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

Revision as of 11:59, 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 $1$ .

Because $\frac{AE}{EC} = \frac{6}{7}, A$ will have a mass of $\frac{7}{6}$ .

Similarly, $B$ will have a mass of $\frac{4}{3}$ .

$mE = mA + mC = \frac{13}{6}$ .

Similarly, $mD = mC + mB = \frac{7}{3}$ .

The mass of $F$ is the sum of the masses of $E$ and $B$ .

$mF = mE + mB = \frac{7}{2}$ .

This can be checked with $mD + mA$ , which is also $\frac{7}{2}$ .

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

@@ Line 10: / Line 10: @@
 Assign point <math>C</math> a mass of <math>1</math>.
-Because <math>\frac{AE}/{EC} = \frac{6}{7}, A</math> will have a mass of <math>\frac{7}{6}</math>.
+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>.