2019A AMC 12 真题（P版）

1 答案 E

圆面积与半径的平方成正比，所以较大面积与较小面积的比为 \(\dfrac{16}{9}\)。

百分比增幅为 \[ \left(\dfrac{16}{9} - 1\right) \times 100 = \dfrac{700}{9} \approx 77.8\text{。} \] 最接近的整数是 \(78\)。

所以正确答案是 E。

English

The areas are proportional to the squares of the radii, so the ratio of the larger area to the smaller is \(\dfrac{16}{9}.\)

The percent increase is \[ \left(\dfrac{16}{9} - 1\right) \times 100 = \dfrac{700}{9} \approx 77.8. \] The closest integer is \(78.\)

Thus, the correct answer is E.

2 答案 D

因为 \(a = 1.5b\)，所以 \[ \dfrac{3b}{a} = \dfrac{3b}{1.5b} = 2\text{。}\]

化成百分数，\(3b\) 是 \(a\) 的 \(200\%\)。

所以正确答案是 D。

English

Since \(a = 1.5b,\) we have \[ \dfrac{3b}{a} = \dfrac{3b}{1.5b} = 2. \]

As a percentage, \(3b\) is \(200\%\) of \(a.\)

Thus, the correct answer is D.

3 答案 B

最坏情况下，我们取红、绿、黄各 \(14\) 个，再取完所有蓝球 \((13),\) 白球 \((11),\) 和黑球 \((9),\) 仍没有任何一种颜色达到 \(15\) 个。

这共有 \[ 14 + 14 + 14 + 13 + 11 + 9 = 75 \] 个球。

再取下一个球就一定会凑成某种颜色的 \(15\) 个，所以需要 \(76\) 个球。

所以正确答案是 B。

English

In the worst case, we draw \(14\) each of red, green, and yellow, plus all of the blue \((13),\) white \((11),\) and black \((9),\) without reaching \(15\) of any color.

That is \[ 14 + 14 + 14 + 13 + 11 + 9 = 75 \] balls.

The next ball must complete a set of \(15,\) so \(76\) balls are needed.

Thus, the correct answer is B.

4 答案 D

可以使用负整数。从 \(-44\) 到 \(44\) 的整数和为 \(0\)，所以从 \(-44\) 到 \(45\) 的整数和为 \(45\)。

这段共有 \(45 - (-44) + 1 = 90\) 个整数。反过来，如果 \(L\) 个连续整数的和为 \(45\)，那么它们和的两倍等于 \(L\) 乘以某个整数，所以 \(90\) 是 \(L\) 的倍数。因此 \(L\le90\)，证明这段最长。

因此，正确答案是 D。

English

Negative integers are allowed. The integers from \(-44\) to \(44\) sum to \(0,\) so the integers from \(-44\) to \(45\) sum to \(45.\)

This run has \(45 - (-44) + 1 = 90\) integers. Conversely, if \(L\) consecutive integers have sum \(45,\) then twice their sum is \(L\) times an integer, so \(90\) is a multiple of \(L.\) Hence \(L\le90,\) proving that this run is longest.

Thus, the correct answer is D.

5 答案 C

两条直线分别为 \(y = \tfrac{1}{2}x + 1\) 和 \(y = 2x - 2\)。分别与 \(x + y = 10\) 联立，得到点 \((6, 4)\) 和 \((4, 6)\)。

三角形顶点为 \((2, 2)\)，\((6, 4)\)，和 \((4, 6)\)。由鞋带公式，

\[ \begin{aligned} &\small \tfrac{1}{2}\left| 2(4 - 6) + 6(6 - 2) + 4(2 - 4) \right| \\ &= \tfrac{1}{2}\left| -4 + 24 - 8 \right| \\ &= 6 \end{aligned}\text{。}\]

所以正确答案是 C。

English

The two lines are \(y = \tfrac{1}{2}x + 1\) and \(y = 2x - 2.\) Intersecting each with \(x + y = 10\) gives the points \((6, 4)\) and \((4, 6).\)

The triangle has vertices \((2, 2),\) \((6, 4),\) and \((4, 6).\) By the shoelace formula,

\[ \begin{aligned} &\small \tfrac{1}{2}\left| 2(4 - 6) + 6(6 - 2) + 4(2 - 4) \right| \\ &= \tfrac{1}{2}\left| -4 + 24 - 8 \right| \\ &= 6. \end{aligned} \]

Thus, the correct answer is C.

