This question was already asked by Martin Gardner in 1976. [2]
Until now such a tetrad was not found, but also the non-existence of such an object was not proven.
For a convex shape every point P of the straight line segment between two arbitrary points A and B of the shape also has to belong to the shape.
A tetrad with hole is not convex, because the points in the hole do not belong to the tetrad.
Only a tetrad with no hole can be convex.
The known tetrads where two tiles surround the others (see question 17) are not convex.
There the tiles have two sides with point symmetric boundaries which are not straight.
The straight line segment from one point of the boundary to the symmetrical point does not entirely belong to the shape.
Can you contribute arguments which clearly prove that these types of tetrads cannot be convex.
A convex tetrad would have to have exactly three vertices A, B, C on its outer boundary which belong to exactly two tiles.
And the boundary of the inner tile has to have three vertices D, E, F which belong to exactly three tiles.
In the points A, B, C the sum of the two angles is at most 180°.
Here are a few convex tetrad fakes. Try to find out what is wrong.
See what is wrong.
See what is wrong.
See what is wrong.