) be the fraction of blue knights who are magical. Then the fraction of red knights who are magical is \(2x.\)

The total magical fraction is a weighted average over the red and blue groups: \[\frac27(2x)+\frac57(x)=\frac16.\]

Thus \(\frac97x=\frac16,\) so \(x=\frac7{54}.\) The red magical fraction is \(2x=\frac7{27}.\)

Thus, the answer is C .

如果 Alice 抽到 \(1\) 或 \(40\)，她会知道谁的数字更大，因此她的数字不是 \(1\) 或 \(40\)。

Bob 由此知道 Alice 的数字不是端点。此时只有当 Bob 的数字属于 \(1,2,39\) 或 \(40\) 时，他才能判断大小。又因为他的数字是质数，所以只能是 \(2\)。

现在 \(100\cdot2\) 加上 Alice 的数字是介于 \(201\) 和 \(240\) 之间的平方数。唯一可能是 \(225\)，所以 Alice 的数字为 \(25\)，两数之和为 \(2+25=27\)。

所以正确答案是 A。

English

If Alice had drawn \(1\) or \(40,\) she would know who had the larger number. Her first statement tells Bob that Alice has neither \(1\) nor \(40.\)

Bob can then know who has the larger number only if his number is \(1,2,39,\) or \(40.\) Since Bob says his number is prime, his number must be \(2.\)

Now \(100\cdot2\) plus Alice’s number is a square between \(201\) and \(240.\) The only square in that interval is \(225,\) so Alice’s number is \(25.\) The sum is \(2+25=27.\)

Thus, the answer is A .

原圆由正六边形和 \(6\) 个相同的圆弓形组成。将每条小弧关于对应边反射后，这 \(6\) 个圆弓形都移到六边形内部。

因此原圆面积与反射弧围成区域面积的平均数等于正六边形面积。正六边形面积为 \(6\cdot\frac{\sqrt3}{4}=\frac{3\sqrt3}{2}\)，原圆半径为 \(1\)，所以面积为 \(\pi\)。

设所求面积为 \(A\)，则 \(\frac{A+\pi}{2}=\frac{3\sqrt3}{2}\)，所以 \(A=3\sqrt3-\pi\)。

所以正确答案是 B。

English

The original circle is made from the regular hexagon plus \(6\) equal circular segments. Reflecting each minor arc over its side puts those same \(6\) segments inside the hexagon instead.

Therefore the average of the circle’s area and the reflected-arc region’s area is the area of the regular hexagon. The hexagon has area \(6\cdot\frac{\sqrt3}{4}=\frac{3\sqrt3}{2},\) and the circle has radius \(1,\) so its area is \(\pi.\)

If the desired area is \(A,\) then \(\frac{A+\pi}{2}=\frac{3\sqrt3}{2},\) so \(A=3\sqrt3-\pi.\)

Thus, the answer is B .

设 \(E=x(x-y)+y(y-z)\) \(+z(z-x)\)。展开得 \[ \begin{aligned} 2E={}&(x-y)^2+(y-z)^2\\ &+(z-x)^2 \end{aligned}\text{。}\]

要使值为 \(1\)，三个非负平方项的和必须为 \(2\)。因为 \(x,y,z\) 是整数，这三个平方项只能是 \(1,1,0\)。

因此两个变量必须相等，第三个变量与它们相差 \(1\)。条件 \(x=z\) 且 \(y-1=x\) 正好保证这一点。

所以正确答案是 D。

English

Let \(E=x(x-y)+y(y-z)\) \(+z(z-x).\) Expanding gives \[ \begin{aligned} 2E={}&(x-y)^2+(y-z)^2\\ &+(z-x)^2. \end{aligned} \]

For the value to be \(1,\) the three nonnegative square terms must sum to \(2.\) Since \(x,y,z\) are integers, this means the squared differences are \(1,1,0.\)

Thus two of the variables must be equal, and the third must differ from them by \(1.\) The condition \(x=z\) and \(y-1=x\) guarantees exactly that.

Thus, the answer is D .

设等腰三角形的高为 \(H\)，底为 \(B\)。由相似性，三角形内水平方向的宽度与到顶点的距离成正比。

大正方形的上边长为 \(3\)，距顶点 \(H-3\)。小正方形的上边长为 \(2\)，距顶点 \(H-5\)。因此 \[\frac{3}{H-3}=\frac{2}{H-5}\text{。}\]

解得 \(H=9\)。又 \(\frac{B}{H}=\frac{3}{H-3}\)，所以 \(B=\frac{9}{2}\)。面积为 \(\frac12\cdot\frac92\cdot9=\frac{81}{4}=20\frac14\)。

所以答案是 B。

English

Let the isosceles triangle have height \(H\) and base \(B.\) By similarity, horizontal widths in the triangle are proportional to distance from the top vertex.

