Criterion on a region to be a triangle in the three dimensional space.

Question

Given the real numbers $a_1,a_2,b_1,b_2,c_1,c_2\in [0,1]$. Suppose we know that the set $$\begin{cases}x+y+z=1 \\a_1\le x \le a_2\\ b_1\le y \le b_2\\ c_1\le z \le c_2\\ \end{cases}$$ is a surface (not empty and not reduced to a point).

Is there a criterion that enables us to know if the surface is a triangle?

For example, if $a_1=b_1=c_1=0$ and $a_2=b_2=c_2=1$, then the surface is a triangle (the standard 2-simplex).

But, for example, the plot of $$\begin{cases}x+y+z=1 \\0\le x \le 0.2\\ 0.1\le y \le 0.6\\ 0.6\le z \le 0.7\\ \end{cases}$$ is not a triangle.

Answer 1

Let me use $S$ to denote the rectangular solid defined by your three inequalities.

Also let $F(x,y,z) = x+y+z$. We are interested in plane defined by the equation $F(x,y,z)=1$ which I'll denote $F_1$

So your question comes down to whether $S \cap F_1$ is a plane. To determine the answer, evaluate the function $$F(x,y,z)=x+y+z $$ on each of the eight points forming the vertices of $S$, namely $$P_{i,j,k} = (a_i,b_j,c_k), \quad i,j,k \in \{1,2\} $$ Now count: let $N_+$ be the count of how many of those eight values are $\ge 1$; and let $N_-$ be the count of how many are $\le 1$.

The intersection $S \cap F_1$ will be a triangle if and only if $N_- = 7$ or $N_+ = 7$.

Here's an intuitive explanation of why this is true; I'll leave out detailed proofs, which should involve nothing more than simple case analysis involving inequalities.

Imagine $S$ starts above the plane $F(x,y,z)=1$ and then moves downwards, passing through the plane. At the first moment that $S$ touches the plane, it touches it at the point $(a_1,b_1,c_1)$, and at that moment we have $N_-=1$ and $N_+ = 8$ and the intersection is just that point. Then, as $S$ moves a little further, $N_-=1$ and $N_+=7$ and the intersection is a triangle.

As $S$ continues to pass further through the plane, the intersection continues to be a triangle up until the first moment that one of the three points $(a_2,b_1,c_1)$ or $(a_1,b_2,c_1)$ or $(a_1,b_1,c_2)$ touches the plane. If $S$ passes just a bit further, one obtains a polygon with 4 to 6 sides, and both of $N_-,N_+$ are $\le 6$. This continues up until the last moment that one of the three points $(a_1,b_2,c_2)$ or $(a_2,b_1,c_2)$ or $(a_2,b_2,c_1)$ touches the plane. At that moment the intersection becomes a triangle again, and continues until the moment that $(a_2,b_2,c_2)$ touches the plane, at which moment we have $N_- = 8$ and $N_+ = 1$.

Answer 2

In fact, your problem can be transformed into a 2D issue by considering $(x,y,z)$ as barycentric coordinates (abbreviated as "b.c.") inside a triangle $ABC$ as displayed in the following figure (made with Geogebra):

Interactive image here : one can move vertices $A,B,C$ and as well blue points on the sides.

where for example $a_1=0.7, \ a_2=0.9, \ b_1=0.4, \ b_2=0.6, \ c_1=0.3, \ c_2=0.6.$

The different constraints are materialized as trapezoidal stripes.

Their intersection can take different forms that can range from triangular (this is the case here) to hexagonal.

There exists evidently no unique criteria for having a triangular intersection ; one is obliged to find sufficient cases for this phenomena to occur.

For example, if we consider in the figure the intersection of the blue and the green trapezoids, we are in a case where their intersection is a parallelogram (with red vertices), a case that is fulfilled if

$$\text{Condition 1:} \ \ b_2 \le a_1$$

The b.c. of the vertices of this parallelogram are

$$\begin{cases}S:(a_1,b_1,1-a_1-b_1),\\ P:(a_1,b_2,1-a_1-b_2),\\ R:(a_2,b_1,1-a_2-b_1),\\ Q:(a_2,b_2,1-a_2-b_2), \end{cases}$$

Till now, all has been done by considering the two first b.c. It is time to see how the third b.c. can be constrained. The constraint on the red trapezoid is

• its longest "base" must leave three of the red points (points $P,Q,S$) below it, and one above it (point $R$)

• its shortest base must leave all red points below it.

This double condition gives rise to:

$$\text{Condition 2:} \ \ \begin{cases}1-a_2-b_1 \ge c_1 \ge \max(1-a_2-b_2, 1-a_1-b_1)\\ \text{and} \ c_2 \ge 1-a_2-b_1\end{cases}$$

(is it clear ?).

To sum up, here is a sufficient condition :

$$\text{If Cond. 1 and Cond? 2 are fullfilled, then we have a triangle}$$

But there are maybe dozen of such conditions...

At least, we have an methodology to find them all...