Monday, February 15, 2010

Problem 426: Triangle, Circumradius, Circumcenter, concurrent Cevians

Proposed Problem
Click the figure below to see the complete problem 426 about Triangle, Circumradius, Circumcenter, concurrent Cevians.

Problem 426: Triangle, Circumradius, Circumcenter, concurrent Cevians.
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Complete Problem 426

Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 7:50 AM
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2 comments:

  1. c .t . e. oFebruary 16, 2010 at 3:54 AM

    from ceva's theorem

    OD/AD + OE/BE + OF/FC = 1

    (AD-R)/AD + (BE-R)/BE + (FC-R)Fc = 1

    1 - R/AD + 1 - R/BE + 1 - R/FC = 1

    2 = R/AD + R/BE + R/FC

    2 = R ( 1/AD + 1/BE + 1/FC )

    2/R = 1/AD + 1/BE + 1/FC
    -----------------------------------------

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  2. AnonymousMarch 2, 2010 at 5:29 AM

    This is a special case of Ceva:
    For any point P in ABC, if we draw three cevians APD, BPE, and CPF, then
    AP/AD + BP/BE + CP/CF = 2

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