6 答案 C

平移一个完整周期会把图形映到自身，所以平移可行。

绕 \(\ell\) 上适当的点旋转 \(180^\circ\) 会把直线上方的每个正方形送到直线下方的正方形，并且对角线段也对应，因此这个旋转可行。

关于 \(\ell\) 的反射会把上方朝右上的对角线送到下方朝右上的对角线，但实际下方的对角线朝左下，因此失败。关于垂直直线的反射也因同样原因失败。四类变换中只有 \(2\) 类可行。

所以正确答案是 C。

English

A translation by one full period maps the figure to itself, so translation works.

A \(180^\circ\) rotation about a suitable point on \(\ell\) sends each square above the line to the square below it, with the diagonal segments matching, so this rotation works.

Reflection across \(\ell\) sends the top-right diagonals to top-right diagonals below the line, but the actual below-line diagonals point to the bottom-left, so it fails. A reflection across a perpendicular line fails for the same reason. Only \(2\) of the four transformations work.

Thus, the correct answer is C.

7 答案 E

数值 \(1\) 到 \(28\) 各出现 \(12\) 次，都是众数，所以 \(d = \dfrac{14 + 15}{2} = 14.5\)。

排序后 \(365\) 个值的第 \(183\) 个是中位数。数值 \(1\) 到 \(15\) 占据前 \(180\) 个位置，所以第 \(183\) 个位置是 \(16\)；因此 \(M = 16\)。

所有数值的总和是 \(12(1 + \cdots + 28)\) \(+ 11(29 + 30)\) \(+ 7 \cdot 31 = 5738\)，所以 \(\mu = \dfrac{5738}{365} \approx 15.7\)。

因此 \(d \lt \mu \lt M\)。

所以正确答案是 E。

English

The values \(1\) through \(28\) each appear \(12\) times and are the modes, so \(d = \dfrac{14 + 15}{2} = 14.5.\)

The \(183\)rd of the \(365\) ordered values is the median. Values \(1\) through \(15\) fill the first \(180\) positions, so position \(183\) is \(16;\) thus \(M = 16.\)

The total of all values is \(12(1 + \cdots + 28)\) \(+ 11(29 + 30)\) \(+ 7 \cdot 31 = 5738,\) so \(\mu = \dfrac{5738}{365} \approx 15.7.\)

Therefore \(d \lt \mu \lt M.\)

Thus, the correct answer is E.

8 答案 D

当四条直线全部平行或全部共点时，交点数分别为 \(0\) 和 \(1\)。三条平行线被第四条截时有 \(3\) 个交点。三条共点直线加上一条不经过公共点的直线时有 \(4\) 个交点。一对平行且没有三线共点时有 \(5\) 个交点；四条直线处于一般位置时有 \(6\) 个交点。

数值 \(2\) 不可能。两条直线在 \(P\) 点相交后，第三条若不经过 \(P\)，就至少产生一个新交点；第四条不同的直线还必须产生另一个新交点。如果其余直线都经过 \(P\)，则只有一个交点。因此可实现的值为 \(0,1,3,4,5,6\)，它们的和为 \(19\)。

因此，正确答案是 D。

English

The values \(0\) and \(1\) occur when all four lines are parallel or all four are concurrent. Three parallel lines crossed by a fourth give \(3\) points. Three concurrent lines together with a fourth line not through their common point give \(4.\) One parallel pair with no three concurrent gives \(5,\) and four lines in general position give \(6.\)

The value \(2\) is impossible. Once two lines meet at \(P,\) a third line not through \(P\) creates at least one new intersection; a fourth distinct line must then create another new point. If every remaining line passes through \(P,\) there is only one intersection point instead. Thus the achievable values are \(0,1,3,4,5,6,\) whose sum is \(19.\)

Thus, the correct answer is D.

