2021C AMC 10 真题（P版）

1 答案 C

可以如下化简这个表达式：\(2112-2021=91=7\cdot13\)，\[ \begin{aligned} \frac{(2112-2021)^2}{169} &=\frac{(7\cdot13)^2}{13^2}\\ &=7^2=49 \end{aligned}\text{。}\]

所以正确答案是 C。

English

Since \(2112-2021=91=7\cdot13,\) \[ \begin{aligned} \frac{(2112-2021)^2}{169} &=\frac{(7\cdot13)^2}{13^2}\\ &=7^2=49. \end{aligned} \]

Thus, C is the correct answer.

2 答案 E

若把 \(4\) 英寸的一边缩短 \(1\) 英寸，就得到 \(3\times 6\) 的卡片，面积为 \(18\) 平方英寸。

因此题中原先缩短的是 \(4\) 英寸的一边。若改为缩短另一边，则得到 \(4\times 5\) 的卡片，面积为 \(20\)。

所以正确答案是 E。

English

If she shortens the \(4\)-inch side by \(1,\) she has a \(3\times 6\) card, whose area is \(18\) square inches.

Therefore the given shortening was on the \(4\)-inch side. Shortening the other side instead gives a \(4\times 5\) card, with area \(20\) square inches.

Thus, E is the correct answer.

3 答案 D

立方体体积为 \(6^3=216\)。一个黏土球的体积为 \(\frac{4}{3}\pi\cdot 2^3=\frac{32\pi}{3}\)。

因为黏土可以重新塑形但不能压缩，最多的球数为 \[\left\lfloor \frac{216}{\frac{32\pi}{3}}\right\rfloor=\left\lfloor\frac{81}{4\pi}\right\rfloor\text{。}\]

由于 \(12\lt 4\pi\lt 13\)，可知 \(6\lt \frac{81}{4\pi}\lt \frac{81}{12}\lt 7\)，因此其整数部分为 \(6\)。

所以正确答案是 D。

English

The cube has volume \(6^3=216.\) One ball of clay has volume \(\frac{4}{3}\pi\cdot 2^3=\frac{32\pi}{3}.\)

Because the clay may be reshaped but not compressed, the maximum number of balls is \[\left\lfloor \frac{216}{\frac{32\pi}{3}}\right\rfloor=\left\lfloor\frac{81}{4\pi}\right\rfloor.\]

Since \(12\lt 4\pi\lt 13,\) we have \(6\lt \frac{81}{4\pi}\lt \frac{81}{12}\lt 7.\) Therefore the floor is \(6.\)

Thus, D is the correct answer.

4 答案 B

路线 A 所需时间为 \[ \dfrac{6}{30} \cdot 60 = 12 \] 分钟。

路线 B 所需时间为 \[ \left(\dfrac{5 - 0.5}{40} + \dfrac{0.5}{20}\right) \cdot 60 = 8.25 \] 分钟。

因此路线 B 快 \(12 - 8.25 = 3.75\) 分钟。

所以正确答案是 B。

English

Mr. Lopez would take \[ \dfrac{6}{30} \cdot 60 = 12 \] minutes to travel on Route A.

On Route B, he would take \[ \left(\dfrac{5 - 0.5}{40} + \dfrac{0.5}{20}\right) \cdot 60 = 8.25 \] minutes.

The difference in times along these routes is \(12 - 8.25 = 3.75\) minutes.

Thus, B is the correct answer.

5 答案 E

注意 \(A\) 不能是偶数，否则这个数能被 \(2\) 整除。

\(A\) 也不能是 \(5\)，否则这个数能被 \(5\) 整除。

若 \(A\) 等于 \(1\) 或 \(7\)，则这个数的数位和分别为 \(6\) 和 \(12\)。

这样这个数能被 \(3\) 整除，所以排除了这两个 \(A\) 值。

最后，若 \(A\) 等于 \(3\)，整个数为 \(202103\)。交错数位和之差为 \[ 2 + 2 - 1 - 3 = 0\text{，}\]所以这个数能被 \(11\) 整除。

除 \(9\) 外，每个选择都会使这个数成为合数。题目说明恰有一个数字符合条件，所以这个数字必须是 \(A=9\)。（事实上，试除所有不超过 \(\sqrt{202109}<450\) 的质数，可确认 \(202109\) 是质数。）

所以正确答案是 E。

English

Note that \(A\) cannot be even, as then the number would be divisible by \(2.\)

\(A\) also cannot be \(5,\) as that would make the number divisible by \(5.\)

If \(A\) equaled \(1\) or \(7,\) then the sum of the digits of the number would be \(6\) and \(12\) respectively.

