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After seeing this question, I was wondering if there is some way to find the area of the region enclosed by three circles that are made with the three sides of a triangle as diameters. Let the sides are $a,b,c$ so there can be a unique such triangle and a unique formula for the area shaded in the image.

We want the light blue region from the image. enter image description here

This is a similar question involving a square might help.

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    $\begingroup$ can you send us an image of the problem? $\endgroup$ – Dr. Sonnhard Graubner Dec 26 '17 at 10:43
  • $\begingroup$ In principle, one could compute the area of the region as the sum of the areas of three circular segments and a triangle. But I don't think there is a nice a formula. $\endgroup$ – Luca Bressan Dec 26 '17 at 11:23
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Update. (There was a mistake in the first version.)

Assume that the given triangle $\triangle$ is acute. The three circles then intersect at the vertices $A$, $B$, $C$ of $\triangle$ and at the pedal points $H_a$, $H_b$, $H_c$ of the heights of $\triangle$. It follows that the vertices of the shape $S$ in question form the orthic triangle $\triangle_{\rm orth}$ of $\triangle$. According to this link the area of $\triangle_{\rm orth}$ is given by $${\rm area}(\triangle_{\rm orth})=2{\rm area}(\triangle)\cos\alpha\cos\beta\cos\gamma\ .$$ The area of $S$ is equal to ${\rm area}(\triangle_{\rm orth})$ plus the areas of three circular segments.

The circular segment $S_A$ with center on $BC$ has radius ${a\over2}$ and central angle $\pi-2\alpha$. (Note that the corresponding peripheral angle is ${\pi\over2}-\alpha$, since the triangle $AH_cC$ has a right angle at $H_c$.) Its area then computes to $${\rm area}(S_A)={a^2\over8}\bigl(\pi-2\alpha-\sin(\pi-2\alpha)\bigr)={a^2\over8}\bigl(\pi-2\alpha-\sin(2\alpha)\bigr)\ .$$

Put all together, and you obtain a formula for ${\rm area}(S)$in terms of the data $a$, $b$,$c$, $\alpha$, $\beta$, $\gamma$.

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