Proof. For the “if” implication, let be a tetrahedron with all three angles at right. We show that bd is trt-convex.
Let . We have four essentially different situations.
Case 1. .
This case follows from the right convexity of any right triangle.
Case 2. .
Similarly, being an acute triangle, it must be rightly convex.
Case 3.
, .
Assume without loss of generality that
y is not closer than
x from
. Then,
Denote by z this intersection. Put . The points u, and z are the vertices of a thin right triangle included in bd.
Case 4.
, .
Consider the point
u from Case 3 and
. Clearly,
If , then . If , then .
So, if , then lie in a thin right triangle with vertices , included in . Analogously, for .
Now, let us prove the “only if” implication.
Let .
Claim 1. If two crossing line segments belong to bdP, then they lie in a facet of P.
Indeed, let be the two line segments, and . Being a convex body, P is not included in the plane , supposed horizontal. Let , below , say. Take and assume that for some . Then, , which contradicts . Hence, there exist no such points , and consequently .
Analogously, and is a facet of P. Claim 1 is proven.
For any , we have . Indeed, otherwise there is no thin right triangle in bdP containing .
We now prove that P has a facet. Assume that P has no facet.
Consider three extreme points . We have . Put . Then, .
Since P contains no facet, is not a supporting plane of P. Hence, divides P into two parts .
Choose . We have .
Consider the non-degenerate polytope . By Radon’s theorem, either one vertex is in the tetrahedron determined by the other four, or one line segment joining two vertices meets the triangle formed by the other three. The first possibility is excluded, the vertices being in Q. Since , avoids intP, whence meets bd, and Claim 1 provides a facet of P, which contradicts our assumption.
Let E be a facet of P.
Claim 2. is nowhere dense in bdE.
Suppose, in contrast, there exists a non-degenerate arc . Choose and put . For close to each other, but different from x, is small. If , then and . Then, choose close to z, such that still . Since are not collinear, being in Q, , whence .
Hence, either the triangle is not right, or we find the triangle which is not right. Obviously, the midpoints of the two long sides do not have the -property.
Claim 3. There are no disjoint line segments in bdE.
Suppose, on the contrary, are such line segments, assumed maximal (with respect to inclusion). Let be the component of not containing T, and the component of not containing S. Since points in and have the -property, we must have and .
Thus, the triangles and are facets of P.
Now, we easily find , , such that . Put and . We can arrange the triangle not to be right. Indeed, if, for example , or not. In the first case, for any choice of a point very close to on , the triangle is not right. In the second case, for any choice of a point very close to q on , the triangle is not right.
Hence, we may suppose the triangle not to be right. However, then p and q do not enjoy the -property. Claim 3 is verified.
From Claims 2 and 3, it follows that E is a triangle . In fact, every facet of P is a triangle. Moreover, are edges of P. Suppose is not a facet of P. Then, some point is separated from k by . However, then meets , which means that , as all four points belong to Q and . By Claim 1, P has a quadrilateral facet, and a contradiction is found. Therefore, .
It remains to prove that the tetrahedron P is right.
We call a tetrahedron quasiright, if for any pair of opposite edges, such as , , we have or or or . Notice that a right tetrahedron is quasiright. We first show that P is quasiright.
Choose arbitrarily the pair of opposite edges and of P. Choose and . Then, the only triangle boundaries in bdP containing are bd and bd.
If or , then the condition for P to be quasiright is fulfilled at , . Otherwise, card and card, so is not right for any . Fix such a point y. Since have the -property in bdP, is right for any . So or for all . Suppose without loss of generality and . Choose close to x such that . We also have . So . Again, the condition for P to be quasiright is fulfilled at , .
Hence,
P is quasiright. If it is not right, it must look like in
Figure 2. However, then, the midpoint of
and a point close to
w on
do not enjoy the
-property. □