Geometry Problem
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Wednesday, March 2, 2011
Problem 587: Triangle, Incenter, Incircle, Tangency Point, Midpoint, Distance, Half the Difference
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Name E, F tgpoints on AB, BC
ReplyDeleteAD+x = DC-x => 2x = DC-AD = CF-AE = (CF+BF)-(AE+BE)
2x = BC-AB = a - c
x = (a - c)/2
E,F tangency points of AB, AC with incircle, say.
ReplyDeleteThen
2MD
=CD-AD
=CF-AE
=(CF+BF)-(AE+BE)
=BC-AB
=a-c
i shall use the usual notation for the sides of the triangle and the semi perimeter.
ReplyDeleteAB = 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.
Name tangency point on BC as P and the tangency point on AB as R
ReplyDeleteLet 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
Let the tangency point of BC and AB be E & F respectively
ReplyDeleteBE=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