Show that given a convex quadrilateral Q of area 1, there is a rectangle R of area 2 such that Q can be drawn inside R

Question

The question is:

Show that given a convex quadrilateral Q of area 1, there is a rectangle R of area 2 such that Q can be drawn inside R.

How do you even approach a problem like this?

I was only able to do this by taking Q to be a square and a rectangle, it is then easy to find 2 sides of a rectangle with area 2 such that the given statement true depending on the sides of the original rectangle..

Also, I saw that Q need not be a rectangle, it can be any convex polygon..how do we prove it then? Can anyone please provide the complete formal proof of the above statements?

PS:I wasn't able to find any relevant answers on Approach0 or while reviewing this question..I am looking for a neat way to approach such problems..

Thanks for any answers!!

Answer 1

Hint: Consider the longest edge/diagonal of the polygon. Consider perpendiculars from those vertices.

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Note: This is true for any convex polygon, not just quadrilaterals.

If you want the answer...

From Cut The Knot