2015A AMC 10 真题（P版）

1 答案 C

可以这样计算。\[\begin{aligned} &(2^0-1+5^2+0)^{-1} \times 5 \\&=\dfrac{1}{1 - 1 + 25} \times 5 \\ &= \dfrac{5}{25}\\ &= \dfrac{1}{5} \end{aligned}\]

所以正确答案是 C。

English

We can evaluate it as follows. \[\begin{aligned} &(2^0-1+5^2+0)^{-1} \times 5 \\&=\dfrac{1}{1 - 1 + 25} \times 5 \\ &= \dfrac{5}{25}\\ &= \dfrac{1}{5} \end{aligned}\]

Thus, C is the correct answer.

2 答案 D

设三角形瓷砖有 \(x\) 块，正方形瓷砖有 \(y\) 块，则 \[ x + y = 25 \] 且 \[ 3x + 4y = 84\text{。}\]把第一个方程乘以 \(3\)，得到 \[ 3x + 3y = 75\text{，}\]两式相减得 \(y = 9\)。

所以正确答案是 D。

English

Let \(x\) be the number of triangular tiles and \(y\) be the number of square tiles. We then have that \[ x + y = 25 \] and \[ 3x + 4y = 84. \] Multiplying the first equation by \(3\) gives us \[ 3x + 3y = 75 \] and subtracting this from the other equation gives us \(y = 9.\)

Thus, D is the correct answer.

3 答案 D

我们试着找出楼梯级数与所需牙签数之间的规律。

对于 \(1\) 级楼梯，只需要 \(4\) 根牙签，也就是一个正方形。

对于 \(2\) 级楼梯，根据图形需要 \(10\) 根牙签。

同理，\(3\) 级楼梯需要 \(18\) 根牙签。

\(2\) 级楼梯比 \(1\) 级楼梯多用 \(10 - 4 = 6\) 根牙签。\(3\) 级楼梯比 \(2\) 级楼梯多用 \(18 - 10 = 8\) 根牙签。

按照这个规律，\(4\) 级楼梯需要 \(18 + 10 = 28\) 根牙签，\(5\) 级楼梯需要 \(28 + 12 = 40\) 根牙签。

这说明 Ann 还需要添加 \(40 - 18 = 22\) 根牙签。

所以正确答案是 D。

English

Let us try to find a pattern between the number of toothpicks needed for the staircases.

For a \(1\)-step staircase, we would only need \(4\) toothpicks (just a square).

For a \(2\)-step staircase, we would need \(10\) toothpicks according to the diagram.

Similarly, we would need \(18\) toothpicks for a \(3\)-step staircase.

A \(2\)-step staircase needs \(10 - 4 = 6\) more toothpicks than a \(1\)-step staircase. A \(3\)-step staircase needs \(18 - 10 = 8\) more toothpicks than a \(2\)-step staircase.

Following this pattern, we can see that a \(4\)-step staircase will need \(18 + 10 = 28\) toothpicks, and a \(5\)-step staircase will need \(28 + 12 = 40\) toothpicks.

This means that Ann would need to add \(40 - 18 = 22\) more toothpicks.

Thus, D is the correct answer.

4 答案 B

设 Mia 有 \(m\) 个糖果蛋，则 Sofia 有 \(2m\) 个，Pablo 有 \(6m\) 个。

糖果蛋总数为 \[ m + 2m + 6m = 9m\text{。}\]三人数量相同时，每人应有 \(9m \div 3 = 3m\) 个。

Sofia 还需要 \(3m - 2m = m\) 个糖果蛋，因此 Pablo 应给她自己原有糖果蛋的 \(\dfrac{m}{6m} = \dfrac{1}{6}\)。

所以正确答案是 B。

English

Let \(m\) be the number of candy eggs that Mia had. Then Sofia had \(2m\) eggs and Pablo had \(6m\) eggs.

The total number of eggs is then \[ m + 2m + 6m = 9m. \] For all of them to have the same number of eggs, they each must have \(9m \div 3 = 3m\) eggs.

Sofia needs \(3m - 2m = m\) more eggs. This means Pablo must give \(\dfrac{m}{6m} = \dfrac{1}{6}\) of his eggs to Sofia.

