2015A AMC 12 真题（P版）

在括号内，\[ \begin{aligned} &2^0-1+5^2+0 \\ &\quad = 1-1+25+0 = 25 \end{aligned}\text{。}\]

于是 \[(25)^{-1}\times 5 = \dfrac{5}{25} = \dfrac{1}{5}\text{。}\]

因此，正确答案是 C。

English

Inside the parentheses, \[ \begin{aligned} &2^0-1+5^2+0 \\ &\quad = 1-1+25+0 = 25. \end{aligned} \]

Then \[(25)^{-1}\times 5 = \dfrac{5}{25} = \dfrac{1}{5}.\]

Thus, the correct answer is C.

根据三角形不等式，第三边 \(s\) 满足 \(20-15 \lt s \lt 20+15\)，也就是 \(5 \lt s \lt 35\)。

周长为 \(35 + s\)，所以它严格介于 \(40\) 和 \(70\) 之间。选项中只有 \(72\) 落在这个范围之外。

因此，正确答案是 E。

English

By the Triangle Inequality, the third side \(s\) satisfies \(20-15 \lt s \lt 20+15,\) that is \(5 \lt s \lt 35.\)

The perimeter is \(35 + s,\) so it lies strictly between \(40\) and \(70.\) Among the choices, only \(72\) falls outside this range.

Thus, the correct answer is E.

其他 \(14\) 个分数的总和是 \(14\cdot 80 = 1120\)。全部 \(15\) 个分数的总和是 \(15\cdot 81 = 1215\)。

因此 Payton 的分数是 \(1215 - 1120 = 95\)。

因此，正确答案是 E。

English

The sum of the \(14\) other scores was \(14\cdot 80 = 1120.\) The sum of all \(15\) scores was \(15\cdot 81 = 1215.\)

Therefore Payton’s score was \(1215 - 1120 = 95.\)

Thus, the correct answer is E.

设这两个数为 \(x\) 和 \(y\)，其中 \(x \gt y \gt 0\)，则 \(x+y = 5(x-y)\)。

展开得 \(x+y = 5x-5y\)，所以 \(6y = 4x\)，于是 \[\dfrac{x}{y} = \dfrac{3}{2}\text{。}\]

因此，正确答案是 B。

English

Let \(x\) and \(y\) be the numbers with \(x \gt y \gt 0.\) Then \(x+y = 5(x-y).\)

Expanding gives \(x+y = 5x-5y,\) so \(6y = 4x\) and \[\dfrac{x}{y} = \dfrac{3}{2}.\]

Thus, the correct answer is B.

要使 \(\dfrac{a}{b}\) 变大，应把分子 \(a\) 向上取整，把分母 \(b\) 向下取整。要使 \(-c\) 变大，应把 \(c\) 向下取整。

只有选项 \(\text{(D)}\) 同时做到这三点：把 \(a\) 向上取整，同时把 \(b\) 和 \(c\) 向下取整，因此每个变化都会使估算值高于精确值。其他选项中至少有一个变化方向相反，所以不能保证估算值更大。

因此，正确答案是 D。

English

To make \(\dfrac{a}{b}\) larger, round the numerator \(a\) up and the denominator \(b\) down. To make \(-c\) larger, round \(c\) down.

Only choice \(\text{(D)}\) does all three: it rounds \(a\) up while rounding \(b\) and \(c\) down, so every change pushes the estimate above the exact value. In the other choices at least one change works the wrong way, so the estimate is not guaranteed to be larger.

Thus, the correct answer is D.

设 \(p\) 和 \(c\) 分别为 Pete 和 Claire 现在的年龄。则 \(p-2 = 3(c-2)\) 且 \(p-4 = 4(c-4)\)。

解得 \(p = 20\) 且 \(c = 8\)，所以 Pete 比 Claire 大 \(12\) 岁。

当 Claire 为 \(12\) 岁时，年龄比为 \(2 : 1\)，这距离现在还有 \(12 - 8 = 4\) 年。

因此，正确答案是 B。

English

Let \(p\) and \(c\) be Pete’s and Claire’s current ages. Then \(p-2 = 3(c-2)\) and \(p-4 = 4(c-4).\)

Solving these gives \(p = 20\) and \(c = 8,\) so Pete is \(12\) years older than Claire.

The ratio is \(2 : 1\) when Claire is \(12,\) which is \(12 - 8 = 4\) years from now.

Thus, the correct answer is B.