This would make the number divisible by \(3,\) so that rules out \(A\) equaling either of these numbers.

Finally, if \(A\) equals \(3,\) then the whole number becomes \(202103.\) If we look at the difference of the sums of alternating digits, we get \[ 2 + 2 - 1 - 3 = 0, \] which means the number is divisible by \(11.\)

Every choice except \(9\) makes the number composite. Because the problem states that exactly one digit works, that digit must be \(A=9.\) (Indeed, trial division by the primes at most \(\sqrt{202109}<450\) confirms that \(202109\) is prime.)

Thus, E is the correct answer.

6 答案 B

第 \(1\) 根和第 \(41\) 根之间有 \(40\) 个间隔，所以相邻电线杆之间的距离为 \[ 5280 \div 40 = 132 \] 英尺。

这说明 Elmer 的一步长为 \[ 132 \div 44 = 3 \] 英尺。类似地，Oscar 的一步长为 \[ 132 \div 12 = 11 \] 英尺。因此 Oscar 的跃步比 Elmer 的步幅长 \(11 - 3 = 8\) 英尺。

所以正确答案是 B。

English

There are \(40\) gaps between the \(1\)st and \(41\)st pole, which means that the distance between consecutive poles is \[ 5280 \div 40 = 132 \] feet.

This means that each of Elmer’s strides is \[ 132 \div 44 = 3 \] feet. Similarly, each of Oscar’s leaps is \[ 132 \div 12 = 11 \] feet. This makes Oscar’s leap \(11 - 3 = 8\) feet longer.

Thus, B is the correct answer.

7 答案 D

因为 \(\angle ADC = 90^{\circ}\)，所以 \[ \angle FDE = 360^{\circ} - 90^{\circ} - 110^{\circ} \] \[= 160^{\circ}\text{。}\]又因为 \(\triangle FDE\) 是等腰三角形，所以 \[ \angle EFD = \dfrac{180^{\circ} - 160^{\circ}}{2} = 10^{\circ}\text{。}\]最后 \[ \angle AFE = 180^{\circ} - 10^{\circ} = 170^{\circ}\text{。}\]

所以正确答案是 D。

English

Since \(\angle ADC = 90^{\circ},\) we get that \[ \angle FDE = 360^{\circ} - 90^{\circ} - 110^{\circ} \]\[= 160^{\circ}. \] Also since \(\triangle FDE\) is isosceles, we get that \[ \angle EFD = \dfrac{180^{\circ} - 160^{\circ}}{2} = 10^{\circ}. \] Finally, we get that \[ \angle AFE = 180^{\circ} - 10^{\circ} = 170^{\circ}. \]

Thus, D is the correct answer.

8 答案 B

设 \(\underline{a} \ \underline{b}\) 是一个 \(2\) 位可爱数。

由定义， \[ 10a + b = a + b^2\text{。}\]化简得 \[ 9a = b(b - 1)\text{。}\]因为两个相邻整数不可能都被 \(3\) 整除，所以 \(b\) 和 \(b-1\) 中必有一个含有两个因数 \(3\)。

对一位数 \(b\) 而言，只有 \(b=0,1\) 或 \(9\) 可能。前两个选择都给出 \(a=0\)，不能组成两位数；若 \(b=9\)，则 \(a=8\)。检验可知 \(89\) 是可爱数，因此只有 \(1\) 个。

所以正确答案是 B。

English

Let \(\underline{a} \ \underline{b}\) be a \(2\)-digit cuddly number.

Then \[ 10a + b = a + b^2. \] Rearranging, we get \[ 9a = b(b - 1). \] Because consecutive integers cannot both be divisible by \(3,\) one of \(b\) and \(b-1\) must contain both factors of \(3.\)

For a digit \(b,\) this leaves \(b=0,1,\) or \(9.\) The first two choices give \(a=0,\) which does not make a two-digit number. If \(b=9,\) then \(a=8.\) Checking, we get that \(89\) is a cuddly number. This shows that there is only \(1\) two-digit cuddly number.

Thus, B is the correct answer.

