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There are many questions on this side asking to compute the area of the largest square that fits in a triangle (equilateral or more general). All otherwise excellent answers (and some of the questions) seem however to take for granted that this square should have one of its side lying entirely on one of the sides of the triangle, while the two corners of the square not on this side should each touch one of the other sides of the triangle.

When we accept this, calculating the area's is not that hard, but how do we know that the largest square does indeed have this position? Why can a square 'balancing' on one of its corners never be largest? How would you prove this?

Vincent
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1 Answers1

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Intuition is that if no side of the square lies along a side of the triangle then there is room left to "jiggle" the square until one of its sides lines up with a side of the triangle, then "inflate" it larger. Therefore the maximal square must have one side on a side of the triangle, and the other two vertices on the other sides of the triangle.

Proving that rigorously is not trivial, though. John M. Sullivan's paper Polygon in Triangle: Generalizing a Theorem of Post proves the following more general statements for arbitrary convex polygons inside a triangle, not just squares:

If a convex polygon P is contained in a triangle T, it can be rigidly moved, staying inside T , until one of its edges lies along an edge of T.

Among the polygons inside a triangle T which are similar to a given polygon P, any largest one sits rigidly inside T.

dxiv
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