9 答案 E

取倒数，\[ \begin{aligned} \dfrac{1}{a_n} &= \dfrac{2a_{n-2} - a_{n-1}}{a_{n-2}a_{n-1}} \\ &= \dfrac{2}{a_{n-1}} - \dfrac{1}{a_{n-2}} \end{aligned}\text{。}\]

令 \(b_n = \dfrac{1}{a_n}\) 则 \(b_n = 2b_{n-1} - b_{n-2}\)，所以 \(b_n\) 是等差数列，且 \(b_1 = 1\)，\(b_2 = \dfrac{7}{3}\)，公差为 \(\dfrac{4}{3}\)。

因此 \(b_{2019} = 1 + 2018 \cdot \dfrac{4}{3} = \dfrac{8075}{3}\)，所以 \(a_{2019} = \dfrac{3}{8075}\)。由于它们互质，\(p + q = 8078\)。

所以正确答案是 E。

English

Taking reciprocals, \[ \begin{aligned} \dfrac{1}{a_n} &= \dfrac{2a_{n-2} - a_{n-1}}{a_{n-2}a_{n-1}} \\ &= \dfrac{2}{a_{n-1}} - \dfrac{1}{a_{n-2}}. \end{aligned} \]

Let \(b_n = \dfrac{1}{a_n}.\) Then \(b_n = 2b_{n-1} - b_{n-2},\) so \(b_n\) is arithmetic with \(b_1 = 1,\) \(b_2 = \dfrac{7}{3},\) and common difference \(\dfrac{4}{3}.\)

Thus \(b_{2019} = 1 + 2018 \cdot \dfrac{4}{3} = \dfrac{8075}{3},\) so \(a_{2019} = \dfrac{3}{8075}.\) Since these are relatively prime, \(p + q = 8078.\)

Thus, the correct answer is E.

10 答案 A

在中心放一个单位圆，六个圆围绕它且圆心距中心为 \(2\)（形成一个六边形），再在外侧空隙中放六个圆，其圆心距中心为 \(2\sqrt{3}\)。这共有 \(1 + 6 + 6 = 13\) 个圆。

最外层圆与大圆相切，所以大圆半径为 \(2\sqrt{3} + 1\)。它的面积为 \[ \pi(2\sqrt{3} + 1)^2 = \pi(13 + 4\sqrt{3})\text{。}\]

减去 \(13\) 个单位圆的面积，剩下 \(\pi(13 + 4\sqrt{3}) - 13\pi = 4\pi\sqrt{3}\)。

所以正确答案是 A。

English

Place a unit circle at the center, six around it with centers at distance \(2\) (a hexagon), and six more with centers at distance \(2\sqrt{3}\) in the outer gaps. That is \(1 + 6 + 6 = 13\) circles.

The outermost circles are tangent to the big circle, whose radius is therefore \(2\sqrt{3} + 1.\) Its area is \[ \pi(2\sqrt{3} + 1)^2 = \pi(13 + 4\sqrt{3}). \]

Subtracting the \(13\) unit circles leaves \(\pi(13 + 4\sqrt{3}) - 13\pi = 4\pi\sqrt{3}.\)

Thus, the correct answer is A.

11 答案 D

循环节给出 \[ 0.\overline{23}_k = \dfrac{2k + 3}{k^2 - 1} = \dfrac{7}{51}\text{。}\]

交叉相乘得 \(51(2k + 3) = 7(k^2 - 1)\)，所以 \(7k^2 - 102k - 160 = 0\)。

由二次公式，\[ \begin{aligned} k &= \dfrac{102 + \sqrt{14884}}{14} \\ &= \dfrac{102 + 122}{14} = 16 \end{aligned}\text{。}\]

所以正确答案是 D。

English

The repeating block gives \[ 0.\overline{23}_k = \dfrac{2k + 3}{k^2 - 1} = \dfrac{7}{51}. \]

Cross-multiplying, \(51(2k + 3) = 7(k^2 - 1),\) so \(7k^2 - 102k - 160 = 0.\)

The quadratic formula gives \[ \begin{aligned} k &= \dfrac{102 + \sqrt{14884}}{14} \\ &= \dfrac{102 + 122}{14} = 16. \end{aligned} \]

Thus, the correct answer is D.

12 答案 B

令 \(a = \log_2 x\) 且 \(b = \log_2 y\)，则 \(\log_y 16 = \dfrac{4}{b}\)，所以 \(a = \dfrac{4}{b}\)，从而 \(ab = 4\)。

因为 \(xy = 64\)，所以 \(a + b = 6\)。

因此 \[ \begin{aligned} \left(\log_2 \tfrac{x}{y}\right)^2 &= (a - b)^2 \\ &= (a + b)^2 - 4ab \\ &= 36 - 16 = 20 \end{aligned}\text{。}\]

所以正确答案是 B。

English

Let \(a = \log_2 x\) and \(b = \log_2 y.\) Then \(\log_y 16 = \dfrac{4}{b},\) so \(a = \dfrac{4}{b},\) giving \(ab = 4.\)

Since \(xy = 64,\) we have \(a + b = 6.\)

Therefore \[ \begin{aligned} \left(\log_2 \tfrac{x}{y}\right)^2 &= (a - b)^2 \\ &= (a + b)^2 - 4ab \\ &= 36 - 16 = 20. \end{aligned} \]

Thus, the correct answer is B.