Thus, B is the correct answer.

5 答案 E

前 \(14\) 份已批改试卷的总分为 \(14\cdot 80=1120\)。

加入 Payton 的试卷后，总分为 \(15\cdot 81=1215\)，所以 Payton 的分数为 \(1215-1120=95\)。

所以正确答案是 E。

English

The total for the first \(14\) graded tests was \(14\cdot 80=1120\).

After Payton’s test was included, the total became \(15\cdot 81=1215\). Therefore Payton’s score was \(1215-1120=95\).

Thus, E is the correct answer.

6 答案 B

设两个数为 \(x\) 和 \(y\)，并假设 \(x \gt y\)。则 \[ x + y = 5(x - y)\text{。}\]化简得到 \[ x + y = 5x - 5y \] \[ 6y = 4x\text{。}\]

两边相除，得到 \[\dfrac{x}{y} = \dfrac{6}{4} = \dfrac{3}{2}\text{。}\]

所以正确答案是 B。

English

Let \(x\) and \(y\) be the two numbers. Then we have that \[ x + y = 5(x - y). \] Note that we are assuming \(x \gt y.\) This gives us \[ x + y = 5x - 5y \]\[ 6y = 4x. \]

Dividing through yields \[\dfrac{x}{y} = \dfrac{6}{4} = \dfrac{3}{2}.\]

Thus, B is the correct answer.

7 答案 B

回忆等差数列的第 \(n\) 项为 \(a + d(n - 1)\)，其中 \(a\) 是首项，\(d\) 是公差。

这里 \(a = 13\)，\(d = 3\)。代入可得 \[\begin{aligned} 73 &= 13 + 3(n - 1) \\ 20&= n - 1 \\ n &= 21 \end{aligned}\text{。}\]所以正确答案是 B。

English

Recall that the \(n\)th term of an arithmetic sequence is \(a + d(n - 1),\) where \(a\) is the first term and \(d\) is the common difference.

For us, \(a = 13\) and \(d = 3.\) Plugging these in, we get that \[\begin{aligned} 73 &= 13 + 3(n - 1) \\ 20&= n - 1 \\ n &= 21. \end{aligned}\] Thus, B is the correct answer.

8 答案 B

设 Pete 和 Claire 当前年龄分别为 \(p\) 和 \(c\)。

题意给出 \[ p - 2 = 3(c - 2) \] 和 \[ p - 4 = 4(c - 4)\text{。}\]化简为 \[ p = 3c - 4 \] 和 \[ p = 4c - 12\text{。}\]令两式相等，得到 \[ 3c - 4 = 4c - 12 \] \[ c = 8\text{。}\]

因此 \[ p = 3 \cdot 8 - 4 = 20\text{。}\]

设还需 \(y\) 年，则 \[ 20 + y = 2(8 + y)\text{，}\]化简得 \[ 20 + y = 16 + 2y \] \[ y = 4\text{。}\]所以正确答案是 B。

English

Let \(p\) and \(c\) be Pete’s and Claire’s current ages respectively.

Then we have that \[ p - 2 = 3(c - 2) \] and \[ p - 4 = 4(c - 4). \] Simplifying both equations gives us \[ p = 3c - 4 \] and \[ p = 4c - 12. \] Setting them equal, we have \[ 3c - 4 = 4c - 12 \]\[ c = 8. \]

This means that \[ p = 3 \cdot 8 - 4 = 20. \]

Now, we need to find the number of years (\(y\)) until \[ 20 + y = 2(8 + y), \] which gives us \[ 20 + y = 16 + 2y \]\[ y = 4. \] Thus, B is the correct answer.

9 答案 D

设两个圆柱的半径和高分别为 \(r_1\)、\(h_1\) 与 \(r_2\)、\(h_2\)。

已知 \[ r_2 = \dfrac{11}{10}r_1 \] 且 \[ \pi r_1^2h_1 = \pi r_2^2h_2\text{。}\]

代入并化简，得到 \[ r_1^2h_1 = \dfrac{121}{100}r_1^2h_2\text{，}\]这告诉我们 \[ h_1 = \dfrac{121}{100}h_2\text{。}\]

所以正确答案是 D。

English

Let \(r_1\) and \(h_1\) be the radius and height of the first cylinder and similarly define \(r_2\) and \(h_2\) for the second cylinder.