设第一个和第二个圆柱的半径、高分别为 \(r,h\) 和 \(R,H\)。体积相等，所以 \(\pi r^2 h = \pi R^2 H\)，且 \(R = 1.1r\)。

则 \(\pi r^2 h = \pi(1.1r)^2 H\) \(= \pi(1.21 r^2) H\)。除以 \(\pi r^2\) 得 \(h = 1.21H\)，所以第一个高比第二个多 \(21\%\)。

因此，正确答案是 D。

English

Let \(r,h\) and \(R,H\) be the radii and heights of the first and second cylinders. The volumes are equal, so \(\pi r^2 h = \pi R^2 H,\) and \(R = 1.1r.\)

Then \(\pi r^2 h = \pi(1.1r)^2 H\) \(= \pi(1.21 r^2) H.\) Dividing by \(\pi r^2\) yields \(h = 1.21H,\) so the first height is \(21\%\) more than the second.

Thus, the correct answer is D.

设长方形的两边为 \(4a\) 和 \(3a\)。根据勾股定理，对角线为 \(5a = d\)，所以 \(a = \dfrac{d}{5}\)。

面积为 \(4a\cdot 3a = 12a^2\) \(= 12\left(\dfrac{d}{5}\right)^2\) \(= \dfrac{12}{25}d^2\)，所以 \(k = \dfrac{12}{25}\)。

因此，正确答案是 C。

English

Let the sides of the rectangle be \(4a\) and \(3a.\) By the Pythagorean Theorem the diagonal is \(5a = d,\) so \(a = \dfrac{d}{5}.\)

The area is \(4a\cdot 3a = 12a^2\) \(= 12\left(\dfrac{d}{5}\right)^2\) \(= \dfrac{12}{25}d^2,\) so \(k = \dfrac{12}{25}.\)

Thus, the correct answer is C.

因为留给 Cheryl 的弹珠是随机决定的，她的两个弹珠等可能是任意一对。固定她的第一个弹珠，第二个弹珠等可能是剩下 \(5\) 个弹珠中的任意一个。

这 \(5\) 个中恰好有一个与第一个弹珠颜色相同，所以概率为 \(\dfrac{1}{5}\)。

因此，正确答案是 C。

English

Because the marbles left for Cheryl are determined at random, her two marbles are equally likely to be any pair. Fixing her first marble, the second is equally likely to be any of the \(5\) remaining marbles.

Exactly one of those \(5\) matches the first marble in color, so the probability is \(\dfrac{1}{5}.\)

Thus, the correct answer is C.

两边同时加 \(1\) 并因式分解，得 \[(x+1)(y+1) = 81 = 3^4\text{。}\]

因为 \(x\) 和 \(y\) 是不同的正整数且 \(x \gt y\)，唯一可能是 \(x+1 = 3^3 = 27\) 且 \(y+1 = 3\)。因此 \(x = 26\)。

因此，正确答案是 E。

English

Adding \(1\) to both sides and factoring gives \[(x+1)(y+1) = 81 = 3^4.\]

Because \(x\) and \(y\) are distinct positive integers with \(x \gt y,\) the only possibility is \(x+1 = 3^3 = 27\) and \(y+1 = 3.\) Therefore \(x = 26.\)

Thus, the correct answer is E.

公切线的数量取决于两个圆的相对位置：

如果小圆在大圆内部，则有 \(0\) 条切线。如果它们内切，则有 \(1\) 条。如果两个圆相交于两点，则有 \(2\) 条。如果它们外切，则有 \(3\) 条。如果它们分离，则有 \(4\) 条。

因此 \(k\) 可以是 \(0, 1, 2, 3, 4\) 中任意一个，共有 \(5\) 个可能值。

因此，正确答案是 D。

English

The number of common tangent lines depends on the relative position of the two circles:

If the smaller circle is inside the larger, there are \(0\) tangents. If it is internally tangent, there is \(1.\) If the circles intersect at two points, there are \(2.\) If they are externally tangent, there are \(3.\) If they are separated, there are \(4.\)

Thus \(k\) can be any of \(0, 1, 2, 3, 4,\) which gives \(5\) possible values.

Thus, the correct answer is D.

