This problem has been posted many times with different numerical examples. For instance, see easy problem (equilateral triangle). Again, the distances from the point to the vertices of the equilateral triangle form a pythagorean triple. See the attachment below.

Attachments:

note the point as $P$ and manage $12^2+5^2=13^2$ let $PA=12,PB=5,PC=13$ rotate the $\triangle BPC$ on $60$ degrees around $C$ so $P$ goes to $X$ $B$ goes to $A$ so we have $PX=XC=PC=12$ and $AX=BP=5$ so we have $\angle AXP=90$ so we have $AX=12,XC=5,\angle APC=150$ so $AC^2=AX^2+XC^2+2AX\cdot XC\cdot cos60$ so $AC=\sqrt 229$ so we are done P.S.sorry for my first solution it was completely incorect hope now it is allright

frenchy wrote: note the point as $P$ and manage $12^2+5^2=13^2$ let $PA=12,PB=5,PC=13$ now on the extension of $PC$ take a point $X$ such that $XC=13$ now we have $XA=BP=5$ and we also have $BX=PA=12$ so $BXAP$ is a rectangle ($\angle XAP=\angle XBP=90$ and $BP\parallel XA$) so we get $AB=13$ analogues we get $BC=AC=13$ so we are done I think your solution is wrong!

3 concentric circles $\mathcal X, \mathcal Y, \mathcal Z$ have given radii $a=5, b=12, c=13.$ A ray from the common center $P$ cuts $\mathcal X, \mathcal Z$ at $X, Z_0.$ Let $\triangle OPX, \triangle XY_0Z_0$ be equilateral, their vertices on the opposite sides of the ray $(OX.$ Rotate/scale $\triangle XY_0Z_0$ around $X$ into $\triangle XYZ,$ while keeping $Z \in \mathcal Z.$ $\triangle OXY \cong \triangle PXZ$ are congruent by SAS $\Longrightarrow$ $OY = PZ = c,$ $Y$ is on a circle $\mathcal O \cong \mathcal Z$ with center $O$ and radius $c.$ Let $Y_1, Y_2 \in \mathcal O \cap \mathcal Y$ and let $Z_1, Z_2 \in \mathcal Z$ be the corresponding vertices of equilateral $\triangle XY_1Z_2, \triangle XY_2Z_2.$ Then $PX : PY_1 : PZ_1 = PX : PY_2 : PZ_2 = a: b: c.$ Triangles $\triangle OPY_1 \cong \triangle OPY_2$ have sides $a, b, c$ and if these are Pythagorean triples, the calculation is particularly easy. Suppose they are not. $XY^2 = PX^2 + PY^2 - 2\ PX \cdot PY \cos (\widehat{OPY} \pm 60^\circ) = \frac{_1}{^2}(a^2 + b^2 + c^2) \pm 2 \sqrt 3 [abc]$ Using Heron formula $16[abc]^2 = 2(b^2c^2 + c^2a^2 + a^2b^2) - (a^4 + b^4 + c^4),$ we get $x^2 = XY^2$ as roots of quadratic equation $x^4 - (a^2 + b^2 + c^2) x^2 + (a^4 + b^4 + c^4) - (b^2c^2 + c^2a^2 + a^2b^2) = 0$ For Pythagorean triple $a=5, b=12, c=13$, $x = \sqrt{c^2 \pm ab \sqrt 3} = \sqrt{169 \pm 60 \sqrt 3}.$ The larger root corresponds to $P$ inside of the equilateral $\triangle XYZ.$

A more general fact, which is easily established using analytical geometry: let $a$, $b$, $c$ be the distances of the point from the vertices and $L$ the sidelenght of the triangle. Then we always have \[ 3(a^4+b^4+c^4+L^4)=(a^2+b^2+c^2+L^2)^2 .\] In this particular case, $L=\sqrt {169+60 \sqrt 3}$.

FACT Given the distances of a point from the three corners of an equilateral triangle, a, b, and c, the length of a side s is given by $3(a^4+b^4+c^4+s^4)=(a^2+b^2+c^2+s^2)^2$