We know that \[ r_2 = \dfrac{11}{10}r_1 \] and \[ \pi r_1^2h_1 = \pi r_2^2h_2. \]

Substituting and simplifying gives us \[ r_1^2h_1 = \dfrac{121}{100}r_1^2h_2, \] which tells us that \[ h_1 = \dfrac{121}{100}h_2. \]

Thus, D is the correct answer.

10 答案 C

禁用的相邻字母对为 \(ab,ba,bc,cb,cd,dc\)。

若排列以 \(a\) 开头，第二个字母只能是 \(c\) 或 \(d\)。

在 \(ac\) 之后，无论怎样排列剩余字母，都会出现 \(cb\) 或 \(cd\)。在 \(ad\) 之后，剩余的 \(b\) 和 \(c\) 必须相邻。因此不存在以 \(a\) 开头的合格排列。

由对称性，也不存在以 \(d\) 开头的合格排列。

若以 \(b\) 开头，后面必须依次为 \(d\)、\(a\)、\(c\)，得到 \(bdac\)。

同理，以 \(c\) 开头时只能得到 \(cadb\)。因此共有 \(2\) 种合格排列。

所以正确答案是 C。

English

The forbidden adjacent pairs are \(ab,ba,bc,cb,cd,dc.\)

If an arrangement starts with \(a,\) its second letter must be \(c\) or \(d.\)

After \(ac,\) either remaining order contains \(cb\) or \(cd.\) After \(ad,\) the remaining \(b\) and \(c\) must be adjacent. Thus no valid arrangement starts with \(a.\)

By symmetry, no valid arrangement starts with \(d.\)

If an arrangement starts with \(b,\) it must continue with \(d,\) then \(a,\) then \(c,\) giving \(bdac.\)

Similarly, starting with \(c\) gives only \(cadb.\) Therefore there are \(2\) valid rearrangements.

Thus, C is the correct answer.

11 答案 C

设长方形边长为 \(4x\) 和 \(3x\)。则对角线长为 \[ \sqrt{(4x)^2 + (3x)^2} = \sqrt{25x^2} = 5x\text{。}\]

长方形面积为 \[ 4x \cdot 3x = 12x^2\text{。}\] 因此 \[ kd^2 = 12x^2 \] \[ k = \dfrac{12x^2}{25x^2} = \dfrac{12}{25}\text{。}\]

所以正确答案是 C。

English

Let the side lengths be \(4x\) and \(3x.\) Then the diagonal has length \[ \sqrt{(4x)^2 + (3x)^2} = \sqrt{25x^2} = 5x. \]

The area of the rectangle is \[ 4x \cdot 3x = 12x^2. \] Then we get that \[ kd^2 = 12x^2 \]\[ k = \dfrac{12x^2}{25x^2} = \dfrac{12}{25}. \]

Thus, C is the correct answer.

12 答案 C

代入 \(x=\sqrt{\pi}\)。于是 \(x^2=\pi\)，\(x^4=\pi^2\)，所以方程变为 \[y^2-2\pi y+\pi^2=1\text{。}\]

因此 \((y-\pi)^2=1\)，从而得到两个可能取值 \(y=\pi+1\) 和 \(y=\pi-1\)。它们的距离为 \(2\)。

所以正确答案是 C。

English

Substitute \(x=\sqrt{\pi}\). Then \(x^2=\pi\) and \(x^4=\pi^2\), so the equation becomes \[y^2-2\pi y+\pi^2=1.\]

Hence \((y-\pi)^2=1\), giving the two possible values \(y=\pi+1\) and \(y=\pi-1\). Their distance is \(2\).

Thus, C is the correct answer.

