# Difference between revisions of "2016 AMC 12A Problems/Problem 12"

(Created page with "== Solution == By the angle bisector theorem, <math>\frac{AB}{AE} = \frac{CB}{CE}</math> <math>\frac{6}{AE} = \frac{7}{8 - AE}</math>, so <math>AE = \frac{48}{13}</math> Sim...") |
(→Solution) |
||

Line 10: | Line 10: | ||

Assign point <math>C</math> a mass of <math>1</math>. | Assign point <math>C</math> a mass of <math>1</math>. | ||

− | Because <math>\frac{AE} | + | Because <math>\frac{AE}{EC} = \frac{6}{7}, A</math> will have a mass of <math>\frac{7}{6}</math>. |

Similarly, <math>B</math> will have a mass of <math>\frac{4}{3}</math>. | Similarly, <math>B</math> will have a mass of <math>\frac{4}{3}</math>. |

## Revision as of 11:59, 4 February 2016

## Solution

By the angle bisector theorem,

, so

Similarly,

Now, we use mass points.

Assign point a mass of .

Because will have a mass of .

Similarly, will have a mass of .

.

Similarly, .

The mass of is the sum of the masses of and .

.

This can be checked with , which is also .

So