13 答案 E

质数 \(5\) 和 \(7\) 在这里没有真因数，因此各有 \(3\) 种选择。

沿着链 \(2 \to 4 \to 8\) 有 \(3 \cdot 2 \cdot 1 = 6\) 种涂法。数 \(9\) 必须与 \(3\) 不同色，所以一旦 \(3\) 的颜色确定，它就有 \(2\) 种选择。

数 \(6\) 必须与 \(2\) 和 \(3\) 都不同色。对 \(2\) 和 \(3\) 的颜色求和（同色有 \(3\) 对，异色有 \(6\) 对），\(4, 8, 9, 6\) 合起来的因子为 \(2 \cdot 2 \cdot (3 \cdot 2 + 6 \cdot 1) = 48\)。

再乘上 \(5\) 和 \(7\) 的涂法数 \(9\)，得到 \(48 \cdot 9 = 432\)。

所以正确答案是 E。

English

The primes \(5\) and \(7\) have no proper divisors here, giving \(3\) choices each.

Along the chain \(2 \to 4 \to 8,\) there are \(3 \cdot 2 \cdot 1 = 6\) colorings. Number \(9\) must differ from \(3,\) giving \(2\) choices once \(3\) is set.

Number \(6\) must differ from both \(2\) and \(3.\) Summing over the colors of \(2\) and \(3\) (equal in \(3\) pairs, unequal in \(6\) pairs), the combined factor for \(4, 8, 9, 6\) totals \(2 \cdot 2 \cdot (3 \cdot 2 + 6 \cdot 1) = 48.\)

Multiplying by the \(9\) ways for \(5\) and \(7\) gives \(48 \cdot 9 = 432.\)

Thus, the correct answer is E.

14 答案 E

因式 \(x^2 - 2x + 2\) 和 \(x^2 - 4x + 8\) 的根分别为 \(1 \pm i\) 和 \(2 \pm 2i\)，这 \(4\) 个值互不相同。

要使 \(P\) 恰有 \(4\) 个不同根，\(x^2 - cx + 4\) 的根必须在这些根中。它们的乘积必须等于 \(4\)，唯一可行的配对是从两个因式各取一个根，例如 \((1 + i)(2 - 2i) = 4\)。

此时 \(c = (1 + i) + (2 - 2i) = 3 - i\)，所以 \(|c| = \sqrt{3^2 + 1^2} = \sqrt{10}\)。

所以正确答案是 E。

English

The factors \(x^2 - 2x + 2\) and \(x^2 - 4x + 8\) have roots \(1 \pm i\) and \(2 \pm 2i,\) which are \(4\) distinct values.

For \(P\) to have exactly \(4\) distinct roots, the roots of \(x^2 - cx + 4\) must lie among these. Their product must equal \(4,\) and the only such pair is one root from each factor, for example \((1 + i)(2 - 2i) = 4.\)

Then \(c = (1 + i) + (2 - 2i) = 3 - i,\) so \(|c| = \sqrt{3^2 + 1^2} = \sqrt{10}.\)

Thus, the correct answer is E.

15 答案 D

令 \(\sqrt{\log a} = p\) 且 \(\sqrt{\log b} = q\)，则 \(\log a = p^2\) 且 \(\log\sqrt{a} = \dfrac{p^2}{2}\)。要使它为整数，\(p\) 必须为偶数；同理 \(q\) 也必须为偶数。

写成 \(p = 2m\)，\(q = 2n\)，方程 \(p + q + \dfrac{p^2}{2} + \dfrac{q^2}{2} = 100\) 变为 \(m(m+1) + n(n+1) = 50\)。

唯一解为 \(\{m, n\} = \{4, 5\}\)，因此 \(\log(ab) = p^2 + q^2\) \(= 4(16 + 25) = 164\)。

所以 \(ab = 10^{164}\)。

所以正确答案是 D。

English

Let \(\sqrt{\log a} = p\) and \(\sqrt{\log b} = q,\) so \(\log a = p^2\) and \(\log\sqrt{a} = \dfrac{p^2}{2}.\) For this to be an integer, \(p\) is even; likewise \(q.\)

Writing \(p = 2m,\) \(q = 2n,\) the equation \(p + q + \dfrac{p^2}{2} + \dfrac{q^2}{2} = 100\) becomes \(m(m+1) + n(n+1) = 50.\)

The only solution is \(\{m, n\} = \{4, 5\},\) giving \(\log(ab) = p^2 + q^2\) \(= 4(16 + 25) = 164.\)

Therefore \(ab = 10^{164}.\)

Thus, the correct answer is D.

