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Math Help - Solutions of Parts of a triangle?

  1. #1
    Super Member fardeen_gen's Avatar
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    Solutions of Parts of a triangle?

    Three circles whose radii are a,b and c touch one other externally and the tangents at their points of contact meet in a point. Prove that the distance of this point from either of their points of contact is √[(abc)/(a + b + c)].
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  2. #2
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    Quote Originally Posted by fardeen_gen View Post
    Three circles whose radii are a,b and c touch one other externally and the tangents at their points of contact meet in a point. Prove that the distance of this point from either of their points of contact is √[(abc)/(a + b + c)].
    The tangent lines are perpendicular to the radii of the 3 circles correspondingly.

    From an external point, the two tangents to a circle are equal. So, the 3 "distance of this point to from either of their points of contact" are equal in lengths. Lets call any of these 3 distances as r.

    Draw line segments connecting the centers of the 3 circles.
    A triangle is formed, whose perimeter is 2a +2b +2c

    By Heron's formula, the area of this triangle whose 3 sides are known is given by:
    A = sqrt[s(s-u)(s-v)(s-w)]
    where s = !/2)perimeter, and u,v,w are the 3 sides.

    So, for our particular triangle,
    s = (1/2)(2a +2b +2c) = (a+b+c)
    s -u = (a+b+c) -(a+b) = c
    s -v = (a+b+c) -(b+c) = a
    s -w = (a+b+c) -(a+c) = b
    And the area, A = sqrt[(a+b+c)(a)(b)c)] = sqrt[(a+b+c)(abc)]

    Back to the drawn triangle,
    since r is perpendicular to all the 3 sides, then r is the radius of the inscribed circle...the inradius.

    There is a formula
    inradius = 2(area of triangle) / (perimeter of triangle)
    {{Or, Area of triangle = (perimeter /2)*(inradius or apothem) }}
    So,
    r = 2*sqrt[(a+b+c)(abc)] / ((2a +2b +2c)
    r = sqrt[(a+b+c)(abc)] / (a+b+c)
    r = sqrt[{a+b+c)(abc) / (a+b+c)^2]
    r = sqrt[abc /(a+b+c)] --------------------------proven.
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