两条抛物线的 \(y\)-截距为 \(-2\) 和 \(4\)。为了与 \(x\)-轴相交，第一条抛物线开口向上，第二条开口向下，所以它们的 \(x\)-截距为 \(\pm t\)，其中 \(t \gt 0\)。

这个风筝形沿 \(y\)-轴的一条对角线长度为 \(4 - (-2) = 6\)，另一条长度为 \(2t\)。面积为 \(\dfrac{1}{2}\cdot 6\cdot 2t = 6t = 12\)，所以 \(t = 2\)。

因此 \(x\)-截距为 \(\pm 2\)。对第一条抛物线，\(0 = a(2)^2 - 2\) 得 \(a = \dfrac{1}{2}\)；对第二条，\(0 = 4 - b(2)^2\) 得 \(b = 1\)。因此 \(a + b = 1.5\)。

因此，正确答案是 B。

English

The \(y\)-intercepts of the two parabolas are \(-2\) and \(4.\) To intersect the \(x\)-axis, the first parabola opens upward and the second opens downward, so their \(x\)-intercepts are \(\pm t\) for some \(t \gt 0.\)

The kite has one diagonal of length \(4 - (-2) = 6\) along the \(y\)-axis and the other of length \(2t.\) Its area is \(\dfrac{1}{2}\cdot 6\cdot 2t = 6t = 12,\) so \(t = 2.\)

Thus the \(x\)-intercepts are \(\pm 2.\) For the first parabola, \(0 = a(2)^2 - 2\) gives \(a = \dfrac{1}{2};\) for the second, \(0 = 4 - b(2)^2\) gives \(b = 1.\) Therefore \(a + b = 1.5.\)

Thus, the correct answer is B.

每支队伍都打 \(11\) 场，因此 \(12\) 支队伍共进行 \(\dfrac{12\cdot 11}{2} = 66\) 场比赛。每场比赛给得分列表增加 \(2\) 分，所以所有得分的总和为 \(66\cdot 2 = 132\)。

如果每场比赛都是平局，每支队伍得 \(11\) 分，所以最高得分不必达到 \(12\)；因此命题 \(\text{(E)}\) 可能不成立。其他命题总成立：总和 \(132 \ge 100\)；总和为偶数，迫使奇数得分的个数为偶数，因而偶数得分的个数也为偶数；两支队伍不可能都得 \(0\) 分，因为它们之间的比赛至少会给其中一队一分。

因此，正确答案是 E。

English

Each of the \(12\) teams plays \(11\) games, so \(\dfrac{12\cdot 11}{2} = 66\) games are played, and each game adds \(2\) points to the list. The total of all scores is \(66\cdot 2 = 132.\)

If every game is a draw, each team scores \(11,\) so the highest score need not reach \(12;\) thus statement \(\text{(E)}\) can fail. The other statements always hold: the sum \(132 \ge 100,\) the sum being even forces an even number of odd scores and hence an even number of even scores, and two teams cannot both score \(0\) because their mutual game gives at least one of them a point.

Thus, the correct answer is E.

根据换底公式，\(\dfrac{1}{\log_b a} = \log_a b\)。因此 \[ \begin{aligned} &1 = \log_a 2 + \log_a 3 \\ &\quad {}+ \log_a 4 = \log_a 24 \end{aligned}\text{。}\]

由此可得 \(a = 24\)。

因此，正确答案是 D。

English

By the change-of-base formula, \(\dfrac{1}{\log_b a} = \log_a b.\) Therefore \[ \begin{aligned} &1 = \log_a 2 + \log_a 3 \\ &\quad {}+ \log_a 4 = \log_a 24. \end{aligned} \]

It follows that \(a = 24.\)

Thus, the correct answer is D.

分子和分母没有公因数。要把这个分数写成小数，就把它改写成分母为 \(10\) 的幂的形式；最小可用的是 \(10^{26}\)：\[\dfrac{123456789}{2^{26}\cdot 5^4} = \dfrac{123456789\cdot 5^{22}}{10^{26}}\text{。}\]

因为分子 \(123456789\cdot 5^{22}\) 不能被 \(10\) 整除，所以小数点后恰好有 \(26\) 位。

因此，正确答案是 C。

English

The numerator and denominator share no common factors. To write the fraction as a decimal, rewrite it with a power of \(10\) in the denominator; the smallest that works is \(10^{26}:\) \[\dfrac{123456789}{2^{26}\cdot 5^4} = \dfrac{123456789\cdot 5^{22}}{10^{26}}.\]

Since the numerator \(123456789\cdot 5^{22}\) is not divisible by \(10,\) the decimal has exactly \(26\) places after the point.

Thus, the correct answer is C.

