Difference between revisions of "1963 AHSME Problems/Problem 39"

Latest revision as of 15:06, 27 December 2024

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

Solution 3 (Menelaus’s Theorem)

By using Menelaus’s Theorem on triangle BCE and points A, P, D, we can substitute in the known values to find that the answer is 5. -pengu14

@@ Line 93: / Line 93: @@
 ==Solution 3 (Menelaus’s Theorem)==
 By using Menelaus’s Theorem on triangle BCE and points A, P, D, we can substitute in the known values to find that the answer is 5.
+-pengu14
 ==See Also==

1963 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 38	Followed by Problem 40
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