2023 年 AMC 8 真题（P版）

1 答案 D

可以按运算顺序化简如下。

\[\begin{align*} (32 + 2) - (8 + 8) &= 34 - 16 \\ &= 18 \end{align*}\]

所以正确答案是 D。

English

We can simplify this as follows.

\[\begin{align*} (32 + 2) - (8 + 8) &= 34 - 16 \\ &= 18 \end{align*}\]

Thus, D is the correct answer.

2 答案 E

把剪开后的纸展开，就得到下面的图形。

所以正确答案是 E。

English

We can unfold the cut up paper to achieve the following figure.

Thus, E is the correct answer.

3 答案 B

代入公式，风寒温度为

\[\begin{align*} 36 - 0.7 \cdot 18 &= 36 - 12.6 \\ &=23.4\text{。} \end{align*}\]

最接近的选项是 \(23\)。

所以正确答案是 B。

English

Using the formula, the wind chill is

\[\begin{align*} 36 - 0.7 \cdot 18 &= 36 - 12.6 \\ &=23.4. \end{align*}\]

The closest choice is \(23.\)

Thus, B is the correct answer.

4 答案 D

把其余的数填进去，就得到完整的网格。

由此可以看出，阴影方格中只有 \(19\)、\(23\) 和 \(47\) 是质数。

所以正确答案是 D。

English

We can fill in the other numbers to get the complete grid.

From this, we can see that the only prime numbers in the shaded boxes are \(19,\) \(23,\) and \(47.\)

Thus, D is the correct answer.

5 答案 B

样本中鳟鱼比例为 \(\dfrac{30}{180} = \dfrac{1}{6}\)。因此湖中所有鱼数是鳟鱼数的 \(6\) 倍。

湖中鱼的总数为 \(6 \cdot 250 = 1500\)。

所以正确答案是 B。

English

Note that \(\dfrac{30}{180} = \dfrac{1}{6}.\) This means that a sixth of the fish in the lake are trout, or in other words, the total number of fish is \(6\) times the number of trout.

Therefore, there are \(6 \cdot 250 = 1500\) fish in the lake.

Thus, B is the correct answer.

6 答案 C

不要把 \(0\) 放在底数位置，因为那样会使整个表达式等于 \(0\)。

所以 \(0\) 必须放在指数位置。指数为 \(0\) 的那一项会自动等于 \(1\)。

把两个 \(2\) 中的一个用于 \(2^0\)，就能把另一个 \(2\) 和 \(3\) 留给另一项。

另一项可以是 \(2^3\) 或 \(3^2\)。因为 \(3^2\) 更大，所以取 \(3^2\)。

于是最大值为 \[ 2^0 \times 3^2 = 1 \times 9 = 9\text{。}\]

所以正确答案是 C。

English

Note that we do not want \(0\) as a base, since that would make the expression equal to \(0.\)

This means that \(0\) must be an exponent. The number whose exponent is \(0\) will automatically evaluate to \(1.\)

Using one of the two copies of \(2\) in the factor \(2^0\) leaves \(2\) and \(3\) for the other factor.

The other factor can then be \(2^3\) or \(3^2.\) Since \(3^2\) is larger, we use \(3^2.\)

This gives us a final value of \[ 2^0 \times 3^2 = 1 \times 9 = 9. \]

Thus, C is the correct answer.

7 答案 B

画出这两条直线。

从图中可见，只有矩形左上角落在其中一条直线上。

所以正确答案是 B。

English

We can graph the two lines.

From this, we see that only the top left corner of the rectangle intersects either line.

Thus, B is the correct answer.

8 答案 A

每一列代表同一轮的两场比赛，因此恰好包含两个胜场记录 \(1\) 和两个负场记录 \(0\)。

逐列填入使该列达到两个 \(1\) 和两个 \(0\) 所需的数，得到蒂约的胜负记录为 \(000101\)。

所以正确答案是 A。

English

Each column represents one round of two matches, so every column must contain exactly two \(1\)’s and two \(0\)’s.

Completing each column to contain two \(1\) entries and two \(0\) entries gives Tiyo’s record \(000101.\)

Thus, A is the correct answer.

