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| The perimeter of a convex quadrilateral is 2004, and one of the diagonals is 1001. Can the second diagonal be a) 1; b) 2; c) 1001?

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a) The doubled sum of the diagonals of a quadrilateral is greater than its perimeter (see problem 5 5152). However, $2(1 + 1001) = 2004$.

b) Consider a quadrilateral $ABCD$ where the diagonals are perpendicular, and diagonal $AC=1001$ bisects diagonal $BD=2$. Its perimeter depends on the position of the intersection point $K$ on $AC$. As $K$ moves from point $A$ to the midpoint of segment $AC$, this perimeter continuously changes from

$2 + 2 \sqrt{1001^2 + 1} > 2004$ to $2 \sqrt{1001^2 + 4} < 2004$. By the Intermediate Value Theorem, there exists an angle at which the perimeter equals 2004.

## Answer

a) It cannot; b), c) it can.