2022B AMC 12 真题（P版）

1 答案 A

因为 \(2 \diamond 3 = |2-3| = 1\)，所以 \(1 \diamond (2 \diamond 3) = 1 \diamond 1 = 0\)。

因为 \(1 \diamond 2 = |1-2| = 1\)，所以 \((1 \diamond 2) \diamond 3 = 1 \diamond 3 = |1-3| = 2\)。

所求值为 \(0 - 2 = -2\)。

所以正确答案是 A。

English

Since \(2 \diamond 3 = |2-3| = 1,\) we get \(1 \diamond (2 \diamond 3) = 1 \diamond 1 = 0.\)

Since \(1 \diamond 2 = |1-2| = 1,\) we get \((1 \diamond 2) \diamond 3 = 1 \diamond 3 = |1-3| = 2.\)

The value is \(0 - 2 = -2.\)

Thus, the correct answer is A.

2 答案 D

边长 \(AD = AP + PD = 5\)，所以 \(AB = 5\)。在直角三角形 \(APB\) 中，\[ \begin{aligned} BP &= \sqrt{AB^2 - AP^2} \\ &= \sqrt{25 - 9} = 4 \end{aligned}\text{。}\]

取 \(AD\) 为底，\(BP\) 为高，面积为 \(AD \cdot BP = 5 \cdot 4 = 20\)。

所以正确答案是 D。

English

The side length is \(AD = AP + PD = 5,\) so \(AB = 5.\) In right triangle \(APB,\) \[ \begin{aligned} BP &= \sqrt{AB^2 - AP^2} \\ &= \sqrt{25 - 9} = 4. \end{aligned} \]

Taking \(AD\) as the base and \(BP\) as the height, the area is \(AD \cdot BP = 5 \cdot 4 = 20.\)

Thus, the correct answer is D.

3 答案 A

第 \(n\) 项由 \(n\) 个一、一个 \(2\)，再接 \(n\) 个一组成。它可分解为循环单位数与形如 \(10^n + 1\) 的数之积。例如 \[ \begin{aligned} &121 = 11 \cdot 11, \\ &\quad 11211 = 111 \cdot 101, \\ &\quad 1112111 = 1111 \cdot 1001\text{，} \end{aligned} \] 一般地，第 \(n\) 项等于 \(\underbrace{1\cdots1}_{n+1} \cdot (10^n + 1)\)。

对每个 \(n \ge 1\)，两个因子都大于 \(1\)，所以每一项都是合数。前十项中没有质数。

所以正确答案是 A。

English

The \(n\)th term consists of \(n\) ones, then a \(2,\) then \(n\) ones. It factors as a repunit times a number of the form \(10^n + 1:\) \[ \begin{aligned} &121 = 11 \cdot 11, \\ &\quad 11211 = 111 \cdot 101, \\ &\quad 1112111 = 1111 \cdot 1001, \end{aligned} \] and in general the \(n\)th term equals \(\underbrace{1\cdots1}_{n+1} \cdot (10^n + 1).\)

For every \(n \ge 1\) both factors exceed \(1,\) so every term is composite. None of the ten numbers is prime.

Thus, the correct answer is A.

4 答案 B

若整数根为 \(p\) 和 \(q\)，则 \(pq = 36\)，且 \(k = -(p+q)\)。两个不同的根必须同号，所以列出 \(36\) 的因子对，并要求 \(p \ne q\)。

正因子对为 \((1,36), (2,18), (3,12), (4,9)\)，负因子对为 \((-1,-36)\)、\((-2,-18)\)、\((-3,-12)\)、\((-4,-9)\)。因为要求两根不同，排除 \((6,6)\)。

这 \(8\) 个因子对各给出一个不同的 \(k\) 值。

所以正确答案是 B。

English

If the roots are integers \(p\) and \(q,\) then \(pq = 36\) and \(k = -(p+q).\) Distinct roots must have the same sign, so we list factor pairs of \(36\) with \(p \ne q.\)

The positive pairs are \((1,36), (2,18), (3,12), (4,9),\) and the negative pairs are \((-1,-36),\) \((-2,-18),\) \((-3,-12),\) \((-4,-9).\) The pair \((6,6)\) is excluded since the roots must be distinct.

Each of these \(8\) pairs gives a different value of \(k.\)

Thus, the correct answer is B.

5 答案 B

相对于旋转中心 \((3, 1)\)，该点的坐标为 \((-1 - 3,\, -2 - 1) = (-4, -3)\)。

逆时针 \(270^\circ\) 旋转将 \((x, y)\) 变为 \((y, -x)\)，所以 \((-4, -3)\) 变为 \((-3, 4)\)。

平移回去得 \((3 - 3,\, 1 + 4) = (0, 5)\)。

所以正确答案是 B。

English

Relative to the center \((3, 1),\) the point is at \((-1 - 3,\, -2 - 1) = (-4, -3).\)

A \(270^\circ\) counterclockwise rotation sends \((x, y)\) to \((y, -x),\) so \((-4, -3)\) becomes \((-3, 4).\)

Translating back gives \((3 - 3,\, 1 + 4) = (0, 5).\)

Thus, the correct answer is B.

