Maximal area of a convex region bounded by hyperbolas

[Xevarion]

This problem was suggested by Gokul43201, based on this year's Putnam A2.

Suppose that K is a convex set in \mathbb{R}^2 which is contained in the region bounded by the graphs of the hyperbolas xy=1, xy=-1 (so the set is in the inner + shaped region which contains the origin also). What is the maximum possible area of K?

 

[coomast]

The only thing you can use is symmetry. I would say that due to the hyperbolas being available with 4 axes of symmetry, the convex area in between the curves must have this property. That leaves us with a circle and a square. Not a rectangular or an ellipse because the area must be maximally convex. The square can have a larger area than the circle if you draw it with it's corners on the systems axes. It touches the four curves of the hyperbolas with the middle of it's edges. The area is then calculated as 4. The area of the circle was \pi.

Whether this is sufficient as a proof remains to be seen...

 

[rs1n]

The only thing you can use is symmetry. I would say that due to the hyperbolas being available with 4 axes of symmetry, the convex area in between the curves must have this property. That leaves us with a circle and a square. Not a rectangular or an ellipse because the area must be maximally convex. The square can have a larger area than the circle if you draw it with it's corners on the systems axes. It touches the four curves of the hyperbolas with the middle of it's edges. The area is then calculated as 4. The area of the circle was \pi.

Whether this is sufficient as a proof remains to be seen...

Well, you can get an area of 4 using any rectangle with vertices on the two hyperbolas. Take for example the rectangle whose vertices are:

(a,1/a), (a, -1/a), (-a, 1/a), and (-a, -1/a)

The are is (2a)(2/a) = 4. For a=1/a=1, this gives a square contained within your region. So the area should be bigger than 4. Perhaps your calculation is off? I'm getting 8 as the area of the region you described.

 

[coomast]

You are absolutely right rs1n, the area of the figure I described is indeed 8. Thank you for noticing it.

 

[coomast]

And the area of the circle I mentioned is 2\cdot \pi

 

[Xevarion]

I think coomast's square region is right because it seems the optimal region should have edges which are tangent to all branches of the hyperbolas. But it doesn't seem obvious to me how one would prove this.

 

[coomast]

Using the tangent property coming from the convex requirement, gives you one type of area which needs to be maximized. This is a regular parallelogram, with corners at the axes of the system and also tangent to the hyperbolas. Calculating the tangent of the hyperbolic should give an equation for the edges and thus also the area can be calculated. This needs to be maximized by taking the derivative and putting it equal to zero.

I will have a look on it later today and will come back on it.

 

[rs1n]

It seems that by symmetry, one could simply consider what happens in say the first quadrant (and then multiply the area by 4). That is, consider the maximal area of a right triangle whose legs are on the axes, and whose hypotenuse is tangent to xy=1. Let the point of tangency be (a,1/a). The slope of the hypotenuse is -1/a^2. Using the point-slope formula of a line, the hypotenuse lies on the line

y = -\frac{x}{a^2} + \frac{2}{a}

The intercepts of this line are (0,2/a) and (2a,0). The area of the corresponding triangle is 2; hence the area of the convex region is 8.

 

[coomast]

Indeed, rs1n, that is the solution. I calculated it in the same way but wanted to be sure that this area is not depending on the "a" value you use. So, the maximal convex area is not necessarily a square, but a regular parallelogram is also valid, which was not obvious to me in the beginning. It is now proven.