# Difference between revisions of "2013 AMC 12A Problems/Problem 11"

(Created page with "Let AD = x, and AG = y. We want to find DE + FG, which is nothing but x+y. Based on the fact that ∆ADE, DEFG, and BCFG have the same perimeters, we can say the following: 3x ...") |
|||

Line 1: | Line 1: | ||

− | Let AD = x, and AG = y. We want to find DE + FG, which is nothing but x+y. | + | Let <math>AD = x</math>, and <math>AG = y</math>. We want to find <math>DE + FG</math>, which is nothing but <math>x+y</math>. |

− | Based on the fact that | + | Based on the fact that <math>ADE</math>, <math>DEFG</math>, and <math>BCFG</math> have the same perimeters, we can say the following: |

− | 3x = x + 2(y-x) + y = y + 2(1-y) + 1 | + | <math>3x = x + 2(y-x) + y = y + 2(1-y) + 1</math> |

Simplifying, we can find that | Simplifying, we can find that | ||

− | 3x = 3y-x = 3-y | + | <math>3x = 3y-x = 3-y</math> |

− | Since 3-y = 3x, y = 3-3x. | + | Since <math>3-y = 3x</math>, <math>y = 3-3x</math>. |

− | After substitution, we find that 9- | + | After substitution, we find that <math>9-10x = 3x</math>, and <math>x</math> = <math>\frac{9}{13}</math>. |

− | Again substituting, we find y = 12/ | + | Again substituting, we find <math>y</math> = <math>\frac{12}{13}</math>. |

− | Therefore, x+y = 21/ | + | Therefore, <math>x+y</math> = <math>\frac{21}{13}</math>, which is <math>C</math> |

## Revision as of 23:55, 6 February 2013

Let , and . We want to find , which is nothing but .

Based on the fact that , , and have the same perimeters, we can say the following:

Simplifying, we can find that

Since , .

After substitution, we find that , and = .

Again substituting, we find = .

Therefore, = , which is