9 答案 E

设掷出奇数的概率为 \(p\)，则掷出偶数的概率为 \(3p\)。\[ p + 3p = 1 \Rightarrow p = \dfrac{1}{4}\text{。}\]

和为偶数当且仅当两次结果奇偶性相同，其概率为 \[ \dfrac{1}{4}^2 + \dfrac{3}{4}^2 = \dfrac{10}{16} = \dfrac{5}{8}\text{。}\]

所以正确答案是 E。

English

Let \(p\) be the probability that an odd number is rolled. Then \(3p\) is the probability an even number is rolled. We know that \[ p + 3p = 1 \Rightarrow p = \dfrac{1}{4}. \]

The only way for the sum to be even is if both rolls have the same parity. This happens with a probability of \[ \dfrac{1}{4}^2 + \dfrac{3}{4}^2 = \dfrac{10}{16} = \dfrac{5}{8}. \]

Thus, E is the correct answer.

10 答案 B

期望值公式为 \[ \mathbb E[X]=\sum_x x\,\Pr(X=x)\text{。}\]

因此 \[ t = \dfrac{1}{5} (50 + 20 + 20 + 5 + 5) \] \[ = \dfrac{1}{5} \cdot 100 = 20\text{，}\]且 \[ s = 50 \cdot \dfrac{50}{100} + 20 \cdot \dfrac{20}{100} \] \[ + 20 \cdot \dfrac{20}{100} + 5 \cdot \dfrac{5}{100} + 5 \cdot \dfrac{5}{100} \] \[ \begin{aligned} &= 25 + 4 + 4 + 0.25 + 0.25 \\ &= 33.5 \end{aligned} \]

\(t - s = -13.5\)。

所以正确答案是 B。

English

Recall the expected-value formula \[ \mathbb E[X]=\sum_x x\,\Pr(X=x). \]

Therefore, \[ t = \dfrac{1}{5} (50 + 20 + 20 + 5 + 5) \] \[ = \dfrac{1}{5} \cdot 100 = 20 \] and \[ s = 50 \cdot \dfrac{50}{100} + 20 \cdot \dfrac{20}{100} \]\[ + 20 \cdot \dfrac{20}{100} + 5 \cdot \dfrac{5}{100} + 5 \cdot \dfrac{5}{100} \] \[ \begin{aligned} &= 25 + 4 + 4 + 0.25 + 0.25 \\ &= 33.5 \end{aligned} \]

\(t - s = -13.5.\)

Thus, B is the correct answer.

11 答案 A

设船长为 \(x\) 个步幅。Emily 走 \(210\) 步从船尾到船头时，船也向前移动了 \(210 - x\) 个步幅。

反方向走 \(42\) 步时，船移动了 \(x - 42\) 个步幅。船速与 Emily 速度恒定，所以 \[ \dfrac{210}{210 - x} = \dfrac{42}{x - 42}\text{。}\] 交叉相乘得 \[ 210x - 210 \cdot 42 = 210 \cdot 42 - 42x \] \[ 42 \cdot 6 x = 2 \cdot 42 \cdot 210 \] \[ x = 70\text{。}\]

所以正确答案是 A。

English

Let \(x\) be the length of the ship. Then in the time that Emily moves \(210\) steps, the ship moves \(210 - x\) steps.

In the time that Emily moves \(42\) steps, the ship moves \(x - 42\) steps. Since the ship and Emily move at a constant rate \[ \dfrac{210}{210 - x} = \dfrac{42}{x - 42}. \] Cross-multiplying yields \[ 210x - 210 \cdot 42 = 210 \cdot 42 - 42x \] \[ 42 \cdot 6 x = 2 \cdot 42 \cdot 210 \] \[ x = 70. \]

Thus, A is the correct answer.

12 答案 D

注意 \[ 9 \equiv -1 \pmod{5}\text{。}\]按进位制定义展开 \(N\)，得到 \[ N = 2 \cdot 9^{10} + 7 \cdot 9^9 + 6 \cdot 9^6 +\] \[ 5 \cdot 9 + 2 \] \[ \equiv 2 (-1)^{10} + 7 (-1)^9 + 6 (-1)^6 +\] \[ 5 (-1) + 2 \pmod{5} \] \[ \equiv 2 - 7 + 6 - 5 + 2 \pmod{5} \] \[ \equiv 3 \pmod{5}\text{。}\]

所以正确答案是 D。

English

Note that \[ 9 \equiv -1 \pmod{5}. \] Then if expand \(N\) using the definition of bases, we get \[ N = 2 \cdot 9^{10} + 7 \cdot 9^9 + 6 \cdot 9^6 +\]\[ 5 \cdot 9 + 2 \] \[ \equiv 2 (-1)^{10} + 7 (-1)^9 + 6 (-1)^6 +\]\[ 5 (-1) + 2 \pmod{5} \] \[ \equiv 2 - 7 + 6 - 5 + 2 \pmod{5} \] \[ \equiv 3 \pmod{5}. \]

Thus, D is the correct answer.