6 答案 B

从 \(1\) 到 \(1000\) 中有 \(\left\lfloor \tfrac{1000}{7} \right\rfloor = 142\) 个 \(7\) 的倍数。由于 \(10 \gt 7\)，每个长度为 \(10\) 的连续整数区间含有一个或两个 \(7\) 的倍数。

设有 \(x\) 个区间含两个倍数，其余 \(100 - x\) 个区间各含一个，则 \(2x + (100 - x) = 142\)，所以 \(x = 42\)。

所以正确答案是 B。

English

Among \(1\) to \(1000\) there are \(\left\lfloor \tfrac{1000}{7} \right\rfloor = 142\) multiples of \(7.\) Because \(10 \gt 7,\) each block of \(10\) consecutive integers contains one or two multiples of \(7.\)

If \(x\) blocks contain two and the remaining \(100 - x\) contain one, then \(2x + (100 - x) = 142,\) so \(x = 42.\)

Thus, the correct answer is B.

7 答案 D

将五个数按递增排列，设中位数为 \(m\)。众数为 \(m + 2 \gt m\)，所以它只能出现在两个最大数中；要成为唯一众数，这两个数必须都等于 \(m + 2\)。

平均数为 \(m - 2\)，所以五个数之和为 \(5(m-2)\)。除去两个 \(m + 2\) 和中位数 \(m\)，两个最小数之和为 \(5(m-2) - m - 2(m+2)\) \(= 2m - 14\)。

两个最小数是不同的正整数，所以 \(2m - 14 \ge 1 + 2 = 3\)，得 \(m \ge 9\)。当 \(m = 9\) 时，\(1, 3, 9, 11, 11\) 满足条件，因此最小众数为 \(m + 2 = 11\)。

所以正确答案是 D。

English

List the numbers in increasing order with median \(m.\) The mode is \(m + 2 \gt m,\) so it can only occur among the two largest entries; for it to be the unique mode, both of them must equal \(m + 2.\)

The mean is \(m - 2,\) so the total is \(5(m-2).\) With the two largest equal to \(m + 2\) and the median \(m,\) the two smallest sum to \(5(m-2) - m - 2(m+2)\) \(= 2m - 14.\)

The two smallest are distinct positive integers, so \(2m - 14 \ge 1 + 2 = 3,\) giving \(m \ge 9.\) With \(m = 9\) the list \(1, 3, 9, 11, 11\) works, so the least mode is \(m + 2 = 11.\)

Thus, the correct answer is D.

8 答案 D

整理得 \(y^4 - 2y^2 + 1 = x^4\)，即 \((y^2 - 1)^2 = (x^2)^2\)。作平方差分解，得 \[ (y^2 - 1 - x^2)(y^2 - 1 + x^2) = 0\text{。}\]

因此或者 \(y^2 - x^2 = 1\)，这是一条双曲线；或者 \(x^2 + y^2 = 1\)，这是一个圆。

所以正确答案是 D。

English

Rearranging, \(y^4 - 2y^2 + 1 = x^4,\) so \((y^2 - 1)^2 = (x^2)^2.\) This factors as \[ (y^2 - 1 - x^2)(y^2 - 1 + x^2) = 0. \]

Thus either \(y^2 - x^2 = 1,\) which is a hyperbola, or \(x^2 + y^2 = 1,\) which is a circle.

Thus, the correct answer is D.

9 答案 B

两边除以 \(2^{27}\)，得 \(2^{a_7 - 27} = a_7\)。因此 \(a_7=2^j\)，并且 \(2^j-j=27\)，其中 \(j=a_7-27\ge0\)。当 \(j\ge1\) 时，左边严格递增，而 \(j=5\) 满足方程，所以唯一解是 \(a_7=32\)。

设公差 \(d \ge 1\)，则 \(a_7 = a_0 + 7d = 32\)，并且 \(a_2 = a_0 + 2d = 32 - 5d\)。要使 \(a_2\) 最小，就要使 \(d\) 尽可能大。因为 \(a_0 = 32 - 7d \ge 1\)，最大选择为 \(d = 4\)（此时 \(a_0 = 4\)）。

于是 \(a_2 = 32 - 20 = 12\)。

因此，正确答案是 B。

English

Dividing by \(2^{27},\) we need \(2^{a_7 - 27} = a_7.\) Thus \(a_7=2^j\) and \(2^j-j=27,\) where \(j=a_7-27\ge0.\) The left side is strictly increasing for \(j\ge1,\) and \(j=5\) works, so uniquely \(a_7=32.\)

With common difference \(d \ge 1,\) we have \(a_7 = a_0 + 7d = 32\) and \(a_2 = a_0 + 2d = 32 - 5d.\) To minimize \(a_2\) we maximize \(d;\) since \(a_0 = 32 - 7d \ge 1,\) the largest choice is \(d = 4\) (giving \(a_0 = 4\)).