13 答案 C

设 \(5\) 美分硬币有 \(x\) 枚，则 \(10\) 美分硬币有 \(12 - x\) 枚。

若 \(x=0\)，则只能组成 \(10,20,\ldots,120\) 美分，共 \(12\) 种金额。因此 \(x>0\)。只要至少有一枚 \(5\) 美分硬币，就可以组成从 \(5\) 美分到总金额 \[5x+10(12-x)=120-5x\] 之间的每个 \(5\) 的倍数。

这样的 \(5\) 的正倍数共有 \(24 - x\) 种。题目给出这个数量为 \(17\)，所以 \(x = 7\)。

因此 \(10\) 美分硬币有 \(12 - 7 = 5\) 枚。

所以正确答案是 C。

English

Let the number of \(5\)-cent coins be \(x\) and the number of \(10\)-cent coins be \(12 - x.\)

If \(x=0,\) the only possible values are \(10,20,\ldots,120\) cents, giving just \(12\) values. Hence \(x>0.\) Having at least one \(5\)-cent coin then makes every multiple of \(5\) from \(5\) through the total value \[5x+10(12-x)=120-5x\] obtainable.

There are \(24 - x\) such multiples of \(5,\) which means that \(x = 7\) to get \(17\) possible different values.

The number of \(10\)-cent coins is therefore \(12 - 7 = 5.\)

Thus, C is the correct answer.

14 答案 C

半径 \(10\) 的圆盘在半径 \(20\) 的钟面外侧滚动。若接触点绕钟面转过圆心角 \(\theta\)，圆盘相对于原方向转过 \(\frac{20+10}{10}\theta=3\theta\)。

箭头再次竖直向上时，\(3\theta\) 等于 \(2\pi\) 的正整数倍。第一次发生在 \(\theta=\frac{2\pi}{3}\)，即从 \(12\) 点钟方向绕一圈的三分之一，对应 \(4\) 点钟方向。

所以正确答案是 C。

English

The disk of radius \(10\) rolls externally around the clock face of radius \(20\). If the point of tangency moves through central angle \(\theta\) around the clock, the disk rotates through \(\frac{20+10}{10}\theta=3\theta\) relative to its original direction.

The arrow next points upward when \(3\theta\) is a positive multiple of \(2\pi\). The first time this happens is \(\theta=\frac{2\pi}{3}\), which is one-third of the way around the clock from \(12\) o’clock, namely \(4\) o’clock.

Thus, C is the correct answer.

15 答案 B

条件为 \[\frac{x+1}{y+1}=\frac{11}{10}\cdot\frac{x}{y}\text{。}\] 交叉相乘得 \(10y(x+1)=11x(y+1)\)，即 \(xy+11x-10y=0\)。

从两边减去 \(110\) 并因式分解，得 \[(x-10)(y+11)=-110\text{。}\] 因为 \(x,y\) 是正整数，可用的负因数对为 \((-1,110)\)、\((-2,55)\) 和 \((-5,22)\)，分别给出 \((x,y)=(9,99),(8,44),(5,11)\)。

其中只有 \(\frac{5}{11}\) 的分子分母互质，所以恰好有一个分数满足条件。

所以正确答案是 B。

English

The condition is \[\frac{x+1}{y+1}=\frac{11}{10}\cdot\frac{x}{y}.\] Cross-multiplying gives \(10y(x+1)=11x(y+1)\), or \(xy+11x-10y=0\).

Factoring by grouping after subtracting \(110\) gives \[(x-10)(y+11)=-110.\] Since \(x,y\) are positive, the useful negative factor pairs are \((-1,110)\), \((-2,55)\), and \((-5,22)\), producing \((x,y)=(9,99),(8,44),(5,11)\).

Only \(\frac{5}{11}\) has relatively prime numerator and denominator, so exactly one fraction works.

Thus, B is the correct answer.

