# Polygon in a Rectangle

Every convex polygon of area 1 is contained in a rectangle of area 2.

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Copyright © 1996-2017 Alexander Bogomolny

Every convex polygon of area 1 is contained in a rectangle of area 2.

### Proof

The claim is certainly true for a triangle. Any triangle can be embedded in a rectangle of twice its area.

Interestingly, the more general fact follows from this simple observation. Let K be a convex polygon of area 1. Pick two most distant vertices of the polygon.

These are vertices A and E in the diagram. Through A and E draw lines L_{A} and L_{E} perpendicular to AE. Since the distance between A and E is the largest of all the distances between any two points of K, neither L_{A} or L_{E} meet K in points other than A and E. Draw a line perpendicular to L_{A} and L_{E} away from K on the right. Move towards K until it first touches K. The line will then share with K either a side of a vertex. In the diagram this is vertex C and the line is called L_{C}. Similarly, find a line perpendicular to L_{A} and L_{E} to the left from K and move it towards K until the first contact. This is line L_{G} in the diagram.

The four lines form a rectangle. Call it R. Line AE separates R into two smaller rectangles, say R_{G} and R_{C}.

Join A and E to C and G. The area of ΔAEG is half that of rectangle R_{G}. The area of ΔAEC is half that of rectangle R_{C}. The area of the quadrilateral ACEG is half that of R. Since K is convex, ACEG lies entirely within K and hence its area is at most 1. It follows that the area of R is at most 2. If need be, to answer the requirement it may be enlarged to enclose the area of 2.

### References

- B. Bollobás,
*The Art of Mathematics: Coffee Time in Memphis*, Cambridge University Press, 2006, p. 55.

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Copyright © 1996-2017 Alexander Bogomolny

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