A circle is inscribed in a triangle ABC with sides a,b,c. Tangents to the circle parallel to the sides of the triangle are constructed. Each of these tangents cuts off a triangle from △ABC. In each of these triangles, a circle is inscribed. Find the sum of the areas of all four inscribed circles (in terms of a,b,c).
This does not solve the problem, but if the triangle were equilateral ...
The inradius of an equilateral triangle is 1/3 times the height. So the three small triangles cut off from △ABC are each equilateral with dimensions 1/3 of the parent triangle. So each incircle of the smaller equilaterals has area 1/9 of △ABC.
So the sum of areas of all 4 incircles is 4/3 that of △ABC
If a=b=c, the area of the incircle of △ABC is πa^2/12, so area of all 4 is πa^2/9
Since many triangle identities are related to the semiperimeter, the answer might be (and this is just a guess):
π(2s/3)^2 / 9 or 4πs^2/81 where s is the semiperimeter, or
(a+b+c)π/81. Again, just a guess.
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Posted by Larry
on 2024-06-19 11:16:55 |
If oversimplified to an equilateral