三角形 \(ABC\) 和 \(ABD\) 都是 \(3\)-\(4\)-\(5\) 直角三角形，面积为 \(6\)，并共用斜边 \(AB\)。设 \(E\) 为从 \(C\) 向 \(AB\) 所作高的垂足，则 \(CE = \dfrac{3\cdot 4}{5} = \dfrac{12}{5}\)。同样，从 \(D\) 向 \(AB\) 所作的高也落在同一点 \(E\)，并且 \(DE = \dfrac{12}{5}\)。

三角形 \(CDE\) 的边长为 \(\dfrac{12}{5}\)，\(\dfrac{12}{5}\)，和 \(CD = \dfrac{12}{5}\sqrt{2}\)，因此它是以 \(E\) 为直角顶点的等腰直角三角形。所以 \(DE \perp CE\) 且 \(DE \perp AB\)，从而 \(DE\) 垂直于平面 \(ABC\)。

四面体的体积为 \(\dfrac{1}{3}\cdot [ABC]\cdot DE = \dfrac{1}{3}\cdot 6\cdot \dfrac{12}{5}\) \(= \dfrac{24}{5}\)。

因此，正确答案是 C。

English

Triangles \(ABC\) and \(ABD\) are \(3\)-\(4\)-\(5\) right triangles with area \(6\) and common hypotenuse \(AB.\) Let \(E\) be the foot of the altitude from \(C\) to \(AB;\) then \(CE = \dfrac{3\cdot 4}{5} = \dfrac{12}{5}.\) Likewise the altitude from \(D\) meets \(AB\) at the same point \(E\) with \(DE = \dfrac{12}{5}.\)

Triangle \(CDE\) has sides \(\dfrac{12}{5},\) \(\dfrac{12}{5},\) and \(CD = \dfrac{12}{5}\sqrt{2},\) so it is an isosceles right triangle with the right angle at \(E.\) Thus \(DE \perp CE\) and \(DE \perp AB,\) making \(DE\) perpendicular to the plane of \(ABC.\)

The tetrahedron’s volume is \(\dfrac{1}{3}\cdot [ABC]\cdot DE = \dfrac{1}{3}\cdot 6\cdot \dfrac{12}{5}\) \(= \dfrac{24}{5}.\)

Thus, the correct answer is C.

共有 \(2^8 = 256\) 个等可能结果。按站起来的人数分类，数出圆周上 \(8\) 个座位中没有两个相邻站立者 （正面）的安排数。

从 \(n\) 个圆周座位中选 \(k\) 个不相邻座位的方法数为 \(\dfrac{n}{n-k}\dbinom{n-k}{k}\)。当 \(n = 8\) 时，对 \(k = 0,1,2,3,4\) 分别得到 \[1,\; 8,\; 20,\; 16,\; 2\] 种；超过 \(4\) 个站立者则不可能避免相邻。

总数为 \(1 + 8 + 20 + 16 + 2 = 47\)， 所以概率是 \(\dfrac{47}{256}\)。

因此，正确答案是 A。

English

There are \(2^8 = 256\) equally likely outcomes. Count the arrangements of standers (heads) with no two adjacent around the circle of \(8\) seats, grouped by how many people stand.

The number of ways to choose \(k\) non-adjacent seats from a circle of \(n\) is \(\dfrac{n}{n-k}\dbinom{n-k}{k}.\) For \(n = 8\) this gives \[1,\; 8,\; 20,\; 16,\; 2\] for \(k = 0,1,2,3,4,\) and more than \(4\) standers is impossible without an adjacency.

The total is \(1 + 8 + 20 + 16 + 2 = 47,\) so the probability is \(\dfrac{47}{256}.\)

Thus, the correct answer is A.

设整数零点为 \(p\) 和 \(q\)。由韦达定理，\(p + q = a\) 且 \(pq = 2a\)，所以 \(pq = 2(p+q)\)。整理得 \[(p-2)(q-2) = 4\text{。}\]

\(4\) 的整数因子对为 \((1,4)\)、\((2,2)\)、\((4,1)\)、\((-1,-4)\)、\((-2,-2)\)、\((-4,-1)\)。由此得到的 \((p,q)\) 对应的和为 \(a = 9, 8, -1\) 和 \(0\)。

\(a\) 的不同可能值为 \(9, 8, 0, -1\)，它们的和为 \(16\)。

因此，正确答案是 C。

English

Let the integer zeros be \(p\) and \(q.\) By Vieta’s formulas \(p + q = a\) and \(pq = 2a,\) so \(pq = 2(p+q).\) Rearranging gives \[(p-2)(q-2) = 4.\]

The integer factor pairs of \(4\) are \((1,4),\) \((2,2),\) \((4,1),\) \((-1,-4),\) \((-2,-2),\) \((-4,-1),\) which yield \((p,q)\) pairs summing to \(a = 9, 8, -1,\) and \(0.\)

The distinct possible values of \(a\) are \(9, 8, 0, -1,\) whose sum is \(16.\)

Thus, the correct answer is C.

