Question

Area of greatest square inscribed in a triangle

Answer 1

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Answer 2

The inscribed square is clearly unique once we choose the triangle side where two vertices of the square lie. If we suppose that the square has two vertices on AB and side l, then:

l+lcotA+lcotB=c,

so:

l=c1+cotA+cotB=2RsinAsinBsinCsinC+sinAsinB=abc2Rc+ab,

where R is the circumradius of ABC. In order to maximize l, you only need to minimize 2Rc+ab=2R(c+2Δc), or "land" the square on the side whose length is as close as possible to 2Δ−−−√, where Δ is the area of ABC.

Say x be the side of the largest square within a triangle with sides a,b,c. Say one side of the square is on the side BC. So we get a square and 3 triangles. Equating the area of the bigger triangle with the sum of area of the 3 small triangle and the square we get:

12ch=x2[area of the square]+12x(c−x)[area of two small base triangle]+12x(h−x)[area of the upper triangle]

Upon solving:

x=ch(c+h)