In triangle $A B C, B R=R C, C S=3 S A$, and $\frac{A T}{T B}=\frac{p}{q}$. If the area of $\triangle R S T$ is twice the area of $\triangle T B R$, then $\frac{p}{q}$ is equal to
(A) $\frac{2}{1}$
(B) $\frac{8}{3}$
(C) $\frac{5}{2}$
(D) $\frac{7}{4}$
(E) $\frac{7}{3}$

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-155.jpg?height=363&width=556&top_left_y=1412&top_left_x=1381)
Solution

Join $A$ to $R$ and $C$ to $T$. Label the diagram as shown. Let the area of $\mathrm{D} A B C$ be $k$.

Since triangles CRS and CRA have equal heights and bases that are in the ratio $3 b: 4 b=3: 4$, then

$$
\text { area of DCRS }=\frac{3}{4}(\text { area of DCRA })
$$

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-293.jpg?height=301&width=542&top_left_y=1232&top_left_x=1431)

However, since triangles $C R A$ and $A B R$ also have equal heights and bases that are in the ratio $a: a=1: 1$, then

$$
\begin{aligned}
& \text { DCRS }=\frac{3}{8}(\text { area of } \mathrm{D} A B C) \\
&=\frac{3 k}{8} \\
& \text { area of } \\
& \text { Similarly, area of } \mathrm{D} T B R=\frac{q}{p+q}(\text { area of } \mathrm{D} A B R) \\
&=\frac{q k}{2(p+q)}
\end{aligned}
$$

and area of DATS $=\frac{1}{4}($ area of DATC $)$

$$
=\frac{p k}{4(p+q)}
$$

Since (area of DRST) $=2($ area of $\mathrm{D} T B R)$, then

$$
k-\frac{3 k}{8}-\frac{q k}{2(p+q)}-\frac{p k}{4(p+q)}=\frac{2 q k}{2(p+q)}
$$

$$
1-\frac{3}{8}-\frac{q}{2(p+q)}-\frac{p}{4(p+q)}=\frac{2 q}{2(p+q)}(\text { since } k\{0)
$$

Simplify this expression to get $\frac{p}{q}=\frac{7}{3}$.

ANSWER: (E)