16 答案 B

有 \(5\) 个奇数和 \(4\) 个偶数。每一行和每一列都必须包含奇数个奇数项。

要放置 \(5\) 个奇数项并使每一行和每一列的奇数项个数都为奇数，唯一的形状是填满一整行和一整列（一个由 \(3 + 3 - 1 = 5\) 个格子组成的加号形）。这样的图案有 \(3 \cdot 3 = 9\) 种。

每种图案中，奇数有 \(5!\) 种放法，偶数有 \(4!\) 种放法，所以概率为 \[ \dfrac{9 \cdot 5! \cdot 4!}{9!} = \dfrac{1}{14}\text{。}\]

所以正确答案是 B。

English

There are \(5\) odd and \(4\) even numbers. Each row and column must contain an odd number of odd entries.

The only way to place \(5\) odd entries with every row and column odd is to fill one complete row and one complete column (a plus shape of \(3 + 3 - 1 = 5\) cells). There are \(3 \cdot 3 = 9\) such patterns.

Each pattern admits \(5!\) placements of the odd numbers and \(4!\) of the even numbers, so the probability is \[ \dfrac{9 \cdot 5! \cdot 4!}{9!} = \dfrac{1}{14}. \]

Thus, the correct answer is B.

17 答案 D

每个根 \(r\) 都满足 \(r^3 = 5r^2 - 8r + 13\)，因此 \(r^{k+1} = 5r^k - 8r^{k-1} + 13r^{k-2}\)。

对三个根求和，得到 \(s_{k+1} = 5s_k - 8s_{k-1} + 13s_{k-2}\)，所以 \(a = 5\)，\(b = -8\)，\(c = 13\)。

因此 \(a + b + c = 5 - 8 + 13 = 10\)。

所以正确答案是 D。

English

Every root \(r\) satisfies \(r^3 = 5r^2 - 8r + 13,\) so \(r^{k+1} = 5r^k - 8r^{k-1} + 13r^{k-2}.\)

Summing over the three roots gives \(s_{k+1} = 5s_k - 8s_{k-1} + 13s_{k-2},\) so \(a = 5,\) \(b = -8,\) \(c = 13.\)

Therefore \(a + b + c = 5 - 8 + 13 = 10.\)

Thus, the correct answer is D.

18 答案 D

球与三角形所在平面相交成一个半径为 \(\sqrt{36 - d^2}\) 的圆，其中 \(d\) 是 \(O\) 到该平面的距离。因为每条边都与球相切，这个圆就是三角形的内切圆。

该三角形面积为 \(\tfrac{1}{2} \cdot 24 \cdot 9 = 108\)，半周长为 \(27\)，所以内切圆半径为 \(\dfrac{108}{27} = 4\)。

因此 \(\sqrt{36 - d^2} = 4\)，得 \(d^2 = 20\)，所以 \(d = 2\sqrt{5}\)。

所以正确答案是 D。

English

The sphere intersects the triangle’s plane in a circle of radius \(\sqrt{36 - d^2},\) where \(d\) is the distance from \(O\) to the plane. Since each side is tangent to the sphere, this circle is the triangle’s incircle.

The triangle has area \(\tfrac{1}{2} \cdot 24 \cdot 9 = 108\) and semiperimeter \(27,\) so its inradius is \(\dfrac{108}{27} = 4.\)

Thus \(\sqrt{36 - d^2} = 4,\) giving \(d^2 = 20\) and \(d = 2\sqrt{5}.\)

Thus, the correct answer is D.

