AMC12 2009 B

AMC12 2009 B · Q14

AMC12 2009 B · Q14. It mainly tests Area & perimeter, Coordinate geometry.

Five unit squares are arranged in the coordinate plane as shown, with the lower left corner at the origin. The slanted line, extending from $(c,0)$ to $(3,3)$, divides the entire region into two regions of equal area. What is $c$?

如图，坐标平面上摆放了五个单位正方形，左下角在原点。连接 $(c,0)$ 与 $(3,3)$ 的斜线将整个区域分成面积相等的两部分。求 $c$。

(A) \frac{1}{2} \frac{1}{2}

(B) \frac{3}{5} \frac{3}{5}

(D) \frac{3}{4} \frac{3}{4}

(E) \frac{4}{5} \frac{4}{5}

Answer

Correct choice: (C)

正确答案：（C）

Solution

For $c\geq 1.5$ the shaded area is at most $1.5$, which is too little. Hence $c<1.5$, and therefore the point $(2,1)$ is indeed inside the shaded part, as shown in the picture. Then the area of the shaded part is one less than the area of the triangle with vertices $(c,0)$, $(3,0)$, and $(3,3)$. The area of the entire triangle is $\frac{3(3-c)}2$, therefore the area of the shaded part is $\frac{7-3c}{2}$. The entire figure has area $5$, hence we want the shaded part to have area $\frac 52$. Solving for $c$, we get $c=\boxed{\frac 23}$. The answer is $\mathrm{(C)}$ ( what a coincidence ;) ).

当 $c\geq 1.5$ 时，阴影部分面积至多为 $1.5$，太小了。因此 $c<1.5$，从而点 $(2,1)$ 的确在阴影部分内，如图所示。此时阴影部分的面积比顶点为 $(c,0)$、$(3,0)$、$(3,3)$ 的三角形面积少 1。整个三角形的面积为 $\frac{3(3-c)}2$，因此阴影部分面积为 $\frac{7-3c}{2}$。整个图形面积为 $5$，因此我们希望阴影部分面积为 $\frac 52$。解得 $c=\boxed{\frac 23}$。答案是 $\mathrm{(C)}$。

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