2014B AMC 12 真题（P版）

设 \(n\) 为五美分硬币的数量，则 Leah 有 \(13-n\) 枚一美分硬币。多一枚五美分硬币后，她会有 \(n+1\) 枚五美分硬币，这等于一美分硬币的数量：\[ n+1 = 13-n\text{。} \] 解得 \(n=6\)，所以她有 \(6\) 枚五美分硬币和 \(7\) 枚一美分硬币。

总价值为 \(6\cdot5 + 7 = 37\) 美分。

所以正确答案是 C。

English

Let \(n\) be the number of nickels, so Leah has \(13-n\) pennies. One more nickel would give her \(n+1\) nickels, and this equals the number of pennies: \[ n+1 = 13-n. \] Solving gives \(n=6,\) so there are \(6\) nickels and \(7\) pennies.

The total value is \(6\cdot5 + 7 = 37\) cents.

Thus, the correct answer is C.

在特价下，一组两个气球的费用是一个原价气球的 \(1 + \tfrac23 = \tfrac53\) 倍。

Orvin 的钱按原价可以买 \(30\) 个气球，而按特价可以买 \[ 30 \div \tfrac53 = 18 \] 组，也就是 \(36\) 个气球。

所以正确答案是 C。

English

Under the sale, a pair of balloons costs \(1 + \tfrac23 = \tfrac53\) times the regular price of one balloon.

Orvin’s money buys \(30\) balloons at the regular price, so he can afford \[ 30 \div \tfrac53 = 18 \] pairs, which is \(36\) balloons.

Thus, the correct answer is C.

铺装路占全程的比例为 \[ 1 - \tfrac13 - \tfrac15 = \tfrac{7}{15}\text{。} \]

因为这部分等于 \(20\) 英里，所以全程为 \[ 20 \div \tfrac{7}{15} = \tfrac{300}{7} \] 英里。

所以正确答案是 E。

English

The fraction of the trip on pavement is \[ 1 - \tfrac13 - \tfrac15 = \tfrac{7}{15}. \]

Since this equals \(20\) miles, the whole trip is \[ 20 \div \tfrac{7}{15} = \tfrac{300}{7} \] miles.

Thus, the correct answer is E.

设一个松饼的价格为 \(m\)，一根香蕉的价格为 \(b\)，则 \[ 2(4m+3b) = 2m+16b\text{。}\]

展开得 \(8m+6b = 2m+16b\)，所以 \(6m = 10b\)，即 \(m = \tfrac53 b\)。

所以正确答案是 B。

English

Let a muffin cost \(m\) and a banana cost \(b.\) Then \[ 2(4m+3b) = 2m+16b. \]

Expanding gives \(8m+6b = 2m+16b,\) so \(6m = 10b\) and \(m = \tfrac53 b.\)

Thus, the correct answer is B.

设每块玻璃片宽 \(2x\)，高 \(5x\)。窗户横向有 \(4\) 块玻璃片和 \(5\) 条竖向边框，所以宽为 \(4(2x) + 5\cdot2 = 8x+10\)。

窗户纵向有 \(2\) 块玻璃片和 \(3\) 条横向边框，所以高为 \(2(5x) + 3\cdot2 = 10x+6\)。

令宽等于高，得到 \(8x+10 = 10x+6\)，所以 \(x=2\)，边长为 \(10\cdot2+6 = 26\)。

所以正确答案是 A。

English

Let each pane have width \(2x\) and height \(5x.\) The window is \(4\) panes wide with \(5\) vertical borders, so its width is \(4(2x) + 5\cdot2 = 8x+10.\)

It is \(2\) panes tall with \(3\) horizontal borders, so its height is \(2(5x) + 3\cdot2 = 10x+6.\)

Setting width equal to height gives \(8x+10 = 10x+6,\) so \(x=2\) and the side length is \(10\cdot2+6 = 26.\)

Thus, the correct answer is A.

设普通杯柠檬水为 \(a\) 盎司，则 Ann 的大杯为 \(\tfrac32 a\)。 每人喝掉 \(\tfrac34\) 后，Ann 剩下 \(\tfrac14\cdot\tfrac32 a = \tfrac38 a\)，她给 Ed \(\tfrac13\cdot\tfrac38 a + 2 = \tfrac18 a + 2\) 盎司。

Ed 喝了自己的全部 \(a\) 盎司再加上这份转给他的饮料，Ann 喝了 \(\tfrac32 a\) 减去这份饮料。令二者相等： \[ a + \left(\tfrac18 a + 2\right) = \tfrac32 a - \left(\tfrac18 a + 2\right)\text{，} \] 得 \(4 = \tfrac14 a\)，所以 \(a = 16\)。

于是 Ed 喝了 \(16\) 盎司，Ann 喝了 \(24\) 盎司，总共 \(40\) 盎司。

所以正确答案是 D。

English

Let a regular lemonade hold \(a\) ounces, so Ann’s large holds \(\tfrac32 a.\) After each drinks \(\tfrac34,\) Ann has \(\tfrac14\cdot\tfrac32 a = \tfrac38 a\) left, and she gives Ed \(\tfrac13\cdot\tfrac38 a + 2 = \tfrac18 a + 2\) ounces.

