2018B AMC 12 真题（P版）

1 答案 A

烤盘面积为 \(20\cdot18=360\) 平方英寸，每块面积为 \(2\cdot2=4\) 平方英寸。

块数为 \[ \dfrac{360}{4}=90\text{。}\]

所以正确答案是 A。

English

The pan has area \(20\cdot18=360\) square inches, and each piece has area \(2\cdot2=4\) square inches.

The number of pieces is \[ \dfrac{360}{4}=90. \]

Thus, the correct answer is A.

2 答案 D

前 \(30\) 分钟 Sam 行驶 \(60\cdot\tfrac12=30\) 英里，接着又行驶 \(65\cdot\tfrac12=32.5\) 英里。

最后 \(30\) 分钟行驶 \(96-30-32.5=33.5\) 英里，所以速度为 \[ \dfrac{33.5}{\frac{1}{2}}=67\text{ 英里/小时}\text{。}\]

所以正确答案是 D。

English

In the first \(30\) minutes Sam covered \(60\cdot\tfrac12=30\) miles, and in the second he covered \(65\cdot\tfrac12=32.5\) miles.

The last \(30\) minutes covered \(96-30-32.5=33.5\) miles, so the speed was \[ \dfrac{33.5}{\frac{1}{2}}=67\text{ mph}. \]

Thus, the correct answer is D.

3 答案 B

斜率为 \(2\) 的直线为 \(y-30=2(x-40)\)；令 \(y=0\)，得 \(x=25\)。斜率为 \(6\) 的直线为 \(y-30=6(x-40)\)；令 \(y=0\)，得 \(x=35\)。

两个截距之间的距离为 \(|35-25|=10\)。

所以正确答案是 B。

English

The line of slope \(2\) is \(y-30=2(x-40);\) setting \(y=0\) gives \(x=25.\) The line of slope \(6\) is \(y-30=6(x-40);\) setting \(y=0\) gives \(x=35.\)

The distance between the intercepts is \(|35-25|=10.\)

Thus, the correct answer is B.

4 答案 B

从圆心向弦作垂线会平分弦，形成直角三角形，其两条直角边分别为 \(5\)（半弦长）和 \(5\)（圆心到弦距离），斜边为 \(r\)。

因此 \(r^2=5^2+5^2=50\)，圆面积为 \(\pi r^2=50\pi\)。

所以正确答案是 B。

English

Dropping a perpendicular from the center to the chord bisects it, forming a right triangle with legs \(5\) (half the chord) and \(5\) (the distance), and hypotenuse \(r.\)

Then \(r^2=5^2+5^2=50,\) so the area is \(\pi r^2=50\pi.\)

Thus, the correct answer is B.

5 答案 D

该集合有 \(8\) 个元素，共有 \(2^8=256\) 个子集。不含质数的子集只能使用四个非质数 \(\{4,6,8,9\}\)，共有 \(2^4=16\) 个。

因此至少含一个质数的子集数为 \(256-16=240\)。

所以正确答案是 D。

English

The set has \(8\) elements, giving \(2^8=256\) subsets. The subsets with no prime use only the four non-primes \(\{4,6,8,9\},\) and there are \(2^4=16\) of these.

So the number containing at least one prime is \(256-16=240.\)

Thus, the correct answer is D.

6 答案 B

一罐汽水需要 \(\tfrac{Q}{S}\) 枚二十五美分硬币，也就是 \(\tfrac{Q}{4S}\) 美元。用 \(D\) 美元可购买的罐数是

\[ \dfrac{D}{\tfrac{Q}{4S}}=\dfrac{4DS}{Q}\text{。}\]

所以正确答案是 B。

English

One can costs \(\tfrac{Q}{S}\) quarters, which is \(\tfrac{Q}{4S}\) dollars. The number of cans that \(D\) dollars can buy is

\[ \dfrac{D}{\tfrac{Q}{4S}}=\dfrac{4DS}{Q}. \]

Thus, the correct answer is B.