The top side of the larger square has width \(3\) and is \(H-3\) units from the top. The top side of the smaller square has width \(2\) and is \(H-5\) units from the top. Hence \[\frac{3}{H-3}=\frac{2}{H-5}.\]

Solving gives \(H=9.\) Also \(\frac{B}{H}=\frac{3}{H-3},\) so \(B=\frac{9}{2}.\) The area is \(\frac12\cdot\frac92\cdot9=\frac{81}{4}=20\frac14.\)

Thus, the answer is B .

计算补事件，也就是乘积不能被 \(4\) 整除。这发生在乘积中没有因子 \(2\)，或恰好有一个因子 \(2\) 时。

六个骰子全为奇数的概率为 \((\frac12)^6=\frac1{64}\)。恰好有一个因子 \(2\)，意味着唯一一个骰子掷出 \(2\) 或 \(6\)，其余五个为奇数，概率为 \[6\cdot\frac26\cdot\left(\frac12\right)^5=\frac4{64}\text{。}\]

补事件的概率为 \(\frac5{64}\)，所以所求概率为 \(1-\frac5{64}=\frac{59}{64}\)。

所以正确答案是 C。

English

Count the complement, where the product is not divisible by \(4.\) This happens if the product has no factor of \(2,\) or exactly one factor of \(2.\)

All dice odd has probability \((\frac12)^6=\frac1{64}.\) Exactly one factor of \(2\) means exactly one die is \(2\) or \(6,\) and the other five dice are odd. This has probability \[6\cdot\frac26\cdot\left(\frac12\right)^5=\frac4{64}.\]

The complement has probability \(\frac5{64},\) so the desired probability is \(1-\frac5{64}=\frac{59}{64}.\)

Thus, the answer is C .

因为 \(BR=6\)、\(PR=7\)，所以 \(BP=13\)。由正方形中的对应直角三角形可得 \(\triangle PAB\cong\triangle QBC\)，于是 \(CQ=BP=13\)。

因为 \(BP\perp CQ\)，点 \(R\) 是从 \(B\) 向直角三角形 \(BQC\) 的斜边 \(CQ\) 所作高的垂足。因此 \[QR\cdot RC=BR^2=36\] 且 \[QR+RC=CQ=13\text{。}\]

于是 \(QR\) 与 \(RC\) 是 \(4\) 和 \(9\)。因为 \(P\) 在 \(\overline{AD}\) 上，所以 \(AP\le AB\)。由全等可得 \(BQ=AP\)，而直角三角形的射影定理给出 \(BQ^2=CQ\cdot QR\) 和 \(BC^2=CQ\cdot RC\)。因此 \(QR\le RC\)，从而 \(QR=4\)，\(RC=9\)。于是 \[ \begin{aligned} BC^2&=BR^2+RC^2\\ &=6^2+9^2\\ &=117 \end{aligned}\text{。}\]

正方形面积为 \(117\)。

所以正确答案是 D。

English

Since \(BR=6\) and \(PR=7,\) we have \(BP=13.\) The right-angle and square-angle chasing gives \(\triangle PAB\cong\triangle QBC,\) so \(CQ=BP=13.\)

Because \(BP\perp CQ,\) point \(R\) is the foot of the altitude from \(B\) to the hypotenuse \(CQ\) of right triangle \(BQC.\) Thus \[QR\cdot RC=BR^2=36\] and \[QR+RC=CQ=13.\]

So \(QR\) and \(RC\) are \(4\) and \(9.\) Because \(P\) lies on \(\overline{AD},\) we have \(AP\le AB.\) Congruence gives \(BQ=AP,\) while the right-triangle projection formulas give \(BQ^2=CQ\cdot QR\) and \(BC^2=CQ\cdot RC.\) Hence \(QR\le RC,\) so \(QR=4\) and \(RC=9.\) Therefore, \[ \begin{aligned} BC^2&=BR^2+RC^2\\ &=6^2+9^2\\ &=117. \end{aligned} \]

The area of the square is \(117.\)

Thus, the answer is D .

