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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

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 

Grade:11

1 Answers

mycroft holmes
272 Points
7 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}}

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