19 答案 A

每个正弦值为 \(\sqrt{1 - \cos^2}\)：\(\sin A = \dfrac{3\sqrt{15}}{16}\)，\(\sin B = \dfrac{2\sqrt{15}}{16}\)，\(\sin C = \dfrac{4\sqrt{15}}{16}\)。

由正弦定理，边长比为 \(3 : 2 : 4\)。最小的整数边长为 \(3, 2, 4\)，它们满足三角形不等式。

最小周长为 \(3 + 2 + 4 = 9\)。

所以正确答案是 A。

English

Each sine is \(\sqrt{1 - \cos^2}:\) \(\sin A = \dfrac{3\sqrt{15}}{16},\) \(\sin B = \dfrac{2\sqrt{15}}{16},\) \(\sin C = \dfrac{4\sqrt{15}}{16}.\)

By the Law of Sines the sides are in ratio \(3 : 2 : 4.\) The smallest integer sides are \(3, 2, 4,\) which satisfy the triangle inequality.

The least perimeter is \(3 + 2 + 4 = 9.\)

Thus, the correct answer is A.

20 答案 B

每个变量以概率 \(\tfrac14\) 等于 \(0\)，以概率 \(\tfrac14\) 等于 \(1\)，并以概率 \(\tfrac12\) 在 \([0, 1]\) 上均匀分布。

考虑九种类型组合：数对 \((0, 1)\) 和 \((1, 0)\) 各贡献 \(\tfrac{1}{16}\)。四种“定点对均匀”的情况各贡献 \(\tfrac{1}{16}\)。“均匀对均匀”的情况贡献 \(\tfrac14 \cdot \tfrac14 = \tfrac{1}{16}\)。

总和为 \(\dfrac{2 + 4 + 1}{16} = \dfrac{7}{16}\)。

所以正确答案是 B。

English

Each variable equals \(0\) with probability \(\tfrac14,\) equals \(1\) with probability \(\tfrac14,\) and is uniform on \([0, 1]\) with probability \(\tfrac12.\)

Considering the nine combinations of types: the pairs \((0, 1)\) and \((1, 0)\) each contribute \(\tfrac{1}{16}.\) Each of the four point-versus-uniform cases contributes \(\tfrac{1}{16}.\) The uniform-versus-uniform case contributes \(\tfrac14 \cdot \tfrac14 = \tfrac{1}{16}.\)

The total is \(\dfrac{2 + 4 + 1}{16} = \dfrac{7}{16}.\)

Thus, the correct answer is B.

21 答案 C

因为 \(z = e^{\frac{i\pi}{4}}\)，所以 \(z^{k^2} = e^{\frac{i\pi k^2}{4}}\)，只取决于 \(k^2 \bmod 8\)。

对 \(k = 1\) 到 \(12\)，余数 \(k^2 \bmod 8\) 为 \(1\)（给出 \(z\)）出现六次，为 \(4\)（给出 \(-1\)）出现三次，为 \(0\)（给出 \(1\)）出现三次。所以第一个和为 \(6z - 3 + 3 = 6z\)。

第二个和同理为 \(\dfrac{6}{z} - 3 + 3 = \dfrac{6}{z}\)。两者乘积为 \(6z \cdot \dfrac{6}{z} = 36\)。

所以正确答案是 C。

English

Since \(z = e^{\frac{i\pi}{4}},\) we have \(z^{k^2} = e^{\frac{i\pi k^2}{4}},\) depending only on \(k^2 \bmod 8.\)

For \(k = 1\) to \(12,\) the residue \(k^2 \bmod 8\) is \(1\) (giving \(z\)) six times, \(4\) (giving \(-1\)) three times, and \(0\) (giving \(1\)) three times. So the first sum is \(6z - 3 + 3 = 6z.\)

The second sum is likewise \(\dfrac{6}{z} - 3 + 3 = \dfrac{6}{z}.\) Their product is \(6z \cdot \dfrac{6}{z} = 36.\)

Thus, the correct answer is C.

22 答案 E

令 \(s = AB\)。因为 \(\dfrac{BP}{CP} = 3\)，所以 \(BP = \dfrac{3s}{4}\) 且 \(CP = \dfrac{s}{4}\)。把 \(P\) 放在原点，令 \(BC\) 在 \(x\)-轴上，\(B = \left(-\tfrac{3s}{4}, 0\right)\)，\(C = \left(\tfrac{s}{4}, 0\right)\)，顶点 \(A = \left(-\tfrac{s}{4}, \tfrac{s\sqrt{3}}{2}\right)\)。

点 \(P, O, A\) 共线，所以 \(O = t \cdot A\) 对某个标量 \(t\) 成立。两个条件确定它：\(O\) 到直线 \(BC\) 的距离为 \(17\) 得 \(|t| \cdot \dfrac{s\sqrt{3}}{2} = 17\)，且 \(A\) 在 \(\omega\) 上，得 \(|t - 1| \cdot \dfrac{s\sqrt{13}}{4} = 20\)，因为 \(|A| = \dfrac{s\sqrt{13}}{4}\)。

