1975 AHSME Problems/Problem 28
Revision as of 21:58, 12 February 2021 by Justinlee2017 (talk | contribs) (Created page with "==Solution== Here, we use Mass Points. Let <math>AF = x</math>. We then have <math>AE = 2x</math>, <math>EC = 16-2x</math>, and <math>FB = 12 - x</math> Let <math>B</math> hav...")
Solution
Here, we use Mass Points.
Let 
. We then have 
, 
, and 
Let 
have a mass of 
. Since 
is the midpoint, 
also has a mass of .
Looking at segment 
, we have
![\[2 \cdot (12-x) = \text{m}A_{AB} \cdot x\]](http://latex.artofproblemsolving.com/4/b/a/4ba774b38e31c896d6ce326d8894d30f044b678c.png)
So
![\[\text{m}A_{AB} = \frac{24-2x}{x}\]](http://latex.artofproblemsolving.com/8/d/3/8d382b4f509b07eada5d63c958f7bbe276ce9864.png)
Looking at segment 
,we have
![\[2 \cdot (16-2x) = \text{m}A_{AC} \cdot 2x\]](http://latex.artofproblemsolving.com/e/5/f/e5f09fc3ca91bd652624d54b497038a8c825ae79.png)
So
![\[\text{m}A_{AC} = \frac{16-2x}{x}\]](http://latex.artofproblemsolving.com/e/e/4/ee46c27c93a0096e8d92e20e5e26533acbe81f33.png)
From this, we get
![\[\text{m}E = \frac{16-2x}{x} + 2 \Rrightarrow \text{m}E = \frac{16}{x}\]](http://latex.artofproblemsolving.com/6/7/5/6750c2537d2c1e4f2b54e0738816a122cf7c9de1.png)
and
![\[\text{m}F = \frac{24-2x}{x} + 2 \Rrightarrow \text{m}F = \frac{24}{x}\]](http://latex.artofproblemsolving.com/4/2/7/427690de5b1ef03a3c02999e0b9c5128905fb4b0.png)
We want the value of 
. This can be written as
![\[\frac{EG}{GF} = \frac{\text{m}F}{\text{m}E}\]](http://latex.artofproblemsolving.com/2/e/b/2eb4e0c2c0cdf8da12dec1cc35c5e82889aece8d.png)
Thus
![\[\frac{\text{m}F}{\text{m}E} = \frac {\frac{24}{x}}{\frac{16}{x}} = \frac{3}{2}\]](http://latex.artofproblemsolving.com/3/3/a/33a1e41fb23cb18b4ad96e10c73244f39dbad55b.png)
~JustinLee2017