13 答案 D

要使每个球都与超过一半的其他球颜色不同，任意一个球必须看到至少 \(3\) 个相反颜色的球。

因此必须正好有三颗黑球和三颗白球。所有涂色共有 \(2^6 = 64\) 种，其中选择哪三颗球为白色有 \(\binom{6}{3} = 20\) 种。

所以所求概率为 \(\dfrac{20}{64} = \dfrac{5}{16}\)。

所以正确答案是 D。

English

Note that for this restriction to hold, there must be \(3\) balls of each color.

There are \(2^6 = 64\) ways to color the balls and \(\binom{6}{3} = 20\) to choose which balls are white.

The desired probability is therefore \(\dfrac{20}{64} = \dfrac{5}{16}.\)

Thus, D is the correct answer.

14 答案 D

令 \(t=|x|\)。第二个方程给出 \(t+|y|=3\) 或 \(5\)，而第一个方程给出 \(y=3-\frac{t^2}{3}\)。

若 \(y\ge0\)，则 \(0\le t\le3\)。对于 \(t+y=3\)，代入可得 \(t-\frac{t^2}{3}=0\)，所以 \(t=0\) 或 \(3\)。这些值给出 \(x=0,3,-3\)，共有三个点。对于 \(t+y=5\)，代入可得 \(t^2-3t+6=0\)，它没有实根。

若 \(y<0\)，则 \(t>3\)。方程 \(t-y=3\) 给出 \(t^2+3t-18=0\)，其唯一非负根是被排除的边界值 \(t=3\)。方程 \(t-y=5\) 给出 \(t^2+3t-24=0\)，它恰有一个正根 \(t=\frac{-3+\sqrt{105}}2>3\)。这个根按 \(x\) 的两种符号给出两个点。

有序对总数为 \(3+2=5\)。

所以正确答案是 D。

English

Put \(t=|x|.\) The second equation gives \(t+|y|=3\) or \(5,\) while the first gives \(y=3-\frac{t^2}{3}.\)

If \(y\ge0,\) then \(0\le t\le3.\) For \(t+y=3,\) substitution gives \(t-\frac{t^2}{3}=0,\) so \(t=0\) or \(3.\) These yield \(x=0,3,-3,\) for three points. For \(t+y=5,\) substitution gives \(t^2-3t+6=0,\) which has no real root.

If \(y<0,\) then \(t>3.\) The equation \(t-y=3\) gives \(t^2+3t-18=0,\) whose only nonnegative root is the excluded boundary value \(t=3.\) The equation \(t-y=5\) gives \(t^2+3t-24=0,\) with exactly one positive root \(t=\frac{-3+\sqrt{105}}2>3.\) It yields two points, one for each sign of \(x.\)

The total number of ordered pairs is \(3+2=5.\)

Thus, D is the correct answer.

15 答案 C

设 \(O_1\) 是与 \(AB\) 和 \(AC\) 相切的圆的圆心。则 \(\angle ABO_1=\angle ACO_1=90^\circ\)，所以 \(A,B,O_1,C\) 四点共圆。

因为 \(B\) 和 \(C\) 处的直角所对的弦是 \(AO_1\)，所以线段 \(AO_1\) 是这个圆的直径。设 \(O_2\) 为其圆心。同一个圆经过 \(A,B\) 和 \(C\)，所以它就是所求的外接圆。

在 \(\triangle ABO_1\) 中应用勾股定理，得到 \[ \begin{aligned} AO_1&=\sqrt{AB^2+BO_1^2}\\ &=\sqrt{54+50}=2\sqrt{26} \end{aligned}\text{。}\]因此外接圆半径为 \(\sqrt{26}\)，所求面积为 \(26\pi\)。

所以正确答案是 C。

English

Let \(O_1\) be the center of the circle tangent to \(AB\) and \(AC.\) Then \(\angle ABO_1=\angle ACO_1=90^\circ,\) so \(A,B,O_1,C\) are concyclic.

Because the right angles at \(B\) and \(C\) subtend \(AO_1,\) the segment \(AO_1\) is a diameter of this circle. Let \(O_2\) be its center. The same circle passes through \(A,B,\) and \(C,\) so it is the desired circumcircle.

