An odd wrapping

A present fits inside a circle, and we tape the orange ribbons to the green midpoints of the sides, choosing them perpendicular to the opposite sides at the purple points. Show that the four ribbons have a point in common.

Scroll down for a solution to this problem.


If the circle is centred at 0, the vertices of the cyclic quadrilateral are the vectors A, B, C and D, all of the same length. The common point is E=(A+B+C+D)/2. Then (E – (A+B)/2) = (C+D)/2, which is perpendicular to D-C because C and D have the same length.

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