Point inside a quadrilateral ABCD.

[Math Tutor]

Suppose a point [tex]O[/tex] is inside a quadrilateral [tex]ABCD[/tex] and [tex]2A=OA^2+OB^2+OC^2+OD^2[/tex], where [tex]A[/tex] is the area of ABCD.
Prove that ABCD is a square.

[Guest]

Spoiler: show

Let the distances from [tex]O[/tex] be [tex]a,b,c,d[/tex] (to save typing). The maximum area of triangle [tex]OAB[/tex] is [tex]ab/2[/tex] which occurs when the angle [tex]AOB[/tex] is [tex]90[/tex] degrees (or [tex]a=0[/tex] or [tex]b=0[/tex]), we get similar results for triangles [tex]OBC[/tex], [tex]OCD[/tex], and [tex]ODA[/tex]. So the maximum value of the area [tex]A[/tex] is [tex](ab+bc+cd+da)/2[/tex]. Which means [tex]a^2 + b^2 + c^2 + d^2 \leq ab+bc+cd+da[/tex], which rearranges to [tex]((a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2)/2 \leq 0[/tex]. We must have equality which implies [tex]a=b=c=d[/tex] and the angles [tex]AOB=BOC=COD=DOA=90[/tex] degrees, consequently [tex]ABCD[/tex] must be a square (and [tex]O[/tex] must be at its centre).

R. Baber.

[Math Tutor]

Bravo