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

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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>. |

− | + | <math>mC \cdot CD = mB \cdot DB</math> , so <math>mB = \frac{4}{3}</math> | |

− | + | Similarly, <math>A</math> will have a mass of <math>\frac{7}{6}</math> | |

− | + | <math>mD = mC + mB = 1 + \frac{4}{3} = \frac{7}{3}</math> | |

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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 14:11, 4 February 2016

## Solution

By the angle bisector theorem,

so

Similarly,

Now, we use mass points. Assign point a mass of .

, so

Similarly, will have a mass of

So