7 答案 C

这些因子可以分成两条逐项相消的链。奇数位置上的因子构成\[ \begin{gathered} \log_3 7\cdot\log_7 11 \\ {}\cdot\log_{11} 15\cdots\log_{23} 27 \\ =\log_3 27=3 \end{gathered}\text{，}\] 而偶数位置上的因子构成\[ \begin{gathered} \log_5 9\cdot\log_9 13\cdots\log_{21} 25 \\ =\log_5 25=2 \end{gathered}\text{。}\]

这两条链的乘积为 \(3\cdot2=6\)。

所以正确答案是 C。

English

The factors split into two telescoping chains. The odd-position factors form \[ \begin{gathered} \log_3 7\cdot\log_7 11 \\ {}\cdot\log_{11} 15\cdots\log_{23} 27 \\ =\log_3 27=3, \end{gathered} \] and the even-position factors form \[ \begin{gathered} \log_5 9\cdot\log_9 13\cdots\log_{21} 25 \\ =\log_5 25=2. \end{gathered} \]

The product is \(3\cdot2=6.\)

Thus, the correct answer is C.

8 答案 C

令 \(O\) 为圆心。\(\triangle ABC\) 的重心是 \(A\)、\(B\)、\(C\) 坐标的平均；由于 \(O\) 是 \(\overline{AB}\) 的中点，重心位于从 \(O\) 到 \(C\) 的三分之一处。

当 \(C\) 沿半径为 \(12\) 的圆运动时，重心沿半径为 \(\tfrac13\cdot12=4\) 的圆运动。该圆面积为 \(16\pi\approx50\)。

所以正确答案是 C。

English

Let \(O\) be the center of the circle. The centroid of \(\triangle ABC\) is the average of \(A,\) \(B,\) and \(C;\) since \(O\) is the midpoint of \(\overline{AB},\) the centroid lies one-third of the way from \(O\) to \(C.\)

As \(C\) traces the circle of radius \(12,\) the centroid traces a circle of radius \(\tfrac13\cdot12=4.\) Its area is \(16\pi\approx50.\)

Thus, the correct answer is C.

9 答案 E

拆开求和：\[ \begin{gathered} \sum_{i=1}^{100}\sum_{j=1}^{100}(i+j) \\ =\sum_{i=1}^{100}\sum_{j=1}^{100}i \\ {}+\sum_{i=1}^{100}\sum_{j=1}^{100}j \\ =100\sum_{i=1}^{100}i \\ {}+100\sum_{j=1}^{100}j \end{gathered}\text{。}\]

因为 \(\sum_{k=1}^{100}k=5050\)，结果为 \(100\cdot5050\) \(+100\cdot5050\) \(=1{,}010{,}000\)。

所以正确答案是 E。

English

Splitting the sum, \[ \begin{gathered} \sum_{i=1}^{100}\sum_{j=1}^{100}(i+j) \\ =\sum_{i=1}^{100}\sum_{j=1}^{100}i \\ {}+\sum_{i=1}^{100}\sum_{j=1}^{100}j \\ =100\sum_{i=1}^{100}i \\ {}+100\sum_{j=1}^{100}j. \end{gathered} \]

Since \(\sum_{k=1}^{100}k=5050,\) this equals \(100\cdot5050\) \(+100\cdot5050\) \(=1{,}010{,}000.\)

Thus, the correct answer is E.

10 答案 D

众数占 \(10\) 项，剩下 \(2008\) 项。因为众数唯一，其他每个值最多出现 \(9\) 次，所以至少需要 \(\left\lceil\tfrac{2008}{9}\right\rceil=224\) 个非众数值。

再加上众数本身，共有 \(224+1=225\) 个不同值。这可以实现：让从 \(1\) 到 \(223\) 的每个数各出现 \(9\) 次，再让 \(224\) 出现十次，让 \(225\) 出现一次。

所以正确答案是 D。

English

The mode uses \(10\) of the entries, leaving \(2008.\) Because the mode is unique, every other value appears at most \(9\) times, so at least \(\left\lceil\tfrac{2008}{9}\right\rceil=224\) distinct non-mode values are needed.

