The excentral triangle of a reference triangle $ABC$ is the triangle with vertices corresponding to the excenters of $ABC$. Denote with $D$, $E$, $F$ the $A$−, $B$−, $C$− excenters, respectively. Denote with $U$, $V$, $W$ the midpoints of $BC$, $AC$, $AB$, respectively. Let $D'$, $E'$, $F'$ be the reflections of the points $D$, $E$, $F$ with respect to the midpoints $U$, $V$,$W$, respectively. Then $D'E'F'$ has the *same* area as the reference triangle $ABC$ (see Proposition 4 of [Dalcín and Kiss - Some Properties of the García Reflection Triangles](https://www.heldermann.de/JGG/JGG25/JGG251/jgg25010.htm)).


According to the [WOLFRAM Demonstrations Project](https://demonstrations.wolfram.com/InsphereAndFourExspheresOfATetrahedron/#more), a tetrahedron has four ex-spheres, but it does not show how to construct these spheres. I am interested in knowing if reflecting the centers of these 4 ex-spheres with respect to the four centroids of the four faces of the tetrahedron respectively, we would obtain a tetrahedron with the *same* volume as the reference tetrahedron. If so, how to prove it?



Looking for the 3D-analog/extension of a 2D theorem
It's definitely not of the same volume. If you apply your operation to the right tetrahedron you will get a right tetrahedron with smaller edges.


Although it's still possible that the volume is proportional to the volume of original. I haven't checked it.


If you want to check it you should use barycentric coordinates with respect to vertices of your tetrahedron. Let's call it $x\_1 x\_2 x\_3 x\_4$. Non-normalized coordinates of some point $p$ are the signed volumes of tetrahedrons $V(p,x\_2,x\_3,x\_4), V(x\_1,p,x\_3,x\_4), V(x\_1,x\_2, p,x\_4), V(x\_1,x\_2,x\_2,p)$. The sign of the first coordinate depends on which side of hyperplane spanned by $x\_2,x\_3,x\_4$ point $p$ sits. And the same for other coordinates.


If $p$ is excenter then the distance from it to any hyperplane spanned by a face is the same since it's the radius of ex-sphere. So you have $|V(x\_1,p,x\_3,x\_4)| = rS(x\_1,x\_3,x\_4)|$, where $S$ is a usual non-signed area.


Putting this all together we get that that non-normalized barycentric coordinates of an excenter $I\_1$ are $(-S(x\_2,x\_3,x\_4): S(x\_1,x\_3,x\_4): S(x\_1,x\_2,x\_4): S(x\_1,x\_2,x\_3)).$ And to get normalized you should divide each by the sum of coordinates so the new sum equals to $1$.


If you do it for a right tetrahedron you get $(-1/2, 1/2, 1/2, 1/2)$. I will not write the general case.


The centroid of $(x\_2, x\_3, x\_4)$ has coordinates $(0, 1/3, 1/3, 1/3)$. So if we reflect $I\_1$ with respect to it you get $(1/2, 1/6, 1/6, 1/6)$ lets call this point $I^\*\_1$. $I^\*\_1$ is closer to the center of the original tetrahedron than the vertices of the original polyhedron. So the right polyhedron with vertices $I^\*\_1, I^\*\_2, I^\*\_3, I^\*\_4$ will be smaller than the original one.


If you want to compute volume of $I^\*\_1, I^\*\_2, I^\*\_3, I^\*\_4$ in the general case than you just have to organize normalized coordinates of $I^\*\_1, I^\*\_2, I^\*\_3, I^\*\_4$ in a matrix, compute determinant and I think this will be $V(I^\*\_1, I^\*\_2, I^\*\_3, I^\*\_4)/V(x\_1,x\_2,x\_3,x\_4)$.