Chris 选定一对相邻球后，Silva 有 \(5\) 对相邻球可等可能选择。

如果 Silva 选择同一对，则所有 \(5\) 个球都回到原位，概率为 \(\frac15\)。

如果 Silva 选择与 Chris 的那对恰好共享一个球的相邻对，则有 \(2\) 个球在原位。这样的对有 \(2\) 个，所以概率为 \(\frac25\)。

如果 Silva 选择与 Chris 的那对不相交的相邻对，则有 \(1\) 个球在原位。概率同样为 \(\frac25\)。

期望为 \[ \begin{aligned} 5\cdot\frac15+2\cdot\frac25+1\cdot\frac25&=\frac{11}{5}\\ &=2.2 \end{aligned}\text{。}\]

所以答案是 D。

English

After Chris chooses an adjacent pair, Silva has \(5\) equally likely adjacent pairs to choose.

If Silva chooses the same pair, all \(5\) balls return to their original positions. This has probability \(\frac15.\)

If Silva chooses a pair sharing exactly one ball with Chris’s pair, then \(2\) balls are in their original positions. There are \(2\) such pairs, so this has probability \(\frac25.\)

If Silva chooses a disjoint adjacent pair, then \(1\) ball is in its original position. This also has probability \(\frac25.\)

The expected number is \[ \begin{aligned} 5\cdot\frac15+2\cdot\frac25+1\cdot\frac25&=\frac{11}{5}\\ &=2.2. \end{aligned} \]

Thus, the answer is D .

两次关于过原点直线的反射，等价于旋转两倍的有向夹角，从第一条反射线转到第二条反射线。

点 \((-1,4)\) 被送到 \((4,1)\)，这是顺时针 \(90^\circ\) 的旋转。因此直线 \(m\) 是将第一条反射线顺时针旋转 \(45^\circ\) 后得到的，而第一条反射线就是 \(\ell\)。

直线 \(\ell\) 的斜率为 \(5\)。若 \(\theta_m=\theta_\ell-45^\circ\)，则 \[\tan\theta_m=\frac{5-1}{1+5}=\frac23\text{。}\]

所以直线 \(m\) 为 \(y=\frac23x\)，即 \(2x-3y=0\)。

所以答案是 D。

English

Two reflections across lines through the origin are equivalent to a rotation by twice the angle from the first reflecting line to the second.

The point \((-1,4)\) is sent to \((4,1),\) which is a \(90^\circ\) clockwise rotation. Therefore line \(m\) is \(45^\circ\) clockwise from line \(\ell.\)

The slope of \(\ell\) is \(5.\) If \(\theta_m=\theta_\ell-45^\circ,\) then \[\tan\theta_m=\frac{5-1}{1+5}=\frac23.\]

Thus line \(m\) is \(y=\frac23x,\) or \(2x-3y=0.\)

Thus, the answer is D .

边界可分成 \(24\) 个全等三角形，角分别为 \(15^\circ\)、\(45^\circ\)、\(120^\circ\)。

每个三角形的高为 \(3\)，即正方形边长的一半。相邻直角三角形有一个 \(30^\circ\) 角，所以从长度为 \(3\) 的底边上截去的部分为 \(3\tan30^\circ=\sqrt3\)。

对其中一个三角形作高，其底为 \(3-\sqrt3\)，高为 \(3\)，面积为 \(\frac{3(3-\sqrt3)}{2}=\frac{9-3\sqrt3}{2}\)。

所以总面积为 \[24\cdot\frac{9-3\sqrt3}{2}=108-36\sqrt3\text{。}\] 因此 \(a+b+c=108+36+3=147\)。

所以答案是 E。

English

The boundary can be split into \(24\) congruent triangles. Each has angles \(15^\circ,\) \(45^\circ,\) and \(120^\circ.\)

For one such triangle, draw the altitude from the center-side direction. The altitude is \(3,\) half the side length of a square. The adjacent right triangle has a \(30^\circ\) angle, so the part cut off from a length \(3\) base is \(3\tan30^\circ=\sqrt3.\)

Thus each small triangle has base \(3-\sqrt3\) and height \(3,\) giving area \(\frac{3(3-\sqrt3)}{2}=\frac{9-3\sqrt3}{2}.\)

The total area is \[24\cdot\frac{9-3\sqrt3}{2}=108-36\sqrt3.\] Hence \(a+b+c=108+36+3=147.\)

Thus, the answer is E .