9 答案 B

她第一次达到 \(7\) 米是在 \(2\) 秒时。

随后她在 \(4\) 秒后低于 \(4\) 米，这一段贡献 \(4 - 2 = 2\) 秒。

接着她在 \(6\) 秒时重新高于 \(4\) 米，并在 \(10\) 秒时再次达到 \(7\) 米。

这一段贡献 \(10 - 6 = 4\) 秒。最后她在 \(12\) 秒时低于 \(7\) 米。

再到 \(14\) 秒时她低于 \(4\) 米，最后一段贡献 \(14 - 12 = 2\) 秒。

三个时间段的总长度为 \[ 2 + 4 + 2 = 8 \] 所以正确答案是 B。

English

The first time that she hits an elevation of \(7\) meters is at \(2\) seconds.

She then dips below \(4\) meters after \(4\) seconds. This adds \(4 - 2 = 2\) seconds to the total answer.

Malaika then goes above \(4\) meters at \(6\) seconds. She hits \(7\) meters again at \(10\) seconds.

This adds \(10 - 6 = 4\) more seconds to the total. She finally dips below \(7\) meters for the last time at \(12\) seconds.

She then falls below \(4\) meters at \(14\) seconds, finally adding \(14 - 12 = 2\) seconds to the total time.

The desired total is \[ 2 + 4 + 2 = 8 \] Thus, B is the correct answer.

10 答案 D

哈罗德把整个派的 \(1 - \frac{1}{4} = \frac{3}{4}\) 留给了朋友。

驼鹿吃掉其中 \(\frac{1}{3} \cdot \frac{3}{4} = \frac{1}{4}\)，所以剩下 \(\frac{3}{4} - \frac{1}{4} = \frac{1}{2}\)。

豪猪再吃掉剩余的 \(\frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6}\)，所以最后剩下 \(\frac{1}{2} - \frac{1}{6} = \frac{1}{3}\)。

所以正确答案是 D。

English

Harold left \(1 - \frac{1}{4} = \frac{3}{4}\) of the pie for his friends.

The moose ate \(\frac{1}{3} \cdot \frac{3}{4} = \frac{1}{4}\) of the pie, leaving \(\frac{3}{4} - \frac{1}{4} = \frac{1}{2}\) of the pie.

Finally, the porcupine ate \(\frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6}\) of the pie. This leaves \(\frac{1}{2} - \frac{1}{6} = \frac{1}{3}\) of the pie.

Thus, D is the correct answer.

11 答案 C

把距离近似为 \(300{,}000{,}000\) 英里，并把 \(6.5\) 个月近似为 \[ 30 \times 6.5 = 195 \approx 200 \] 天。

于是每天约行进 \[ 300{,}000{,}000 \div 200 = 1{,}500{,}000 \] 英里。

再除以 \(24\)，得到约 \(62{,}500\) 英里每小时，最接近 \(60{,}000\)。

所以正确答案是 C。

English

We can round the distance to \(300{,}000{,}000\) miles and approximate \(6.5\) months as \[ 30 \times 6.5 = 195 \approx 200 \] days.

This gives \[ 300{,}000{,}000 \div 200 = 1{,}500{,}000 \] miles per day.

Dividing by \(24\) gives about \(62{,}500\) miles per hour, which is closest to \(60{,}000.\)

Thus, C is the correct answer.

12 答案 B

不妨设每个最小圆的半径为 \(1\)。

这样有 \(3\) 个阴影单位圆，总面积为 \(3 \cdot 1^2 \pi = 3 \pi\)。

另外还有一个半径为 \(4\) 的阴影圆，其中挖去了两个半径为 \(2\) 的未涂色圆。

这部分额外阴影面积为 \[\begin{align*} 4^2 \pi - 2 \cdot 2^2 \pi &= 16 \pi - 8 \pi \\ &= 8 \pi\text{。} \end{align*}\]

阴影总面积为 \[8\pi + 3\pi = 11\pi\text{。}\]大未涂色圆的面积为 \[6^2\pi = 36\pi\text{。}\]所以所求分数为 \[\dfrac{11\pi}{36\pi} = \dfrac{11}{36}\text{。}\]

所以正确答案是 B。

English

Let each smallest circle have radius \(1.\)

This means that there are \(3\) shaded unit circles, which total to \(3 \cdot 1^2 \pi = 3 \pi\) area.

