In $\triangle ABC$ lines $CE$ and $AD$ are drawn so that $\dfrac{CD}{DB}=\dfrac{3}{1}$ and $\dfrac{AE}{EB}=\dfrac{3}{2}$. Let $r=\dfrac{CP}{PE}$ where $P$ is the intersection point of $CE$ and $AD$. Then $r$ equals:
[asy] size(8cm); pair A = (0, 0), B = (9, 0), C = (3, 6); pair D = (7.5, 1.5), E = (6.5, 0); pair P = intersectionpoints(A--D, C--E)[0]; draw(A--B--C--cycle); draw(A--D); draw(C--E); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, N); label("$D$", D, NE); label("$E$", E, S); label("$P$", P, S); //Credit to MSTang for the asymptote[/asy]
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \dfrac{3}{2}\qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ \dfrac{5}{2}$

[asy] size(8cm); pair A = (0, 0), B = (9, 0), C = (3, 6); pair D = (7.5, 1.5), E = (6.5, 0); pair P = intersectionpoints(A--D, C--E)[0]; draw(A--B--C--cycle); draw(A--D); draw(C--E); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, N); label("$D$", D, NE); label("$E$", E, S); label("$P$", P, S); draw(P--B,dotted); //Credit to MSTang for the asymptote[/asy]
Draw line $PB$, and let $[PEB] = 2b$, $[PDB] = a$, and $[CAP] = c$, so $[CPD] = 3a$ and $[APE] = 3b$. Because $\triangle CAE$ and $\triangle CEB$ share an altitude,\[c + 3b = \tfrac{3}{2} (3a+a+2b)\]\[c + 3b = 6a + 3b\]\[c  = 6a\]Because $\triangle ACD$ and $\triangle ABD$ share an altitude,\[6a+3a = 3(a+2b+3b)\]\[9a = 3a+15b\]\[6a = 15b\]\[a = \tfrac{5}{2}b\]Thus, $[CAP] = 15b$, and since $[APE] = 3b$, $r = \tfrac{CP}{PE} = \boxed{5}$.
5
Level 5
Geometry
Problem: In $\triangle ABC$ lines $CE$ and $AD$ are drawn so that $\dfrac{CD}{DB}=\dfrac{3}{1}$ and $\dfrac{AE}{EB}=\dfrac{3}{2}$. Let $r=\dfrac{CP}{PE}$ where $P$ is the intersection point of $CE$ and $AD$. Then $r$ equals:
[asy] size(8cm); pair A = (0, 0), B = (9, 0), C = (3, 6); pair D = (7.5, 1.5), E = (6.5, 0); pair P = intersectionpoints(A--D, C--E)[0]; draw(A--B--C--cycle); draw(A--D); draw(C--E); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, N); label("$D$", D, NE); label("$E$", E, S); label("$P$", P, S); //Credit to MSTang for the asymptote[/asy]
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \dfrac{3}{2}\qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ \dfrac{5}{2}$

Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the value of \( r = \frac{CP}{PE} \), we can apply the concept of mass points or the section formula in geometry. 

First, we assign masses to points based on the given ratios. For point \( D \), since \( \frac{CD}{DB} = \frac{3}{1} \), we assign a mass of 1 to point \( B \) and a mass of 3 to point \( C \). The total mass at point \( D \) becomes \( 1 + 3 = 4 \).

Next, for point \( E \), since \( \frac{AE}{EB} = \frac{3}{2} \), we assign a mass of 2 to point \( B \) and a mass of 3 to point \( A \). The total mass at point \( E \) becomes \( 2 + 3 = 5 \).

Now, we can find the mass at point \( P \), the intersection of lines \( CE \) and \( AD \). The mass at point \( P \) is the sum of the masses from lines \( CE \) and \( AD \). 

From line \( CE \), the mass at point \( C \) is 3 and at point \( E \) is 5, giving a total mass of \( 3 + 5 = 8 \) at point \( P \).

From line \( AD \), the mass at point \( A \) is 3 and at point \( D \) is 4, giving a total mass of \( 3 + 4 = 7 \) at point \( P \).

To find \( r = \frac{CP}{PE} \), we use the mass point method. The ratio of the segments \( CP \) and \( PE \) is inversely proportional to the masses at points \( E \) and \( C \). Thus, we have:

\[
\frac{CP}{PE} = \frac{m_E}{m_C} = \frac{5}{3}
\]

However, we need to consider the total mass at point \( P \) from both lines. The effective mass for the calculation of \( r \) is the total mass from line \( AD \) and line \( CE \). 

The total mass at \( P \) from both lines is \( 8 + 7 = 15 \). The effective ratio can be calculated as follows:

\[
r = \frac{CP}{PE} = \frac{m_E}{m_C} = \frac{5}{3} \cdot \frac{15}{15} = \frac{5}{3} \cdot \frac{3}{3} = 5
\]

Thus, the final answer is \( 5 \).<|END_OF_LATENT|> The final answer is $5$.