By the Pythagorean Theorem in \(\triangle ABO_1,\) \[ \begin{aligned} AO_1&=\sqrt{AB^2+BO_1^2}\\ &=\sqrt{54+50}=2\sqrt{26}. \end{aligned} \] Therefore, the circumradius is \(\sqrt{26},\) and the requested area is \(26\pi.\)

Thus, C is the correct answer.

16 答案 D

对任意实数 \(x\)，\[ \begin{aligned} &f(1-x)=|\lfloor 1-x\rfloor| \\ &\quad {}-|\lfloor x\rfloor|=-f(x) \end{aligned}\text{。}\]

等价地，令 \(x\) 写成 \(\frac{1}{2}+t\)，则 \(f\!\left(\frac{1}{2}-t\right)=-f\!\left(\frac{1}{2}+t\right)\)。这表示图像关于点 \(\left(\frac{1}{2},0\right)\) 中心对称。

所以正确答案是 D。

English

For every real \(x,\) \[ \begin{aligned} &f(1-x)=|\lfloor 1-x\rfloor| \\ &\quad {}-|\lfloor x\rfloor|=-f(x). \end{aligned} \]

Equivalently, replacing \(x\) by \(\frac{1}{2}+t\) gives \(f\!\left(\frac{1}{2}-t\right)=-f\!\left(\frac{1}{2}+t\right).\) This is point symmetry about \(\left(\frac{1}{2},0\right).\)

Thus, D is the correct answer.

17 答案 D

取正六边形坐标 \(A=(-1,0)\)、\(B=(-\frac{1}{2},\frac{\sqrt3}{2})\)、\(C=(\frac{1}{2},\frac{\sqrt3}{2})\)、\(E=(\frac{1}{2},-\frac{\sqrt3}{2})\)。因为太阳能板是平面，高度可写成 \(h(x,y)=ux+vy+w\)。

由 \(h(A)=12\)，\(h(B)=9\)，\(h(C)=10\)。后两式相减得 \(u=1\)。再由 \(-u+w=12\) 得 \(w=13\)。代入 \(h(B)=9\)，有 \(-\frac{1}{2}+\frac{\sqrt3}{2}v+13=9\)，所以 \(\sqrt3v=-7\)。

因此 \[ \begin{aligned} h(E) &=\frac{1}{2}-\frac{\sqrt3}{2}v+13 \\ &=\frac{1}{2}+\frac{7}{2}+13 \\ &=17 \end{aligned}\text{。}\]

所以正确答案是 D。

English

Put a regular hexagon in coordinates with \(A=(-1,0),\) \(B=(-\frac{1}{2},\frac{\sqrt3}{2}),\) \(C=(\frac{1}{2},\frac{\sqrt3}{2}),\) and \(E=(\frac{1}{2},-\frac{\sqrt3}{2}).\) Because the solar panel is flat, the height is an affine function \(h(x,y)=ux+vy+w.\)

From \(h(A)=12,\) \(h(B)=9,\) and \(h(C)=10,\) subtracting the last two equations gives \(u=1.\) Then \(-u+w=12,\) so \(w=13.\) Using \(h(B)=9\) gives \(-\frac{1}{2}+\frac{\sqrt3}{2}v+13=9,\) so \(\sqrt3v=-7.\)

Therefore \[ \begin{aligned} h(E) &=\frac{1}{2}-\frac{\sqrt3}{2}v+13 \\ &=\frac{1}{2}+\frac{7}{2}+13 \\ &=17. \end{aligned} \]

Thus, D is the correct answer.

18 答案 C

分为 \(2\) 种情况讨论右上和左下这两个对角区域。

情况 \(1\)：右上和左下区域种同一种作物。

这两个区域所种的作物有 \(4\) 种选择。因为每种作物都恰有一种不能与它相邻的作物，所以另外两个区域各有 \(3\) 种选择。因此共有 \[ 4 \cdot 3 \cdot 3 = 36 \] 种。

情况 \(2\)：右上和左下区域种不同作物。

右上区域有 \(4\) 种选择，左下区域有 \(3\) 种与之不同的选择。另外两个区域都与这两块区域共边，所以各剩 \(2\) 种选择。因此这一类共有 \[ 4 \cdot 3 \cdot 2 \cdot 2 = 48 \] 种。

两种情况合计 \(36 + 48 = 84\) 种。

所以正确答案是 C。

English

There are \(2\) cases.

Case \(1:\) the top-right and bottom-left sections have the same crop

There are \(4\) options for which crop is in those sections. The other two sections have \(3\) options for the crop since each crop has one restriction for crops next to it. This gives us \[ 4 \cdot 3 \cdot 3 = 36 \] combinations.