Ed drinks his full \(a\) ounces plus that gift, and Ann drinks her \(\tfrac32 a\) minus the gift. Setting these equal, \[ a + \left(\tfrac18 a + 2\right) = \tfrac32 a - \left(\tfrac18 a + 2\right), \] which gives \(4 = \tfrac14 a,\) so \(a = 16.\)

Then Ed drank \(16\) ounces and Ann drank \(24\) ounces, for a total of \(40\) ounces.

Thus, the correct answer is D.

写成 \[ \dfrac{n}{30-n} = \dfrac{30}{30-n} - 1\text{。}\]

要使它为正整数，\(30-n\) 必须是 \(30\) 的正因数，并且 \(\dfrac{30}{30-n} \ge 2\)，即 \(30-n \le 15\)。

\(30\) 的不超过 \(15\) 的因数为 \(1, 2, 3, 5, 6, 10, 15\)，给出 \(7\) 个 \(n\) 的值（即 \(15, 20, 24, 25, 27, 28, 29\)）。

所以正确答案是 D。

English

Write \[ \dfrac{n}{30-n} = \dfrac{30}{30-n} - 1. \]

For this to be a positive integer, \(30-n\) must be a positive divisor of \(30\) with \(\dfrac{30}{30-n} \ge 2,\) i.e. \(30-n \le 15.\)

The divisors of \(30\) that are at most \(15\) are \(1, 2, 3, 5, 6, 10, 15,\) giving \(7\) values of \(n\) (namely \(15, 20, 24, 25, 27, 28, 29\)).

Thus, the correct answer is D.

最左列显示 \(A+B = D\)，且没有向外进位，所以 \(A+B \le 9\)。查看十位和千位（每列都形如 \(C + \text{数字} + \text{进位}\) 仍得到同一个数字）可知 \(C = 0\)，并排除所有进位。

每一列于是都化为 \(A+B = D\)，且 \(A, B, C=0\) 互不相同。因为 \(A\) 和 \(B\) 是不同的正数字，\(D = A+B\) 可以是从 \(3\) 到 \(9\) 的任意值，共 \(7\) 种可能，例如 \((A,B,C,D) = (1,2,0,3)\)，\((1,3,0,4), \ldots\)，\((2,7,0,9)\)。

所以正确答案是 C。

English

The leftmost column shows \(A+B = D\) with no carry out, so \(A+B \le 9.\) Examining the tens and thousands columns (each of the form \(C + \text{digit} + \text{carry}\) producing the same digit) forces \(C = 0\) and eliminates all carries.

Every column then reduces to \(A+B = D,\) with \(A, B, C=0\) distinct. Since \(A\) and \(B\) are distinct positive digits, \(D = A+B\) can be any value from \(3\) up to \(9,\) giving \(7\) possibilities, for example \((A,B,C,D) = (1,2,0,3),\) \((1,3,0,4), \ldots,\) \((2,7,0,9).\)

Thus, the correct answer is C.

由直角三角形 \(ABC\) 中的勾股定理，\(AC = \sqrt{3^2+4^2} = 5\)。

因为 \(5^2 + 12^2 = 13^2\)，勾股定理逆定理说明 \(\angle DAC = 90^\circ\)，所以 \(\triangle DAC\) 是直角三角形。

\(\triangle ABC\) 的面积是 \(\tfrac12\cdot3\cdot4 = 6\)，\(\triangle DAC\) 的面积是 \(\tfrac12\cdot5\cdot12 = 30\)。四边形面积为 \(6 + 30 = 36\)。

所以正确答案是 B。

English

By the Pythagorean Theorem in right triangle \(ABC,\) \(AC = \sqrt{3^2+4^2} = 5.\)

Since \(5^2 + 12^2 = 13^2,\) the converse of the Pythagorean Theorem shows \(\angle DAC = 90^\circ,\) so \(\triangle DAC\) is right.

