Checkers, Anyone?

This solution doesn't explain how to calculate the area, but instead focuses on why we tend to get circles.

Here's the best explanation that I can come up with

1) Since $\csc ( x + 2\pi ) = \csc x$ , it suffices to look at the the region that $0 \leq x \leq 2 \pi$ . Since $\csc x \csc y > 0$ , we will look at $0 \leq x , y \leq \pi$ .

2) We want to understand why the circles seem to form about the points $(\frac{ \pi}{2 }, \frac{ \pi }{2} )$ . To do so, we will simply take the Taylor expansion about that point. This gives us:

$( 1 + \frac{1}{2} ( x - \frac{\pi}{2} ) ^2 + \frac{5}{24} ( x - \frac{ \pi }{2} ) ^ 4 + \ldots ) ( 1 + \frac{1}{2} ( y - \frac{\pi}{2} ) ^2 + \frac{5}{24} ( y - \frac{ \pi }{2} ) ^ 4 + \ldots ) = \phi$

Observe that the RHS only has positive coefficients and even exponents, hence is an increasing function on $x \geq \frac { \pi } {2}$ . (It is also an "increasing function" on $x \leq \frac{\pi}{2}$ if we trace our steps backwards, but that doesn't really matter ).

3) For small values of $a, b$ , and $k \gtrsim 1$ the solution to

$( 1 + a^2 ) ( 1 + b^2 ) = k$

looks very much like the circle $a^2 + b^2 = (k-1)$ . This is obvious, because we are dropping the "very small term of $a^2 b^2$ ".

Thus, this explains why with $\phi$ , we obtain the circles.

We'll focus on the region that is centered at $O=\big(\frac{\pi}{2},\frac{\pi}{2}\big)$ .

First find the radius $r_1$ of the largest circle centered at O contained inside the region. When $x=\frac{\pi}{2}$ , the smallest possible positive value of y is $\sin^{-1}\big(\frac{1}{\phi}\big)$ . So the radius of the largest circle centered at O contained entirely within our region is $r_1=\frac{\pi}{2}-\sin^{-1}\big(\frac{1}{\phi}\big)$ .

Now find the radius $r_2$ of the smallest circle centered at O that contains the region. When $x=y$ , the equation becomes $\csc^2(x)=\phi$ , and the smallest possible positive solution is $\Big(\sin^{-1}\big(\sqrt{\frac{1}{\phi}}\big),\sin^{-1}\big(\sqrt{\frac{1}{\phi}}\big)\Big).$ So $r_2=\Big(\frac{\pi}{2}-\sin^{-1}\big(\sqrt{\frac{1}{\phi}}\big)\Big)\sqrt{2}.$

Since $\pi r_1^2 \approx 2.57$ and $\pi r_2^2 \approx 2.79$ , we're looking for an answer in that range. Of the options, $\frac{\pi^2}{\phi^e}$ is the only one.

Dunno how to solve this without using a graph. The circle is what I approximate the area to, The integral is stupid too, wolfram gave up tryna compute it. In the picture above, the little circles with dots in them are intersections of the real graph with the approximate circle

Does anyone know how to do the integral? Or is it just too ugly.

I used this approach, but I'm sure it's wrong because the value I get is too far from the right answer (although the right answer was still the closest value to my answer). Does anybody know what's the problem? Let us think about the function $z=sin(x)sin(y)$ the plot of this function looks like alternating mountains and valleys. If we take the cross section of that for a certain value of $z$ , we obtain the circles shown in the picture. For any value of $z$ , the center of those circles should be on the point $(x,y)$ that gives us the maximum $z$ for each mountain/valley. Looking at the partial derivative of the function, it's easy to see that this happens first at $(\frac{\pi}{2},\frac{\pi}{2})$ . Now, using this in the function given in the problem, we can see for which values of $x$ it is true that $sin(x)sin(y)=\frac{1}{\phi}$ , when $y=\frac{\pi}{2}$ , and the radius of the circle will be $\frac{\pi}{2} - x$ . Doing that we get that $x=arcsin(\frac{1}{\phi})$ and the area is about $2.57$ .