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  "body": "\n## Statement\n\nLet $ABC$ be a [[triangle|0CTNFGYMM2]] in the Euclidean plane.\nOn each side of $ABC$ erect an equilateral triangle, either all three outside $ABC$ or all three overlapping $ABC$ inwards.\nLet $X$, $Y$ and $Z$ be the [[centroids|0X7EEWES5G]] of the equilateral triangles on $BC$, $CA$ and $AB$.\nThen $XYZ$ is an equilateral triangle.\n\n## Remarks\n\nA short proof uses rotations.\nFor the outer case, the rotations by $120^\\circ$ about $Z$, $X$ and $Y$, in the same sense, send $A$ to $B$, $B$ to $C$ and $C$ to $A$.\nTheir composition has total angle $360^\\circ$, so it is a translation, and it fixes $A$, so it is the identity.\nThree rotations by $120^\\circ$ whose composition is the identity have centres that form an equilateral triangle.\nThe triangle $XYZ$ is called the outer or the inner Napoleon triangle.\nBoth Napoleon triangles have the same centroid as $ABC$, and the area of the outer one minus the area of the inner one equals the area of $ABC$.\nFor an equilateral triangle $ABC$ the inner Napoleon triangle degenerates to a single point, its centre.\n\n## Explanations\n\n### L1 human\n\nTake any triangle, even a very lopsided one.\nDraw an equilateral triangle on each of its sides, pointing outwards, and mark the centre of each.\nThe three centres always form an equilateral triangle.\nThe irregular shape of the starting triangle disappears completely.\n\n### L2 human\n\nThe theorem is a standard exercise for complex numbers or for compositions of rotations in the plane.\nThe same equilateral triangles appear in the construction of the [[Fermat point|QM0CGSQ4TF]].\nThe lines joining each vertex to the apex of the outer equilateral triangle on the opposite side meet in one point.\nIt is one of the first examples where a [[regular polygon|01KJJ9A400]] is produced from an arbitrary triangle by a fixed recipe.\n",
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