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

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+ | ==Problem 12== | ||

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+ | In <math>\triangle ABC</math>, <math>AB = 6</math>, <math>BC = 7</math>, and <math>CA = 8</math>. Point <math>D</math> lies on <math>\overline{BC}</math>, and <math>\overline{AD}</math> bisects <math>\angle BAC</math>. Point <math>E</math> lies on <math>\overline{AC}</math>, and <math>\overline{BE}</math> bisects <math>\angle ABC</math>. The bisectors intersect at <math>F</math>. What is the ratio <math>AF</math> : <math>FD</math>? | ||

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+ | <pre style="color: gray">TODO: Diagram</pre> | ||

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+ | <math>\textbf{(A)}\ 3:2\qquad\textbf{(B)}\ 5:3\qquad\textbf{(C)}\ 2:1\qquad\textbf{(D)}\ 7:3\qquad\textbf{(E)}\ 5:2</math> | ||

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== Solution == | == Solution == | ||

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> so <math>AE = \frac{48}{13}</math> | <math>\frac{6}{AE} = \frac{7}{8 - AE}</math> so <math>AE = \frac{48}{13}</math> | ||

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Similarly, <math>CD = 4</math>. | Similarly, <math>CD = 4</math>. |

## Revision as of 14:16, 4 February 2016

## Problem 12

In , , , and . Point lies on , and bisects . Point lies on , and bisects . The bisectors intersect at . What is the ratio : ?

TODO: Diagram

## 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