16 答案 B

展开并相加两个方程，得到 \[ x^2 + y^2 - 4x - 4y + 8 \] \[ = x + y + 8\text{。}\] 整理得 \[ x^2 + y^2 = 5(x + y)\text{。}\] 两式相减，得到 \[ x^2 - y^2 - 4x + 4y = y - x\text{。}\] 再整理得 \[ x^2 - y^2 = 3(x - y)\text{。}\] 因为 \(x \neq y\)，两边可除以 \(x - y\)，得到 \[ x + y = 3 \]\[ x^2 + y^2 = 15\text{。}\]

所以正确答案是 B。

English

Adding the two equations gives us \[ x^2 + y^2 - 4x - 4y + 8 \] \[ = x + y + 8. \] We can rearrange this equation to get \[ x^2 + y^2 = 5(x + y). \] We can then subtract them to get \[ x^2 - y^2 - 4x + 4y = y - x. \] Once again rearranging, we can find \[ x^2 - y^2 = 3(x - y). \] We have that \(x \neq y,\) which means that we can divide both sides by \(x - y.\) This gives us \[ x + y = 3 \]\[ x^2 + y^2 = 15. \]

Thus, B is the correct answer.

17 答案 D

因为等边三角形的一边是竖直线，所以其对称轴为水平线。

这说明第三条边的斜率必须是第二条边斜率的相反数，也就是 \(-\dfrac{\sqrt{3}}{3}\)。

要求周长，只需求出这个三角形一条边的长度。

把 \(x = 1\) 代入另外两个方程，就得到竖直边上的两个顶点。

两个 \(y\) 值分别为 \(1 + \dfrac{\sqrt{3}}{3}\) 和 \(-\dfrac{\sqrt{3}}{3}\)。它们的差为 \(1 + \dfrac{2\sqrt{3}}{3}\)，所以周长为 \[ 3 \cdot \left(1 + \dfrac{2\sqrt{3}}{3}\right) = 3 + 2\sqrt{3}\text{。}\] 因此正确答案是 D。

English

Since one of the sides of the equilateral triangle is a vertical line, the line of symmetry perpendicular to this side must be horizontal.

This means that the slope of the third side must be opposite the slope of the second side, which would be \(-\dfrac{\sqrt{3}}{3}.\)

To find the perimeter, we only need to find the length of one of the sides of the triangle.

We can plug in \(x = 1\) into the two other equations to get the two vertices on the vertical line.

The two \(y\)-values are \(1 + \dfrac{\sqrt{3}}{3}\) and \(-\dfrac{\sqrt{3}}{3}.\) Their difference is \(1 + \dfrac{2\sqrt{3}}{3},\) which makes the perimeter \[ 3 \cdot \left(1 + \dfrac{2\sqrt{3}}{3}\right) = 3 + 2\sqrt{3}. \] Thus, D is the correct answer.

18 答案 E

在十六进制中，\(1000=3E8_{16}\)。把较短的表示在前面补零，补成三位数。从 \(000_{16}\) 到 \(399_{16}\) 的每个纯数字字符串都不超过 \(3E8_{16}\)，而且这些就是全部可能。

第一位有 \(4\) 种选择，另外两位各有 \(10\) 种选择。因此共有 \[4\cdot10\cdot10=400\] 个字符串，其中包括 \(000_{16}\)。去掉零后得到 \(n=399\)。

\(399\) 的数位和为 \(21\)。

所以正确答案是 E。

English

In hexadecimal, \(1000=3E8_{16}.\) Pad each shorter representation to three digits with leading zeros. Every numeric-only string from \(000_{16}\) through \(399_{16}\) is at most \(3E8_{16},\) and these are all the possibilities.

The first digit has \(4\) choices, and each of the other two digits has \(10\) choices. This gives \[4\cdot10\cdot10=400\] strings, including \(000_{16}.\) Excluding zero leaves \(n=399.\)

The sum of the digits in \(399\) is \(21.\)

Thus, E is the correct answer.

19 答案 D

因为 \(\triangle ABC\) 是面积为 \(12.5\) 的等腰直角三角形，所以两条直角边长为 \(5\)；三等分线给出 \(\angle ACD=30^\circ\) 和 \(\angle BCE=30^\circ\)，并且 \(\triangle ACD\) 与 \(\triangle BCE\) 的面积相等。

从 \(D\) 向 \(AC\) 作垂线，垂足为 \(F\)。因为 \(D\) 在 \(AB\) 上且 \(\angle A=45^\circ\)，所以 \(\triangle AFD\) 是等腰直角三角形。设 \(AF=DF=h\)，则 \(CF=5-h\)。由 \(30^\circ\) 直角三角形关系， \[\frac{CF}{DF}=\sqrt{3}\text{。}\] 因此 \(5-h=h\sqrt{3}\)，所以 \(h=\frac{5}{1+\sqrt{3}}=\frac{5\sqrt{3}-5}{2}\)。