The area of \(\triangle ABC\) is \(\tfrac12\cdot3\cdot4 = 6\) and the area of \(\triangle DAC\) is \(\tfrac12\cdot5\cdot12 = 30.\) The quadrilateral has area \(6 + 30 = 36.\)

Thus, the correct answer is B.

行驶距离为 \(cba - abc = 99(c-a)\)，是 \(9\) 的倍数。以每小时 \(55\) 英里行驶整数小时，也使它是 \(55\) 的倍数，因此它是 \(495\) 的倍数。

由于里程表差值至多是一个 \(3\) 位数且 \(a \ge 1\)，距离只能是 \(495\)，所以 \(c - a = 5\)。

在 \(a \ge 1\) 且 \(a+b+c \le 7\) 下，唯一可能是 \(a=1\)，\(c=6\)，\(b=0\)。因此 \(a^2+b^2+c^2 = 1 + 0 + 36 = 37\)。

所以正确答案是 D。

English

The distance driven is \(cba - abc = 99(c-a),\) a multiple of \(9.\) Driving a whole number of hours at \(55\) mph makes it a multiple of \(55\) too, hence a multiple of \(495.\)

Since the odometer difference is at most a \(3\)-digit number and \(a \ge 1,\) the distance must be \(495,\) so \(c - a = 5.\)

With \(a \ge 1\) and \(a+b+c \le 7,\) the only choice is \(a=1,\) \(c=6,\) \(b=0.\) Then \(a^2+b^2+c^2 = 1 + 0 + 36 = 37.\)

Thus, the correct answer is D.

列表总和为 \(11 \cdot 10 = 110\)。 要最大化一个数，应最小化另外十个数的和。

排序后，第六个数必须是 \(9\)（中位数），且 \(8\) 出现次数必须比其他任何值都多。让 \(8\) 出现三次时，最小可能的十个数为 \[ 1,1,8,8,8,9,9,10,10,11\text{，} \] 它们的和为 \(75\)，并且仍使 \(8\) 是唯一众数。

最大项于是为 \(110 - 75 = 35\)。

所以正确答案是 E。

English

The list sums to \(11 \cdot 10 = 110.\) To maximize one entry, minimize the sum of the other ten.

Sorted, the sixth number must be \(9\) (the median), and \(8\) must appear more often than any other value. Trying \(8\) three times, the smallest possible ten numbers are \[ 1,1,8,8,8,9,9,10,10,11, \] which sum to \(75\) and keep \(8\) the unique mode.

The largest entry is then \(110 - 75 = 35.\)

Thus, the correct answer is E.

按非递增顺序写出每个三角形的边长。等边三角形只能选一个（它们都相似），并且相似的一对 \(2,2,1\) 和 \(4,4,2\) 中也只能选一个。

其余有效且两两不相似的三角形为 \[ \begin{gathered} 4\,4\,3,\ 4\,4\,1,\ 4\,3\,3,\ 4\,3\,2,\ \\ 3\,3\,2,\ 3\,3\,1,\ 3\,2\,2\text{，} \end{gathered} \] 一共七个。再加上一个等边三角形和相似对中的一个，\(S\) 最多有 \(9\) 个元素。

所以正确答案是 B。

English

Write each triangle by its side lengths in nonincreasing order. Only one equilateral triangle is allowed (all are similar), and of the similar pair \(2,2,1\) and \(4,4,2\) only one may appear.

The remaining valid, pairwise non-similar triangles are \[ \begin{gathered} 4\,4\,3,\ 4\,4\,1,\ 4\,3\,3,\ 4\,3\,2,\ \\ 3\,3\,2,\ 3\,3\,1,\ 3\,2\,2, \end{gathered} \] seven in all. Together with one equilateral and one of the similar pair, \(S\) has at most \(9\) elements.

Thus, the correct answer is B.

因为 \(b\) 是 \(1, a, b\) 中最大的边，不存在这样的三角形当且仅当 \(b \ge a+1\)。因为 \(1\) 是 \(\tfrac1b, \tfrac1a, 1\) 中最大的边，不存在这样的三角形当且仅当 \(1 \ge \tfrac1a + \tfrac1b\)，即 \(a \le \tfrac{b}{b-1}\)。

当 \(a+1 = b\) 与 \(a = \tfrac{b}{b-1}\) 相交时，同时满足两个条件的 \(b\) 最小，得到 \(b - 1 = \tfrac{b}{b-1}\)，即 \(b^2 - 3b + 1 = 0\)。

大于 \(1\) 的根为 \(b = \dfrac{3+\sqrt5}{2}\)。

所以正确答案是 C。

English

Since \(b\) is the largest of \(1, a, b,\) no such triangle exists exactly when \(b \ge a+1.\) Since \(1\) is the largest of \(\tfrac1b, \tfrac1a, 1,\) no such triangle exists exactly when \(1 \ge \tfrac1a + \tfrac1b,\) that is \(a \le \tfrac{b}{b-1}.\)

Both conditions hold with \(b\) smallest when \(a+1 = b\) and \(a = \tfrac{b}{b-1}\) meet, giving \(b - 1 = \tfrac{b}{b-1},\) or \(b^2 - 3b + 1 = 0.\)

The root larger than \(1\) is \(b = \dfrac{3+\sqrt5}{2}.\)

Thus, the correct answer is C.

