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

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<math>\frac{6}{AE} = \frac{7}{8 - AE}</math> so <math>AE = \frac{48}{13}</math>
 
<math>\frac{6}{AE} = \frac{7}{8 - AE}</math> so <math>AE = \frac{48}{13}</math>
  
Similarly, <math>CD = 4</math>
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Similarly, <math>CD = 4</math>.
  
 
Now, we use mass points. Assign point <math>C</math> a mass of <math>1</math>.
 
Now, we use mass points. Assign point <math>C</math> a mass of <math>1</math>.

Revision as of 14:13, 4 February 2016

Solution

By the angle bisector theorem, $\frac{AB}{AE} = \frac{CB}{CE}$

$\frac{6}{AE} = \frac{7}{8 - AE}$ so $AE = \frac{48}{13}$


Similarly, $CD = 4$.

Now, we use mass points. Assign point $C$ a mass of $1$.

$mC \cdot CD = mB \cdot DB$ , so $mB = \frac{4}{3}$

Similarly, $A$ will have a mass of $\frac{7}{6}$

$mD = mC + mB = 1 + \frac{4}{3} = \frac{7}{3}$

So $\frac{AF}{AD} = \frac{mD}{mA} = \boxed{\textbf{(C)}\; 2 : 1}$

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