See the problem as posed originally here.
Triangle T has area A and sides of length a,b, and c.
Given 3 concentric circles of radius a,b,c, respectively, find the areas of the largest and smallest equilateral triangles with a vertex on each circle in terms of the given variables.
A,B,C are vertex and O is center of the circles, with a rotation of 60° degrees of the segment OA about C you get a bigger equilateral triangle OO'C where O'B is equal to OA.
Then also OBO' is an abc triangle.
From there using two times theorem of cosine I got a formula for the asked Area
A=(1/4)*[a^2+b^2+c^2]-sqr3*AreaT(abc)
Edited on January 13, 2019, 8:35 am
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Posted by armando
on 2019-01-13 06:35:15 |
Smallest one