设三条边长为 \(x, y, z\)，则 \(xy+yz+zx = 47\)，且 \(x+y+z = 12\)。因此 \[ \begin{gathered} x^2+y^2+z^2 = (x+y+z)^2 \\ {}- 2(xy+yz+zx) \\ = 144 - 94 \\ = 50 \end{gathered}\text{。}\]

\(4\) 条体对角线中的每一条长度为 \(\sqrt{x^2+y^2+z^2} = \sqrt{50} = 5\sqrt2\)，所以总长度为 \(4 \cdot 5\sqrt2 = 20\sqrt2\)。

所以正确答案是 D。

English

Let the edges be \(x, y, z.\) Then \(xy+yz+zx = 47\) and \(x+y+z = 12.\) Therefore \[ \begin{gathered} x^2+y^2+z^2 = (x+y+z)^2 \\ {}- 2(xy+yz+zx) \\ = 144 - 94 \\ = 50. \end{gathered} \]

Each of the \(4\) interior diagonals has length \(\sqrt{x^2+y^2+z^2} = \sqrt{50} = 5\sqrt2,\) so their total length is \(4 \cdot 5\sqrt2 = 20\sqrt2.\)

Thus, the correct answer is D.

因为 \(k \ln k = \ln(k^k)\)，这个和给出 \(p = \ln\left(\prod_{k=1}^{6} k^k\right)\)，所以 \[ e^p = 1^1 \cdot 2^2 \cdot 3^3 \cdot 4^4 \cdot 5^5 \cdot 6^6\text{。}\]

因子 \(2\) 来自 \(2^2\)（贡献 \(2\)），\(4^4 = 2^8\)（贡献 \(8\)），以及 \(6^6 = 2^6 \cdot 3^6\)（贡献 \(6\)）。因此 \(2\) 的指数总共为 \(2 + 8 + 6 = 16\)。

所以正确答案是 C。

English

Since \(k \ln k = \ln(k^k),\) the sum gives \(p = \ln\left(\prod_{k=1}^{6} k^k\right),\) so \[ e^p = 1^1 \cdot 2^2 \cdot 3^3 \cdot 4^4 \cdot 5^5 \cdot 6^6. \]

The factors of \(2\) come from \(2^2\) (giving \(2\)), \(4^4 = 2^8\) (giving \(8\)), and \(6^6 = 2^6 \cdot 3^6\) (giving \(6\)). In total the exponent of \(2\) is \(2 + 8 + 6 = 16.\)

Thus, the correct answer is C.

因为 \(P(0) = k\)， 写作 \(P(x) = ax^3 + bx^2 + cx + k\)。

于是 \(P(1) = a+b+c+k = 2k\)，且 \(P(-1) = -a+b-c+k = 3k\)。 两式相加得 \(2b + 2k = 5k\)， 所以 \(2b = 3k\)。

在和中奇次项抵消： \[ \begin{gathered} P(2)+P(-2) \\ = (8a+4b+2c+k) \\ {}+ (-8a+4b-2c+k) \\ = 8b + 2k\text{。} \end{gathered} \] 因为 \(8b = 4(2b) = 12k\)，所以这个和等于 \(12k + 2k = 14k\)。

所以正确答案是 E。

English

Since \(P(0) = k,\) write \(P(x) = ax^3 + bx^2 + cx + k.\)

Then \(P(1) = a+b+c+k = 2k\) and \(P(-1) = -a+b-c+k = 3k.\) Adding these gives \(2b + 2k = 5k,\) so \(2b = 3k.\)

The odd-power terms cancel in the sum: \[ \begin{gathered} P(2)+P(-2) \\ = (8a+4b+2c+k) \\ {}+ (-8a+4b-2c+k) \\ = 8b + 2k. \end{gathered} \] Since \(8b = 4(2b) = 12k,\) this equals \(12k + 2k = 14k.\)

Thus, the correct answer is E.