在每个这样的四边形中，\(CD \ge AB\)。设 \(E\) 为从 \(A\) 向 \(\overline{CD}\) 所作垂线的垂足，则 \(CE = 2\) 且 \(AE = BC\)。令 \(x = AE\)、\(y = DE\)，所以 \(AD = 2 + y\)。

根据勾股定理，\(x^2 + y^2 = (2+y)^2\)，所以 \(x^2 = 4 + 4y\)，且 \(x\) 为偶数。写作 \(x = 2z\) 得 \(y = z^2 - 1\)，周长为 \[x + 2y + 6 = 2z^2 + 2z + 4\text{。}\]

递增的 \(z = 1, 2, 3, \dots\) 给出满足条件且周长递增的四边形。当 \(z = 31\) 时周长为 \(1988\)，当 \(z = 32\) 时为 \(2116\)。因此 \(p \lt 2015\) 的可能值有 \(31\) 个。

因此，正确答案是 B。

English

In every such quadrilateral \(CD \ge AB.\) Let \(E\) be the foot of the perpendicular from \(A\) to \(\overline{CD};\) then \(CE = 2\) and \(AE = BC.\) Let \(x = AE\) and \(y = DE,\) so \(AD = 2 + y.\)

By the Pythagorean Theorem \(x^2 + y^2 = (2+y)^2,\) so \(x^2 = 4 + 4y\) and \(x\) is even. Writing \(x = 2z\) gives \(y = z^2 - 1,\) and the perimeter is \[x + 2y + 6 = 2z^2 + 2z + 4.\]

Increasing values \(z = 1, 2, 3, \dots\) give the required quadrilaterals with increasing perimeter. For \(z = 31\) the perimeter is \(1988,\) and for \(z = 32\) it is \(2116.\) Therefore there are \(31\) possible values of \(p \lt 2015.\)

Thus, the correct answer is B.

\(T\) 到底边 \(8\) 的高为 \(\sqrt{5^2 - 4^2} = 3\)，所以 \(T\) 的面积为 \(\dfrac{1}{2}\cdot 8\cdot 3 = 12\)，周长为 \(18\)。

对 \(T'\)，需要 \(2a + b = 18\)，且面积 \(\dfrac{1}{4}b\sqrt{4a^2 - b^2} = 12\)。代入 \(a = \dfrac{18 - b}{2}\) 并平方，可得 \[(b - 8)(b^2 - b - 8) = 0\text{。}\]

因为 \(T\) 和 \(T'\) 不全等，\(b \ne 8\)，所以 \(b^2 - b - 8 = 0\)，且 \(b = \dfrac{1 + \sqrt{33}}{2}\)。因为 \(25 \lt 33 \lt 36\)，这个数介于 \(3\) 和 \(3.5\) 之间，所以最接近的整数是 \(3\)。

因此，正确答案是 A。

English

The altitude of \(T\) to its base of length \(8\) is \(\sqrt{5^2 - 4^2} = 3,\) so \(T\) has area \(\dfrac{1}{2}\cdot 8\cdot 3 = 12\) and perimeter \(18.\)

For \(T'\) we need \(2a + b = 18\) and area \(\dfrac{1}{4}b\sqrt{4a^2 - b^2} = 12.\) Substituting \(a = \dfrac{18 - b}{2}\) and squaring leads to \[(b - 8)(b^2 - b - 8) = 0.\]

Since \(T\) and \(T'\) are not congruent, \(b \ne 8,\) so \(b^2 - b - 8 = 0\) and \(b = \dfrac{1 + \sqrt{33}}{2}.\) Because \(25 \lt 33 \lt 36,\) this is between \(3\) and \(3.5,\) so the closest integer is \(3.\)

Thus, the correct answer is A.

