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

Revision as of 13:13, 4 February 2016

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

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