1965 AHSME Problems/Problem 37

Revision as of 09:38, 20 July 2024 by Thepowerful456 (talk | contribs) (diagram)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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

[asy]  import geometry;  point A = (0,0); point B = (16,0); point C = (3, 10); point D, E, F; real d;  // Triangle ABC draw(A--B--C--A); dot(A); label("A", A, SW); dot(B); label("B", B, SE); dot(C); label("C", C, NW);  // Segments AD and CE D = 2/3*C+1/3*B; dot(D); label("D", D, NE); draw(A--D); E = midpoint(A--midpoint(A--B)); dot(E); label("E", E, S); draw(C--E);  // Point F pair[] f=intersectionpoints((A--D), (C--E)); F=f[0]; dot(F); label("F", F, SE);  [/asy]

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 $\fbox{\textbf{(C)}}$.

~JustinLee2017

See Also

1965 AHSC (ProblemsAnswer KeyResources)
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. AMC logo.png