Question

If l,m,n are the distances of vertices of a triangle from the corresponding points of contact with the incircle, then prove that lmn/(l+m+n)=r^2
(r is inradius)

Answer 1

A

Question:

If $l, m, n$ are the distances of vertices of a triangle from the corresponding points of contact with the incircle, then prove that $\lim_{n \to \infty} (l + m + n) = r^{\wedge} 2$ ( $r$ is inradius)

Solution:

For proof using next formula for radius incircle:

$$r = \sqrt {\frac {(p - a) (p - b) (p - c)}{p}} (1)$$

where

$$a = A B = n + l,$$ $$b = B C = l + m,$$ $$c = A C = n + m,$$ $$p = \frac {1}{2} (a + b + c) = \frac {1}{2} (n + l + l + m + n + m) = l + n + m.$$

Insert $a, b, c, p$ into (1):

$$r = \sqrt {\frac {(l + n + m - (n + l)) (l + n + m - (l + m)) (l + n + m - (n + m))}{l + n + m}}$$ $$r = \sqrt {\frac {m \cdot n \cdot l}{l + n + m}}$$ $$r ^ {2} = \frac {m n l}{l + n + m}$$