因此 \[ \begin{aligned} &[ACD]=\frac12\cdot 5\cdot h \\ &=\frac{25\sqrt{3}-25}{4} \end{aligned}\text{。}\] 从 \(\triangle ABC\) 的面积中减去两个全等的角上三角形，得到 \[ \begin{aligned} &[CDE]=\frac{25}{2} \\ &\quad {}-2\cdot\frac{25\sqrt{3}-25}{4} \\ &=\frac{50-25\sqrt{3}}{2} \end{aligned}\text{。}\]

所以正确答案是 D。

English

Since \(\triangle ABC\) is isosceles right with area \(12.5\), its legs have length \(5\). The trisectors make \(\angle ACD=30^\circ\) and \(\angle BCE=30^\circ\), so \(\triangle ACD\) and \(\triangle BCE\) have equal area.

Drop a perpendicular from \(D\) to \(AC,\) with foot \(F.\) Since \(D\) lies on \(AB\) and \(\angle A=45^\circ,\) \(\triangle AFD\) is isosceles right. Let \(AF=DF=h.\) Then \(CF=5-h,\) and the \(30^\circ\) angle gives \[\frac{CF}{DF}=\sqrt{3}.\] Thus \(5-h=h\sqrt{3}\), so \(h=\frac{5}{1+\sqrt{3}}=\frac{5\sqrt{3}-5}{2}\).

Therefore \[ \begin{aligned} &[ACD]=\frac12\cdot 5\cdot h \\ &=\frac{25\sqrt{3}-25}{4}. \end{aligned} \] Subtracting the two congruent corner triangles from \(\triangle ABC\), \[ \begin{aligned} &[CDE]=\frac{25}{2} \\ &\quad {}-2\cdot\frac{25\sqrt{3}-25}{4} \\ &=\frac{50-25\sqrt{3}}{2}. \end{aligned} \]

Thus, D is the correct answer.

20 答案 B

设长方形的正整数边长为 \(x\) 和 \(y\)，则 \[ \begin{aligned} &A+P=xy+2x+2y \\ &=(x+2)(y+2)-4 \end{aligned}\text{。}\] 因此 \(A+P+4\) 必须分解成两个都至少为 \(3\) 的整数之积。

各选项加 \(4\) 后得到 \(104,106,108,110,112\)。除 \(106\) 外，其余数都有两个不小于 \(3\) 的因数： \[104=4\cdot26\text{，}\] \[108=9\cdot12\text{，}\] \[110=10\cdot11\text{，}\] \[112=7\cdot16\text{。}\] 但 \(106=2\cdot53\)，不能等于 \((x+2)(y+2)\)。

所以正确答案是 B。

English

Let the side lengths be positive integers \(x\) and \(y\). Then \[ \begin{aligned} &A+P=xy+2x+2y \\ &=(x+2)(y+2)-4. \end{aligned} \] Hence \(A+P+4\) must factor into two integers both at least \(3\).

The answer choices plus \(4\) are \(104,106,108,110,112\). All except \(106\) have a factorization with both factors at least \(3\): \[104=4\cdot26,\] \[108=9\cdot12,\] \[110=10\cdot11,\] \[112=7\cdot16.\] But \(106=2\cdot53\), so it cannot equal \((x+2)(y+2)\).

Thus, B is the correct answer.

21 答案 C

我们断言三角形 \(ABC\) 和 \(ABD\) 所在平面互相垂直。

分别从 \(C\) 和 \(D\) 向 \(AB\) 作垂线，即可证明这一点。

因为 \(AC = AD\) 且 \(BC = BD\)，这两条高的垂足重合于点 \(P\)。

于是 \[ CP = DP = \dfrac{3 \cdot 4}{5} = \dfrac{12}{5}\text{。}\] 又有 \(CD = CP\sqrt{2}\)，所以 \(\triangle CPD\) 是等腰直角三角形，且 \(CP\perp DP\)。

因为 \(CP\perp AB\) 且 \(CP\perp DP\)，所以线段 \(CP\) 垂直于平面 \(ABD\)。因此四面体的体积为 \[ \dfrac{1}{3}[ABD] \cdot CP = \dfrac{6}{3} \cdot \dfrac{12}{5} = \dfrac{24}{5}\text{。}\]

所以正确答案是 C。

English

We claim that the planes \(ABC\) and \(ABD\) are perpendicular to each other.

