Circles and squares and polygon

Question

If i replace the square with circle, i can very easily find the area of circle and get some approximation for the area of the square.
Even the square is doable, but i was thinking that can we find the area of the regular polygon which is made to sit in place of the square ?
Since, i don't have the answer to that problem, i don't know if approximating that with circle would work.
Would a regular polygon of arbitrarily large number of sides fit into that space and can we find its area (i know only upto undergrad. maths)?
I am trying to solve it by coordinate geometry but i am not getting as many equations as many variables i have.
$a^2$+b(b-7)=49=$d^2$+c(c-7)=($e^2$+$f^2$)=($g^2$+$h^2$)=(f-b)(g-c)49/(e-a)(h-d)=$(f-b)^2$+$(e-a)^2$+49-$(g-c)^2$-$(h-d)^2$=$(a-c)^2$+$(b-d)^2$+49-$(e-g)^2$-$(f-h)^2$ Thank you.

Answer 1

A solution for a regular polygon with one edge a chord of the large circle and the vertices of the opposite edge on the smaller circles

Consider coordinates with $O$ as origin and the line of symmetry as the $x$-axis. Let the edge length of the regular polygon be $2L$ and let the distance between opposite edges be $D$.

$D$ is $2L$ for a square, $2\sqrt 3L$ for a hexagon and $(2+2\sqrt 2)L$ for an octagon.

The large circle has equation $x^2+y^2=49$.

The upper small circle has equation $x^2-\frac{7}{\sqrt 2}x+y^2-\frac{7}{\sqrt 2}y=0$.

The line $y=L$ intersects the upper circle at $(x,L)$ and the large circle at $(x+D,L)$. Eliminating $x$ from the two equations for these points gives us a polynomial equation for $L$ in terms of $D$.

(1) $x^2+2Dx+D^2+L^2=49$.

(2) $x^2-\frac{7}{\sqrt 2}x+L^2-\frac{7}{\sqrt 2}L=0$.

First eliminate $x^2$.

$(7+2\sqrt 2 D)x=49\sqrt 2-\sqrt 2 D^2-7L$

Substituting into Equation 1 then gives us the required polynomial. We shall now do this for the square.

$(7+4\sqrt 2 L)x=49\sqrt 2-7L-4\sqrt 2 L^2$

$(7+4\sqrt 2 L)(x+D)=4\sqrt 2 L^2+7L+49\sqrt 2$

$(4\sqrt 2 L^2+7L+49\sqrt 2)^2=(49-L^2)(4\sqrt 2 L+7)^2$

$64L^4+112\sqrt 2 L^3-686L^2-2058\sqrt 2 L+2401=0$

The relevant solution is $L=0.7274$ and so the length of a side is approximately $1.455$.

Answer 2

I would argue that even with the square it is a very tough problem. At some point, you have to solve either quartic equation or the intersecting ellipses problem (which is again boils down to quartic equation). And the final answer is the smallest real root of $64x^4-144x^3+177x^2-100x+4=0$. You can express the root with radicals, but it's nasty as hell.

Good news is the problem with $n=6$ or $n=8$ isn't harder, since you get the same quadratic equation just with other coefficients.

True problem will appear at $n=10$, since the inscribed dodecagon touch the inner circles with its sides rather than vertices.

Although the calculation of this scenario isn't harder per se, it makes the general formula totally uncomprehensible, since you need to determine first whether the n-gon touches inner circles with sides or vertices and which sides vertices and then use the corresponding equation.

This problem is not that interesting to go through so much hassle.

Answer 3

The intersection of the two small circles is 7/2,7/2.

With the transformation for the equation of circles at the point of intersection, we have

(x-7/2)^2+(y-7/2)^2=7^2 (1) big quadrant

(x-7/2)^2+y^2=(7/2)^2 (2) semicircle with horizontal diameter

x^2+(y-7/2)^2=(7/2)^2 (3) semicircle with vertical diameter Now, the square has sides inclined at 45 degrees with some intercept.

Let c be the intercept.

Equation of straight line with slope as 1, and the intercept c is given by

y=x+c which is away from the line passing through the origin with unit slope equal to c/sqrt(2).

With points connecting on either side of the line through the origin, the side of the square is csqrt(2).

We need to solve for the intersesction of the straight line y=x+c with the big quadrant.

Followed by the evaluation of the intersection point of the straight line y=x+c and semicircle with vertical diameter.

It is a quadratic equation in each case.

Considering the positive values, we can equate the distance to csqrt(2) to get the dimensions of the expected square.