数 \(N\) 满足 \[7\cdot10^{312}\lt N\lt 8\cdot10^{312}\text{。}\] 乘除 \(10\) 的幂只会移动小数点，因此每次只需关注取出合适 \(10\) 的幂后剩余的首因子。

对 \(r=2\)，\(\sqrt N\) 的首因子在 \(\sqrt7\) 和 \(\sqrt8\) 之间，所以 \(f(2)=2\)。

对 \(r=3\)，首因子在 \(\sqrt[3]{7}\) 和 \(\sqrt[3]{8}\) 之间，所以 \(f(3)=1\)。对 \(r=4\)，首因子在 \(\sqrt[4]{7}\) 和 \(\sqrt[4]{8}\) 之间，所以 \(f(4)=1\)。

对 \(r=5\)，因为 \(312=5\cdot62+2\)，首因子在 \(\sqrt[5]{700}\) 和 \(\sqrt[5]{800}\) 之间。由于 \(3^5\lt700\lt800\lt4^5\)，得 \(f(5)=3\)。

对 \(r=6\)，首因子在 \(\sqrt[6]{7}\) 和 \(\sqrt[6]{8}\) 之间，所以 \(f(6)=1\)。总和为 \(2+1+1+3+1=8\)。

所以答案是 A。

English

The number \(N\) satisfies \[7\cdot10^{312}\lt N\lt 8\cdot10^{312}.\] Multiplying or dividing by a power of \(10\) only shifts the decimal point, so we only need the leading factor left after taking out the largest convenient power of \(10.\)

For \(r=2,\) \(\sqrt N\) has leading factor between \(\sqrt7\) and \(\sqrt8,\) so \(f(2)=2.\)

For \(r=3,\) the leading factor is between \(\sqrt[3]{7}\) and \(\sqrt[3]{8},\) so \(f(3)=1.\) For \(r=4,\) the leading factor is between \(\sqrt[4]{7}\) and \(\sqrt[4]{8},\) so \(f(4)=1.\)

For \(r=5,\) since \(312=5\cdot62+2,\) the leading factor is between \(\sqrt[5]{700}\) and \(\sqrt[5]{800}.\) Since \(3^5\lt700\lt800\lt4^5,\) \(f(5)=3.\)

For \(r=6,\) the leading factor is between \(\sqrt[6]{7}\) and \(\sqrt[6]{8},\) so \(f(6)=1.\) The sum is \(2+1+1+3+1=8.\)

Thus, the answer is A .

由对称性，\(P(\text{Hugo 获胜})=\frac14\)。因此所求条件概率等于 Hugo 第一次掷出 \(5\) 并最终获胜的概率的四倍。

若 Hugo 第一次掷出 \(5\)，其他人不能掷出 \(6\)。按其他三人中也掷出 \(5\) 的人数分类；若有 \(t\) 人与 Hugo 并列，则 Hugo 后续获胜概率为 \(\frac1{t+1}\)。

按并列人数分类，得到 \[ \begin{gathered} P(\text{Hugo 掷出 }5\text{ 且获胜})\\ {}=\frac{1}{6^4}\sum_{t=0}^3 \binom3t\frac{4^{3-t}}{t+1}\\ {}=\frac{64+24+4+\frac14}{6^4}\\ {}=\frac{369}{4\cdot1296} \end{gathered}\text{。}\]

乘以 \(4\)，得到 \(\frac{369}{1296}=\frac{41}{144}\)。

所以正确答案是 C。

English

By symmetry, \(P(\text{Hugo wins})=\frac14.\) Therefore, the requested conditional probability is four times the probability that Hugo first rolls \(5\) and eventually wins.

If Hugo rolls \(5,\) then no other player can roll \(6.\) Case on how many of the other three players also roll \(5.\) If \(t\) other players tie Hugo, then Hugo wins the eventual tiebreaker with probability \(\frac1{t+1}.\)

Thus \[ \begin{gathered} P(\text{Hugo rolls }5\text{ and wins})\\ {}=\frac{1}{6^4}\sum_{t=0}^3 \binom3t\frac{4^{3-t}}{t+1}\\ {}=\frac{64+24+4+\frac14}{6^4}\\ {}=\frac{369}{4\cdot1296}. \end{gathered} \]

Multiplying by \(4\) gives \(\frac{369}{1296}=\frac{41}{144}.\)

Thus, the answer is C .