We can show this by dropping the perpendiculars from \(C\) and \(D\) to \(AB\).

Since \(AC = AD\) and \(BC = BD,\) we have that the feet of these altitudes will coincide at point \(P.\)

Then we have that \[ CP = DP = \dfrac{3 \cdot 4}{5} = \dfrac{12}{5}. \] We also have \(CD = CP\sqrt{2},\) so \(\triangle CPD\) is an isosceles right triangle and \(CP\perp DP.\)

Since \(CP\perp AB\) and \(CP\perp DP\), the segment \(CP\) is perpendicular to the plane \(ABD\). Finally, the volume of the tetrahedron is \[ \dfrac{1}{3}[ABD] \cdot CP = \dfrac{6}{3} \cdot \dfrac{12}{5} = \dfrac{24}{5}. \]

Thus, C is the correct answer.

22 答案 A

统计站起来的人构成的集合。站起 \(0\) 人和 \(1\) 人时，分别有 \(1\) 种和 \(8\) 种。

站起 \(2\) 人时，从任意两人的组合中减去 \(8\) 对相邻者，共有 \(\binom82-8=20\) 种。

站起 \(3\) 人时，先选一人；其余五个非相邻座位有 \(10\) 对，其中 \(4\) 对相邻，剩下 \(6\) 对。每个最终集合被计数三次，所以共有 \(\frac{8\cdot6}{3}=16\) 种。

站起 \(4\) 人时，只有两种交替站立的情况。因此有利结果共有 \(1+8+20+16+2=47\) 种。全部 \(2^8=256\) 种结果等可能，所以概率为 \(\frac{47}{256}\)。

所以正确答案是 A。

English

Count the possible sets of people who stand. For \(0\) and \(1\) people standing, there are \(1\) and \(8\) possibilities.

For \(2\) people standing, choose any pair and subtract the \(8\) adjacent pairs: \(\binom82-8=20\).

For \(3\) people standing, first choose one standing person. Among the remaining five non-neighbor seats, \(10\) pairs are possible, but \(4\) of those pairs are adjacent, leaving \(6\). This counts each final set three times, so there are \(\frac{8\cdot6}{3}=16\) possibilities.

For \(4\) people standing, the only possibilities are the two alternating sets. Thus the number of favorable coin-flip outcomes is \(1+8+20+16+2=47\). Since all \(2^8=256\) outcomes are equally likely, the probability is \(\frac{47}{256}\).

Thus, A is the correct answer.

23 答案 C

设整数零点为 \(r\) 和 \(s\)。由韦达定理，\(a = r + s\)，且 \(2a = rs\)。

因此 \[ rs = 2(r + s)\text{，}\] 整理并配方得到 \[ rs - 2r - 2s = 0 \] \[ rs - 2r - 2s + 4 = 4 \] \[ (r - 2)(s - 2) = 4\text{。}\]

可行的 \((r - 2, s - 2)\) 只有 \[ (1, 4), (-1, -4), (4, 1), (-4, -1)\text{，}\] \[ (2, 2), (-2, -2)\text{。}\] 对每一对，都有 \[ a = r - 2 + s - 2 + 4\text{。}\] 所有不同的 \(a\) 值为 \[ -1, 0, 8, 9\text{。}\] 它们的和为 \(-1 + 8 + 9 = 16\)。

所以正确答案是 C。

English

Let the zeros be \(r\) and \(s.\) Using Vieta’s formulas, we have that \(a = r + s\) and \(2a = rs.\)

Then we get that \[ rs = 2(r + s), \] which rearranges to \[ rs - 2r - 2s = 0 \] \[ rs - 2r - 2s + 4 = 4 \] \[ (r - 2)(s - 2) = 4. \]

The only possible pairs \((r - 2, s - 2)\) that work are \[ (1, 4), (-1, -4), (4, 1), (-4, -1), \] \[ (2, 2), (-2, -2). \] For any of these pairs, we have that \[ a = r - 2 + s - 2 + 4. \] We want all the such unique values of \(a.\) We get that they are \[ -1, 0, 8, 9. \] The sum of these values is \(-1 + 8 + 9 = 16.\)

Thus, C is the correct answer.