解得 \(|t| s = \dfrac{34}{\sqrt{3}}\) 且 \(|t - 1| s = \dfrac{80}{\sqrt{13}}\)。有效构型中 \(O\) 与 \(A\) 位于 \(P\) 的两侧，所以 \(|t-1|s=|t|s+s\)。因此 \[ AB = s = \dfrac{80}{\sqrt{13}} - \dfrac{34}{\sqrt{3}}\text{。}\]

因此 \(m + n + p + q = 80 + 13\) \(+ 34 + 3 = 130\)。

所以正确答案是 E。

English

Let \(s = AB.\) Since \(\dfrac{BP}{CP} = 3,\) we have \(BP = \dfrac{3s}{4}\) and \(CP = \dfrac{s}{4}.\) Put \(P\) at the origin with \(BC\) on the \(x\)-axis, \(B = \left(-\tfrac{3s}{4}, 0\right),\) \(C = \left(\tfrac{s}{4}, 0\right),\) and apex \(A = \left(-\tfrac{s}{4}, \tfrac{s\sqrt{3}}{2}\right).\)

Points \(P, O, A\) are collinear, so \(O = t \cdot A\) for some scalar \(t.\) Two conditions pin it down: \(O\) is at distance \(17\) from line \(BC,\) giving \(|t| \cdot \dfrac{s\sqrt{3}}{2} = 17,\) and \(A\) is on \(\omega,\) giving \(|t - 1| \cdot \dfrac{s\sqrt{13}}{4} = 20\) since \(|A| = \dfrac{s\sqrt{13}}{4}.\)

Solving, \(|t| s = \dfrac{34}{\sqrt{3}}\) and \(|t - 1| s = \dfrac{80}{\sqrt{13}}.\) The valid configuration has \(O\) and \(A\) on opposite sides of \(P,\) so \(|t-1|s=|t|s+s.\) Therefore \[ AB = s = \dfrac{80}{\sqrt{13}} - \dfrac{34}{\sqrt{3}}. \]

Then \(m + n + p + q = 80 + 13\) \(+ 34 + 3 = 130.\)

Thus, the correct answer is E.

23 答案 D

令 \(L(x) = \log_7 x\)。则 \(L(a \diamondsuit b) = L(a)L(b)\)，且 \(L(a \heartsuit b) = \dfrac{L(a)}{L(b)}\)。

所以 \(L(a_3) = \dfrac{L(3)}{L(2)}\)，且 \(L(a_n) = \dfrac{L(n)}{L(n-1)} \cdot L(a_{n-1})\)。该乘积会裂项相消：\[ \begin{aligned} L(a_N) &= \dfrac{L(3)}{L(2)} \cdot \dfrac{L(N)}{L(3)} \\ &= \dfrac{L(N)}{L(2)} \end{aligned}\text{。}\]

因此 \(L(a_{2019}) = \dfrac{\log_7 2019}{\log_7 2}\) \(= \log_2 2019 \approx 10.98\)，四舍五入为 \(11\)。

所以正确答案是 D。

English

Let \(L(x) = \log_7 x.\) Then \(L(a \diamondsuit b) = L(a)L(b)\) and \(L(a \heartsuit b) = \dfrac{L(a)}{L(b)}.\)

So \(L(a_3) = \dfrac{L(3)}{L(2)},\) and \(L(a_n) = \dfrac{L(n)}{L(n-1)} \cdot L(a_{n-1}).\) The product telescopes: \[ \begin{aligned} L(a_N) &= \dfrac{L(3)}{L(2)} \cdot \dfrac{L(N)}{L(3)} \\ &= \dfrac{L(N)}{L(2)}. \end{aligned} \]

Hence \(L(a_{2019}) = \dfrac{\log_7 2019}{\log_7 2}\) \(= \log_2 2019 \approx 10.98,\) which rounds to \(11.\)

Thus, the correct answer is D.