Then \(a_2 = 32 - 20 = 12.\)

Thus, the correct answer is B.

10 答案 D

将中心置于原点，取 \(A = (-1, \sqrt3)\)，\(B = (1, \sqrt3)\)，\(C = (2, 0)\)，\(D = (1, -\sqrt3)\)，\(E = (-1, -\sqrt3)\)，\(F = (-2, 0)\)。

则 \(G = (0, \sqrt3)\)，\(H = (0, -\sqrt3)\)。由对称性，四边形 \(GCHF\) 的四条边相等，并且 \[ GC = \sqrt{2^2 + (\sqrt3)^2} = \sqrt7\text{。}\]

所以周长为 \(4\sqrt7\)。

所以正确答案是 D。

English

Place the hexagon with center at the origin: \(A = (-1, \sqrt3),\) \(B = (1, \sqrt3),\) \(C = (2, 0),\) \(D = (1, -\sqrt3),\) \(E = (-1, -\sqrt3),\) \(F = (-2, 0).\)

Then \(G = (0, \sqrt3)\) and \(H = (0, -\sqrt3).\) By symmetry all four sides of \(GCHF\) are equal, and \[ GC = \sqrt{2^2 + (\sqrt3)^2} = \sqrt7. \]

The perimeter is \(4\sqrt7.\)

Thus, the correct answer is D.

11 答案 E

两个底数是本原三次单位根 \(\omega = e^{\frac{2\pi i}{3}}\) 及其共轭 \(\omega^2 = e^{-\frac{2\pi i}{3}}\)。因此 \(f(n) = \omega^n + \omega^{-n} = 2\cos\dfrac{2\pi n}{3}\)。

由于 \(2022\) 是 \(3\) 的倍数，\(\omega^{2022} = 1\)，所以 \(f(2022) = 1 + 1 = 2\)。

所以正确答案是 E。

English

The two bases are the primitive cube roots of unity, \(\omega = e^{\frac{2\pi i}{3}}\) and its conjugate \(\omega^2 = e^{-\frac{2\pi i}{3}}.\) So \(f(n) = \omega^n + \omega^{-n} = 2\cos\dfrac{2\pi n}{3}.\)

Since \(2022\) is a multiple of \(3,\) \(\omega^{2022} = 1,\) so \(f(2022) = 1 + 1 = 2.\)

Thus, the correct answer is E.

12 答案 D

将每个骰子分类为低档 \(\{1,2\}\)、中档 \(\{3,4\}\) 或高档 \(\{5,6\}\)；每类概率都是 \(\tfrac13\)，所以 \(3^4 = 81\) 个类别模式等可能。

需要至少一个高骰，即骰面大于 \(4\)，并且至少两个骰面大于 \(2\)，也就是中档或高档。两个坏事件分别是没有高骰，以及至多一个非低档骰。

没有高骰的模式有 \(2^4 = 16\) 个，至多一个非低骰的模式有 \(9\) 个，两种坏条件同时发生的模式有 \(5\) 个。由容斥，可行模式数为 \(81 - 16 - 9 + 5 = 61\)。

概率为 \(\dfrac{61}{81}\)。

所以正确答案是 D。

English

Sort each die into low \(\{1,2\},\) mid \(\{3,4\},\) or high \(\{5,6\};\) each has probability \(\tfrac13,\) so the \(3^4 = 81\) category patterns are equally likely.

We need at least one high die (a number greater than \(4\)) and at least two dice that are greater than \(2\) (mid or high). The two bad events are having no high die and having at most one non-low die.

There are \(2^4 = 16\) patterns of the first kind, \(9\) of the second kind, and \(5\) in their intersection. By inclusion-exclusion the count of good patterns is \(81 - 16 - 9 + 5 = 61.\)

The probability is \(\dfrac{61}{81}.\)

Thus, the correct answer is D.

