Sunday, August 23, 2009

Problem 345. Equal circles, Tangents, Concurrency, Hexagon, Semiperimeter

Proposed Problem
Click the figure below to see the complete problem 345 about Equal circles, Tangents, Concurrent lines, Hexagon, Semiperimeter.

 Problem 345. Equal circles, Tangents, Concurrency, Hexagon, Semiperimeter.
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Complete Problem 345
Level: High School, SAT Prep, College geometry

1 comment:

  1. 1. Connect BC3 and CB3
    note that BC3=CB3 and angle( BC3E)= angle (CB3E)=90 , angle (BEC3)= angle (B3EC)
    Triangle BEC3 congruence with triangle B3EC ( case ASA)
    so BE=CE
    Similarly we also have BH=HC >> HE is the perpendicular bisector of BC
    With the same logic DG, FM are the perpendicular bisectors of AB and AC
    So 3 perpen. bisectors DG, EH, FM are concurrent at the center of circumcircle of triangle ABC

    2. from part 1 we have DA=DB, BE=CE, CF=AF
    so s= half of perimeter of polygon ADBECF = DA+EB+CF=DB+EC+FA

    Peter Tran

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