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

m (Solution)
(Solution)
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Similarly, <math>CD = 4</math>
 
Similarly, <math>CD = 4</math>
  
Now, we use mass points.
+
Now, we use mass points. Assign point <math>C</math> a mass of <math>1</math>.
  
Assign point <math>C</math> a mass of <math>1</math>.
+
<math>mC \cdot CD = mB \cdot DB</math> , so <math>mB = \frac{4}{3}</math>
  
Because <math>\frac{AE}{EC} = \frac{6}{7}, A</math> will have a mass of <math>\frac{7}{6}</math>
+
Similarly, <math>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>
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<math>mD = mC + mB = 1 + \frac{4}{3} = \frac{7}{3}</math>
 
 
<math>mE = mA + mC = \frac{13}{6}</math>.
 
 
 
Similarly, <math>mD = mC + mB = \frac{7}{3}</math>
 
 
 
The mass of <math>F</math> is the sum of the masses of <math>E</math> and <math>B</math>.
 
 
 
<math>mF = mE + mB = \frac{7}{2}</math>
 
 
 
This can be checked with <math>mD + mA</math>, which is also <math>\frac{7}{2}</math>
 
  
 
So <math>\frac{AF}{AD} = \frac{mD}{mA} = \boxed{\textbf{(C)}\; 2 : 1}</math>
 
So <math>\frac{AF}{AD} = \frac{mD}{mA} = \boxed{\textbf{(C)}\; 2 : 1}</math>

Revision as of 13:11, 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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