13 答案 D

将长方形放在 \([0,8] \times [0,4]\)。利用 \(3\)-\(4\)-\(5\) 直角三角形，倾斜正方形的顶点可取为 \((4, 0)\)、\((0, 3)\)、\((3, 7)\) 和 \((7, 4)\)。

除了伸出顶边 \(y = 4\) 的三角形外，整个正方形都在长方形内。该三角形的顶点为 \((0.75, 4)\)、\((3, 7)\) 和 \((7, 4)\)，面积为 \[ \dfrac12 \cdot (7 - 0.75) \cdot (7 - 4) = \dfrac{75}{8}\text{。}\]

重叠区域的面积为 \(25 - \dfrac{75}{8} = \dfrac{125}{8} = 15\dfrac58\)。

所以正确答案是 D。

English

Place the rectangle as \([0,8] \times [0,4].\) The tilted square, using the \(3\)-\(4\)-\(5\) right triangles, has vertices \((4, 0),\) \((0, 3),\) \((3, 7),\) and \((7, 4).\)

The entire square lies inside the rectangle except for the triangle poking above the top edge \(y = 4.\) That triangle has vertices \((0.75, 4),\) \((3, 7),\) and \((7, 4),\) with area \[ \dfrac12 \cdot (7 - 0.75) \cdot (7 - 4) = \dfrac{75}{8}. \]

The region inside both is \(25 - \dfrac{75}{8} = \dfrac{125}{8} = 15\dfrac58.\)

Thus, the correct answer is D.

14 答案 E

因式分解得 \(x^2 + 2x - 15 = (x+5)(x-3)\)，所以 \(A = (-5, 0)\)，\(C = (3, 0)\)，且 \(y\)-截距为 \(B = (0, -15)\)。

于是 \(\vec{BA} = (-5, 15)\)，\(\vec{BC} = (3, 15)\)。利用叉积和点积，\[ \begin{gathered} \tan(\angle ABC) = \scriptsize \dfrac{|(-5)(15) - (15)(3)|}{(-5)(3) + (15)(15)} \\ = \dfrac{120}{210} \\ = \dfrac47 \end{gathered}\text{。}\]

所以正确答案是 E。

English

Factoring, \(x^2 + 2x - 15 = (x+5)(x-3),\) so \(A = (-5, 0)\) and \(C = (3, 0),\) and the \(y\)-intercept is \(B = (0, -15).\)

Then \(\vec{BA} = (-5, 15)\) and \(\vec{BC} = (3, 15).\) Using the cross and dot products, \[ \begin{gathered} \tan(\angle ABC) = \scriptsize \dfrac{|(-5)(15) - (15)(3)|}{(-5)(3) + (15)(15)} \\ = \dfrac{120}{210} \\ = \dfrac47. \end{gathered} \]

Thus, the correct answer is E.

15 答案 C

所有选项都是奇数，所以只需检查质数 \(3, 5, 7\)。

选项 A 中，\(2^{606} \equiv 1 \pmod 3\)，所以 \(2^{606} - 1\) 能被 \(3\) 整除。选项 B 中，\(2^{606} \equiv 4 \pmod 5\)，所以 \(2^{606} + 1\) 能被 \(5\) 整除。选项 D 中，\(2^{607} \equiv 2 \pmod 3\)，所以 \(2^{607} + 1\) 能被 \(3\) 整除。选项 E 对 \(5\) 取模时，\(2^{607} + 3^{607} \equiv 3 + 2 = 5 \equiv 0\)。

对于 \(2^{607} - 1\)，它对三取模时 \(\equiv 1 \pmod 3\)，对五取模时 \(\equiv 2 \pmod 5\)；而且由 \(2^3 \equiv 1 \pmod 7\) 和 \(607 \equiv 1 \pmod 3\) 可得它对七取模时 \(\equiv 1 \pmod 7\)。因此它不能被任何小于 \(10\) 的质数整除。

所以正确答案是 C。

English

Every option is odd, so only the primes \(3, 5, 7\) need checking.

Option A: \(2^{606} \equiv 1 \pmod 3,\) so \(2^{606} - 1\) is divisible by \(3.\) Option B: \(2^{606} \equiv 4 \pmod 5,\) so \(2^{606} + 1\) is divisible by \(5.\) Option D: \(2^{607} \equiv 2 \pmod 3,\) so \(2^{607} + 1\) is divisible by \(3.\) Option E: modulo \(5,\) \(2^{607} + 3^{607} \equiv 3 + 2 = 5 \equiv 0.\)

For \(2^{607} - 1:\) it is \(\equiv 1 \pmod 3,\) \(\equiv 2 \pmod 5,\) and (since \(2^3 \equiv 1 \pmod 7\) and \(607 \equiv 1 \pmod 3\)) \(\equiv 1 \pmod 7.\) So it is not divisible by any prime below \(10.\)

Thus, the correct answer is C.

