There is this famous problem:
Consider a stick of length 1. Pick two points uniformly at random on the stick, and break the stick at those points. What is the probability that the three segments obtained in this way form a triangle?
The answer is $1/4$ and there are many different ways one can solve it. For a comprehensive list, see: https://mathoverflow.net/questions/2014/if-you-break-a-stick-at-two-points-chosen-uniformly-the-probability-the-three-r
However, I would like to ask about two particular solutions of this problem. Frankly saying, I have some problems with understanding them...
It seems natural to rephrase the question in terms of barycentric coordinates in a triangle. These coordinates are numbers x, y, z in the interval [0,1] satisfying the equation x+y+z=1. We are looking for triples (x,y,z) of such numbers satisfying the three triangle inequalities x≤y+z, y≤x+z, and z≤x+y. Replacing the relations "≤" by "=", we get line segments joining the midpoints of the edges of the triangle. These line segments cut the triangle into four congruent subtriangles. The central one of these four subtriangles is the region where all three triangle inequalities hold, and this region has area equal to one quarter of the area of the big triangle.
Say that our radnom points are $U_1, U_2 \sim \mathcal{U}(0,1)$. As I understand, $x,y,z$ from this solution are the so called "spacings", defined as: $$x:= \min(U_1,U_2),$$ $$y:=\max(U_1, U_2)-\min(U_1, U_2),$$ $$z:=1-\max(U_1, U_2).$$ For the cited solution to hold, we should require $(x,y,z)$ to be uniformly distributed on $$\Delta \ = \ \{(a,b,c)\in \mathbb{R}_+^3: a+b+c=1 \}.$$ I know that it is true, but I do not find it anyhow obvious... Can this statement be demonstrated in an elementary fashion?
Additionally, there is another solution in this spirit:
(with the same notation) we have $(x,y)\in \tilde{\Delta}$, where $$\tilde{\Delta} \ = \ \{(a,b)\in \mathbb{R}_+^2: a+b\le 1 \}.$$ In order to make a triangle, we must have $$a\le 1/2,$$ $$b\le 1/2,$$ $$a+b\ge 1/2.$$ Hence probability is $1/8$ divided by $1/2$ (measure of $\tilde{\Delta}$) which results in $1/4$.
My question again is: how do we know that $(x,y)\sim \mathcal{U}(\tilde{\Delta})$. Is it by any means obvious?
EDIT: My question is about proving those two little facts (used in cited solutions without any justification) in an elementary way. But, what is important, I know that both those facts are true (e.g. via dypheomorphic transformation formula for densities) - I just believe that proving them is harder than the actual triangle problem, and I am wondering if I do not miss any simple argument...
Lets write $\max(U_1,U_2)=U_1\vee U_2$ and $\min(U_1,U_2)=U_1\wedge U_2\,.$
Proof. For all $a,b\in[0,1],$ with $a+b\le 1\,,$ \begin{align}\tag{1} &\mathbb P\Big\{x\le a,y\le b\Big\}\\&=\mathbb P\Big\{x\le a,y\le b,U_1\le U_2\Big\}+\mathbb P\Big\{x\le a,y\le b,U_1> U_2\Big\}\\ &=\underbrace{\mathbb P\Big\{U_1\le a, U_2-U_1\le b,U_1\le U_2\Big\}}_{(*)}+\underbrace{\mathbb P\Big\{U_2\le a, U_1-U_2\le b,U_1>U_2\Big\}}_{(**)}\,. \end{align} The terms (*) and (**) are \begin{align} \int_0^a\int_{u_1}^{u_1+b}\,du_2\,du_1&=\int_0^ab\,du_1=a\,b\,,\\ \int_0^a\int_{u_2}^{u_2+b}\,du_1\,du_2&=\int_0^ab\,du_2=a\,b\,. \end{align} Therefore $$ \mathbb P\Big\{x\le a,y\le b\Big\}=2\,a\,b\,,\quad a+b\le 1\,, $$ which shows that $(x,y)$ is uniform in $\tilde{\Delta}\,$.
Clearly, all $(x,y,z)$ are in $\Delta$. The function $$ \{(a,b)\in\mathbb R^2_+:a+b<1\}\ni\left(\begin{matrix}a\\b\end{matrix}\right)\mapsto\left(\begin{matrix}a\\b\\1-a- b\end{matrix}\right) $$ is a parametrization of $\Delta$. If $a$ and $b$ move by small amounts $da,db$ respectively, then the $3$-vector on $\Delta$ moves by $$ \left(\begin{matrix}da\\0\\-da\end{matrix}\right)\text{ resp. by }\left(\begin{matrix}0\\db\\-db\end{matrix}\right)\,. $$ The area of the parallelogram on $\Delta$ that is spanned by those two "tangent" vectors is the norm of the their cross product: $$ \left(\begin{matrix}da\,db\\da\,db\\da\,db\end{matrix}\right) $$ which is $\sqrt{3}\,da\,db\,.$ This surface measure on $\Delta$ is obviously uniform (it is the pushforward of the Lebesgue measure on the two dimensional triangle $\{(a,b)\in\mathbb R^2_+:a+b<1\})$. We have shown in 1. that the latter is the distribution of $(x,y)$ therefore, the uniform surface measure on $\Delta$ is the distribution of $(x,y,z)$.