Monday, November 17, 2014

Geometry Problem 1061: Triangle, Inradius (r), Circumradius (R), Circumcircle, Angle Bisector, Distance from a point to a Line, Perpendicular

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 1061: Triangle, Inradius (r), Circumradius (R), Circumcircle, Angle Bisector, Distance from a point to a Line, Perpendicular.
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2 comments:

  1. Jacob HA (EMK2000)November 17, 2014 at 7:37 AM

    Join A₁B, A₁C. Then A₁B=A₁C=2Rsin(A/2).

    By Ptolemy Theorem on ABA₁C,
    a×AA₁ = 2Rsin(A/2)×(b+c)
    a×AA₁sin(A/2) = 2Rsin²(A/2)×(b+c)
    a×a₁ = 2Rsin²(A/2)×(b+c)
    sinA×a₁ = 2Rsin²(A/2)(sinB+sinC)
    2sin(A/2)cos(A/2)×a₁ = 4Rsin²(A/2) cos(A/2)cos((B-C)/2)
    a₁ = 2Rsin(A/2)cos((B-C)/2) = R(cosB+cosC)

    Similarly,
    b₁ = R(cosA+cosC)
    c₁ = R(cosA+cosB)

    a₁+b₁+c₁
    = 2R (cosA + cosB + cosC)
    = 2R + 8Rsin(A/2)sin(B/2)sin(C/2)
    = 2R + 2r
    = 2(R+r)

    ReplyDelete
  2. Ivan BazarovDecember 27, 2014 at 1:52 PM

    From property of incenter I, A1, B1, and C1 are circumcenters of BIC, AIC, and AIB respectively.
    Let C' and A' be perpendiculars from C1 and A1 onto AB and BC respectively.
    Due to first statement,
    c1-r=IC1*sin(<C/2)=C1C'
    a1-r=IA1*sin(<A/2)=A1A'
    Let P be perpendicular from O to B1B2.
    <OB1P=<C=<BOC', and OB1=OB=R, so triangles B1OP and OBC' are congruent. Therefore,
    b1-OA'=B1P=OC',
    b1=OA'+OC'
    Summing up previous equations gives
    (c1-r)+(a1-r)+(b1)=(C1C')+(A1A')+(OA'+OC')=2R.

    ReplyDelete

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