16 答案 C

令 \(a = \log_2 x\)、\(b = \log_2 y\)。对第一个方程取 \(\log_2\)，由 \(x^y = 2^{64}\) 得 \(y \log_2 x = 64\)，即 \(a \cdot 2^b = 2^6\)。

对第二个方程取 \(\log_2\) 得到 \(b \log_2 a = 7\)，所以 \(a = 2^{\frac{7}{b}}\)。代入前式得 \(2^{\frac{7}{b}} \cdot 2^b = 2^6\)，也就是 \(b + \dfrac7b = 6\)，即 \(b^2 - 6b + 7 = 0\)。

解得 \(b = 3 \pm \sqrt2\)，因此 \(\log_2 y\) 的最大可能值为 \(3 + \sqrt2\)。

所以正确答案是 C。

English

Let \(a = \log_2 x\) and \(b = \log_2 y.\) Taking \(\log_2\) of \(x^y = 2^{64}\) gives \(y \log_2 x = 64,\) i.e. \(a \cdot 2^b = 2^6.\)

Taking \(\log_2\) of the second equation gives \(b \log_2 a = 7,\) so \(a = 2^{\frac{7}{b}}.\) Substituting, \(2^{\frac{7}{b}} \cdot 2^b = 2^6,\) so \(b + \dfrac7b = 6,\) i.e. \(b^2 - 6b + 7 = 0.\)

Thus \(b = 3 \pm \sqrt2,\) and the greatest value of \(\log_2 y\) is \(3 + \sqrt2.\)

Thus, the correct answer is C.

17 答案 D

行和为 \(4\) 的行全是 \(1\)，列和为 \(4\) 的列也全是 \(1\)。将行和 \(1, 2, 3, 4\) 分配给四行有 \(4!\) 种方式，选择哪一列的和为 \(4\) 有 \(4\) 种方式。

删除该列后，剩下的 \(4 \times 3\) 数组的行和为 \(0, 1, 2, 3\)，列和必须为 \(1, 2, 3\)。全零行和全一行已确定；行和为 \(1\) 和 \(2\) 的两行有 \(6\) 种排法，使列和按某种顺序成为 \(1, 2, 3\)。

因此总数为 \(24 \cdot 4 \cdot 6 = 576\)。

所以正确答案是 D。

English

The row with sum \(4\) is all \(1\)s and the column with sum \(4\) is all \(1\)s. There are \(4!\) ways to assign the row sums \(1, 2, 3, 4\) to the four rows, and \(4\) choices for which column has sum \(4.\)

Delete that column. The remaining \(4 \times 3\) array has row sums \(0, 1, 2, 3\) and must have column sums \(1, 2, 3.\) The all-zero and all-one rows are forced; the rows of reduced sum \(1\) and \(2\) can be placed in \(6\) ways to produce column sums \(1, 2, 3\) in some order.

The total is \(24 \cdot 4 \cdot 6 = 576.\)

Thus, the correct answer is D.

18 答案 C

只有内部 \(3 \times 3\) 子网格可以初始填充。要使中心在变换后填充，若它原为空，则需恰有 \(3\) 个填充邻居；若它原已填充，则需有 \(2\) 或 \(3\) 个填充邻居。

其他所有方格在变换后都必须为空。关键限制是任何边界方格都不能恰好获得三个填充邻居，这就排除了把 \(3 \times 3\) 子网格某条外边上的三格全部填充的情况。

若中心初始为空，则它的 \(8\) 个邻居中必须恰有 \(3\) 个被填充。在正方形的对称下检查这些三元组，只剩四种类型。以中心为 \((0,0)\) 建立环坐标，代表元为 \[ \begin{gathered} \{(-1,-1),(-1,0),(1,1)\},\\ \{(-1,-1),(-1,1),(1,-1)\},\\ \{(-1,-1),(-1,1),(1,0)\},\\ \{(-1,-1),(0,1),(1,0)\} \end{gathered}\text{。}\] 它们的对称轨道大小依次为 \(8,4,4,4\)，共给出 \(20\) 种构型。

若中心初始已填充，同样的邻居检查只留下两种构型，其中另外两个填充格是一对相对的角邻居。因此总数为 \(20+2=22\)。

所以正确答案是 C。

English

Only the inner \(3 \times 3\) squares can start filled. For the center to be filled afterward, if it began empty it needs exactly \(3\) filled neighbors, and if it began filled it needs \(2\) or \(3.\)

Every other square must end empty. The key restriction is that no border square may acquire exactly three filled neighbors, which rules out filling all three squares along an outer edge of the \(3 \times 3.\)

If the center starts empty, exactly \(3\) of its \(8\) neighbors must be filled. Checking these triples up to square symmetry leaves four types. With ring coordinates centered at \((0,0),\) representatives are \[ \begin{gathered} \{(-1,-1),(-1,0),(1,1)\},\\ \{(-1,-1),(-1,1),(1,-1)\},\\ \{(-1,-1),(-1,1),(1,0)\},\\ \{(-1,-1),(0,1),(1,0)\}. \end{gathered} \] Their symmetry-orbit sizes are \(8,4,4,4,\) giving \(20\) configurations.