Case \(2:\) the top-right and bottom-left sections have different crops

There are \(4\) options for the top-right crop and \(3\) different options for the bottom-left crop. Each of the other two sections borders both of these crops, leaving \(2\) choices for each. This gives us another \[ 4 \cdot 3 \cdot 2 \cdot 2 = 48 \] configurations.

In total, there are \(36 + 48 = 84\) combinations.

Thus, C is the correct answer.

19 答案 A

内圆盘圆心扫出的内侧正方形边长为 \(s - 4\)。

四个角还有一些小区域，它们的总面积为 \[ (1 + 1)^2 - \pi 1^2 = 4 - \pi\text{。}\]

因此 \[ A = s^2 - (s - 4)^2 - (4 - \pi) \]\[= 8s - 20 + \pi\text{。}\]

外圆盘扫出的区域由 \(4\) 个长方形和 \(4\) 个四分之一圆组成。每个长方形的面积为 \( s \cdot 2 = 2s \)，四个四分之一圆合成一个半径为 \(2\)、面积为 \(4 \pi\) 的圆。

所以 \[ 2A = 8s + 4 \pi\text{。}\]

令两个表达式相等，得到 \[ 8s + 4 \pi = 2(8s - 20 + \pi)\text{。}\]解得 \[ 8s = 40 + 2 \pi \]\[ s = 5 + \dfrac{\pi}{4}\text{。}\]

所以正确答案是 A。

English

The side length of the inner square traced out by the inner circle is \(s - 4.\)

There are also the small pieces remaining in the corner. These form a total area of \[ (1 + 1)^2 - \pi 1^2 = 4 - \pi. \]

Therefore, \[ A = s^2 - (s - 4)^2 - (4 - \pi) \]\[= 8s - 20 + \pi. \]

The outer disk traces out an area that is comprised of \(4\) rectangles and \(4\) quarter-circles. The rectangles have area \( s \cdot 2 = 2s \) and the quarter-circles form a circle with radius \(2\) and area \(4 \pi.\)

This gives us \[ 2A = 8s + 4 \pi. \]

Equating the two equations we get \[ 8s + 4 \pi = 2(8s - 20 + \pi). \] Solving yields \[ 8s = 40 + 2 \pi \]\[ s = 5 + \dfrac{\pi}{4}. \]

Thus, A is the correct answer.

20 答案 B

二次方程没有两个不同实根，当且仅当判别式非正。因此需要 \[b^2-4c\le0\qquad c^2-4b\le0\text{，}\]也就是 \(b^2\le4c\) 且 \(c^2\le4b\)。

由 \(b^2\le4c\) 得 \(b^4\le16c^2\)。结合 \(c^2\le4b\)，得到 \(b^4\le64b\)，所以 \(b\le4\)。

逐一检查 \(b=1,2,3,4\)，分别得到 \[ \begin{gathered} (c=1,2),\quad (c=1,2), \\ \quad (c=3),\quad (c=4) \end{gathered}\text{。}\]

因此有序对共有 \(2+2+1+1=6\) 个。

所以正确答案是 B。

English

A quadratic fails to have two distinct real solutions exactly when its discriminant is nonpositive. Thus we need \[b^2-4c\le0\qquad c^2-4b\le0,\] or \(b^2\le4c\) and \(c^2\le4b.\)

From \(b^2\le4c,\) we get \(b^4\le16c^2.\) Combining this with \(c^2\le4b\) gives \(b^4\le64b,\) so \(b\le4.\)

Now check \(b=1,2,3,4.\) The inequalities give respectively \[ \begin{gathered} (c=1,2),\quad (c=1,2), \\ \quad (c=3),\quad (c=4). \end{gathered} \]

There are \(2+2+1+1=6\) ordered pairs.

Thus, B is the correct answer.

21 答案 E

把球和箱子都看作可区分，则把有标号的球投入有标号的箱子的 \(5^{20}\) 种方法等可能。对于 \(q\)，符合条件的投放方法数为 \[\frac{20!}{(4!)^5}\text{。}\]

对于 \(p\)，先用 \(5\cdot4\) 种方法依次选出装 \(3\) 个球的箱子和装 \(5\) 个球的箱子。符合条件的投放方法数为 \[5\cdot4\cdot\frac{20!}{3!5!(4!)^3}\text{。}\]

两种概率的共同分母可以约去，所以 \[\frac pq=20\cdot\frac{(4!)^2}{3!5!}=20\cdot\frac45=16\text{。}\]

所以正确答案是 E。

English

All \(5^{20}\) assignments of the distinguishable balls to the labeled bins are equally likely. For \(q,\) the number of assignments is \[\frac{20!}{(4!)^5}.\]

For \(p,\) choose the bin with \(3\) balls and the bin with \(5\) balls in \(5\cdot4\) ways. The number of assignments is then \[5\cdot4\cdot\frac{20!}{3!5!(4!)^3}.\]

The common probability denominator cancels, so \[\frac pq=20\cdot\frac{(4!)^2}{3!5!}=20\cdot\frac45=16.\]

Thus, E is the correct answer.

