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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)[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

[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 $[APE] = 3b$, $r = \tfrac{CP}{PE} = 5$, 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

See Also

1963 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 38 		Followed by
Problem 40
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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