In the diagram, two straight lines are to be drawn through $O(0,0)$ so that the lines divide the figure $O P Q R S T$ into 3 pieces of equal area. The sum of the slopes of the lines will be
(A) $\frac{35}{24}$
(B) $\frac{7}{6}$
(C) $\frac{5}{4}$
(D) $\frac{4}{3}$
(E) $\frac{11}{8}$

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-088.jpg?height=344&width=572&top_left_y=2251&top_left_x=1145)
Extend $Q R$ downwards to meet the $x$-axis at $U(6,0)$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-175.jpg?height=350&width=585&top_left_y=2270&top_left_x=824)

The area of figure $O P Q R S T$ equals the sum of the areas of square $O P Q U$ (which has side length 6 , so area 36) and rectangle $R S T U$ (which has height 2 and width 6 , so area 12).

Thus, the area of figure $O P Q R S T$ is 48 . If we are to divide the figure into three pieces of equal area, then each piece has area 16.

Let $V$ be the first point on the perimeter (measured clockwise from $P$ ) so that the line through $O$ and $V$ cuts off an area of 16 .

Note that the area of $\triangle O P Q$ is half of the area of square $O P Q U$, or 18 , so $V$ is to the left of $Q$ on $P Q$.

Thus, $V$ has coordinates $(v, 6)$ for some number $v$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-176.jpg?height=344&width=569&top_left_y=674&top_left_x=838)

Consider $\triangle O P V$ having base $O P$ of length 6 and height $P V$ of length $v$.

Since the area of $\triangle O P V$ is 16 , then $\frac{1}{2}(6)(v)=16$ or $3 v=16$ or $v=\frac{16}{3}$.

Therefore, the slope of $O V$ is $\frac{6}{\frac{16}{3}}=\frac{9}{8}$.

Let $W$ be the second desired point.

Since the area of $\triangle O T S$ is $\frac{1}{2}(O T)(T S)=\frac{1}{2}(12)(2)=12$ (less than $\frac{1}{3}$ of the total area) and the area of trapezoid $O R S T$ is $\frac{1}{2}(R S+O T)(S T)=\frac{1}{2}(6+12)(2)=18$ (more than $\frac{1}{3}$ of the total area), then $W$ lies on $R S$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-176.jpg?height=344&width=566&top_left_y=1538&top_left_x=839)

Suppose that $W$ has coordinates $(w, 2)$ for some number $w$. We want the area of trapezoid $W S T O$ to be 16 .

Therefore,

$$
\begin{aligned}
\frac{1}{2}(W S+O T)(S T) & =16 \\
\frac{1}{2}(12-w+12)(2) & =16 \\
24-w & =16 \\
w & =8
\end{aligned}
$$

Thus, the coordinates of $W$ are $(8,2)$, and so the slope of $O W$ is $\frac{2}{8}=\frac{1}{4}$.

Thus, the sum of the two required slopes is $\frac{9}{8}+\frac{1}{4}=\frac{11}{8}$.

ANswer: (E)