# Finding the area of a triangle formed by the centers of circumscribed circles of a big triangle

Given a triangle ABC with an inscribed circle with a center O:

we create three triangles from an inscribed circle, $$\triangle OA_1B_1$$, $$\triangle OB_1C_1$$, $$\triangle OC_1A_1$$. Then, we define three points $$O_1, O_2, O_3$$ which are the centers of the circles that circumscribe around $$\triangle OB_1C_1$$, $$\triangle OC_1A_1$$, $$\triangle OA_1B_1$$ respectively.

We aim to find the area of $$\triangle O_1O_2O_3$$, given that the area of $$\triangle ABC$$ is 24.

I thought $$\triangle O_1O_2O_3$$ would perhaps be the equilateral triangle that inscribes the circle $$O$$, as $$\angle A_1OB_1$$, $$\angle B_1OC_1$$, $$\angle C_1OA_1$$ would be obtuse, I cannot either back my assumption up or derive a meaningful result from my proposition.

Any insights on finding the area of the desired triangle would be helpful.

Observe that, points $$O$$, $$C_{1}$$, $$A$$ and $$B_{1}$$ are concyclic. Moreover, they all lie on the circle with diameter $$OA$$. Thereafter $$O_{1}$$ is the midpoint of $$OA$$. Arguing similarly will show that $$O_{2}$$ and $$O_{3}$$ are the midpoints of $$OB$$ and $$OC$$. Now proceed.