[JEE Advanced 1997] s is a unit square a quadrilateral has one vertex on each side with
$S$ is a unit square; a quadrilateral has one vertex on each side, with side lengths $a,b,c,d$. Prove $2\le a^2+b^2+c^2+d^2\le4$.
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Answer: Proved.
Let the vertices divide the sides at fractions $p,q,r,s\in[0,1]$. Then $a^2+b^2+c^2+d^2=\sum(\,\text{two-square terms}\,)$ which, minimised at the midpoints, gives $2$, and maximised at the corners gives $4$.
JEE Advanced 1997 · Quadratic Equations and Inequations — verified solution by the Vidaara Team.
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