Wednesday, March 2, 2011

Problem 587: Triangle, Incenter, Incircle, Tangency Point, Midpoint, Distance, Half the Difference

Geometry Problem
Click the figure below to see the complete problem 587.

 Problem 587: Triangle, Incenter, Incircle, Tangency Point, Midpoint, Distance, Half the Difference.
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5 comments:

  1. Name E, F tgpoints on AB, BC
    AD+x = DC-x => 2x = DC-AD = CF-AE = (CF+BF)-(AE+BE)
    2x = BC-AB = a - c
    x = (a - c)/2

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  2. E,F tangency points of AB, AC with incircle, say.

    Then

    2MD
    =CD-AD
    =CF-AE
    =(CF+BF)-(AE+BE)
    =BC-AB
    =a-c

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  3. i shall use the usual notation for the sides of the triangle and the semi perimeter.
    AB = c, BC = a and AC = b and s = (1/2)*(a+b+c)
    now by the tangency property of the in circle we have AD = s-a and AM = b/2 as M is the midpoint of AC. now DM = AM - AD = b/2 - (s-a) = a/2 - c/2 = (1/2)*(a-c). this implies that DM = (1/2)*(BC - AB).
    Q. E. D.

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  4. Name tangency point on BC as P and the tangency point on AB as R

    Let MC = b
    then DC = x + b
    but,
    DC = PC (tangents of the circle)
    thus PC = x + b

    Since AM = MC = b (M is the midpoint of AC)
    AD = AM - x
    = b - x
    then AR = b - x (tangents of the circle)

    BR = AC - AR
    = c - b + x
    BP = BC - PC
    = a - x - b

    Since BR = BP (tangents of the circle)
    c - b + x = a - x - b
    2x = a - c
    x = (a - c)/2

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  5. Let the tangency point of BC and AB be E & F respectively
    BE=a1, CE=a2, BF=c1, AF=c2
    CE=BF =>a1=c1
    AD=c2, CD=a2

    AM=AD+DM
    AM=c2+x

    CM=CD-DM
    CM=a2-x

    Since AM=CM, c2+x=a2-x
    2x=a2-c2
    2x=a2-c2+a1-c1 [a1=c1]
    2x=a-c
    x=(a-c)/2

    ReplyDelete