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

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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.

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1 Answer 1

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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.

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