If the center starts filled, the same neighbor check leaves only the two configurations in which the other filled cells are opposite corner neighbors. Hence the total is \(20+2=22.\)

Thus, the correct answer is C.

19 答案 A

令 \(a = BC\)、\(b = CA\)、\(c = AB\)。由于 \(E\) 是 \(AC\) 的中点，\(AE = \tfrac{b}{2}\)。重心性质给出 \(AG = \tfrac23 m_a\) 和 \(GE = \tfrac13 m_b\)，其中 \(m_a, m_b\) 分别是从 \(A\) 和 \(B\) 出发的中线长。

等边三角形 \(\triangle AGE\) 说明 \(AG = GE = AE\)。由 \(\tfrac23 m_a = \tfrac{b}{2}\) 得 \(m_a = \tfrac34 b\)，结合 \(m_a^2 = \tfrac{2b^2 + 2c^2 - a^2}{4}\) 得 \(2c^2 - a^2 = \tfrac{b^2}{4}\)。由 \(\tfrac13 m_b = \tfrac{b}{2}\) 得 \(m_b = \tfrac32 b\)，所以 \(a^2 + c^2 = 5b^2\)。

解这两个关系得 \(c^2 = \tfrac{7b^2}{4}\)、\(a^2 = \tfrac{13b^2}{4}\)。取 \(b = 2\)，则 \(a^2 = 13\)，\(c^2 = 7\)，于是 \[ \begin{aligned} \cos C &= \dfrac{a^2 + b^2 - c^2}{2ab} \\ &= \dfrac{13 + 4 - 7}{2 \cdot \sqrt{13} \cdot 2} \\ &= \dfrac{5\sqrt{13}}{26} \end{aligned}\text{。}\]

因此 \(m + n + p = 5 + 26 + 13 = 44\)。

所以正确答案是 A。

English

Let \(a = BC,\) \(b = CA,\) \(c = AB.\) Since \(E\) is the midpoint of \(AC,\) \(AE = \tfrac{b}{2}.\) The centroid gives \(AG = \tfrac23 m_a\) and \(GE = \tfrac13 m_b,\) where \(m_a, m_b\) are the medians from \(A\) and \(B.\)

Equilateral \(\triangle AGE\) means \(AG = GE = AE.\) From \(\tfrac23 m_a = \tfrac{b}{2}\) we get \(m_a = \tfrac34 b,\) which with \(m_a^2 = \tfrac{2b^2 + 2c^2 - a^2}{4}\) gives \(2c^2 - a^2 = \tfrac{b^2}{4}.\) From \(\tfrac13 m_b = \tfrac{b}{2}\) we get \(m_b = \tfrac32 b,\) giving \(a^2 + c^2 = 5b^2.\)

Solving, \(c^2 = \tfrac{7b^2}{4}\) and \(a^2 = \tfrac{13b^2}{4}.\) Taking \(b = 2\) gives \(a^2 = 13,\) \(c^2 = 7,\) so \[ \begin{aligned} \cos C &= \dfrac{a^2 + b^2 - c^2}{2ab} \\ &= \dfrac{13 + 4 - 7}{2 \cdot \sqrt{13} \cdot 2} \\ &= \dfrac{5\sqrt{13}}{26}. \end{aligned} \]

Then \(m + n + p = 5 + 26 + 13 = 44.\)

Thus, the correct answer is A.

20 答案 E

一次多项式不可能给出这两个不同的余式。若 \(P(x)=ax^2+bx+c\) 是二次的，比较它的两个余式会同时得到 \(b-a=1,\ c-a=2\) 和 \(b=2,\ c-a=1\)，互相矛盾。因此最低可能次数是 \(3\)。写成 \(P(x) = (x + 2)\) \(+ (x^2 + x + 1)(px + q)\)，它除以 \(x^2 + x + 1\) 的余式就是 \(x + 2\)。

模 \(x^2 + 1\) 化简（即令 \(x^2 \equiv -1\)），所得余式为 \((q + 1)x + (2 - p)\)。令它等于 \(2x + 1\)，得到 \(q = 1\) 且 \(p = 1\)。

于是 \(P(x) = x^3 + 2x^2 + 3x + 3\)，各系数的平方和为 \(1 + 4 + 9 + 9 = 23\)。

所以正确答案是 E。

English

No linear polynomial can have the two different remainders. If \(P(x)=ax^2+bx+c\) were quadratic, comparing its remainders would give both \(b-a=1,\ c-a=2\) and \(b=2,\ c-a=1,\) a contradiction. Thus the least possible degree is \(3.\) Write \(P(x) = (x + 2)\) \(+ (x^2 + x + 1)(px + q),\) which has remainder \(x + 2\) upon division by \(x^2 + x + 1.\)

Reducing modulo \(x^2 + 1\) (so \(x^2 \equiv -1\)) gives remainder \((q + 1)x + (2 - p).\) Setting this equal to \(2x + 1\) gives \(q = 1\) and \(p = 1.\)

Then \(P(x) = x^3 + 2x^2 + 3x + 3,\) and the sum of the squares of the coefficients is \(1 + 4 + 9 + 9 = 23.\)

Thus, the correct answer is E.

