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Let ABC be a triangle and a circle T' be drawn lying inside the triangle touching its incircle T externally and also touching the two sides AB and AC. Show that the ratio of the radii of the circles T' and T is equal to tan square((pi-A)/4).

Let ABC be a triangle and a circle T' be drawn lying inside the triangle touching its incircle T externally and also touching the two sides AB and AC. Show that the ratio of the radii of the circles T' and T is equal to tan square((pi-A)/4). Β 

Grade:12th Pass

1 Answers

Ashwin Muralidharan IIT Madras
290 Points
9 years ago

Let I and I' be the incentres respectively of the two circles T and T'

 

Now  AI = r/sin(A/2) -------- draw a triangle and incircle to it and verify (Angle between AI and one of the side is A/2 since angle bisector)

 

and AI' = r'/sin(A/2)

 

We also know that A divides the line joining the centres of the two cirlces externally in the ratio of the radii (since A is the meeting point of the direct common tangets of the two circles) ------ that is the property of direct common tangents (the proof for that is directly by similar traingles)

 

So AI/AI' = r/r'

 

(AI - AI')/AI' = (r-r')/r'

But AI - AI' = II' = r+r'

hence (r+r')/AI' = (r-r')/r'

 

Substitue AI' = r'/sin(A/2)  and simplify, we get

(r-r')/(r+r') = Sin(A/2)

 

which will give r'/r = {1-Sin(A/2)}/{1+Sin(A/2)} ------- (1)

 

Now, Cos2x = [1-sq(tanx)]/[1+sq(tanx)] ------- (2)

 

take x = (pie-A)/4 and put in (2), you will get

Hence Sin(A/2) = substituted value with x value as (pie-A)/4

 

Now Substitue for sin(A/2) in (1),

 

and you will get r'/r = tan^2 [(pie-A)/4]

 

And hence the problem solved.......

 

This Problem requires a combination of concepts of many topics, and this is what precisely IIT JEE does. It will test your concepts, and your ability to apply your concepts to a particular situation.

 

All the Best.

Regards,

Ashwin (IIT Madras)

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