Remarkable property of equilateral triangles

Question

In a discussion with a math friend and colleague, I was asked to find the angle $APB$ in an equilateral triangle if $AP=3$, $BP=4$ and $CP=5$. In solving this question I rediscovered that the same angle is always found when $a$, $b$ and $c$ satisfy the Pythagorean theorem.

But I rediscovered another, very surprising theorem.

In an equilateral triangle, the following is true: $$L^2=\frac{1}{2}(a^2+b^2+c^2)+2{\sqrt{3}}Area_{\triangle _{(PA,PB,PC)}}$$

I was able to prove this by using two times the cosine rule and expressing $Area_{\triangle _{(PA,PB,PC)}}$ by applying Héron's rule. (And using this theorem, the property mentioned first is easy to prove.)

My question is: is this a known theorem? Does it have a name?

Answer 1

The wikipedia article about Pompeiu's theorem mentions this:

Generally, by the point $P$ and the lengths to the vertices of the equilateral triangle - $PA$, $PB$, and $PC$ two equilateral triangles (the larger and the smaller) with sides $a_{1}$ and $a_{2}$ are defined:

$$a_{1,2}^{2}={\frac {1}{2}}\left(PA^{2}+PB^{2}+PC^{2}\pm 4{\sqrt{3}}\triangle _{(PA,PB,PC)}\right)$$

The symbol $\triangle$ denotes the area of the triangle whose sides have lengths $PA$, $PB$, $PC$.

Answer 2

In fact, the following figure, copied from the final state of the very nice animation here explains the connection with pythagorean triples and more generaly with any right triangle :

The explanations are given in the "cut-the-knot" article for the $3-4-5$ triangle, as you have, but are valid for any right triangle.

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Edit : A natural question now : let us consider the identity obtained by summing up all the areas of the figure above, knowing that the area of $A'B'C'$ is four times the area of $ABC$ :

With notations $L=AB=AC=BC$, $a=PA,b=PB,c=PC$, and [...] for the areas, we get :

$$[A'B'C'] = \begin{cases} 3 \times [PBP_A] + & (\text{3 right triangles})\\ [PCP_a]+[PAP_b]+[PBP_c] + & (\text{equilateral triangles})\\ 2 \times ([A'P_aB]+[A'P_bC]+[A'P_cA]) & (\text{various triangles}) \end{cases} $$

Otherwise said :

$$2 \sqrt{3} L^2 = \begin{cases} \frac32 ab + \\ \frac{\sqrt{3}}{2}(a^2+b^2+c^2) + \\ 2 \times \frac14(\delta(L,b,c)+\delta(a,L,c)+\delta(a,b,L)) \end{cases}\tag{1}$$

where $\frac14\delta(x,y,z)$ is the area of a triangle with sides $x,y,z$, that can be expressed under the form :

$$\mathfrak{A}^2=\frac{1}{16}\delta(x,y,z)^2=\frac{1}{16}\begin{vmatrix} 0&1&1&1\\ 1&0&x^2&y^2\\ 1&x^2&0&z^2\\ 1&y^2&z^2&0\end{vmatrix}$$

Besides, we have the relationship given by the Cayley-Menger identity for the mutual distance of 4 points ($A,B,C,P$) in a plane which is :

$$\begin{vmatrix} 0&1&1&1&1\\ 1&0&L^2&L^2&a^2\\ 1&L^2&0&L^2&b^2\\ 1&L^2&L^2&0&c^2\\ 1&a^2&b^2&c^2&0\end{vmatrix}=0\tag{2}$$

It is almost evident that (1) and (2) are two faces of a same result, but till now, I have been unable to make a bridge between them...