Adding the mode gives \(224+1=225.\) This is achievable: use \(9\) copies each of \(1\) through \(223,\) ten copies of \(224,\) and one copy of \(225.\)

Thus, the correct answer is D.

11 答案 A

沿着折线从纸角到盒顶中心看，纸角到纸中心的距离为 \[ \dfrac{w}{2}+h+\dfrac{w}{2}=w+h\text{。}\]

这段是 \(45\)-\(45\)-\(90\) 三角形的一条直角边，其斜边是正方形包装纸的一整条边，所以边长为 \(\sqrt2\,(w+h)\)。

包装纸面积为 \(\left(\sqrt2\,(w+h)\right)^2=2(w+h)^2\)。

所以正确答案是 A。

English

Following a fold from a corner of the paper to the center of the box top, the distance from a corner of the sheet to its center is \[ \dfrac{w}{2}+h+\dfrac{w}{2}=w+h. \]

That segment is a leg of a \(45\)-\(45\)-\(90\) triangle whose hypotenuse is a full side of the square sheet, so the side length is \(\sqrt2\,(w+h).\)

The area of the sheet is \(\left(\sqrt2\,(w+h)\right)^2=2(w+h)^2.\)

Thus, the correct answer is A.

12 答案 C

令 \(q=AC\)，\(r=BD\)。由角平分线定理，\(\tfrac{q}{3}=\tfrac{10}{r}\)，所以 \(r=\tfrac{30}{q}\)。

对三角形三边 \(q\)、\(10\)、\(3+r\) 应用三角形不等式，并代入 \(r=\tfrac{30}{q}\) 得到 \((q-15)(q+2)\lt0\) 和 \((q-3)(q+10)\gt0\)（第三个不等式自动成立）。合起来得到 \(3\lt q\lt15\)。

所以 \((m,n)=(3,15)\)，且 \(m+n=18\)。

所以正确答案是 C。

English

Let \(q=AC\) and \(r=BD.\) The angle bisector theorem gives \(\tfrac{q}{3}=\tfrac{10}{r},\) so \(r=\tfrac{30}{q}.\)

Applying the triangle inequalities to sides \(q,\) \(10,\) and \(3+r\) and substituting \(r=\tfrac{30}{q}\) yields \((q-15)(q+2)\lt0\) and \((q-3)(q+10)\gt0\) (the third inequality holds automatically). Together these force \(3\lt q\lt15.\)

So \((m,n)=(3,15)\) and \(m+n=18.\)

Thus, the correct answer is C.

13 答案 C

取 \(A=(0,30)\)、\(B=(0,0)\)、\(C=(30,0)\)、\(D=(30,30)\)，并令 \(P=(3x,3y)\)。对每组三个顶点的坐标取平均，四个重心为 \[ \begin{gathered} (x,\,y+10),\ \\ (x+10,\,y),\ \\ (x+20,\,y+10),\ \\ (x+10,\,y+20) \end{gathered}\text{。}\]

这些点形成一个正方形，一条水平对角线和一条竖直对角线的长度均为 \(20\)。它的面积为 \(\tfrac12\cdot20\cdot20=200\)，与 \(P\) 的位置无关。

所以正确答案是 C。

English

Place \(A=(0,30),\) \(B=(0,0),\) \(C=(30,0),\) \(D=(30,30),\) and \(P=(3x,3y).\) Averaging the vertices, the four centroids are \[ \begin{gathered} (x,\,y+10),\ \\ (x+10,\,y),\ \\ (x+20,\,y+10),\ \\ (x+10,\,y+20). \end{gathered} \]

These form a square whose diagonals, one horizontal and one vertical, each have length \(20.\) Its area is \(\tfrac12\cdot20\cdot20=200,\) independent of where \(P\) lies.

Thus, the correct answer is C.