过 \(Q\) 的直线为 \(y = m(x-20) + 14\)。代入 \(y = x^2\) 得 \[ x^2 - mx + (20m - 14) = 0\text{。}\]

没有交点当且仅当这个方程没有实根，也就是判别式 \(m^2 - 4(20m-14)\) \(= m^2 - 80m + 56\) 为负。这发生在 \(m^2 - 80m + 56 = 0\) 的两个根 \(r\) 与 \(s\) 之间。

由韦达定理，\(r + s = 80\)。

所以正确答案是 E。

English

The line through \(Q\) is \(y = m(x-20) + 14.\) Substituting into \(y = x^2\) gives \[ x^2 - mx + (20m - 14) = 0. \]

There is no intersection exactly when this has no real root, i.e. when the discriminant \(m^2 - 4(20m-14)\) \(= m^2 - 80m + 56\) is negative. That happens between the two roots \(r\) and \(s\) of \(m^2 - 80m + 56 = 0.\)

By Vieta’s formulas, \(r + s = 80.\)

Thus, the correct answer is E.

任意单个数字都能得到 \(1\) 到 \(5\) 的和。如果某个连续段的和为 \(n\)，剩余数字也组成一个连续段，其和为 \(15 - n\)，所以从 \(10\) 到 \(14\) 的和也自动可以得到。因此，一个排列是坏排列，只可能是因为不能得到 \(6\) 或 \(7\)。

如果不能得到 \(6\)，通过旋转和翻折可设顺序为 \(1bc5e\)。数对 \(\{b,c\}\) 不能是 \(\{2,3\}\) 或 \(\{2,4\}\)，所以 \(e=2\)；再避开连续段 \(213\)，就迫使排列为 \(14352\)。如果不能得到 \(7\)，将顺序写成 \(2bc5e\)。此时 \(\{b,c\}\) 不能是 \(\{3,4\}\) 或 \(\{1,4\}\)，所以 \(e=4\)，避开 \(421\) 就迫使排列为 \(23154\)。

在旋转和翻折意义下，只有这两个坏排列。

因此，正确答案是 B。

English

Any single number covers sums \(1\) through \(5.\) If a consecutive block sums to \(n,\) the remaining numbers form a consecutive block summing to \(15 - n,\) so sums \(10\) through \(14\) are automatically covered as well. Thus an arrangement is bad only if it fails to produce \(6\) or \(7.\)

If \(6\) cannot be formed, rotate and reflect so the order is \(1bc5e.\) The pair \(\{b,c\}\) cannot be \(\{2,3\}\) or \(\{2,4\},\) so \(e=2;\) avoiding the block \(213\) then forces \(14352.\) If \(7\) cannot be formed, write the order as \(2bc5e.\) Now \(\{b,c\}\) cannot be \(\{3,4\}\) or \(\{1,4\},\) so \(e=4,\) and avoiding \(421\) forces \(23154.\)

These are the only two bad arrangements up to rotation and reflection.

Thus, the correct answer is B.

设上底半径为 \(1\)，下底半径为 \(r\)，球半径为 \(a\)。球与两个底面都相切，所以圆台的高为 \(2a\)。在轴截面中把球心放在 \((0,a)\)；过 \((r,0)\) 和 \((1,2a)\) 的右侧腰所在直线的方程为 \(2ax+(r-1)y-2ar=0\)。它到 \((0,a)\) 的距离为 \(a\)，所以 \((r+1)^2=4a^2+(r-1)^2\)，从而 \(r=a^2\)。

圆台体积为 \(\tfrac13 \pi (r^2 + r + 1)(2a)\)。令它等于球体积的两倍 \(\tfrac43 \pi a^3\)，并用 \(r = a^2\)，得到 \[ a^4 - 3a^2 + 1 = 0\text{，}\] 即 \(r^2 - 3r + 1 = 0\)。

正根为 \(r = \dfrac{3+\sqrt5}{2}\)。

所以正确答案是 E。

English

Let the top radius be \(1,\) the bottom radius \(r,\) and the sphere radius \(a.\) The sphere touches both bases, so the frustum height is \(2a.\) In an axial cross-section put the sphere center at \((0,a)\); the right slanted side through \((r,0)\) and \((1,2a)\) has equation \(2ax+(r-1)y-2ar=0.\) Its distance from \((0,a)\) is \(a,\) so \((r+1)^2=4a^2+(r-1)^2,\) giving \(r=a^2.\)

The frustum volume is \(\tfrac13 \pi (r^2 + r + 1)(2a).\) Setting it equal to twice the sphere volume \(\tfrac43 \pi a^3\) and using \(r = a^2\) yields \[ a^4 - 3a^2 + 1 = 0, \] that is \(r^2 - 3r + 1 = 0.\)

The positive root is \(r = \dfrac{3+\sqrt5}{2}.\)

Thus, the correct answer is E.

