Math Is Fun Forum

Identity

Member

Registered: 2007-04-18

Posts: 934

Convex Polygon

What is the area of the smallest convex polygon that intersects each of the four branches of y = -2/x and y = 2/x?

Please help thanks

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: Convex Polygon

by intuition i would guess a rectangle with vertices (1,2) (1,-2) (-1, 2) (-1, -2) which gives an area 8units squared

Well actually, you could say 0 is the answer since as x tends to infinity and -infinity it will intersect each of the 4 branches, and hence you'd have a line which has no area Otherwise the y axis with same argument.

Last edited by luca-deltodesco (2008-05-07 20:10:56)

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Identity

Member

Registered: 2007-04-18

Posts: 934

Re: Convex Polygon

Thanks luca, I got 8 units squared, but it seemed too easy. I like your solution as x tends to +/- infinity, but are you sure it is valid?

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: Convex Polygon

I see no reason why not.

the 4 regions given by y = -2/x and y = 2/x are all symetrical about the x and y axis. so simply consider the top right region with a rectangle formed between the x and y axis with the origin and the point y = 2/x.

its area will be 4/x², giving the total area of polygon 16/x².

as x tends to infinity, the total area tends to 0

rearrange as x = 2/y. so area is now 16/y^2. as y tends to infinity, total area tends to 0.

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mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Convex Polygon

You're calculating the area by 2/x * 2/x, when you should be doing it by x * 2/x. This gives an area of two, regardless of what x is, and so the overall area will be 8.

This is true for any rectangle whose vertices lie on the 4 curves and whose lines are parallel (or perpendicular) to the axes.
There might be a smaller polygon if we start considering angled rectangles or heptagons or something.

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Identity

Member

Registered: 2007-04-18

Posts: 934

Re: Convex Polygon

Ok, I'm confused of what to think now... I understand that if you draw an infinitely long rectangle with width 0 then these conditions would be satisfied at infinity, but then that rectangle has area 8 according to mathsy.... so paradox or what?
Also, isn't the rectangle the smallest possible area for this problem? I mean, generally as you increase the number of vertices, extra space is filled so the area goes up.