14 答案 E

设 Chloe 今天 \(n\) 岁，那么她比 Zoe 大 \(n-1\) 岁。\(y\) 年后，Chloe 的年龄 \(n+y\) 是 Zoe 年龄 \(1+y\) 的倍数，当且仅当 \(1+y\) 整除 \(n-1\)。有 \(9\) 个这样的生日，意味着 \(n-1\) 恰有 \(9\) 个因数。

恰有 \(9\) 个因数的数形如 \(p^2q^2\)（其中 \(p,q\) 是不同质数）或形如 \(p^8\)。因为在所问的未来生日，Joey 的年龄是两位数，所以 \(n-1\lt99\)；唯一可能是 \(2^2\cdot3^2=36\)。因此 Chloe 是 \(37\) 岁，Joey 是 \(38\) 岁。

Joey 的年龄 \(38+y\) 是 \(1+y\) 的倍数，当且仅当 \(1+y\) 整除 \(37\)。下一次发生在 \(y=36\)，此时 Joey 是 \(74\) 岁，数位和为 \(7+4=11\)。

因此，正确答案是 E。

English

Let Chloe be \(n\) today, so she is \(n-1\) years older than Zoe. In \(y\) years Chloe’s age \(n+y\) is a multiple of Zoe’s age \(1+y\) exactly when \(1+y\) divides \(n-1.\) Having \(9\) such birthdays means \(n-1\) has exactly \(9\) divisors.

A number with exactly \(9\) divisors has the form \(p^2q^2\) for distinct primes \(p,q,\) or \(p^8.\) Because Joey’s age at the requested future birthday has two digits, \(n-1\lt99;\) the only possibility is \(2^2\cdot3^2=36.\) So Chloe is \(37\) and Joey is \(38.\)

Joey’s age \(38+y\) is a multiple of \(1+y\) exactly when \(1+y\) divides \(37.\) The next time is \(y=36,\) making Joey \(74,\) with digit sum \(7+4=11.\)

Thus, the correct answer is E.

15 答案 A

写作 \(\overline{abc}\)。百位数字 \(a\) 有 \(8\) 种选择（\(1,2,4,5,6,7,8,9\)），个位数字 \(c\) 有 \(4\) 种选择（\(1,5,7,9\)）。

十位数字 \(b\) 可为 \(\{0,1,2,4,5,6,7,8,9\}\)。这些数字按模 \(3\) 分成三个大小相等的剩余类 \(\{0,6,9\},\{1,4,7\},\{2,5,8\}\)，所以恰有 \(3\) 个 \(b\) 使 \(a+b+c\) 能被 \(3\) 整除。

总数为 \(8\cdot4\cdot3=96\)。

所以正确答案是 A。

English

Write the number as \(\overline{abc}.\) The hundreds digit \(a\) has \(8\) choices (\(1,2,4,5,6,7,8,9\)), and the units digit \(c\) has \(4\) choices (\(1,5,7,9\)).

The tens digit \(b\) may be any of \(\{0,1,2,4,5,6,7,8,9\}.\) These split into three residue classes mod \(3\) of equal size \(\{0,6,9\},\{1,4,7\},\{2,5,8\},\) so exactly \(3\) choices of \(b\) make \(a+b+c\) divisible by \(3.\)

The count is \(8\cdot4\cdot3=96.\)

Thus, the correct answer is A.

16 答案 B

平移 \(6\) 后，\(z^8=81\) 的解是半径为 \(81^{\frac{1}{8}}=\sqrt3\) 的圆上的八个点，构成正八边形。面积最小的三角形使用三个连续顶点。

取 \(A=\left(\tfrac12\sqrt6,\tfrac12\sqrt6\right)\)、\(B=(\sqrt3,0)\) 和 \(C=\left(\tfrac12\sqrt6,-\tfrac12\sqrt6\right)\)。此时 \(AC=\sqrt6\)，高为 \(\sqrt3-\tfrac12\sqrt6\)，所以面积为 \[ \begin{gathered} \tfrac12\cdot\sqrt6\left(\sqrt3-\tfrac12\sqrt6\right) \\ =\tfrac{3}{2}\sqrt2-\tfrac{3}{2} \end{gathered}\text{。}\]

所以正确答案是 B。

English

Translating by \(6,\) the solutions of \(z^8=81\) are eight points on a circle of radius \(81^{\frac{1}{8}}=\sqrt3,\) forming a regular octagon. The minimum-area triangle uses three consecutive vertices.