椭圆 \(\dfrac{x^2}{16} + y^2 = 1\) 的半轴长为 \(4\) 和 \(1\)，所以 \(c^2 = 16 - 1 = 15\)，焦点为 \((\pm\sqrt{15}, 0)\)。

经过两个焦点的圆的圆心在 \(y\)-轴上，设为 \((0, k)\)，半径为 \(\sqrt{k^2 + 15}\)。它的最高点总在椭圆外。要有四个交点，它的最低点 \((0, k - \sqrt{k^2 + 15})\) 必须低于 \(y = -1\)，这恰好在 \(0 \le k \lt 7\) 时发生。

当 \(k\) 在 \([0, 7)\) 中变化时，半径 \(\sqrt{k^2 + 15}\) 的取值范围为 \([\sqrt{15}, 8)\)，所以 \(a + b = \sqrt{15} + 8\)。

因此，正确答案是 D。

English

The ellipse \(\dfrac{x^2}{16} + y^2 = 1\) has semi-axes \(4\) and \(1,\) so \(c^2 = 16 - 1 = 15\) and the foci are \((\pm\sqrt{15}, 0).\)

A circle through both foci has its center on the \(y\)-axis, say \((0, k),\) with radius \(\sqrt{k^2 + 15}.\) Its top point always lies outside the ellipse. For four intersection points, its bottom point \((0, k - \sqrt{k^2 + 15})\) must be below \(y = -1,\) which happens exactly when \(0 \le k \lt 7.\)

As \(k\) ranges over \([0, 7),\) the radius \(\sqrt{k^2 + 15}\) ranges over \([\sqrt{15}, 8),\) so \(a + b = \sqrt{15} + 8.\)

Thus, the correct answer is D.

注意 \(S(1) = 2\)，\(S(2) = 4\)，\(S(3) = 8\)。每个合法序列都以一段长度为一、二或三的同字母段结尾；去掉这段后，剩下长度为 \(n-1\)，\(n-2\)，或 \(n-3\) 的合法序列。\[ \begin{aligned} &S(n) = S(n-1) + S(n-2) \\ &\quad {}+ S(n-3) \end{aligned}\text{。}\]

模 \(3\) 时，前 \(13\) 项为 \[2,1,2,2,2,0,1,0,1,2,0,0,2\text{。}\]接下来的三项是 \(2,1,2\)，与初始状态相同，所以递推式以 \(13\) 为周期重复。因为 \(2015 = 13\cdot 155\)，可得 \(S(2015) \equiv S(13) \equiv 2 \pmod 3\)。

模 \(4\) 时，各项按 \(2,0,0,2\) 循环，因为这四项之后的三项状态又回到 \((2,0,0)\)。所以周期为 \(4\)。因为 \(2015 = 4\cdot 503 + 3\)，所以 \(S(2015) \equiv S(3) \equiv 0 \pmod 4\)。

写 \(S(2015) = 4k\)，条件 \(4k \equiv 2 \pmod 3\) 给出 \(k \equiv 2 \pmod 3\)，因此 \(S(2015) \equiv 8 \pmod{12}\)。

因此，正确答案是 D。

English

Note \(S(1) = 2,\) \(S(2) = 4,\) \(S(3) = 8.\) Every valid sequence ends in a run of one, two, or three equal letters; removing that run leaves a valid sequence of length \(n-1,\) \(n-2,\) or \(n-3.\) Thus \[ \begin{aligned} &S(n) = S(n-1) + S(n-2) \\ &\quad {}+ S(n-3). \end{aligned} \]

Modulo \(3,\) the first \(13\) terms are \[2,1,2,2,2,0,1,0,1,2,0,0,2.\] The next three terms are \(2,1,2,\) which reproduce the initial state, so the recurrence repeats with period \(13.\) Since \(2015 = 13\cdot 155,\) it follows that \(S(2015) \equiv S(13) \equiv 2 \pmod 3.\)

Modulo \(4,\) the terms repeat as \(2,0,0,2,\) because the next three-term state after these four terms is again \((2,0,0).\) Thus the period is \(4.\) As \(2015 = 4\cdot 503 + 3,\) we have \(S(2015) \equiv S(3) \equiv 0 \pmod 4.\)

Writing \(S(2015) = 4k,\) the condition \(4k \equiv 2 \pmod 3\) gives \(k \equiv 2 \pmod 3,\) so \(S(2015) \equiv 8 \pmod{12}.\)

Thus, the correct answer is D.