对内接同圆且无共点顶点的正 \(k\) 边形和正 \(n\) 边形，其中 \(k\lt n\)，它们的边界相交于 \(2k\) 个点。

较小多边形的每条边会被较大多边形边界穿过两次。\(5\) 边形与各较大多边形分别贡献 \(2\cdot5\) 个交点，对应的边数为 \(6\)、\(7\)、\(8\)；\(6\) 边形与各较大多边形分别贡献 \(2\cdot6\) 个交点，对应的边数为 \(7\)、\(8\)；\(7\) 边形贡献 \(2\cdot7\) 个交点，对应边数为 \(8\)。

对所有多边形对求和。\[3(10)+2(12)+14=68\text{。}\]

所以正确答案是 E。

English

For a regular \(k\)-gon and a regular \(n\)-gon inscribed in the same circle with \(k\lt n\) and no shared vertices, their boundaries intersect in \(2k\) points. Each side of the smaller polygon is crossed twice by the boundary of the larger polygon.

Therefore, sum over all pairs of polygons. The \(5\)-gon contributes \(2\cdot5\) intersections with each of the \(6\)-, \(7\)-, and \(8\)-gons. The \(6\)-gon contributes \(2\cdot6\) intersections with each of the \(7\)- and \(8\)-gons. The \(7\)-gon contributes \(2\cdot7\) with the \(8\)-gon.

The total is \[3(10)+2(12)+14=68.\]

Thus, the answer is E .

从 \(S_{n-1}\) 到 \(S_n\)，新加入的项为 \(jn\)，其中 \(1\le j\lt n\)。\[ \begin{gathered} n(1+2+\cdots+(n-1))\\ =\frac{n^2(n-1)}2 \end{gathered}\text{。}\]

模 \(3\) 看，这个增量为 \(0\) 的情形是 \(n\equiv0\) 或 \(1\pmod3\)，而增量为 \(2\) 的情形是 \(n\equiv2\pmod3\)。

因为 \(S_2=2\)，在第 \(3\) 次遇到与 \(2\pmod3\) 同余的数时，也就是 \(n=8\)，\(S_n\) 开始能被 \(3\) 整除。之后 \(n=8,9,10\) 都满足，并且这个模式每隔 \(9\) 个 \(n\) 重复一次。

最小的十个值为 \(8,9,10,17,18,19,26,27,28,35\)。它们的和为 \(197\)。

所以答案是 B。

English

When passing from \(S_{n-1}\) to \(S_n,\) the new terms are \(jn\) for \(1\le j\lt n.\) Their sum is \[ \begin{gathered} n(1+2+\cdots+(n-1))\\ =\frac{n^2(n-1)}2. \end{gathered} \]

Modulo \(3,\) this increment is \(0\) when \(n\equiv0\) or \(1\pmod3,\) and is \(2\) when \(n\equiv2\pmod3.\)

Since \(S_2=2,\) the sequence becomes divisible by \(3\) after the third occurrence of a number congruent to \(2\pmod3,\) namely at \(n=8.\) Then \(S_n\) stays divisible by \(3\) for \(n=8,9,10,\) and the same pattern repeats every \(9\) in \(n.\)

The ten least values are \(8,9,10,17,18,19,26,27,28,35.\) Their sum is \(197.\)

Thus, the answer is B .

计算补事件：将 \(10\) 条边组成的 \(K_5\) 用两种颜色染色，且没有单色三角形。

在任意顶点，如果有 \(3\) 条关联边同色，那么这三条边另一端之间的边都必须是另一种颜色；但这又会形成单色三角形。因此每个顶点恰有 \(2\) 条红边和 \(2\) 条蓝边。

所以红边构成一个 \(2\) 正则图，顶点数为 \(5\)，只能是一个 \(5\) 环。带标号的 \(5\) 环共有 \(\frac{(5-1)!}{2}=12\) 个。

总染色数为 \(2^{10}=1024\)，所以所求概率为 \[1-\frac{12}{1024}=\frac{253}{256}\text{。}\]

所以正确答案是 D。

English

Count the complement: colorings of the \(10\) edges of \(K_5\) with no monochromatic triangle.

At any vertex, if \(3\) incident edges had the same color, then the edges among their other endpoints would all have to be the other color, making a monochromatic triangle. Thus each vertex has exactly \(2\) red and \(2\) blue incident edges.

So the red edges form a \(2\)-regular graph on \(5\) vertices, which must be a \(5\)-cycle. The number of labeled \(5\)-cycles is \(\frac{(5-1)!}{2}=12.\)

There are \(2^{10}=1024\) total colorings, so the desired probability is \[1-\frac{12}{1024}=\frac{253}{256}.\]

Thus, the answer is D .

