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

  • $\begingroup$ Please make use of MathJax $\endgroup$
    – Mayank M.
    Commented Nov 7, 2016 at 18:21
  • $\begingroup$ $\frac{1}{2}<\frac{130}{232}=\frac{65}{116}$ $\endgroup$ Commented Nov 7, 2016 at 18:32
  • $\begingroup$ @above, I have made the corrections. thanks! but how to find that by polynominal equations? & Mayank, kindly tell how to! $\endgroup$ Commented Nov 7, 2016 at 18:39
  • $\begingroup$ I'm fairly certain that it cannot be solved using polynomials. It should be doable with calculus, though $\endgroup$ Commented Nov 7, 2016 at 19:47

1 Answer 1


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 $:)$.

  • $\begingroup$ Your value is actually the value suggested by the intermediate value theorem. The OP's value of $\frac{65}{116}$ Intersects the circle four times, and $\frac{1}{\sqrt{3}}$ only intersects twice. So any value in between your $h$ and the other (yet to be determined) value would intersect the circle four times $\endgroup$ Commented Nov 7, 2016 at 20:06
  • $\begingroup$ @teadawg1337 Oh, I actually misunderstood the question I think, despite talkign about it in chat. It appears OP is looking for a value that minimises the distance of the curves (wrt to some measure like $\sup_{x\in\gamma_1,y\in\gamma_2}d(x,y)$)? $\endgroup$
    – s.harp
    Commented Nov 7, 2016 at 20:10
  • $\begingroup$ No, I believe the OP is asking for values of $n$ such that the two curves intersect more than twice. In fact, this may be the only solution. WolframAlpha appears to return four intersections for any value greater than 1/2. So the solution set may be $$\frac{1}{2}<n\leq \frac{\ln(2)}{\ln(2+\sqrt{2})}$$ I have no idea how to approach it, though. $\endgroup$ Commented Nov 7, 2016 at 20:19
  • $\begingroup$ The user s.harp has surely got it right, as at that value the curve touches the circle and at (0,1) & (1,0) $\endgroup$ Commented Nov 8, 2016 at 18:07
  • $\begingroup$ The user s.harp has surely got it right, as at that value the curve touches the circle and at (0,1) & (1,0) but the most suitable value where the curve should resemble that circle should be at that value of 'n' where the area bounded by the curve and the circle both becomes same or equal to (1 - pi/4) or the value of 'n' at which the curve x^(1/n) +y^(1/n) = 1 bounds the area (1 - pi/4) with x-axis and this 'n' is just greater than ln(2)/(ln(2+sqrt2). I too reached up til here but can not find this 'n' here afterwards, can somebody please find it $\endgroup$ Commented Nov 8, 2016 at 18:13

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