1965 AHSME Problems/Problem 37

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

$\frac{\text{m} C}{\text{m} E} + \frac{\text{m} D}{\text{m} A} = \frac{2}{4} + \frac{3}{3} = \frac{1}{2} + 1 = \frac{3}{2}$ $\boxed{C}$

~JustinLee2017

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1965 AHSME Problems/Problem 37

Solution