问题等价于选择哪 \(4\) 个位置涂蓝，位置总数为 \(8\)，并对立方体的 \(24\) 个旋转使用 Burnside 引理。

恒等旋转固定 \(\binom84=70\) 种染色。\(6\) 个绕面中心轴的四分之一转各固定 \(2\) 种；\(3\) 个绕面中心轴的半转各固定 \(6\) 种；\(8\) 个绕相对顶点轴的旋转各固定 \(4\) 种；\(6\) 个绕相对棱中点轴的半转各固定 \(6\) 种。

因此不等价构造数为 \[ \begin{gathered} \frac{70+6\cdot2+3\cdot6+8\cdot4+6\cdot6}{24}\\ =7 \end{gathered}\text{。}\]

所以正确答案是 A。

English

This is the number of ways to choose which \(4\) of the \(8\) cube vertices are blue, up to cube rotation. Use Burnside’s lemma on the \(24\) rotations of the cube.

The identity fixes \(\binom84=70\) colorings. The \(6\) quarter-turn face rotations each fix \(2.\) The \(3\) half-turn face rotations each fix \(6.\) The \(8\) rotations about opposite vertices each fix \(4.\) The \(6\) half-turn rotations about opposite edges each fix \(6.\)

Thus the number of inequivalent constructions is \[ \begin{gathered} \frac{70+6\cdot2+3\cdot6+8\cdot4+6\cdot6}{24}\\ =7. \end{gathered} \]

Thus, the answer is A .

如图使用相似三角形。大正方形的两条边长分别为 \(4x+2y\) 与 \(3y+x\)，所以 \(3y+x=4x+2y\)，得到 \(y=3x\)，大正方形边长为 \(10x\)。

在图形上部，设标出的水平线段为 \(m\)。矩形 \(R\) 两侧形成的两个直角三角形具有平行的对应边和相等的斜边，因此全等，由此得到下图标出的长度。

由相似三角形，\[\frac{m}{3x}=\frac{4x-\frac m3}{6x-m}\text{。}\] 因而 \(6xm-m^2=12x^2-xm\)，所以 \[ \begin{gathered} m^2-7xm+12x^2=0\\ =(m-3x)(m-4x)\text{。} \end{gathered} \]

若 \(m=3x\)，矩形 \(R\) 的两边均为 \(3\sqrt2x\)，面积为 \(18x^2\)；若 \(m=4x\)，两边为 \(5x\) 与 \(\frac{10}{3}x\)，面积为 \(\frac{50}{3}x^2\)。

两个可能面积之和为 \(\frac{104}{3}x^2\)。由 \(1\times3\) 矩形可知 \(x^2+(3x)^2=1\)，所以 \(x^2=\frac1{10}\)。因此面积和为 \[\frac{104}{3}\cdot\frac1{10}=\frac{52}{15}\text{。}\]

于是 \(m+n=52+15=67\)，正确答案是 E。

English

Use similar triangles as shown in the diagram. The left side of the large square has length \(4x+2y,\) and the bottom side has length \(3y+x.\) Since these are equal, \(3y+x=4x+2y,\) so \(y=3x.\) The side length of the large square is therefore \(10x.\)

In the upper part of the figure, let the marked horizontal segment be \(m.\) The two right triangles formed by the sides of rectangle \(R\) have parallel corresponding sides and equal hypotenuses, so they are congruent. This gives the lengths shown below.

Similar triangles give \[\frac{m}{3x}=\frac{4x-\frac m3}{6x-m}.\] Hence \(6xm-m^2=12x^2-xm,\) so \[ \begin{gathered} m^2-7xm+12x^2=0\\ =(m-3x)(m-4x). \end{gathered} \]

If \(m=3x,\) rectangle \(R\) has side length \(3\sqrt2x\) in both directions, so its area is \(18x^2.\) If \(m=4x,\) its side lengths are \(5x\) and \(\frac{10}{3}x,\) so its area is \(\frac{50}{3}x^2.\)

The two possible areas sum to \(\frac{104}{3}x^2.\) Since the \(1\times3\) rectangle gives \(x^2+(3x)^2=1,\) we have \(x^2=\frac1{10}.\) The sum of the possible areas is \[\frac{104}{3}\cdot\frac1{10}=\frac{52}{15}.\]

Thus \(m+n=52+15=67,\) and the answer is E .