24 答案 B

设 \(BC=x\)，且 \(CD=AD=y\)。从 \(A\) 向 \(CD\) 作高，得到一个直角三角形，其直角边为 \(x\) 和 \(y-2\)，斜边为 \(y\)。因此 \[(y-2)^2+x^2=y^2\text{，}\] 所以 \(x^2=4(y-1)\)。

因为 \(x\) 是整数，写作 \(x=2k\)。于是 \(y=k^2+1\)，周长为 \[ \begin{aligned} &p=2+x+y+y \\ &=2k^2+2k+4 \end{aligned}\text{。}\]

需要 \(2k^2+2k+4<2015\)，也就是 \(k^2+k<1005.5\)。这对 \(k=1,2,\ldots,31\) 成立，而 \(k=32\) 过大，所以共有 \(31\) 个可能的周长。

所以正确答案是 B。

English

Let \(BC=x\) and \(CD=AD=y\). Dropping the altitude from \(A\) to \(CD\) gives a right triangle with legs \(x\) and \(y-2\), and hypotenuse \(y\). Therefore \[(y-2)^2+x^2=y^2, \] so \(x^2=4(y-1)\).

Since \(x\) is an integer, write \(x=2k\). Then \(y=k^2+1\), and the perimeter is \[ \begin{aligned} &p=2+x+y+y \\ &=2k^2+2k+4. \end{aligned} \]

We need \(2k^2+2k+4<2015\), or \(k^2+k<1005.5\). This holds for \(k=1,2,\ldots,31\), while \(k=32\) is too large. Thus there are \(31\) possible perimeters.

Thus, B is the correct answer.

25 答案 A

先固定第一个点所在的边。第二个点在同一边、相邻边、对边上的概率分别为 \(\frac14\)、\(\frac12\)、\(\frac14\)。

同一边上时，两个坐标 \(a,b\in[0,1]\) 的距离至少为 \(\frac12\)，等价于 \(|a-b|\ge\frac12\)。这个区域由两个直角三角形组成，总面积为 \(\frac14\)。

相邻边上时，距离形如 \(\sqrt{a^2+b^2}\)。失败区域是半径为 \(\frac12\) 的四分之一圆，所以成功概率为 \(1-\frac{\pi}{16}\)。

对边上时，距离总是至少为 \(1\)，所以成功概率为 \(1\)。因此所求概率为 \[ \begin{aligned} &\frac14\cdot\frac14 \\ &\quad {}+\frac12\left(1-\frac{\pi}{16}\right) \\ &\quad {}+\frac14=\frac{26-\pi}{32} \end{aligned}\text{。}\] 所以 \(a+b+c=26+1+32=59\)。

所以正确答案是 A。

English

Fix one of the two points. The second point is on the same side with probability \(\frac14\), on an adjacent side with probability \(\frac12\), and on the opposite side with probability \(\frac14\).

On the same side, two coordinates \(a,b\in[0,1]\) are at distance at least \(\frac12\) when \(|a-b|\ge\frac12\). This region consists of two right triangles with total area \(\frac14\).

On adjacent sides, the distance has the form \(\sqrt{a^2+b^2}\). The failing region is a quarter circle of radius \(\frac12\), so the success probability is \(1-\frac{\pi}{16}\).

On opposite sides, the distance is always at least \(1\), so the success probability is \(1\). Therefore the desired probability is \[ \begin{aligned} &\frac14\cdot\frac14 \\ &\quad {}+\frac12\left(1-\frac{\pi}{16}\right) \\ &\quad {}+\frac14=\frac{26-\pi}{32}. \end{aligned} \] Hence \(a+b+c=26+1+32=59\).

Thus, A is the correct answer.