Take \(A=\left(\tfrac12\sqrt6,\tfrac12\sqrt6\right),\) \(B=(\sqrt3,0),\) and \(C=\left(\tfrac12\sqrt6,-\tfrac12\sqrt6\right).\) Then \(AC=\sqrt6\) and the height is \(\sqrt3-\tfrac12\sqrt6,\) so the area is \[ \begin{gathered} \tfrac12\cdot\sqrt6\left(\sqrt3-\tfrac12\sqrt6\right) \\ =\tfrac{3}{2}\sqrt2-\tfrac{3}{2}. \end{gathered} \]

Thus, the correct answer is B.

17 答案 A

由 \(\tfrac59\lt\tfrac pq\) 得 \(9p-5q\ge1\)，由 \(\tfrac pq\lt\tfrac47\) 得 \(4q-7p\ge1\)。于是 \[ \begin{gathered} \dfrac{1}{63}=\dfrac47-\dfrac59 \\ =\dfrac{4q-7p}{7q}+\dfrac{9p-5q}{9q} \\ \ge\dfrac{1}{7q}+\dfrac{1}{9q} \\ =\dfrac{16}{63q} \end{gathered}\text{。}\]

因此 \(q\ge16\)。当 \(q=16\) 时，分数 \(\tfrac{9}{16}\) 严格介于 \(\tfrac59\) 和 \(\tfrac47\) 之间，所以 \(p=9\)，且 \(q-p=16-9=7\)。

所以正确答案是 A。

English

From \(\tfrac59\lt\tfrac pq\) we get \(9p-5q\ge1,\) and from \(\tfrac pq\lt\tfrac47\) we get \(4q-7p\ge1.\) Now \[ \begin{gathered} \dfrac{1}{63}=\dfrac47-\dfrac59 \\ =\dfrac{4q-7p}{7q}+\dfrac{9p-5q}{9q} \\ \ge\dfrac{1}{7q}+\dfrac{1}{9q} \\ =\dfrac{16}{63q}. \end{gathered} \]

Hence \(q\ge16.\) With \(q=16,\) the fraction \(\tfrac{9}{16}\) lies strictly between \(\tfrac59\) and \(\tfrac47,\) so \(p=9\) and \(q-p=16-9=7.\)

Thus, the correct answer is A.

18 答案 B

反复将递推式代入自身，可得 \[ f(n)=f(n-6)+6\text{。} \] 所以每当 \(n\) 增加 \(6\)，\(f\) 增加 \(6\)。

因为 \(2018=2+6\cdot336\)，所以 \(f(2018)=f(2)+6\cdot336\) \(=1+2016=2017\)。

正确答案是 B。

English

Repeatedly substituting the recursion into itself gives \[ f(n)=f(n-6)+6. \] So \(f\) increases by \(6\) every time \(n\) increases by \(6.\)

Since \(2018=2+6\cdot336,\) we have \(f(2018)=f(2)+6\cdot336\) \(=1+2016=2017.\)

Thus, the correct answer is B.

19 答案 C

设 \(d\) 是 \(323\) 后面的下一个因数。若 \(\gcd(d,323)=1\)，则 \(323d\) 整除 \(n\)，迫使 \(n\ge323d\gt323^2\gt9999\)，不可能是 \(4\) 位数。因此 \(d\) 与 \(323=17\cdot19\) 有公共质因数。

于是 \(d-323\ge\gcd(d,323)\ge17\)，所以 \(d\ge340\)。确实，\(d=340=17\cdot20\) 可由 \(n=17\cdot19\cdot20=6460\) 实现，而这是偶数且为 \(4\) 位数。

所以正确答案是 C。

English

Let \(d\) be the next divisor after \(323.\) If \(\gcd(d,323)=1,\) then \(323d\) divides \(n,\) forcing \(n\ge323d\gt323^2\gt9999,\) impossible for a \(4\)-digit number. So \(d\) shares a prime factor with \(323=17\cdot19.\)

Then \(d-323\ge\gcd(d,323)\ge17,\) so \(d\ge340.\) Indeed \(d=340=17\cdot20\) occurs for \(n=17\cdot19\cdot20=6460,\) which is even and \(4\)-digit.