There is also a shaded circle with radius \(4\) with two unshaded circles of radius \(2\) inside.

This gives us an extra shaded area of \[\begin{align*} 4^2 \pi - 2 \cdot 2^2 \pi &= 16 \pi - 8 \pi \\ &= 8 \pi. \end{align*}\]

The total shaded area is \[ 8\pi + 3\pi = 11\pi. \] The area of the large unshaded circle is \[ 6^2\pi = 36\pi. \] Therefore, the desired fraction is \[ \dfrac{11\pi}{36\pi} = \dfrac{11}{36}. \]

Thus, B is the correct answer.

13 答案 D

第 \(3\) 个补水站在全程的 \(\frac{3}{8}\) 处，因为补水站把全程分成 \(8\) 个相等间隔。

第一个维修站在全程的 \(\frac{1}{3}\) 处。两站之间的距离占全程的 \[ \dfrac{3}{8} - \dfrac{1}{3} = \dfrac{1}{24}\text{。}\]这段距离是 \(2\) 英里，所以全程为 \(24 \cdot 2 = 48\) 英里。

所以正确答案是 D。

English

The \(3\)rd water station is located \(\frac{3}{8}\) of the way along the race (the water stations split the race up into \(8\) equal spaces).

The first repair station is located \(\frac{1}{3}\) of the way along the race. The distance between the stations is \[ \dfrac{3}{8} - \dfrac{1}{3} = \dfrac{1}{24} \] of the race length. This distance is \(2\) miles, so the race is \(24 \cdot 2 = 48\) miles long.

Thus, D is the correct answer.

14 答案 E

设 \(a\)、\(b\)、\(c\) 分别为 \(5\) 分、\(10\) 分、\(25\) 分邮票的张数，并设 \(n=a+b+c\)。金额方程除以 \(5\) 得 \[ a+2b+5c=142\text{，}\]所以 \(n+b+4c=142\)。

假设 \(n\ge56\)，则 \(b+4c\le86\)。又因为 \[ a=142-2b-5c\le20\text{，}\]所以 \(2b+5c\ge122\)。合并这两个不等式可得 \(c\le16\)，但这样 \(2b+5c\le2(20)+5(16)=120\)，矛盾。因此最多只能使用 \(55\) 张邮票。

取 \(a=19\)、\(b=19\)、\(c=17\) 时可以达到这个上界：\[ 19(5)+19(10)+17(25)=710\text{。}\]此时邮票总数为 \(19+19+17=55\)。

所以正确答案是 E。

English

Let \(a,\) \(b,\) \(c\) be the numbers of \(5\)-, \(10\)-, and \(25\)-cent stamps, and let \(n=a+b+c.\) Dividing the value equation by \(5\) gives \[ a+2b+5c=142, \] so \(n+b+4c=142.\)

Suppose \(n\ge56.\) Then \(b+4c\le86.\) Also \[ a=142-2b-5c\le20, \] so \(2b+5c\ge122.\) Combining these inequalities gives \(c\le16,\) but then \(2b+5c\le2(20)+5(16)=120,\) a contradiction. Thus at most \(55\) stamps can be used.

The bound is attainable with \(a=19,\) \(b=19,\) \(c=17:\) \[ 19(5)+19(10)+17(25)=710. \] Hence the greatest possible number of stamps is \(19+19+17=55.\)

Thus, E is the correct answer.

15 答案 B

半英里对应 \(10\) 个街区，所以一个街区长 \(\dfrac{1}{2} \div 10 = \dfrac{1}{20}\) 英里。

从绕路开始，维斯瓦姆需要走 \(7 \cdot \dfrac{1}{20} = \dfrac{7}{20}\) 英里。

正常情况下，从这个位置到学校需要 \(5\) 分钟。现在他需要在这 \(5\) 分钟内走完 \(\dfrac{7}{20}\) 英里。

注意 \(5\) 分钟等于 \(\dfrac{5}{60} = \dfrac{1}{12}\) 小时，所以所需速度为 \[\begin{align*} \dfrac{\frac{7}{20}}{\frac{1}{12}} &= 12 \cdot \dfrac{7}{20} \\&= \dfrac{21}{5} \\ &= 4.2 \end{align*}\] 英里每小时。

所以正确答案是 B。

English

If half a mile is the same as \(10\) blocks, then one block is \(\dfrac{1}{2} \div 10 = \dfrac{1}{20}\) miles.