只有当 \(x - 40 \gt 0\) 且 \(60 - x \gt 0\) 时，对数才有定义，所以 \(40 \lt x \lt 60\)。

在这个范围内，不等式变为 \((x-40)(60-x) \lt 100\)，展开得 \(x^2 - 100x + 2500 \gt 0\)，即 \((x-50)^2 \gt 0\)。这对所有 \(x \ne 50\) 成立。

严格介于 \(40\) 和 \(60\) 之间且不等于 \(50\) 的整数为 \(41, \ldots, 49\) 和 \(51, \ldots, 59\)，共 \(18\) 个。

所以正确答案是 B。

English

The logarithms are defined only when \(x - 40 \gt 0\) and \(60 - x \gt 0,\) so \(40 \lt x \lt 60.\)

Within this range the inequality becomes \((x-40)(60-x) \lt 100,\) which expands to \(x^2 - 100x + 2500 \gt 0,\) i.e. \((x-50)^2 \gt 0.\) This holds for every \(x \ne 50.\)

The integers strictly between \(40\) and \(60\) except \(50\) are \(41, \ldots, 49\) and \(51, \ldots, 59,\) which is \(18\) values.

Thus, the correct answer is B.

设 \(x = BE = GH = CF\)，并设 \(\theta = \angle DHG = \angle AGJ\) \(= \angle FKH\)， 其中 \(AD = GJ = HK = 1\)。 在直角三角形 \(GDH\) 中，\(x \sin\theta = DG = 1 - \cos\theta\)， 所以 \(x = \dfrac{1 - \cos\theta}{\sin\theta}\)。

沿边 \(CD\)，有 \[ \begin{gathered} 1 = CF + FH + HD \\ = x + \sin\theta + x\cos\theta\text{。} \end{gathered} \] 代入 \(x\)，得 \[ \begin{gathered} 1 = \dfrac{(1-\cos\theta)(1+\cos\theta)}{\sin\theta} \\ {}+ \sin\theta \\ = \dfrac{\sin^2\theta}{\sin\theta} \\ {}+ \sin\theta \\ = 2\sin\theta\text{。} \end{gathered} \]

因此 \(\sin\theta = \tfrac12\)， 所以 \(\theta = 30^\circ\)，并且 \[ x = \dfrac{1 - \frac{\sqrt3}{2}}{\frac12} = 2 - \sqrt3\text{。} \]

所以正确答案是 C。

English

Let \(x = BE = GH = CF\) and \(\theta = \angle DHG = \angle AGJ\) \(= \angle FKH,\) with \(AD = GJ = HK = 1.\) In right triangle \(GDH,\) \(x \sin\theta = DG = 1 - \cos\theta,\) so \(x = \dfrac{1 - \cos\theta}{\sin\theta}.\)

Along side \(CD,\) \[ \begin{gathered} 1 = CF + FH + HD \\ = x + \sin\theta + x\cos\theta. \end{gathered} \] Substituting for \(x\) gives \[ \begin{gathered} 1 = \dfrac{(1-\cos\theta)(1+\cos\theta)}{\sin\theta} \\ {}+ \sin\theta \\ = \dfrac{\sin^2\theta}{\sin\theta} \\ {}+ \sin\theta \\ = 2\sin\theta. \end{gathered} \]

Hence \(\sin\theta = \tfrac12,\) so \(\theta = 30^\circ\) and \[ x = \dfrac{1 - \frac{\sqrt3}{2}}{\frac12} = 2 - \sqrt3. \]

Thus, the correct answer is C.

设 \(p_j\) 为从标号 \(j\) 的叶子出发最终到达标号 \(10\) 的概率。由中心处跳跃规则的对称性，\(p_5 = \tfrac12\)。

每个内部位置满足 \(p_j = \tfrac{10-j}{10}\,p_{j+1} + \tfrac{j}{10}\,p_{j-1}\)，因此 \[ \begin{gathered} p_4 = \tfrac25 p_3 + \tfrac35 p_5, \\ \quad p_3 = \tfrac{3}{10} p_2 + \tfrac{7}{10} p_4 \end{gathered}\text{，}\]\[ p_2 = \tfrac15 p_1 + \tfrac45 p_3,\quad p_1 = \tfrac{9}{10} p_2\text{。}\]