Thus, the correct answer is C.

20 答案 C

把正六边形放在单位圆上，取 \(A=(1,0)\)、\(C=(-\tfrac12,\tfrac{\sqrt3}{2})\) 和 \(E=(-\tfrac12,-\tfrac{\sqrt3}{2})\)。题中指定的三个中点为 \(X=(\tfrac34,\tfrac{\sqrt3}{4})\)、\(Y=(-\tfrac34,\tfrac{\sqrt3}{4})\) 和 \(Z=(0,-\tfrac{\sqrt3}{2})\)。

求 \(\triangle ACE\) 与 \(\triangle XYZ\) 各边所在直线的交点，就得到公共内部的六个顶点，按循环顺序为 \[ \begin{gathered} (-\tfrac12,0),\quad (-\tfrac18,-\tfrac{3\sqrt3}{8}),\\ (\tfrac14,-\tfrac{\sqrt3}{4}),\quad (\tfrac58,\tfrac{\sqrt3}{8}),\\ (\tfrac14,\tfrac{\sqrt3}{4}),\quad (-\tfrac12,\tfrac{\sqrt3}{4}) \end{gathered}\text{。}\] 对这些顶点使用鞋带公式，得到面积为 \(\dfrac{15\sqrt3}{32}\)。

所以正确答案是 C。

English

Place the regular hexagon on the unit circle with \(A=(1,0),\) \(C=(-\tfrac12,\tfrac{\sqrt3}{2}),\) and \(E=(-\tfrac12,-\tfrac{\sqrt3}{2}).\) The three specified midpoints are \(X=(\tfrac34,\tfrac{\sqrt3}{4}),\) \(Y=(-\tfrac34,\tfrac{\sqrt3}{4}),\) and \(Z=(0,-\tfrac{\sqrt3}{2}).\)

Intersecting the side lines of \(\triangle ACE\) and \(\triangle XYZ\) gives the six vertices of their common interior, in cyclic order: \[ \begin{gathered} (-\tfrac12,0),\quad (-\tfrac18,-\tfrac{3\sqrt3}{8}),\\ (\tfrac14,-\tfrac{\sqrt3}{4}),\quad (\tfrac58,\tfrac{\sqrt3}{8}),\\ (\tfrac14,\tfrac{\sqrt3}{4}),\quad (-\tfrac12,\tfrac{\sqrt3}{4}). \end{gathered} \] The shoelace formula applied to these vertices gives area \(\dfrac{15\sqrt3}{32}.\)

Thus, the correct answer is C.

21 答案 E

因为 \(5^2+12^2=13^2\)，三角形在 \(C\) 处为直角。设 \(C=(0,0)\)、\(A=(12,0)\) 且 \(B=(0,5)\)。则 \(O\) 是 \(\overline{AB}\) 的中点，即 \(O=\left(6,\tfrac52\right)\)，外接圆半径为 \(\tfrac{13}{2}\)。内切圆半径为 \(\tfrac{\text{面积}}{s}=\tfrac{30}{15}=2\)，所以 \(I=(2,2)\)。

因为以 \(M\) 为圆心的圆与两条直角边都相切，所以 \(M=(\rho,\rho)\)。与外接圆内切给出 \(MO=\tfrac{13}{2}-\rho\)。令它等于 \(\sqrt{(\rho-6)^2+\left(\rho-\tfrac52\right)^2}\) 并求解，得到 \(\rho=4\)，所以 \(M=(4,4)\)。

对 \(M=(4,4)\)、\(O=\left(6,\tfrac52\right)\) 和 \(I=(2,2)\) 用鞋带公式，面积为 \(\tfrac72\)。