Starting from the detour, Viswam has to walk \(7 \cdot \dfrac{1}{20} = \dfrac{7}{20}\) miles.

Normally, from this spot Viswam would take \(5\) minutes to walk to school. Now he has to travel \(\dfrac{7}{20}\) miles in \(5\) minutes.

Note that \(5\) minutes is \(\dfrac{5}{60} = \dfrac{1}{12}\) hours. This means his speed must be \[\begin{align*} \dfrac{\frac{7}{20}}{\frac{1}{12}} &= 12 \cdot \dfrac{7}{20} \\&= \dfrac{21}{5} \\ &= 4.2 \end{align*}\] miles per hour.

Thus, B is the correct answer.

16 答案 C

因为 \(20 = 3 \cdot 6 + 2\)，每列底部的 \(2\) 个字母会比第三个字母多出现一次。

所以每列中第三个字母出现 \(6\) 次，另外 \(2\) 个字母各出现 \(7\) 次。

在 \(20\) 列中，有 \(7\) 列的 \(P\) 只出现 \(6\) 次，其余 \(13\) 列中则出现 \(7\) 次。\(R\) 的情况相同。

因此 \(P\) 和 \(R\) 都各出现 \(7\cdot6+13\cdot7=42+91=133\) 次。

因此 \(Q\) 出现了 \[\begin{align*} 20 \cdot 20 - 2 \cdot 133 &=400 - 266 \\ &= 134 \end{align*}\] 次。

所以正确答案是 C。

English

Since \(20 = 3 \cdot 6 + 2,\) the bottom \(2\) letters in each column will occur one more time than the third letter.

This means that the third letter in each column will occur \(6\) times, whereas the other \(2\) will appear \(7\) times.

Across the \(20\) columns, \(P\) is the letter that appears \(6\) times in \(7\) columns and appears \(7\) times in the other \(13\) columns. The same is true of \(R.\)

Thus \(P\) and \(R\) each appear \(7\cdot6+13\cdot7=42+91=133\) times.

\(Q\) will therefore appear \[\begin{align*} 20 \cdot 20 - 2 \cdot 133 &=400 - 266 \\ &= 134 \end{align*}\] times.

Thus, C is the correct answer.

17 答案 A

折叠后，编号 \(2\)、\(3\)、\(4\) 和 \(5\) 的面形成八面体的下半部分，因此剩余四个面构成上半部分。

候选面是 \(1\)、\(6\) 和 \(7\)。从展开图可以看出，\(6\) 会在 \(Q\) 的左侧，而 \(7\) 又在 \(6\) 的左侧。

因此，剩下的面 \(1\) 必须在 \(Q\) 的右侧。

所以正确答案是 A。

English

Begin by observing that when folded, the faces labelled \(2,\) \(3,\) \(4,\) and \(5\) form the bottom half of the octahedron. As such, the remaining four faces must make up the top half of the octahedron.

From here, we have narrowed down our possibilities to \(1,\) \(6,\) and \(7.\) We can see that \(6\) will be the face to the left of the shaded region \(Q.\) This also gives us that \(7\) is to the left of \(6.\)

Therefore, we know that the only remaining face, \(1,\) must be to the right of the shaded region \(Q.\)

Thus, A is the correct answer.

18 答案 D

设向右跳 \(r\) 次，向左跳 \(l\) 次，则需要 \[ 5r-3l=2023\text{。}\]模 \(5\) 化简得 \(-3l\equiv3\pmod5\)，所以 \(l\equiv4\pmod5\)。

为了最少跳跃，取满足条件的最小 \(l\)，也就是 \(l=4\)。此时 \(5r-12=2023\)，所以 \(r=407\)。

总跳数为 \(407+4=411\)。

所以正确答案是 D。

English

Let \(r\) be the number of right jumps and \(l\) be the number of left jumps. We need \[ 5r-3l=2023. \] Reducing modulo \(5\) gives \(-3l\equiv3\pmod5\), so \(l\equiv4\pmod5\).

