AMC12 2018 A

AMC12 2018 A · Q17

AMC12 2018 A · Q17. It mainly tests Area & perimeter, Coordinate geometry.

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直角三角形田地，直角三角形的两条直角边长分别为 3 和 4 单位。在两条直角边相交的直角角上，他留出一个小的未种植正方形 $S$，从空中看像直角符号。田地的其余部分都种植了。从 $S$ 到斜边的最近距离为 2 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上的顶点为 $(4,0)$、$(0,3)$、$(0,0)$，用 $[0,s]\times[0,s]$ 表示未种植的田地区域。斜边的方程为 $3x+4y-12=0$，因此，从 $(s,s)$（$S$ 中最靠近斜边的那个顶点）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。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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