# Difference between revisions of "1982 AHSME Problems/Problem 11"

(Created page with "== Problem 11 Solution == Since <math>BO</math> and <math>CO</math> are angle bisectors of angles <math>B</math> and <math>C</math> respectively, <math>\angleMBO = \angleOBC<...") |
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== Problem 11 Solution == | == Problem 11 Solution == | ||

− | Since <math>BO</math> and <math>CO</math> are angle bisectors of angles <math>B</math> and <math>C</math> respectively, <math>\ | + | Since <math>BO</math> and <math>CO</math> are angle bisectors of angles <math>B</math> and <math>C</math> respectively, <math>\angle MBO = \angle OBC</math> and similarly <math>\angle NCO = \angle OCB</math>. Because <math>MN</math> and <math>BC</math> are parallel, <math>\angle OBC = \angle MOB</math> and <math>\angle NOC = \angle OCB</math> by corresponding angles. This relation makes <math>\triangle MOB</math> and <math>\triangle NOC</math> isosceles. This makes <math>MB = MO</math> and <math>NO = NC</math>. Therefore the perimeter of <math>\triangle AMN</math> is <math>12 + 18 = 30</math>. |

## Revision as of 14:32, 5 August 2017

## Problem 11 Solution

Since and are angle bisectors of angles and respectively, and similarly . Because and are parallel, and by corresponding angles. This relation makes and isosceles. This makes and . Therefore the perimeter of is .