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

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By the angle bisector theorem, <math>\frac{AB}{AE} = \frac{CB}{CE}</math> | By the angle bisector theorem, <math>\frac{AB}{AE} = \frac{CB}{CE}</math> | ||

− | <math>\frac{6}{AE} = \frac{7}{8 - AE}</math> | + | <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> | ||

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Assign point <math>C</math> a mass of <math>1</math>. | 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>. | <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>. | 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> | So <math>\frac{AF}{AD} = \frac{mD}{mA} = \boxed{\textbf{(C)}\; 2 : 1}</math> |

## Revision as of 14:00, 4 February 2016

## Solution

By the angle bisector theorem,

so

Similarly,

Now, we use mass points.

Assign point a mass of .

Because will have a mass of

Similarly, will have a mass of

.

Similarly,

The mass of is the sum of the masses of and .

This can be checked with , which is also

So