Monday, December 1, 2014

Geometry Problem 1065: Triangle, Acute Angle, Orthocenter, Circumradius, Inradius, Exradius, Distance, Diameter

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

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Online Math: Geometry Problem 1065: Triangle, Acute Angle, Orthocenter, Circumradius, Inradius, Exradius, Distance, Diameter.

2 comments:

  1. Since a₁² = 2R²(1+cos2A) = 4R² cos²A, thus a₁ = 2R cosA.
    r = 4R sin(A/2) sin(B/2) sin(C/2)
    r₁ = 4R sin(A/2) cos(B/2) cos(C/2)

    r₁ − r = 4R sin(A/2) cos(B/2 + C/2) = 4R sin²(A/2) = 2R(1−cosA)
    a₁ + r₁ − r = 2R
    r₁ + a₁ = 2R + r

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  2. Let IE cut circumcircle at P. By properties of incenter, P is circumcenter of BICE.
    Let A' be midpoint of BC
    ra-r=IE*sin(<A/2)=2*IP*sin(<A/2)=2*BP*sin(<A/2)=2*A'P,
    (ra-r)/2=A'P
    It is also known that a1/2=OA'
    Summing up previous equations gives
    (ra-r)/2+a1/2=A'P+OA'=R, or
    ra+a1-r=2R

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