所以正确答案是 E。

English

Since \(5^2+12^2=13^2,\) the triangle is right-angled at \(C.\) Set \(C=(0,0),\) \(A=(12,0),\) and \(B=(0,5).\) Then \(O\) is the midpoint of \(\overline{AB},\) namely \(O=\left(6,\tfrac52\right),\) with circumradius \(\tfrac{13}{2}.\) The inradius is \(\tfrac{\text{area}}{s}=\tfrac{30}{15}=2,\) so \(I=(2,2).\)

Because \(M\)’s circle is tangent to both legs, \(M=(\rho,\rho).\) Internal tangency to the circumcircle gives \(MO=\tfrac{13}{2}-\rho.\) Setting this equal to \(\sqrt{(\rho-6)^2+\left(\rho-\tfrac52\right)^2}\) and solving gives \(\rho=4,\) so \(M=(4,4).\)

The shoelace formula on \(M=(4,4),\) \(O=\left(6,\tfrac52\right),\) \(I=(2,2)\) gives area \(\tfrac72.\)

Thus, the correct answer is E.

22 答案 D

写 \(P(x)=ax^3+bx^2+cx+d\)，其中 \(a,b,c,d\) 都在 \(\{0,\ldots,9\}\) 中。条件为 \(-a+b-c+d=-9\)。

令 \(a'=9-a\) 和 \(c'=9-c\)，它们也都在 \([0,9]\) 中。于是 \(a'+b+c'+d=9\)。由隔板法，非负整数解的个数为 \(\binom{9+3}{3}=\binom{12}{3}=220\)，且因为总和是 \(9\)，每个变量自动不超过上界。

所以正确答案是 D。

English

Write \(P(x)=ax^3+bx^2+cx+d\) with each of \(a,b,c,d\) in \(\{0,\ldots,9\}.\) The condition is \(-a+b-c+d=-9.\)

Let \(a'=9-a\) and \(c'=9-c,\) both in \([0,9].\) Then \(a'+b+c'+d=9.\) By stars and bars the number of nonnegative solutions is \(\binom{9+3}{3}=\binom{12}{3}=220,\) and each automatically satisfies the upper bounds since the sum is \(9.\)

Thus, the correct answer is D.

23 答案 C

两地经度差为 \(360^\circ-(110^\circ+115^\circ)=135^\circ\)，且 \(B\) 位于北纬 \(45^\circ\)。把 \(A=(1,0,0)\) 放在单位球面上。

于是 \(B=\tiny\left(\cos45^\circ\cos135^\circ,\ \cos45^\circ\sin135^\circ,\ \sin45^\circ\right)\) \(=\left(-\tfrac12,\tfrac12,\tfrac{\sqrt2}{2}\right)\)。点积为 \(A\cdot B=-\tfrac12\)，所以 \(\cos\angle ACB=-\tfrac12\)，从而 \(\angle ACB=120^\circ\)。

所以正确答案是 C。

English

The longitudes differ by \(360^\circ-(110^\circ+115^\circ)=135^\circ,\) and \(B\) is at latitude \(45^\circ\) N. Place \(A=(1,0,0)\) on the unit sphere.

Then \(B=\tiny\left(\cos45^\circ\cos135^\circ,\ \cos45^\circ\sin135^\circ,\ \sin45^\circ\right)\) \(=\left(-\tfrac12,\tfrac12,\tfrac{\sqrt2}{2}\right).\) The dot product is \(A\cdot B=-\tfrac12,\) so \(\cos\angle ACB=-\tfrac12\) and \(\angle ACB=120^\circ.\)

Thus, the correct answer is C.