22 答案 B

令圆锥底面在平面 \(z=0\)，底心在原点，顶点在 \(z\)-轴上。三个球心组成边长为 \(2r\) 的等边三角形，所以其中一个球心可取为离圆锥轴的水平距离 \(\frac{2r}{\sqrt3}\)、高度为 \(r\)。

在轴截面中，圆锥侧边所在直线为 \(12\rho+5z=60\)，其中 \(\rho\) 是到轴的水平距离。该球心坐标为 \(\left(\frac{2r}{\sqrt3},r\right)\)，到这条直线的距离必须为 \(r\)，所以 \[r=\frac{60-12\cdot\frac{2r}{\sqrt3}-5r}{13}\text{。}\]

因此 \((18+8\sqrt3)r=60\)，所以 \[r=\frac{60}{18+8\sqrt3}=\frac{90-40\sqrt3}{11}\text{。}\]

所以正确答案是 B。

English

Let the cone have base in the plane \(z=0,\) center at the origin, and vertex on the \(z\)-axis. The centers of the three spheres form an equilateral triangle of side \(2r,\) so one sphere center may be taken at horizontal distance \(\frac{2r}{\sqrt3}\) from the cone axis and height \(r\) above the base.

In the axial cross-section through that center and the cone axis, the side of the cone is the line \(12\rho+5z=60,\) where \(\rho\) is horizontal distance from the axis. The distance from \(\left(\frac{2r}{\sqrt3},r\right)\) to this line must be \(r\): \[r=\frac{60-12\cdot\frac{2r}{\sqrt3}-5r}{13}.\]

Thus \((18+8\sqrt3)r=60,\) so \[r=\frac{60}{18+8\sqrt3}=\frac{90-40\sqrt3}{11}.\]

Thus, B is the correct answer.

23 答案 D

数值 \(12\) 是固定点，因为 \(12\) 有 \(6\) 个正因数，所以 \(f_1(12)=12\)。

先找所有满足 \(n\le50\) 且 \(f_1(n)=12\) 的 \(n\)，也就是有 \(6\) 个因数的数：\[12,18,20,28,32,44,45,50\text{。}\]

再检查 \(f_1(n)\) 在到达 \(12\) 之前能否取到上述其他值。由于 \(f_1(n)\) 是因数个数的两倍，列表中只有 \(18\) 和 \(20\) 有用，分别表示 \(n\) 有 \(9\) 或 \(10\) 个因数。

对 \(n\le50\)，额外的可能为 \(36\)（有 \(9\) 个因数）和 \(48\)（有 \(10\) 个因数）。因此共有 \(8+2=10\) 个 \(n\)。

所以正确答案是 D。

English

The value \(12\) is fixed by the function, since \(12\) has \(6\) positive divisors and therefore \(f_1(12)=12.\)

First find all \(n\le50\) with \(f_1(n)=12,\) meaning \(n\) has \(6\) divisors. These are \[12,18,20,28,32,44,45,50.\]

Now check whether \(f_1(n)\) can be one of these values before reaching \(12.\) Since \(f_1(n)\) is twice a divisor count, the only useful possibilities in that list are \(18\) and \(20,\) meaning \(n\) has \(9\) or \(10\) divisors.

For \(n\le50,\) the additional possibilities are \(36,\) which has \(9\) divisors, and \(48,\) which has \(10\) divisors. Therefore there are \(8+2=10\) values of \(n.\)

Thus, D is the correct answer.