21 答案 E

前两个圆同心，半径分别为 \(2\) 和 \(8\)。同时与它们相切的圆，要么半径为 \(3\)，圆心到原点的距离为 \(5\)；要么半径为 \(5\)，圆心到原点的距离为 \(3\)。

第三个圆的圆心为 \((5, 0)\)，半径为 \(\sqrt3\)。对每个候选半径 \(s\in\{3,5\}\)，与第三个圆相切要求圆心到 \((5,0)\) 的距离为 \(s+\sqrt3\) 或 \(s-\sqrt3\)。这两个距离圆各与相应的圆心轨迹交于两个对称的点。因此恰有四个半径为 \(3\) 的圆和四个半径为 \(5\) 的圆满足条件。

面积和为 \(4 \cdot \pi(3)^2 + 4 \cdot \pi(5)^2\) \(= 36\pi + 100\pi = 136\pi\)。

所以正确答案是 E。

English

The first two circles are concentric with radii \(2\) and \(8.\) A circle tangent to both either has radius \(3\) with center at distance \(5\) from the origin, or radius \(5\) with center at distance \(3\) from the origin.

The third circle has center \((5, 0)\) and radius \(\sqrt3.\) For each candidate radius \(s\in\{3,5\},\) tangency requires the center’s distance from \((5,0)\) to be \(s+\sqrt3\) or \(s-\sqrt3.\) Each of these two distance circles intersects the appropriate center-locus in two symmetric points. Hence exactly four radius-\(3\) circles and four radius-\(5\) circles work.

The sum of the areas is \(4 \cdot \pi(3)^2 + 4 \cdot \pi(5)^2\) \(= 36\pi + 100\pi = 136\pi.\)

Thus, the correct answer is E.

22 答案 C

因为每个 \(t_n \lt 1\)，Amelia 一定至少完成两步。她恰好两步后停止当且仅当 \(t_1 + t_2 \gt 1\)，其概率为 \(\tfrac12\)；否则她走满三步。

位移与时间独立。若走两步，位置为 \(x_1 + x_2\)，且 \(P(x_1 + x_2 \gt 1) = \tfrac12\)。若走三步，位置为 \(x_1 + x_2 + x_3\)，且 \(P(x_1 + x_2 + x_3 \gt 1)\) \(= 1 - \tfrac16 = \tfrac56\)。

所求概率为 \(\tfrac12 \cdot \tfrac12 + \tfrac12 \cdot \tfrac56 = \tfrac14 + \tfrac{5}{12} = \tfrac23\)。

所以正确答案是 C。

English

Because each \(t_n \lt 1,\) Amelia always completes at least two steps. She stops after exactly two steps when \(t_1 + t_2 \gt 1,\) which happens with probability \(\tfrac12;\) otherwise she takes all three steps.

The increments are independent of the times. If she takes two steps, her position is \(x_1 + x_2,\) and \(P(x_1 + x_2 \gt 1) = \tfrac12.\) If she takes three, her position is \(x_1 + x_2 + x_3,\) and \(P(x_1 + x_2 + x_3 \gt 1)\) \(= 1 - \tfrac16 = \tfrac56.\)

The answer is \(\tfrac12 \cdot \tfrac12 + \tfrac12 \cdot \tfrac56 = \tfrac14 + \tfrac{5}{12} = \tfrac23.\)

Thus, the correct answer is C.

23 答案 A

因为 \(S_n\) 是由低 \(n\) 位构成的整数，条件 \(7S_n \equiv 1\) 表示 \(S_n \equiv 7^{-1} \pmod{2^n}\) 对每个 \(n\) 都成立。因此 \(x_k\) 是以 \(2\) 为底表示 \(\tfrac17\) 时的数位，也就是它作为 \(2\)-进数时的数位。

用 \(2\) 进制长除可得 \(x_0 = x_1 = x_2 = 1\)，之后以 \(3\) 为周期重复：当 \(k \ge 1\) 时，恰在 \(k\) 是 \(3\) 的倍数时有 \(x_k = 0\)，否则 \(x_k = 1\)。

因为 \(2019\) 和 \(2022\) 都是 \(3\) 的倍数，而 \(2020 \equiv 1\)、\(2021 \equiv 2 \pmod 3\)，所以 \(x_{2019} = 0\)、\(x_{2020} = 1\)、\(x_{2021} = 1\)、\(x_{2022} = 0\)。所求和为 \(0 + 2 + 4 + 0 = 6\)。

所以正确答案是 A。

English

Since \(S_n\) is the integer formed by the low \(n\) bits, the condition \(7S_n \equiv 1\) means \(S_n \equiv 7^{-1} \pmod{2^n}\) for every \(n.\) Thus the digits \(x_k\) are the base-\(2\) digits of \(\tfrac17\) as a \(2\)-adic number.