令 \(d_j=p_j-p_{j-1}\)。递推式等价于 \((10-j)d_{j+1}=jd_j\)，所以 \[ \begin{gathered} d_2=\dfrac{d_1}{9},\quad d_3=\dfrac{d_1}{36}, \\ d_4=\dfrac{d_1}{84},\quad d_5=\dfrac{d_1}{126} \end{gathered}\text{。}\] 又因为 \(p_0=0\) 且 \(p_5=\frac{1}{2}\)，所以 \[ \begin{aligned} \dfrac12&=d_1\biggl(1+\dfrac19+\dfrac1{36} \\ &\qquad+\dfrac1{84}+\dfrac1{126}\biggr) \\ &=d_1\cdot\dfrac{73}{63} \end{aligned}\text{。}\] 因此 \(p_1=d_1=\frac{63}{146}\)。

所以正确答案是 C。

English

Let \(p_j\) be the probability of eventually reaching pad \(10\) starting from pad \(j.\) By the symmetry of the jump rule at the center, \(p_5 = \tfrac12.\)

Each interior pad satisfies \(p_j = \tfrac{10-j}{10}\,p_{j+1} + \tfrac{j}{10}\,p_{j-1},\) which gives \[ \begin{gathered} p_4 = \tfrac25 p_3 + \tfrac35 p_5, \\ \quad p_3 = \tfrac{3}{10} p_2 + \tfrac{7}{10} p_4, \end{gathered} \] \[ p_2 = \tfrac15 p_1 + \tfrac45 p_3,\quad p_1 = \tfrac{9}{10} p_2. \]

Put \(d_j=p_j-p_{j-1}.\) The recurrence is equivalent to \((10-j)d_{j+1}=jd_j,\) so \[ \begin{gathered} d_2=\dfrac{d_1}{9},\quad d_3=\dfrac{d_1}{36}, \\ d_4=\dfrac{d_1}{84},\quad d_5=\dfrac{d_1}{126}. \end{gathered} \] Since \(p_0=0\) and \(p_5=\frac{1}{2},\) \[ \begin{aligned} \dfrac12&=d_1\biggl(1+\dfrac19+\dfrac1{36} \\ &\qquad+\dfrac1{84}+\dfrac1{126}\biggr) \\ &=d_1\cdot\dfrac{73}{63}. \end{aligned} \] Therefore \(p_1=d_1=\frac{63}{146}.\)

Thus, the correct answer is C.

在模 \(2017\) 下，恒等式 \(\binom{2014}{k} \cdot k! \cdot (2014-k)! = 2014!\) 结合 \(2016 \cdot 2015 \cdots (2015-k)\) \(\equiv (-1)^k (k+2)!\)，得到 \[ \begin{gathered} 2\binom{2014}{k} \equiv (-1)^k \\ {}\cdot (k+2)(k+1) \pmod{2017}\text{，} \end{gathered} \] 因此 \(\binom{2014}{k} \equiv (-1)^k \binom{k+2}{2}\)。

于是 \[ \begin{gathered} S \equiv \sum_{k=0}^{62} (-1)^k \binom{k+2}{2} \\ = 1 \\ {}+ \sum_{k=1}^{31}\left[\binom{2k+2}{2} - \binom{2k+1}{2}\right] \\ = 1 + \sum_{k=1}^{31}(2k+1)\text{。} \end{gathered} \]

剩下的和为 \(3 + 5 + \cdots + 63 = 1023\)， 所以 \(S \equiv 1 + 1023\) \(= 1024 \pmod{2017}\)。

所以正确答案是 C。

English

Working modulo \(2017,\) the identity \(\binom{2014}{k} \cdot k! \cdot (2014-k)! = 2014!\) together with \(2016 \cdot 2015 \cdots (2015-k)\) \(\equiv (-1)^k (k+2)!\) leads to \[ \begin{gathered} 2\binom{2014}{k} \equiv (-1)^k \\ {}\cdot (k+2)(k+1) \pmod{2017}, \end{gathered} \] so \(\binom{2014}{k} \equiv (-1)^k \binom{k+2}{2}.\)

Then \[ \begin{gathered} S \equiv \sum_{k=0}^{62} (-1)^k \binom{k+2}{2} \\ = 1 \\ {}+ \sum_{k=1}^{31}\left[\binom{2k+2}{2} - \binom{2k+1}{2}\right] \\ = 1 + \sum_{k=1}^{31}(2k+1). \end{gathered} \]

The remaining sum is \(3 + 5 + \cdots + 63 = 1023,\) so \(S \equiv 1 + 1023\) \(= 1024 \pmod{2017}.\)

Thus, the correct answer is C.

