Area of the region enclosed by the three circles that have sides of a triangle as diameters. 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.

This is a similar question involving a square might help.
 A: 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 $a$, $b$,$c$, $\alpha$, $\beta$, $\gamma$.
