# 1963 AHSME Problems/Problem 39

## Problem 39

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: $[asy] size(8cm); pair A = (0, 0), B = (9, 0), C = (3, 6); pair D = (7.5, 1.5), E = (6.5, 0); pair P = intersectionpoints(A--D, C--E); draw(A--B--C--cycle); draw(A--D); draw(C--E); label("A", A, SW); label("B", B, SE); label("C", C, N); label("D", D, NE); label("E", E, S); label("P", P, S); //Credit to MSTang for the asymptote[/asy]$ $\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \dfrac{3}{2}\qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ \dfrac{5}{2}$

## Solution $[asy] size(8cm); pair A = (0, 0), B = (9, 0), C = (3, 6); pair D = (7.5, 1.5), E = (6.5, 0); pair P = intersectionpoints(A--D, C--E); draw(A--B--C--cycle); draw(A--D); draw(C--E); label("A", A, SW); label("B", B, SE); label("C", C, N); label("D", D, NE); label("E", E, S); label("P", P, S); draw(P--B,dotted); //Credit to MSTang for the asymptote[/asy]$

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 $[APE] = 3b$, $r = \tfrac{CP}{PE} = 5$, which is answer choice $\boxed{\textbf{(D)}}$.

## Solution 2 (Mass Geometry)

Let the mass of point $A$, $B$, $C$, $D$, and $E$ be $mA$, $mB$, $mC$, $mD$, and $mE$ respectively. By mass geometry theorems, we have \begin{align*} \frac{CP}{PE} = \frac{mE}{mC} \end{align*} Focusing on the line segment $AB$, using mass geometry theorems, we have \begin{align*} mE = mA + mB \end{align*} and \begin{align*} \frac{3}{2} &= \frac{mB}{mA} \\ mA &= \frac{2}{3}mB \end{align*} which leads to $mE = \frac{5}{3}mB$. For line segment $CB$, similarly, we got $$mC = \frac{1}{3}mB$$ Substituting $mC$ and $mE$ back to the equation we obtained at the beginning, we got: \begin{align*} \frac{CP}{PE} = \frac{mE}{mC} = \frac{\frac{5}{3}mB}{\frac{1}{3}mB} = 5 \end{align*} which gives us the answer choice $\boxed{\textbf{(D)}}$. -nullptr07

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