因为弧 \(AB, CD\) 相等，弧 \(BC, DE\) 相等，所以弦 \(AC, BD, CE\) 都相等；设 \(x = AC = BD = CE\)， \(y = AD\)， 且 \(z = BE\)。

对 \(ABCD\)、\(BCDE\) 和 \(ABDE\) 使用托勒密定理，得到 \[ \begin{gathered} 10y + 9 = x^2, \\ \quad 100 + 3z = x^2, \\ \quad 30 + 14x = yz\text{。} \end{gathered} \] 由前两个方程解出 \(y\) 和 \(z\)，再代入第三个方程，得到 \[ \begin{gathered} x^3 - 109x - 420 = 0 \\ = (x-12)(x+5)(x+7)\text{。} \end{gathered} \]

所以 \(x = 12\)， \(y = \tfrac{135}{10} = \tfrac{27}{2}\)， 且 \(z = \tfrac{44}{3}\)。 五条对角线为 \(AC, BD, CE, AD, BE\)， 它们的和为 \[ \begin{gathered} 3x + y + z = 36 \\ {}+ \tfrac{27}{2} + \tfrac{44}{3} \\ = \tfrac{385}{6}\text{。} \end{gathered} \]

因此 \(m + n = 385 + 6 = 391\)， 正确答案是 D。

English

Because arcs \(AB, CD\) are equal and arcs \(BC, DE\) are equal, the chords \(AC, BD, CE\) are all equal; let \(x = AC = BD = CE,\) \(y = AD,\) and \(z = BE.\)

Ptolemy’s theorem on \(ABCD,\) \(BCDE,\) and \(ABDE\) gives \[ \begin{gathered} 10y + 9 = x^2, \\ \quad 100 + 3z = x^2, \\ \quad 30 + 14x = yz. \end{gathered} \] Solving the first two for \(y\) and \(z\) and substituting into the third yields \[ \begin{gathered} x^3 - 109x - 420 = 0 \\ = (x-12)(x+5)(x+7). \end{gathered} \]

So \(x = 12,\) \(y = \tfrac{135}{10} = \tfrac{27}{2},\) and \(z = \tfrac{44}{3}.\) The five diagonals are \(AC, BD, CE, AD, BE,\) summing to \[ \begin{gathered} 3x + y + z = 36 \\ {}+ \tfrac{27}{2} + \tfrac{44}{3} \\ = \tfrac{385}{6}. \end{gathered} \]

Thus \(m + n = 385 + 6 = 391,\) and the correct answer is D.

令 \(x = \tfrac{\pi y}{2}\)。两边除以 \(2\)，并使用 \(\tfrac12(1 - \cos(2\pi y)) = \sin^2(\pi y)\)，方程化简为 \[ \cos(\pi y)\cos\left(\dfrac{4028\pi}{y}\right) = 1\text{。}\]

两个余弦必须都等于 \(1\)，或都等于 \(-1\)，因此 \(y\) 和 \(\tfrac{4028}{y}\) 是同奇偶性的整数。因为 \(4028 = 2^2 \cdot 19 \cdot 53\) 是偶数，二者都必须为偶数，所以 \(y = 2a\)，其中 \(a\) 是 \(2014 = 2 \cdot 19 \cdot 53\) 的正奇因数，故 \(a \in \{1, 19, 53, 19 \cdot 53\}\)。

每个这样的 \(a\) 给出 \(x = \tfrac{\pi y}{2} = \pi a\)，所以解的和为 \[ \begin{gathered} \pi(1 + 19 + 53 + 19\cdot53) \\ = \pi(19+1)(53+1) \\ = 1080\pi \end{gathered}\text{。}\]

所以正确答案是 D。

English

Let \(x = \tfrac{\pi y}{2}.\) Dividing by \(2\) and using \(\tfrac12(1 - \cos(2\pi y)) = \sin^2(\pi y),\) the equation simplifies to \[ \cos(\pi y)\cos\left(\dfrac{4028\pi}{y}\right) = 1. \]

Both cosines must equal \(1\) or both equal \(-1,\) so \(y\) and \(\tfrac{4028}{y}\) are integers of the same parity. Since \(4028 = 2^2 \cdot 19 \cdot 53\) is even, both must be even, so \(y = 2a\) with \(a\) a positive odd divisor of \(2014 = 2 \cdot 19 \cdot 53,\) giving \(a \in \{1, 19, 53, 19 \cdot 53\}.\)

Each such \(a\) gives \(x = \tfrac{\pi y}{2} = \pi a,\) so the sum of solutions is \[ \begin{gathered} \pi(1 + 19 + 53 + 19\cdot53) \\ = \pi(19+1)(53+1) \\ = 1080\pi. \end{gathered} \]

Thus, the correct answer is D.