24 答案 C

令 \(\{x\}=x-\lfloor x\rfloor\)。方程化为 \(x^2=10{,}000\{x\}\)，所以 \(\tfrac{x^2}{10{,}000}=\{x\}\)。因为 \(0\le\{x\}\lt1\)，必须有 \(0\le x^2\lt10{,}000\)，即 \(-100\lt x\lt100\)。

在每个区间 \([k,k+1)\) 上，写成 \(x=k+t\)，其中 \(0\le t\lt1\)。方程变为 \((k+t)^2-10{,}000t=0\)。当 \(-100\le k\le98\) 时，左边严格递减：在 \(t=0\) 处它等于 \(k^2\ge0\)，而当 \(t\) 趋于 \(1\) 时它趋于 \((k+1)^2-10{,}000\lt0\)。因此这些区间中每一个都恰好含有一个解。这样的区间共有 \(98-(-100)+1=199\) 个。

所以正确答案是 C。

English

Let \(\{x\}=x-\lfloor x\rfloor.\) The equation becomes \(x^2=10{,}000\{x\},\) so \(\tfrac{x^2}{10{,}000}=\{x\}.\) Since \(0\le\{x\}\lt1,\) we need \(0\le x^2\lt10{,}000,\) i.e. \(-100\lt x\lt100.\)

On each interval \([k,k+1),\) write \(x=k+t\) with \(0\le t\lt1.\) The equation becomes \((k+t)^2-10{,}000t=0.\) For \(-100\le k\le98,\) the left side is strictly decreasing; at \(t=0\) it is \(k^2\ge0,\) while as \(t\) approaches \(1\) it approaches \((k+1)^2-10{,}000\lt0.\) Thus each of these intervals contains exactly one solution. There are \(98-(-100)+1=199\) such intervals.

Thus, the correct answer is C.

25 答案 D

设 \(O_i\) 为 \(\omega_i\) 的圆心，\(K\) 为直线 \(O_1P_1\) 与 \(O_2P_2\) 的交点。因为 \(\angle P_1P_2P_3=60^\circ\)，三角形 \(P_2KP_1\) 是 \(30\)-\(60\)-\(90^\circ\) 三角形。令 \(d=P_1K\)，得 \(P_2K=2d\) 且 \(P_1P_2=\sqrt3\,d\)。

在 \(\triangle O_1KO_2\) 中使用余弦定理，并利用 \(O_1O_2=8\) 得到 \[ \begin{gathered} 8^2=(d+4)^2+(2d-4)^2 \\ {}-2(d+4)(2d-4)\cos60^\circ\text{，} \end{gathered} \] 化简为 \(3d^2-12d-16=0\)，所以 \(d=2+\tfrac23\sqrt{21}\)。

因此 \(P_1P_2=\sqrt3\,d=2\sqrt3+2\sqrt7\)，面积为 \[ \begin{gathered} \dfrac{\sqrt3}{4}\left(2\sqrt3+2\sqrt7\right)^2 \\ =10\sqrt3+6\sqrt7 \\ =\sqrt{300}+\sqrt{252}\text{。} \end{gathered} \]

所以 \(a+b=300+252=552\)。

所以正确答案是 D。

English

Let \(O_i\) be the center of \(\omega_i,\) and let \(K\) be the intersection of lines \(O_1P_1\) and \(O_2P_2.\) Because \(\angle P_1P_2P_3=60^\circ,\) triangle \(P_2KP_1\) is a \(30\)-\(60\)-\(90^\circ\) triangle. With \(d=P_1K,\) we get \(P_2K=2d\) and \(P_1P_2=\sqrt3\,d.\)

The Law of Cosines in \(\triangle O_1KO_2\) (with \(O_1O_2=8\)) gives \[ \begin{gathered} 8^2=(d+4)^2+(2d-4)^2 \\ {}-2(d+4)(2d-4)\cos60^\circ, \end{gathered} \] which simplifies to \(3d^2-12d-16=0,\) so \(d=2+\tfrac23\sqrt{21}.\)

Then \(P_1P_2=\sqrt3\,d=2\sqrt3+2\sqrt7,\) and the area is \[ \begin{gathered} \dfrac{\sqrt3}{4}\left(2\sqrt3+2\sqrt7\right)^2 \\ =10\sqrt3+6\sqrt7 \\ =\sqrt{300}+\sqrt{252}. \end{gathered} \]

So \(a+b=300+252=552.\)

Thus, the correct answer is D.