Sunday, July 12, 2009

Problem 321: Triangle, Incenter, Excircle, Altitude, Collinearity

Proposed Problem
Click the figure below to see the complete problem 321.

 Geometry Problem 321: Triangle, Incenter, Excircle, Midpoint of the Altitude, Collinearity.
See more:
Complete Problem 321
Level: High School, SAT Prep, College geometry


  1. Can you give me Hint????

  2. Solution:

    Note that A, I and E are collinear, since they all lie on the bisector of angle BAC.

    Let AE and BC intersect at N. Since AH || EF, triangles NEF and NAH are similar. Thus, AH/FE = AN/NE, so AM/FE = AN/2NE.

    Let the incircle have radius r, and the excircle have radius R. Then r/R = AI/AE = IN/NE, so we have

    AI * NE = AE * IN
    AI * NE = (AI + IN + NE) * IN

    2 AI * NE =
    = (AI + IN + NE) * IN + AI * NE
    = IN^2 + AI * IN + NE * IN + AI * NE
    = (IN + AI) (IN + NE)
    = AN * IE

    Therefore, AM/FE = AN/2NE = AI/IE. Since the angles IAM and IEF are equal, the triangles IAM and IEF are similar. Thus, the angles AIM and EIF are equal, so F, I, M are collinear.

    see sketch for location of points K, L and N
    denote 2p= perimeter of triangle ABC.
    Note that AK= p-a
    KL= BK+BL= BN+BK= BC= a
    And AM//EF => ∠ (MAI)= ∠ (IEF)
    Calculate area of triangle ABC in 2 ways
    S(ABC)= ½ AH.BC= AM.BC
    S(ABC)= (p-a).EF=AK.EF= AM.BC
    So triangle AMI similar to EFI ( case SAS) => ∠ (AIM)= ∠ (FIE)=> M, I, F are collinear