JEE Advance - Mathematics (1997 - No. 11)
Let $$S$$ be a square of unit area. Consider any quadrilateral which has one vertex on each side of $$S$$. If $$a,\,b,\,c$$ and $$d$$ denote the lengths of the sides of the quadrilateral, prove that $$2 \le {a^2} + {b^2} + {c^2} + {d^2} \le 4.$$
The minimum value of a^2 + b^2 + c^2 + d^2 is achieved when the quadrilateral is a square.
The maximum value of a^2 + b^2 + c^2 + d^2 is achieved when the quadrilateral collapses into a line segment.
The given inequality holds true for any quadrilateral inscribed in a unit square.
If the vertices of the quadrilateral coincide with the midpoints of the sides of the square, then a^2 + b^2 + c^2 + d^2 = 2.
a^2 + b^2 + c^2 + d^2 can take any value between 2 and 4 inclusive.
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