# If the incircle of the triangle ABC touches its sides respectively at L , M , N and if x , y , z be the circumradii of the triangles MIN , NIL , LIM  where I is the incentre then the product xyz is equal to

mycroft holmes
272 Points
6 years ago
Note that since angle ILB = angle INB = 900, INBC is a cyclic quadrilateral, which is also the circumcircle of Tr INC. This circle has IB as its diameter. From IBC we have angle IBC = B/2 and BC = (s-b) and hence radius of the circumcircle of Tr INC is $\frac{(s-b)}{2 \sin \frac{B}{2}}$.

Hence the product of the radii is $\frac{(s-a)(s-b)(s-c)}{8 \sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}}$

Using $\sin \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{bc}}$ etc.,

this product becomes $\frac{abc(s-a)(s-b)(s-c))}{8 (s-a)(s-b)(s-c)}$

$=\boxed{\frac{abc}{8}}$