Long division in base \(2\) gives digits \(x_0 = x_1 = x_2 = 1,\) and thereafter the block repeats with period \(3:\) for \(k \ge 1,\) \(x_k = 0\) exactly when \(k\) is divisible by \(3,\) and \(x_k = 1\) otherwise.

Since \(2019\) and \(2022\) are both divisible by \(3,\) while \(2020 \equiv 1\) and \(2021 \equiv 2 \pmod 3,\) we get \(x_{2019} = 0,\) \(x_{2020} = 1,\) \(x_{2021} = 1,\) \(x_{2022} = 0.\) The sum is \(0 + 2 + 4 + 0 = 6.\)

Thus, the correct answer is A.

24 答案 C

相隔 \(d\) 步的两个顶点之间弦长平方为 \(2 - 2\cos\dfrac{2\pi d}{7}\)，且 \(d = 1, 2, 3\) 时每类有 \(7\) 条弦。因此所求和为 \[ 7 \sum_{d=1}^{3} \left(2 - 2\cos\tfrac{2\pi d}{7}\right)^2\text{。}\]

利用 \(\displaystyle\sum_{d=1}^{3} \cos\tfrac{2\pi d}{7} = -\tfrac12\) 以及 \(\displaystyle\sum_{d=1}^{3} \cos^2\tfrac{2\pi d}{7} = \tfrac54\)，括号内的和展开为 \(4\left(3 + 1 + \tfrac54\right) = 21\)。

因此总和是 \(7 \cdot 21 = 147\)。

所以正确答案是 C。

English

A chord joining two vertices \(d\) steps apart has squared length \(2 - 2\cos\dfrac{2\pi d}{7},\) and there are \(7\) chords for each of \(d = 1, 2, 3.\) The required sum is \[ 7 \sum_{d=1}^{3} \left(2 - 2\cos\tfrac{2\pi d}{7}\right)^2. \]

Using \(\displaystyle\sum_{d=1}^{3} \cos\tfrac{2\pi d}{7} = -\tfrac12\) and \(\displaystyle\sum_{d=1}^{3} \cos^2\tfrac{2\pi d}{7} = \tfrac54,\) the inner sum expands to \(4\left(3 + 1 + \tfrac54\right) = 21.\)

Therefore the total is \(7 \cdot 21 = 147.\)

Thus, the correct answer is C.

25 答案 B

将正方形中心置于原点，顶点为 \(\left(\pm\tfrac12, \pm\tfrac12\right)\)。每个六边形与正方形共用一条边，并向正方形的对侧延伸。例如底边上的六边形，其远端上边从 \(\left(-\tfrac12, \sqrt3 - \tfrac12\right)\) 到 \(\left(\tfrac12, \sqrt3 - \tfrac12\right)\)。

外边界是一个 \(12\) 边形，其平边到中心的距离为 \(\sqrt3 - \tfrac12\)，凸顶点包括 \(\left(\sqrt3 - \tfrac12, \tfrac12\right)\)，而相邻六边形的斜边在 \(\left(\tfrac52 - \sqrt3, \tfrac52 - \sqrt3\right)\) 及其对称点相交，形成四个凹入的顶点。

对这 \(12\) 个顶点使用鞋带公式，面积为 \(16\sqrt3 - 23\)，所以 \(m = 16\)，\(n = 3\)，\(p = -23\)，从而 \(m + n + p = -4\)。

所以正确答案是 B。

English

Center the square at the origin with vertices \(\left(\pm\tfrac12, \pm\tfrac12\right).\) Each hexagon shares one edge with the square and extends across to the opposite side; the hexagon on the bottom edge, for instance, has its far (top) edge from \(\left(-\tfrac12, \sqrt3 - \tfrac12\right)\) to \(\left(\tfrac12, \sqrt3 - \tfrac12\right).\)

The outer boundary is a \(12\)-gon with flat edges at distance \(\sqrt3 - \tfrac12\) from the center, convex vertices such as \(\left(\sqrt3 - \tfrac12, \tfrac12\right),\) and four reflex notches where adjacent hexagons’ slanted edges meet, at \(\left(\tfrac52 - \sqrt3, \tfrac52 - \sqrt3\right)\) and its symmetric images.

Applying the shoelace formula to these \(12\) vertices gives area \(16\sqrt3 - 23,\) so \(m = 16,\) \(n = 3,\) \(p = -23,\) and \(m + n + p = -4.\)

Thus, the correct answer is B.