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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.

George carlin
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3 Answers3

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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.

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

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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.

Vasily Mitch
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  • The polynomial equation you give has two complex roots and two real roots $\approx 0.9867$ and $\approx 0.0432$. It is not the smallest root but the biggest one which is candidate. But visibly the answer should be (slightly) bigger than $1$. – Jean Marie Sep 13 '19 at 11:01
  • The radius of the largest circle is $1$, of course =) – Vasily Mitch Sep 13 '19 at 11:03
  • @VasilyMitch that is really detailed explanation , thanks a lot – George carlin Sep 13 '19 at 11:59
  • @VasilyMitch if i draw a circle in place of the square, and since we know the inradius of the square, we can easily find the area of square, instead of the biquadratic ? m i correct ? – George carlin Sep 16 '19 at 11:21
  • I don't understand you. Can you provide a drawing? – Vasily Mitch Sep 16 '19 at 11:30
  • @VasilyMitch i will try to provide a drawing, i am saying that we can draw a circle tangent to three given circles and we may find its radius, which will be circumcircle to given square possibly,,,,,,and i have one more doubt,,,,,it seems like problem of intersecting circles rather than ellipses to me, contrary to your explanation – George carlin Sep 16 '19 at 11:54
  • @VasilyMitch i have provided in my question the equations that i have got while solving it by coordinates, but the numbers of equations are inadequate – George carlin Sep 16 '19 at 13:16
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    The square inscribed to the inscribed circle won't be touching any of the circles defined in the problem. To your doubt, if the center of polygon is at point $(d,d)$ and has the vertices on the distance $r$ from the center, the equations are of the form: $(d+\alpha r)^2+(d+\beta r)^2=1$ and $(d+\gamma r-1/2)^2+(d+\delta r)^2=1/4$ which are the equations of ellipses in coordinates $(d,r)$ – Vasily Mitch Sep 16 '19 at 13:18
  • @VasilyMitch in your equation, the value of x are .9 and .04 ,which can not be the area of square not can it be the side of the square,,,plz clarify – George carlin Sep 19 '19 at 09:28
  • As I said early, I consider the radius 7 cm to be 1. – Vasily Mitch Sep 19 '19 at 14:55
  • @VasilyMitch see the other solution posted here by donald, i am getting a different answer by his equation. – George carlin Sep 20 '19 at 11:21
  • He got the same answer. His area is $1.455^2=2.18$ which is exactly 49 times larger my 0.0432. – Vasily Mitch Sep 20 '19 at 12:18
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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 $1.455$.

Community
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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.

Shiva Prakash MV
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