# Thread: Solutions of Parts of a triangle?

1. ## 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)].

2. Originally Posted by fardeen_gen
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.