To minimize the total number of jumps, use the smallest possible \(l\), namely \(l=4\). Then \(5r-12=2023\), so \(r=407\).

The fewest number of jumps is \(407+4=411\).

Thus, D is the correct answer.

19 答案 C

内三角形边长是大三角形的 \(\frac{2}{3}\)，所以面积是大三角形的 \(\left(\frac{2}{3}\right)^2 = \frac{4}{9}\)。

三个梯形总面积是大三角形的 \(1 - \frac{4}{9} = \frac{5}{9}\)。

一个梯形面积是大三角形的 \(\frac{5}{9} \div 3 = \frac{5}{27}\)。

它与内三角形的面积比为 \[ \dfrac{\frac{5}{27}}{\frac{4}{9}} = \dfrac{5}{27} \cdot \dfrac{9}{4} = \dfrac{5}{12}\text{。}\]

所以正确答案是 C。

English

Since the inner triangle’s side length is \(\frac{2}{3}\) the side length of the outer triangle, its area is \(\left(\frac{2}{3}\right)^2 = \frac{4}{9}\) the area of the outer triangle.

This means that the three trapezoids are \(1 - \frac{4}{9} = \frac{5}{9}\) the area of the outer triangle.

Therefore, one trapezoid is \(\frac{5}{9} \div 3 = \frac{5}{27}\) the area of the outer triangle.

This makes the ratio of the areas of one trapezoid and the inner triangle \[ \dfrac{\frac{5}{27}}{\frac{4}{9}} = \dfrac{5}{27} \cdot \dfrac{9}{4} = \dfrac{5}{12}. \]

Thus, C is the correct answer.

20 答案 D

原来的极差是 \(28-3=25\)，所以新极差必须是 \(50\)。要保持中位数为 \(8\)，新增的一个数 \(x\) 必须小于 \(8\)，另一个数 \(y\) 必须大于 \(8\)。

若 \(x\ge3\)，最小值仍为 \(3\)，所以新的最大值必须是 \(53\)。取 \(x=3\) 会改变众数，因此 \(x\le7\)。所以 \(x+y\le7+53=60\)。

若 \(x<3\) 且 \(y>28\)，则 \(y-x=50\)，所以 \(x+y=2x+50\le54\)。若 \(y\le28\)，最大值仍为 \(28\)，这会迫使 \(x=-22\)，得到的和更小。因此没有任何情况能超过 \(60\)。

取 \(x=7\) 和 \(y=53\) 时，众数和中位数都不变，极差为 \(50\)，所以最大和是 \(60\)。

所以正确答案是 D。

English

The original range is \(28-3=25,\) so the new range must be \(50.\) To keep the median \(8,\) one added number \(x\) must be less than \(8\) and the other, \(y,\) must be greater than \(8.\)

If \(x\ge3,\) the minimum remains \(3,\) so the new maximum must be \(53.\) The value \(x=3\) would change the mode, so \(x\le7.\) Hence \(x+y\le7+53=60.\)

If \(x<3\) and \(y>28,\) then \(y-x=50,\) so \(x+y=2x+50\le54.\) If \(y\le28,\) the maximum remains \(28,\) forcing \(x=-22\) and giving an even smaller sum. Therefore no case exceeds \(60.\)

The values \(x=7\) and \(y=53\) preserve the mode and median and give range \(50,\) so the maximum sum is \(60.\)

Thus, D is the correct answer.

21 答案 C

所有数字的总和为 \(\dfrac{9 \cdot 10}{2} = 45\)，所以每组和必须为 \(45 \div 3 = 15\)。

含 \(9\) 的组中，另外两个数和为 \(6\)。可能是 \(1\) 和 \(5\)，或 \(2\) 和 \(4\)。

情况 \(1\)：一组是 \(1\)、\(5\) 和 \(9\)。

考虑含 \(8\) 的那组，另外两个数必须和为 \(7\)。唯一选择是 \(3\) 和 \(4\)。

剩下的一组是 \(2\)、\(6\) 和 \(7\)，其和为 \(15\)，所以这种情况给出一种分法。

情况 \(2\)：一组是 \(2\)、\(4\) 和 \(9\)。

同理，含 \(8\) 的那组另外两个数和为 \(7\)。唯一选择是 \(1\) 和 \(6\)。

最后一组是 \(3\)、\(5\) 和 \(7\)，其和为 \(15\)，这又给出一种分法。

所有情况都已检查，共有 \(2\) 种分法。

所以正确答案是 C。

English

The sum of all the numbers is \(\dfrac{9 \cdot 10}{2} = 45.\) This means that the sum of each group is \(45 \div 3 = 15.\)

