1965 AHSME Problems/Problem 37

Problem

Point $E$ is selected on side $AB$ of $\triangle{ABC}$ in such a way that $AE: EB = 1: 3$ and point $D$ is selected on side $BC$ such that $CD: DB = 1: 2$ . The point of intersection of $AD$ and $CE$ is $F$ . Then $\frac {EF}{FC} + \frac {AF}{FD}$ is:

$\textbf{(A)}\ \frac {4}{5} \qquad \textbf{(B) }\ \frac {5}{4} \qquad \textbf{(C) }\ \frac {3}{2} \qquad \textbf{(D) }\ 2\qquad \textbf{(E) }\ \frac{5}{2}$

Solution

We use mass points for this problem. Let $\text{m} A$ denote the mass of point $A$ . Rewrite the expression we are finding as $\frac{EF}{FC} + \frac{AF}{FD} = \frac{FE}{FC} + \frac{FA}{FD} = \frac{\text{m} C}{\text{m} E} + \frac{\text{m} D}{\text{m} A}$ Now, let $\text{m} C = 2$ . We then have $2 \cdot 1 = \text{m} B \cdot 2$ , so $\text{m} B = 1$ , and $\text{m} D = 2+1 = 3$ We can let $\text{m} A = 3$ . We have $\text{m} E = \text{m} A + \text{m} B = 3+1 = 4$ From here, substitute the respective values to get

$\frac{\text{m} C}{\text{m} E} + \frac{\text{m} D}{\text{m} A} = \frac{2}{4} + \frac{3}{3} = \frac{1}{2} + 1 = \frac{3}{2}$ This answer corresponds to answer $\fbox{\textbf{(C)}}$ .

~JustinLee2017

1965 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 36	Followed by Problem 38
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1965 AHSME Problems/Problem 37

Problem

Solution

See Also