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

Revision as of 14:00, 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 2: / Line 2: @@
 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>
+<math>\frac{6}{AE} = \frac{7}{8 - AE}</math> so <math>AE = \frac{48}{13}</math>
 Similarly, <math>CD = 4</math>
@@ 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>.
+Similarly, <math>B</math> will have a mass of <math>\frac{4}{3}</math>
 <math>mE = mA + mC = \frac{13}{6}</math>.
-Similarly, <math>mD = mC + mB = \frac{7}{3}</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>.
+<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>.
+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>