1965 AHSME Problems/Problem 37
Problem
Point 
is selected on side 
of 
in such a way that 
and point 
is selected on side 
such that 
. The point of intersection of 
and 
is 
. Then 
is:
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]](http://latex.artofproblemsolving.com/2/7/1/2714544397c280b137d2054bd7d65d2f867baaa6.png)
We use mass points for this problem. Let 
denote the mass of point 
.
Rewrite the expression we are finding as
![\[\frac{EF}{FC} + \frac{AF}{FD} = \frac{FE}{FC} + \frac{FA}{FD} = \frac{\text{m} C}{\text{m} E} + \frac{\text{m} D}{\text{m} A}\]](http://latex.artofproblemsolving.com/f/3/a/f3a717b11449847a3c77699b0f4bd4dad3cefce6.png)
Now, let 
. We then have 
, so 
, and 
We can let 
. We have 
From here, substitute the respective values to get
![\[\frac{\text{m} C}{\text{m} E} + \frac{\text{m} D}{\text{m} A} = \frac{2}{4} + \frac{3}{3} = \frac{1}{2} + 1 = \frac{3}{2}\]](http://latex.artofproblemsolving.com/e/0/4/e0476efaade33366dd29f64d3c2c9f4ee7095498.png)
This answer corresponds to 
.
~JustinLee2017
See Also
| 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. 
