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2018 AMC 12A 第 17 题

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所有题目均经美国数学协会（MAA）官方合法授权使用。

17.

Farmer Pythagoras 有一块直角三角形田地。这个直角三角形的两条直角边长分别为 33 和 44 个单位。在两边相交的直角角落里，他留下一小块未种植的正方形 SS，使得从空中看像直角标记。田地其余部分都已种植。SS 到斜边的最短距离是 22 个单位。田地中已种植的部分占多少比例？

Farmer Pythagoras has a field in the shape of a right triangle. The right triangle’s legs have lengths of 33 and 44 units. In the corner where those sides meet at a right angle, he leaves a small unplanted square SS so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from SS to the hypotenuse is 22 units. What fraction of the field is planted?

2527\tfrac{25}{27}

2627\tfrac{26}{27}

7375\tfrac{73}{75}

145147\tfrac{145}{147}

7475\tfrac{74}{75}

答案：D
知识点：坐标几何距离公式面积
难度评级：1910
小提示：

把直角放在原点，使斜边为直线 3x+4y=123x + 4y = 12。

Put the right angle at the origin so the hypotenuse is the line 3x+4y=123x + 4y = 12

大提示：

正方形远角 (s,s)(s, s) 到该直线的距离为 22；解 ∣3s+4s−12∣5=2\tfrac{|3s + 4s - 12|}{5} = 2 并保留有效根

The far corner (s,s)(s, s) of the square is distance 22 from that line; solve ∣3s+4s−12∣5=2\tfrac{|3s + 4s - 12|}{5} = 2 and keep the valid root

解答：

将直角放在原点，两条直角边沿坐标轴，则顶点为 (4,0)(4, 0)、(0,3)(0, 3)、(0,0)(0, 0)，正方形 SS 为 [0,s]×[0,s][0, s] \times [0, s]。斜边是 3x+4y−12=03x + 4y - 12 = 0，它到正方形最近的角 (s,s)(s, s) 的距离满足 ∣3s+4s−12∣32+42=∣7s−12∣5=2。 \frac{|3s + 4s - 12|}{\sqrt{3^2 + 4^2}} = \frac{|7s - 12|}{5} = 2\text{。} 这给出 s=227s = \tfrac{22}{7} 或 s=27s = \tfrac27；只有 s=27s = \tfrac27 能使正方形留在三角形内部。

田地面积为 12⋅3⋅4=6\tfrac12 \cdot 3 \cdot 4 = 6，未种植正方形面积为 (27)2=449\left(\tfrac27\right)^2 = \tfrac{4}{49}。 已种植比例为 1−4496=1−2147=145147。 1 - \frac{\frac{4}{49}}{6} = 1 - \frac{2}{147} = \frac{145}{147}\text{。}

所以正确答案是 D。

Place the right angle at the origin with legs on the axes, so the vertices are (4,0),(4, 0), (0,3),(0, 3), (0,0),(0, 0), and the square SS is [0,s]×[0,s].[0, s] \times [0, s]. The hypotenuse is 3x+4y−12=0,3x + 4y - 12 = 0, and the distance from its nearest corner (s,s)(s, s) is ∣3s+4s−12∣32+42=∣7s−12∣5=2. \frac{|3s + 4s - 12|}{\sqrt{3^2 + 4^2}} = \frac{|7s - 12|}{5} = 2. This gives s=227s = \tfrac{22}{7} or s=27;s = \tfrac27; only s=27s = \tfrac27 keeps the square inside the triangle.

The field has area 12⋅3⋅4=6\tfrac12 \cdot 3 \cdot 4 = 6 and the unplanted square has area (27)2=449.\left(\tfrac27\right)^2 = \tfrac{4}{49}. The planted fraction is 1−4496=1−2147=145147. 1 - \frac{\frac{4}{49}}{6} = 1 - \frac{2}{147} = \frac{145}{147}.

Thus, the correct answer is D.

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