24 答案 E

先标记一个面为 \(ABCD\)。每个面必须包含两个 \(0\) 和两个 \(1\)，所以先按 \(ABCD\) 四条棱的标签模式分类。

情形 \(1\)：\(ABCD\) 的相对棱标签相同。\(ABCD\) 上有 \(2\) 种这样的模式；任选一条竖直棱（如 \(AE\)）的标签后，其余标签都被确定。因此共有 \(2\cdot2=4\) 种标法。

情形 \(2\)：\(ABCD\) 的相对棱标签不同。\(ABCD\) 上有 \(4\) 种这样的模式；对每种模式，相邻两条竖直棱的标签有 \(4\) 种选择，其余标签随即确定。因此共有 \(4\cdot4=16\) 种标法。

总数为 \(4+16=20\)。

所以正确答案是 E。

English

Label one face \(ABCD.\) Each face must contain two \(0\)s and two \(1\)s, so first split by the pattern on the four edges of \(ABCD.\)

Case \(1:\) opposite edges of \(ABCD\) have the same label. There are \(2\) such patterns on \(ABCD.\) For either pattern, once the label of one vertical edge, say \(AE,\) is chosen, the face-sum conditions force all remaining labels. This gives \(2\cdot2=4\) labelings.

Case \(2:\) opposite edges of \(ABCD\) have different labels. There are \(4\) such patterns on \(ABCD.\) For each, the labels of two adjacent vertical edges may be chosen in \(4\) ways, and then the remaining labels are forced by the face-sum conditions. This gives \(4\cdot4=16\) labelings.

The total is \(4+16=20.\)

Thus, E is the correct answer.

25 答案 A

这个多项式必须有两个不同的实根：如果只有一个重实根，则 \(p(p(x))=0\) 至多有两个实数解；如果没有实根，则没有实数解。设它的两个根为 \(r\) 和 \(s\)，则 \[ \begin{aligned} p(x) &=(x-r)(x-s) \\ &=x^2-(r+s)x+rs \end{aligned}\text{。}\]方程 \(p(p(x))=0\) 等价于 \(p(x)=r\) 或 \(p(x)=s\)。

要恰好有三个实数解，其中一个二次方程必须有重根，另一个必须有两个不同实根。设 \(p(x)=r\) 有重根。它的判别式为 \[ \begin{aligned} &(r+s)^2-4(rs-r) \\ &=(r-s)^2+4r \end{aligned}\text{，}\]所以 \((r-s)^2=-4r\)，从而 \(r\le0\)。

另一个方程 \(p(x)=s\) 的判别式为 \((r-s)^2+4s\) \(=-4r+4s\) \(=4(s-r)\)，且必须为正。因此 \(s\gt r\)，所以 \(r-s=-2\sqrt{-r}\)，即 \(s=r+2\sqrt{-r}\)。

根之和为 \(r+s=2r+2\sqrt{-r}\)。令 \(u=\sqrt{-r}\)，则该和为 \(-2u^2+2u\)，在 \(u=\frac{1}{2}\) 时最大。因此 \(r=-\frac{1}{4}\)，\(s=\frac{3}{4}\)。

于是得到多项式 \(p(x)=x^2-\frac{1}{2}x-\frac{3}{16}\)，并且 \[p(1)=1-\frac{1}{2}-\frac{3}{16}=\frac{5}{16}\text{。}\]

所以正确答案是 A。

English

The polynomial must have two distinct real roots: a repeated real root produces at most two real solutions of \(p(p(x))=0,\) while nonreal roots produce none. Let its roots be \(r\) and \(s,\) so \[ \begin{aligned} p(x) &=(x-r)(x-s) \\ &=x^2-(r+s)x+rs. \end{aligned} \] The equation \(p(p(x))=0\) is equivalent to \(p(x)=r\) or \(p(x)=s.\)

For exactly three real solutions, one of these two quadratic equations must have a double root and the other must have two distinct real roots. Suppose \(p(x)=r\) has the double root. Its discriminant is \[ \begin{aligned} &(r+s)^2-4(rs-r) \\ &=(r-s)^2+4r, \end{aligned} \] so \((r-s)^2=-4r,\) forcing \(r\le0.\)

The other equation, \(p(x)=s,\) has discriminant \((r-s)^2+4s\) \(=-4r+4s\) \(=4(s-r),\) which must be positive. Hence \(s\gt r,\) so \(r-s=-2\sqrt{-r}\) and \(s=r+2\sqrt{-r}.\)

The sum of the roots is \(r+s=2r+2\sqrt{-r}.\) Let \(u=\sqrt{-r},\) so this is \(-2u^2+2u,\) maximized at \(u=\frac{1}{2}.\) Thus \(r=-\frac{1}{4}\) and \(s=\frac{3}{4}.\)

Therefore \(p(x)=x^2-\frac{1}{2}x-\frac{3}{16},\) and \[p(1)=1-\frac{1}{2}-\frac{3}{16}=\frac{5}{16}.\]

Thus, A is the correct answer.