Consider the group with \(9\) in it. The other two numbers must add to \(6.\) Therefore, the other cards in this group are \(1\) and \(5\) or \(2\) and \(4.\)

Case \(1:\) One group is \(1,\) \(5,\) and \(9.\)

Consider the group with \(8\) in it. The other numbers must add to \(7.\) The only option is \(3\) and \(4\) with the remaining cards.

The other group is then \(2,\) \(6,\) and \(7.\) This adds to \(15,\) so this case contributes one possibility.

Case \(2:\) One group is \(2,\) \(4,\) and \(9.\)

Consider the group with \(8\) in it. As above, the other numbers have to add to \(7.\) The only option is \(1\) and \(6.\)

The final group is \(3,\) \(5,\) and \(7,\) which adds to \(15.\) This is another configuration.

We have gone through all the cases, which revealed that there are only \(2\) possible groupings.

Thus, C is the correct answer.

22 答案 D

设第一项为 \(x\)，第二项为 \(y\)。

于是，序列的前几项为 \[ x, y, xy, xy^2, x^2y^3, x^3y^5, \cdots \] 因此 \(x^3y^5 = 4000\)。

将 \(4000\) 分解质因数，得到 \[ 4000 = 2^5 \cdot 5^3\text{。}\]因为 \(x\) 和 \(y\) 都是正整数，所以能整除 \(4000\) 的完全五次方只有 \(1\) 和 \(32\)。

若 \(y = 1\)，则 \(x^3 = 4000\)，但 \(4000\) 不是完全立方数，不可能成立。因此 \[ y^5 = 32 \] 且 \[ y = 2\text{。}\]

这迫使 \(x\) 等于 \(5\)。

所以正确答案是 D。

English

Let \(x\) be the first term and \(y\) be the second.

Then we get the following sequence: \[ x, y, xy, xy^2, x^2y^3, x^3y^5, \cdots \] Thus \(x^3y^5 = 4000.\)

Factoring \(4000,\) gives \[ 4000 = 2^5 \cdot 5^3. \] Because \(x\) and \(y\) are positive integers, the only fifth powers that divide \(4000\) are \(1\) and \(32.\)

If \(y = 1,\) then \(x^3 = 4000,\) which is impossible because \(4000\) is not a perfect cube. Therefore, \[ y^5 = 32 \] and \[ y = 2. \]

This forces \(x\) to equal \(5.\)

Thus, D is the correct answer.

23 答案 C

总铺法数为 \(4^9\)。可形成大阴影菱形的 \(2\times2\) 小网格有 \(4\) 个。

选定其中一个 \(2\times2\) 小网格后，它内部四块瓷砖的方向就被确定了，其余 \(5\) 个方格可以任意填。所以每选定一个这样的 \(2\times2\) 小网格，都对应 \(4^5\) 种铺法。

两个不同的 \(2\times2\) 小网格不能同时形成大阴影菱形，因为重叠方格会要求不同方向。因此有利铺法数为 \(4\cdot4^5=4^6\)。

所求概率为 \[ \dfrac{4^6}{4^9} = \dfrac{1}{4^3} = \dfrac{1}{64}\text{。}\]

所以正确答案是 C。

English

There are \(4^9\) possible tilings. There are \(4\) possible \(2\times2\) grids where a large shaded diamond could appear.

After one of these \(2\times2\) grids is chosen, the four tile orientations inside it are forced, and the other \(5\) squares can be filled in any way. This gives \(4^5\) tilings for each chosen \(2\times2\) grid.

Two different \(2\times2\) grids cannot both contain a large shaded diamond, because their overlapping squares would require incompatible tile orientations. Therefore the number of favorable tilings is \(4\cdot4^5=4^6.\)

The desired probability is then \[ \dfrac{4^6}{4^9} = \dfrac{1}{4^3} = \dfrac{1}{64}. \]

Thus, C is the correct answer.

