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Problem: A quadrilateral is inscribed in a parabola, then

(A) Quadrilateral may be cyclic.

(B) Diagonals of the quadrilateral may be equal.

(C) All possible pairs of adjacent side may be perpendicular.

(D) None of the above $\quad$ $\blacksquare$

Option (A) can be verified easily. I think both (B) and (C) are the criterion for rectangle, but I am unable to prove or disprove options (B) and (C). Please help.

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  • $\begingroup$ So finally why did you not settle for (D) as the answer? $\endgroup$ – Narasimham Dec 8 '18 at 15:45
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If you place a circle such that it intersects the parabola in four points, those four points will form the corners of a cyclic quadrilateral inscribed in the parabola, satisfying A. If the circle is also placed such that its center lies on the parabola's line of symmetry, these four points form the vertices of an isosceles trapezoid, which is sufficient to satisfy B but not C. Satisfying C would, in fact, require the figure to be a rectangle, which is not possible.

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