AMC10 2013 A

AMC10 2013 A · Q18

AMC10 2013 A · Q18. It mainly tests Linear equations, Area & perimeter.

Let points $A = (0, 0)$, $B = (1, 2)$, $C = (3, 3)$, and $D = (4, 0)$. Quadrilateral $ABCD$ is cut into equal area pieces by a line passing through $A$. This line intersects $CD$ at point $\left( \frac{p}{q}, \frac{r}{s} \right)$, where these fractions are in lowest terms. What is $p + q + r + s$?

设点$A = (0, 0)$、$B = (1, 2)$、$C = (3, 3)$和$D = (4, 0)$。四边形$ABCD$被一条经过$A$的直线切割成相等面积的部分。该直线与$CD$相交于点$\left( \frac{p}{q}, \frac{r}{s} \right)$，其中这些分数为最简形式。求$p + q + r + s$？

(A) 54 54

(B) 58 58

(D) 70 70

(E) 75 75

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Let line $AG$ be the required line, with $G$ on $\overline{CD}$. Divide $ABCD$ into triangle $ABF$, trapezoid $BCEF$, and triangle $CDE$, as shown. Their areas are $1$, $5$, and $\frac{3}{2}$, respectively. Hence the area of $ABCD=\frac{15}{2}$, and the area of triangle $ADG=\frac{15}{4}$. Because $AD=4$, it follows that $GH=\frac{15}{8}=\frac{r}{s}$. The equation of $\overline{CD}$ is $y=-3(x-4)$, so when $y=\frac{15}{8}$, $x=\frac{p}{q}=\frac{27}{8}$. Therefore $p+q+r+s=58$.

答案（B）：设直线 $AG$ 为所求直线，且点 $G$ 在 $\overline{CD}$ 上。如图，将 $ABCD$ 分割为三角形 $ABF$、梯形 $BCEF$ 和三角形 $CDE$。它们的面积分别为 $1$、$5$、$\frac{3}{2}$。因此，$ABCD$ 的面积为 $\frac{15}{2}$，三角形 $ADG$ 的面积为 $\frac{15}{4}$。由于 $AD=4$，可得 $GH=\frac{15}{8}=\frac{r}{s}$。$\overline{CD}$ 的方程为 $y=-3(x-4)$，所以当 $y=\frac{15}{8}$ 时，$x=\frac{p}{q}=\frac{27}{8}$。因此 $p+q+r+s=58$。

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