Monday, August 11, 2008

Elearn Geometry Problem 160



See complete Problem 160
Triangle, Incircle, Incenter, Circumcircle, Circumcenter, Inradius. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.
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3 comments:

  1. AnonymousAugust 13, 2008 at 11:17 PM

    d=R-(r-OI)=R-r+OI,e=R-r-OI.
    So, de=(R-r)²-OI²=(R²-2Rr-OI²)+r².
    But, OI²=R²-2Rr (Problem 155)
    ∴de=r²

    ReplyDelete
  2. AnonymousJuly 7, 2010 at 4:48 AM

    http://ahmetelmas.wordpress.com/2010/05/15/geo-geo/

    ReplyDelete
  3. ArifApril 10, 2018 at 7:48 AM

    Join B with I , then extend BI to meet the circle at point P.
    Join D with P and B with E.

    Triangles DPI and BEI are similar therefore we get :

    BI/IE=DI/IP
    (BI)(IP)=(DI)(IE)

    We have 2R=d+e+2r and from problem 154 we have:

    (BI)(IP) =2Rr
    (BI)(IP) =(d+e+2r)r

    and

    DI= d+r
    IE=e+r

    therefore

    2rR=(e+r)(d+r)
    2r^2+r(d+e)=r^2+r(d+e)+de
    r^2=de

    ReplyDelete

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