24 答案 A

设 \(\triangle ABC\) 的面积为 \(a\)。左图中的未阴影三角形高为 \(11\)，并且与整个三角形相似，所以它的面积是 \(a\left(\frac{11}{h}\right)^2\)。因此，左图的阴影面积是 \(a\left(1-\left(\frac{11}{h}\right)^2\right)\)。

右图中的阴影三角形高为 \(h-5\)，并且与整个三角形相似，所以它的面积是 \(a\left(\frac{h-5}{h}\right)^2\)。令两图的阴影面积相等，再约去 \(a\)，得到 \[ 1-\left(\dfrac{11}{h}\right)^2 =\left(\dfrac{h-5}{h}\right)^2\text{。}\]

化简得 \[ h^2 - 121 = h^2 - 10h + 25\text{，}\]也就是 \[ 10h = 146\text{。}\]

所以 \(h=14.6\)。

所以正确答案是 A。

English

Let \(a\) be the area of \(\triangle ABC.\) The unshaded triangle in the left figure has height \(11\) and is similar to the full triangle, so its area is \(a\left(\frac{11}{h}\right)^2.\) Therefore, the shaded area there is \(a\left(1-\left(\frac{11}{h}\right)^2\right).\)

In the right figure, the shaded triangle has height \(h-5\) and is similar to the full triangle, so its area is \(a\left(\frac{h-5}{h}\right)^2.\) Equating the shaded areas and canceling \(a\) gives \[ 1-\left(\dfrac{11}{h}\right)^2 =\left(\dfrac{h-5}{h}\right)^2. \]

Simplifying yields \[ h^2 - 121 = h^2 - 10h + 25. \] This simplifies to \[ 10h = 146, \]

so \(h=14.6.\)

Thus, A is the correct answer.

25 答案 A

设公差为 \(d\)。将 \(a_1\) 取其最大可能值 \(10\)，将 \(a_{15}\) 取其最小可能值 \(241\)，可得 \[ d \geq \dfrac{241 - 10}{14} = 16.5\text{。}\]因为所有数都是整数，所以 \(d\) 至少为 \(17\)。

将 \(a_1\) 取其最小可能值 \(1\)，将 \(a_{15}\) 取其最大可能值 \(250\)，可得 \[ d \leq \dfrac{250 - 1}{14} \approx 17.8\text{。}\]因为所有数都是整数，所以 \(d\) 至多为 \(17\)。因此 \(d=17\)。

注意 \(17 \cdot 14 = 238\)。由于 \(a_{15}\) 至少为 \(241\)，所以 \(a_1\) 至少为 \(3\)。

另一方面，如果 \(a_1\) 大于 \(3\)，那么 \(a_2=a_1+17\) 就会大于 \(20\)，不符合条件。

所以 \(a_1 = 3\)，且 \(d = 17\)。于是 \[ a_{14} = 3 + 13 \cdot 17 = 224\text{。}\]

数字和为 \[2 + 2 + 4 = 8\text{。}\]

所以正确答案是 A。

English

Let \(d\) be the common difference. Using the largest possible value \(10\) for \(a_1\) and the smallest possible value \(241\) for \(a_{15},\) we have \[ d \geq \dfrac{241 - 10}{14} = 16.5. \] Since all the numbers are integers, \(d\) must be at least \(17.\)

Using the smallest possible value \(1\) for \(a_1\) and the largest possible value \(250\) for \(a_{15},\) we have \[ d \leq \dfrac{250 - 1}{14} \approx 17.8. \] Since all the numbers are integers, \(d\) is at most \(17.\) Therefore, \(d=17.\)

Note that \(17 \cdot 14 = 238.\) Since \(a_{15}\) is at least \(241,\) \(a_1\) must be at least \(3.\)

On the other hand, if \(a_1\) were greater than \(3,\) then \(a_2=a_1+17\) would be greater than \(20,\) which is not allowed.

Now we know that \(a_1 = 3\) and \(d = 17.\) This tells us that \[ a_{14} = 3 + 13 \cdot 17 = 224. \]

Therefore, sum of the digits is \[2 + 2 + 4 = 8.\]

Thus, A is the correct answer.