They Look Like Pythagorean Triple

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In an equivalent triangle side S, if a point inside it is at distances X, Y, Z from the vertices, their relation is given in the link above by,
$3(X^4+Y^4+Z^4+S^4)=(X^2+Y^2+Z^2+S^2)^2 \\ \implies\ 3(3^4+4^4+5^4+S^4)=(3^2+4^2+5^2+S^2)^2.\\ Expanding\ and\ simplifying,\\ 2S^2-100S+386=0\ \ or\ \ S^2 - 50S+193=0.\\ Solving\ the\ quadratic\ \ S^2=45.7846,\ rejecting \ the\ other\ root\ being\ too\ small.\\ \therefore\ S=\color{#D61F06}{6.7664325}$

Refer here to get that twice the area of the equilateral triangle is $25\frac{\sqrt3}{2}+18$ Hence, if the side of the required triangle is $a$ , $a^2 \frac{\sqrt3}{2} = 25\frac{\sqrt3}{2}+18\\\implies a=\sqrt{25+12\sqrt3} \approx 6.766$

Rotate triangle $ABC$ in magnitude of $60^\circ$ in the counterclockwise direction as shown in my figure.

$\sin \angle D’DC=\dfrac{3}{5}$

$\angle D’DC \approx 36.87^\circ$

By cosine rule, we have

$(BC)^2=4^2+5^2-2(4)(5)[\cos (60+36.87)]$

$BC = AB \approx 6.766~\text{units}$

Let generalize this problem for any Pythagorean triple with cathetus $a$ and $b$ , and hypotenuse $c$ .

First moves $\triangle APC$ together $\triangle APB$ as image shows, clearly, $\triangle AP'P$ is an equilateral triangle and $\triangle PP'B$ is an rectangle triangle of sides $a,b$ and $c$ so $\angle BPA = 60° + 90° = 150°$ .

By law of cosines, the side of the equilateral triangle is $\boxed{s=\sqrt{a^2 + b^2 - 2(a)(b)\cos 150°}}$ .

For this triangle $\boxed{AB=\sqrt{3^2+4^2-2(3)(4)\cos150°}\approx6.766}$ .

Similar problem: Equilateral triangle point

Any question, be free of asking in comments.