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Thrown three points into a triangle of unit area, find the mathematical expectation of the area of ​​the triangle formed by them

I already found the solution of similar task, but there was a square of unit area. But I really don’t understand what to do with the outer triangle, because it can be anything. I think the outer shape doesn’t matter, all that matters is shape’s area (in this problem it equals 1), but I could be wrong

Deepest
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  • (I don't know for certain, but I think) It it likely that the expected area depends on the shape of the triangle. – Calvin Lin Nov 15 '20 at 19:04
  • I think it doesn't. And it could be useful to take isosceles right triangle as an outer with side length equal to sqrt(2). But of course, I could be wrong :) – Deepest Nov 15 '20 at 19:11
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    $\frac{1}{12}$, see triangle triangle picking on mathworld. – achille hui Nov 15 '20 at 19:18
  • @achillehui could you explain to me why there is taken an isosceles right triangle with unit leg lengths as an outer triangle? Its area is 1/2, not 1. And why there is integral of area function is divided to the integral of unity over the region? – Deepest Nov 18 '20 at 08:35
  • I don't understand your question. No one (me, Calvin, the question you linked to nor mathworld's article) talks about isosceles right triangle with unit leg lengths.... – achille hui Nov 18 '20 at 08:41
  • @achillehui Since the problem is affine, it can be solved by considering for simplicity an isosceles right triangle with unit leg lengths - mathworld's article. This triangle's area equals 1/2, but the area should be 1 due to the problem's condition. Is not it? – Deepest Nov 18 '20 at 09:33
  • I see what you mean now. Under affine transformation, ratio of areas of different geometric shapes remain invaraint. this mean if you start from any triangle with area $A$, compute the expected area $\Delta$, the ratio $\frac{\Delta}{A}$ will be always the same (independent of shape and area of the outer triangle) – achille hui Nov 18 '20 at 09:37
  • @achillehui So much thanks, now I understand this :) – Deepest Nov 18 '20 at 09:58

1 Answers1

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The answer is $\frac1{12}$, independent of the shape of triangle.

To see the independence, let $T_0$ be a fixed triangle with unit area. For any triangle $T$ with unit area, let $\eta : \mathbb{R}^2 \to \mathbb{R}^2$ be an affine transformation which sends $T$ to $T_0$. Recall affine transformation preserves ratio of area of geometric shapes. If $p_1, p_2, p_3$ are $3$ points in $T$, then $\eta(p_1),\eta(p_2),\eta(p_3)$ are three points in $T_0 = \eta(T)$ with $$\verb/Area/(\triangle_{p_1p_2p_3}) = \verb/Area/(\triangle_{\eta(p_1)\eta(p_2)\eta(p_3)})$$ Furthermore, if $p_k$ are sampled uniformly from $T$, then $\eta(p_k)$ will be uniformly distributed over $T_0$. This implies the expected area of "uniformly sampled" triangles in $T$ equals to that of $T_0$.

Since this is true for all $T$, the answer is independent of shape of triangle $T$.

This problem of finding the expected area of "uniformly sampled" triangles in a triangle is known as Triangle triangle picking, see the entry on mathworld for more information.

There are various generalizations of this problem. Instead of $3$ points, one can uniformly sampled $n$ points from a triangle of unit area and askes: What is the expected area of the convex hull formed by those $n$ points?

This is part of the Triangle Point Picking problem, see entry on mathworld again for more information.

The answer of the expected area of convex hull of $n$ points is given by the formula: $$\Delta_n = 1 - \frac{2}{n+1}\sum_{k=1}^n \frac1k = 1 - \frac{2H_n}{n+1}$$

where $H_n$ is a harmonic number. It is possible to write down a double integral to compute this number. For a derivation, see the second half of my answer to a semi-related question.

In any event, for the problem at hand, the expected area of "uniformly sampled" triangle is

$$\Delta_3 = 1 - \frac{2 H_3}{3 + 1} = 1 - \frac12\left(1 + \frac12 + \frac13\right) = \frac{1}{12}$$

achille hui
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