24 答案 D

固定一个质数 \(p\le n\)。由勒让德公式，分子中 \(p\) 的指数减去分母中该质数的指数为 \[ \begin{aligned} D_p &=\sum_{k\ge1}\left\lfloor\dfrac{n^2-1}{p^k}\right\rfloor\\ &\quad-n\sum_{k\ge1}\left\lfloor\dfrac{n}{p^k}\right\rfloor \end{aligned}\text{。}\]若 \(r_k\) 是 \(n\) 除以 \(p^k\) 的余数，则第 \(k\) 个加项为 \(\left\lfloor\dfrac{nr_k-1}{p^k}\right\rfloor\)。

令 \(a=v_p(n)\)。前 \(a\) 个加项都是 \(-1\)。如果 \(n\) 不是 \(p\) 的幂，写成 \(n=p^a m\)，其中 \(m\ge2\)。当 \(a\ge1\) 时，下一个加项至少为 \(\dfrac{n}{p}-1\ge a\)，其后各项都非负；当 \(a=0\) 时，所有加项本来就非负。因此除非 \(n\) 是 \(p\) 的幂，否则 \(D_p\ge0\)。

当 \(n=p^a\) 时，勒让德公式将条件化为 \(p^a-1\ge2a(p-1)\)。在不超过 \(50\) 的质数幂中，这个条件恰好在 \(a=1\)（即 \(n\) 为质数）以及 \(n=2^2=4\) 时失败。不超过 \(50\) 的质数共有 \(15\) 个，再加上 \(n=4\)，共 \(16\) 个失败值。因此有 \(50 - 16 = 34\) 个 \(n\) 满足条件。

因此，正确答案是 D。

English

Fix a prime \(p\le n.\) By Legendre’s formula, the difference between the exponent of \(p\) in the numerator and its exponent in the denominator is \[ \begin{aligned} D_p &=\sum_{k\ge1}\left\lfloor\dfrac{n^2-1}{p^k}\right\rfloor\\ &\quad-n\sum_{k\ge1}\left\lfloor\dfrac{n}{p^k}\right\rfloor. \end{aligned} \] If \(r_k\) is the remainder of \(n\) modulo \(p^k,\) the \(k\)th summand is \(\left\lfloor\dfrac{nr_k-1}{p^k}\right\rfloor.\)

Let \(a=v_p(n).\) The first \(a\) summands are \(-1.\) If \(n\) is not a power of \(p,\) write \(n=p^a m\) with \(m\ge2.\) When \(a\ge1,\) the next summand is at least \(\dfrac{n}{p}-1\ge a,\) and all later summands are nonnegative; when \(a=0,\) every summand is already nonnegative. Thus \(D_p\ge0\) unless \(n\) is a power of \(p.\)

For \(n=p^a,\) Legendre’s formula reduces the requirement to \(p^a-1\ge2a(p-1).\) Among prime powers at most \(50,\) this fails exactly when \(a=1\) (so \(n\) is prime) and when \(n=2^2=4.\) There are \(15\) primes at most \(50,\) plus \(n=4,\) giving \(16\) failures. Hence \(50 - 16 = 34\) values of \(n\) work.

Thus, the correct answer is D.

25 答案 E

对锐角三角形，垂足三角形（高的垂足构成的三角形）的每个角为 \(180^\circ - 2\alpha\)，其中 \(\alpha\) 是原三角形的对应角。

把一个角写成 \(60^\circ + x\)，新角就是 \(60^\circ - 2x\)，所以每一步都会把相对 \(60^\circ\) 的偏差乘以 \(-2\)。初始偏差为 \(\pm 0.001^\circ\)。

\(n\) 步后，偏差大小为 \(0.001 \cdot 2^n\) 度。三角形第一次变成钝角时这个值超过 \(30^\circ\)，即 \(2^n \gt 30000\)。因为 \(2^{14} = 16384\) 且 \(2^{15} = 32768\)，最小的 \(n\) 是 \(15\)。

所以正确答案是 E。

English

For an acute triangle, the orthic triangle (feet of the altitudes) has angles \(180^\circ - 2\alpha\) for each original angle \(\alpha.\)

Writing an angle as \(60^\circ + x,\) the new angle is \(60^\circ - 2x,\) so each deviation from \(60^\circ\) is multiplied by \(-2.\) The initial deviations are \(\pm 0.001^\circ.\)

After \(n\) steps a deviation has magnitude \(0.001 \cdot 2^n\) degrees. The triangle first becomes obtuse when this exceeds \(30^\circ,\) i.e. \(2^n \gt 30000.\) Since \(2^{14} = 16384\) and \(2^{15} = 32768,\) the least such \(n\) is \(15.\)

Thus, the correct answer is E.