第二个点与第一个点在同一边上的概率为 \(\dfrac14\)， 在对边上的概率为 \(\dfrac14\)， 在相邻边上的概率为 \(\dfrac12\)。

对边： 距离总是至少为 \(1 \ge \dfrac12\)，概率为 \(1\)。

同一边： 对点 \((a, 0)\) 和 \((b, 0)\)， 条件 \(|a - b| \ge \dfrac12\) 的概率为 \(\dfrac14\)。

相邻边： 对点 \((a, 0)\) 和 \((0, b)\)， 条件 \(\sqrt{a^2 + b^2} \ge \dfrac12\) 是半径为 \(\dfrac12\) 的四分之一圆外部的区域， 概率为 \(1 - \dfrac14\pi\left(\dfrac12\right)^2 = 1 - \dfrac{\pi}{16}\)。

总概率为 \[ \begin{aligned} &\dfrac14\cdot 1 + \dfrac14\cdot\dfrac14 \\ &\quad {}+ \dfrac12\left(1 - \dfrac{\pi}{16}\right) \\ &\quad = \dfrac{26 - \pi}{32}\text{。} \end{aligned} \] 因此 \(a + b + c = 26 + 1 + 32 = 59\)。

因此，正确答案是 A。

English

The second point is on the same side as the first with probability \(\dfrac14,\) on the opposite side with probability \(\dfrac14,\) and on an adjacent side with probability \(\dfrac12.\)

Opposite sides: the distance is at least \(1 \ge \dfrac12\) always, probability \(1.\)

Same side: for points \((a, 0)\) and \((b, 0),\) the condition \(|a - b| \ge \dfrac12\) has probability \(\dfrac14.\)

Adjacent sides: for points \((a, 0)\) and \((0, b),\) the condition \(\sqrt{a^2 + b^2} \ge \dfrac12\) is the region outside a quarter-circle of radius \(\dfrac12,\) with probability \(1 - \dfrac14\pi\left(\dfrac12\right)^2 = 1 - \dfrac{\pi}{16}.\)

The total probability is \[ \begin{aligned} &\dfrac14\cdot 1 + \dfrac14\cdot\dfrac14 \\ &\quad {}+ \dfrac12\left(1 - \dfrac{\pi}{16}\right) \\ &\quad = \dfrac{26 - \pi}{32}. \end{aligned} \] Thus \(a + b + c = 26 + 1 + 32 = 59.\)

Thus, the correct answer is A.

各有 \(20\) 个可能的 \(a\) 和 \(b\)。写成最简分数时，分母 \(1,2,3,4,5\) 分别贡献 \(2,2,4,4,8\) 个位于 \([0,2)\) 中的值，总计 \(20\)。

令 \(x = \cos(a\pi)\) 且 \(y = \sin(b\pi)\)，四次方 \((x + iy)^4\) 为实数，当且仅当 \(x = 0\)，\(y = 0\)，或 \(x = \pm y\)。

情形 \(x = 0\) 意味着 \(a \in \left\{\dfrac12, \dfrac32\right\}\)，给出 \(2\cdot 20 = 40\) 对；情形 \(y = 0\) 意味着 \(b \in \{0, 1\}\)，再给出 \(40\) 对，其中 \(4\) 对已经计数。

对于剩余条件 \(\cos(a\pi) = \pm\sin(b\pi)\)，且两边都非零，允许的 \(b\) 值为 \(\dfrac14\)，\(\dfrac12\)，\(\dfrac34\)，\(\dfrac54\)，\(\dfrac32\)，和 \(\dfrac74\)。对应的 \(a\) 值个数依次为 \(4,2,4,4,2,4\)。例如，当 \(b = \dfrac14\) 时，各值为 \(a = \dfrac14\)，\(\dfrac34\)，\(\dfrac54\)，和 \(\dfrac74\)；其他各行同样由四分之一周的恒等式得到。因此此条件再贡献 \(4+2+4+4+2+4=20\) 对。

总共有 \(40 + 40 - 4 + 20 = 96\) 对有效组合，而全部组合共 \(400\) 对，所以概率为 \(\dfrac{96}{400} = \dfrac{6}{25}\)。

因此，正确答案是 D。

English

There are \(20\) possible values for each of \(a\) and \(b.\) In reduced form, denominators \(1,2,3,4,5\) contribute respectively \(2,2,4,4,8\) values in \([0,2),\) for a total of \(20.\)

Writing \(x = \cos(a\pi)\) and \(y = \sin(b\pi),\) the fourth power \((x + iy)^4\) is real if and only if \(x = 0,\) \(y = 0,\) or \(x = \pm y.\)

The case \(x = 0\) means \(a \in \left\{\dfrac12, \dfrac32\right\},\) giving \(2\cdot 20 = 40\) pairs; the case \(y = 0\) means \(b \in \{0, 1\},\) giving another \(40\) pairs, of which \(4\) were already counted.

