Let's suppose each of the mass points has a unit mass. Then, mathematically, the problem is the following.
Let $A$, $B$, $C$ be differentiable function $\mathbb R\to \mathbb R^3:t\mapsto A(t),B(t),C(t)$ with the following properties.
- $v_A+v_B+v_C=0$ where $v_A:=\dot A$, $v_B:=\dot B$, $v_C:=\dot B$ (momentum is constant $0$)
- $A \times v_A+ B\times v_B+ C\times v_C = 0$ (angular momentum is constant $0$)
- $(C-A)\times (C-B)$ nowhere vanishes. (the $ABC$ triangle is never degenerated)
from 1, $A+B+C$ is constant. Without loss of generality, suppose it is $0$, that is,
$$A+B+C=0 \tag{4}$$
I can prove that under these conditions, the 3 points remain forever in the initial plane of the $ABC$ triangle (i.e. $(C(t)-A(t))\times (C(t)-B(t))$ is collinear with $(C(0)-A(0))\times (C(0)-B(0))$ for any $t$). But I feel my proof is too long and complicated and I'd like to see a more elegant proof.
My proof consists of two steps. In the first step, I prove that for every $t$, $v_A(t)$, $v_B(t)$ and $v_C(t)$ are in the 2-dimensional subspace $P(t)$ spanned by $A(t)$ and $B(t)$. In the second step, I prove that the normal vector $n(t)$ to $P(t)$ is constant.
Step 1. By condition (4), $P(t)$ is a 2-dimensional linear subspace of $\mathbb R^3$. Let $n$ be a unit normal vector to $P(t)$. Let $v'_A$, $v'_B$ and $v'_C$ be real numbers such that $v'_An$, $v'_Bn$ and $v'_Cn$ are the components of $v_A(t)$, $v_B(t)$ and $v_C(t)$ orthogonal to $P(t)$. Substituting $v_C = – (v_A + v_B)$ from 1. into 2,
$$0 = A \times v_A + B \times v_B – C \times (v_A + v_B ) = (A – C) \times v_A + (B – C) \times v_B \tag{5}$$
For the normal component of the vector on the right side,
$$(A – C) \times v'_An + (B – C) \times v'_Bn = (v'_A(A – C) + v'_B(B – C)) \times n \tag{6}$$Since the vector $v'_A(A – C) + v'_B(B – C)$ is orthogonal to $n$ and $n \neq 0$, from (6) follows $$ (A – C) v'_A+ (B – C) v'_B = 0.$$
Thus, by the linear independency of $(A -C)$ and $(B – C)$, $v'_A = v'_B = 0$ and this completes step 1.
My proof of step 2. isn't exact enough so I even don't write it down here. Could somebody give a mathematically exact proof for step 2, or for the whole statement?
Best Answer
Here is a short self-contained proof - inspired by the earlier remarks and answers.
By the first condition, the baricenter $(A+B+C)/3$ is constant. The assumptions and the conclusion are invariant under shifting $A$, $B$, $C$ by a constant vector, hence we can assume from the beginning that $A+B+C=0$. Then the first condition is automatic, while the second and third conditions can be rewritten as (after substituting $C=-A-B$): $$(A+2B)\times v_B=v_A\times(2A+B),$$ where $A$ and $B$ are linearly independent. Let us call $X$ the common two sides of this equation. First, $X$ is orthogonal to $A+2B$ and $2A+B$, hence it is also orthogonal to $A$ and $B$. Second, decomposing $X$ as $$X=A\times v_B+2B\times v_B\quad\text{and}\quad X=2 v_A\times A+v_A\times B,$$ and observing that any cross product containing $A$ (resp. $B$) is trivially orthogonal to $A$ (resp. $B$), we infer that both terms on both right-hand sides are orthogonal to $A$ and $B$. In particular, the derivative of $A\times B$ is orthogonal to $A$ and $B$, whence $A\times B$ is constrained to a ray (half-line starting at the origin). Done.