Take a look and try to solve it - circle intersecting a circle like curve Let there be a circle of unit radius centered at $(1,1)$ in Cartesian plane
Another curve
$(x)^{\frac 12} + (y)^{\frac 12} = 1,$ 
if drawn, will meet the circle at only (0,1) & (1,0)
Same goes for $(x)^{\frac 13} + (y)^{\frac 13} = 1,$ but graphically there is some 'n' such that 
$(x)^{\frac 1n} + (y)^{\frac 1n} = 1,$ cuts circle at two more points ( besides $(0,1)$ & $(1,0) )$
This n is just greater than $0.5$ (like $130/232$, see picture), but can it be found out solving a polynomial?
Or can an 'n' be found such that the two curves are most close (or overlapping)!
Also find that value of 'n' such that area enclosed under both curves are same!
Courtesy: graphsketch.com
 A: Here is a small calculation to find a value of $n$ so that $x^{n}+y^{n}=1$ has $3$ intersections with the circle. There should be a value that has $4$ intersections though.
The map
$$[0,1]\times \mathbb R_{>0}\to \mathbb R^2 \qquad (t,h)\mapsto (t^{1/h},(1-t)^{1/h})$$
is continuous and for fixed $h$ the graph of the function is exactly the line $x^h+y^h=1$ in the first quadrant. To find an intersection with $(x-1)^2+(y-1)^2=1$ we could look at the element of the curves on the diagonal. For $x^h+y^h=1$ and $x^2+y^2=1$ respectively this is given by:
$$\left(\frac{1}{2^{1/h}},\frac{1}{2^{1/h}}\right) \qquad\quad \left(1-\frac1{\sqrt 2},1-\frac1{\sqrt 2}\right)$$
When $h\to0$ the left term goes to $(0,0)$ and when $h\to\infty$ it goes to $(1,1)$. From continuity therefore there is an intersection, specifically solving:
$$\frac1{2^{1/h}}=1-\frac1{\sqrt{2}}$$
gives
$$h=-\frac{\ln(2)}{\ln(1-\frac1{\sqrt 2})}\approx 0.564476 $$
As a point that has one intersection with the circle on the diagonal (so a total of $3$).
This is the only possible value of $h$ that has an intersection on the diagonal. There are apparently values of $h$ that have points intersecting outside (and thus have $4$ total intersections with the circle)!
I think the final answer should find such a curve, this is only an inbetween solution until somebody finds that $:)$.