For the remaining condition \(\cos(a\pi) = \pm\sin(b\pi),\) with neither side zero, the allowed values of \(b\) are \(\dfrac14,\) \(\dfrac12,\) \(\dfrac34,\) \(\dfrac54,\) \(\dfrac32,\) and \(\dfrac74.\) The corresponding numbers of allowed values of \(a\) are respectively \(4,2,4,4,2,4.\) For example, when \(b = \dfrac14,\) the values are \(a = \dfrac14,\) \(\dfrac34,\) \(\dfrac54,\) and \(\dfrac74;\) the other rows follow from the same quarter-turn identities. Hence this condition contributes \(4+2+4+4+2+4=20\) more pairs.

In all there are \(40 + 40 - 4 + 20 = 96\) valid pairs out of \(400,\) so the probability is \(\dfrac{96}{400} = \dfrac{6}{25}.\)

Thus, the correct answer is D.

如果半径为 \(r\) 的圆与 \(x\)-轴相切，并嵌在两个半径为 \(r_1\) 和 \(r_2\)、也与该轴相切且彼此外切的圆之间，那么 \[\dfrac{1}{\sqrt{r}} = \dfrac{1}{\sqrt{r_1}} + \dfrac{1}{\sqrt{r_2}}\text{。}\]

令 \(x = \dfrac{1}{\sqrt{70^2}} + \dfrac{1}{\sqrt{73^2}}\) \(= \dfrac{1}{70} + \dfrac{1}{73}\)，这是 \(L_0\) 上的和。\(L_1\) 中唯一的圆也贡献 \(x\)。对 \(k \ge 2\)，每个新圆贡献它两个相邻圆的和；除 \(L_0\) 的两个圆外，每个较早的圆都被计算两次。因此，\(L_k\) 上的和为 \(3^{k-1}x\)。

因此 \[ \begin{gathered} \sum_{C \in S} \dfrac{1}{\sqrt{r(C)}} = x \\ {}+ \sum_{k=1}^{6} 3^{k-1}x \\ = x\left(1 + \dfrac{3^6 - 1}{2}\right) \\ = x\cdot\dfrac{3^6 + 1}{2} \\ = 365x \end{gathered}\text{。}\]

因为 \(x = \dfrac{1}{70} + \dfrac{1}{73}\) \(= \dfrac{143}{70\cdot 73}\) \(= \dfrac{143}{5110}\)，所以总和为 \(365\cdot\dfrac{143}{5110} = \dfrac{143}{14}\)。

因此，正确答案是 D。

English

If a circle of radius \(r\) is tangent to the \(x\)-axis and nestled in the crevice between two circles of radii \(r_1\) and \(r_2\) that are also tangent to the axis and to each other, then \[\dfrac{1}{\sqrt{r}} = \dfrac{1}{\sqrt{r_1}} + \dfrac{1}{\sqrt{r_2}}.\]

Let \(x = \dfrac{1}{\sqrt{70^2}} + \dfrac{1}{\sqrt{73^2}}\) \(= \dfrac{1}{70} + \dfrac{1}{73},\) which is the sum over \(L_0.\) The single circle of \(L_1\) also contributes \(x.\) For \(k \ge 2,\) each new circle contributes the sum of its two neighbors, and every earlier circle is counted twice except the two circles of \(L_0;\) this yields a sum of \(3^{k-1}x\) over \(L_k.\)

Therefore \[ \begin{gathered} \sum_{C \in S} \dfrac{1}{\sqrt{r(C)}} = x \\ {}+ \sum_{k=1}^{6} 3^{k-1}x \\ = x\left(1 + \dfrac{3^6 - 1}{2}\right) \\ = x\cdot\dfrac{3^6 + 1}{2} \\ = 365x. \end{gathered} \]

Since \(x = \dfrac{1}{70} + \dfrac{1}{73}\) \(= \dfrac{143}{70\cdot 73}\) \(= \dfrac{143}{5110},\) the sum is \(365\cdot\dfrac{143}{5110} = \dfrac{143}{14}.\)

Thus, the correct answer is D.