AMC10 2018 A

AMC10 2018 A · Q23

AMC10 2018 A · Q23. It mainly tests Pythagorean theorem, Area & perimeter.

Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths of 3 and 4 units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from $S$ to the hypotenuse is 2 units. What fraction of the field is planted?

毕达哥拉斯农夫有一个形状为直角三角形的田地。该直角三角形的直角边长分别为 3 和 4 个单位。在这两条边构成直角的顶点处，他留出一个小的未种植正方形 $S$，从空中看就像直角符号。田地的其余部分都种植了。从 $S$ 到斜边的最近距离为 2 个单位。田地中有多少分数被种植了？

(A) $\frac{25}{27}$ $\frac{25}{27}$

(B) $\frac{26}{27}$ $\frac{26}{27}$

(D) $\frac{145}{147}$ $\frac{145}{147}$

(E) $\frac{74}{75}$ $\frac{74}{75}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Let the triangle's vertices in the coordinate plane be (4,0), (0,3), and (0,0), with $[0,s]\times[0,s]$ representing the unplanted portion of the field. The equation of the hypotenuse is $3x+4y-12=0$, so the distance from $(s,s)$, the corner of $S$ closest to the hypotenuse, to this line is given by $$ \frac{|3s+4s-12|}{\sqrt{3^2+4^2}}. $$ Setting this equal to 2 and solving for $s$ gives $s=\frac{22}{7}$ and $s=\frac{2}{7}$, and the former is rejected because the square must lie within the triangle. The unplanted area is thus $\left(\frac{2}{7}\right)^2=\frac{4}{49}$, and the requested fraction is $$ 1-\frac{\frac{4}{49}}{\frac{1}{2}\cdot 4\cdot 3}=\frac{145}{147}. $$

答案（D）：设该三角形在坐标平面中的顶点为 $(4,0)$、$(0,3)$ 和 $(0,0)$，并用 $[0,s]\times[0,s]$ 表示未种植的正方形区域。斜边的方程为 $3x+4y-12=0$，因此从 $(s,s)$（正方形 $S$ 距离斜边最近的那个顶点）到这条直线的距离为 $$ \frac{|3s+4s-12|}{\sqrt{3^2+4^2}}。 $$ 令其等于 2 并解 $s$，得 $s=\frac{22}{7}$ 和 $s=\frac{2}{7}$；由于正方形必须位于三角形内部，故舍去前者。于是未种植面积为 $\left(\frac{2}{7}\right)^2=\frac{4}{49}$，所求比例为 $$ 1-\frac{\frac{4}{49}}{\frac{1}{2}\cdot 4\cdot 3}=\frac{145}{147}。 $$

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