Squaring Any Polygon!\Large{\purple{Squaring}\space \blue{Any}\space \red{Polygon!}}

The Problem\large{\green{{\underline{The\space Problem}}}} Given a polygon A1A2A3...An we need to construct a square PQRS such thatGiven\space a \space polygon \space A_1A_2A_3...A_n \space we \space need \space to \space construct \space a \space square\space PQRS \space such \space that ar(A1A2A3...An)=ar(PQRS)\boxed{ar(A_1A_2A_3...A_n)=ar(PQRS)} Using Straight  edge and compass constructionUsing\space Straight \space - \space edge \space and \space compass \space construction


The Solution\large{\purple{\underline{The\space Solution}}}

The problem may seem very hard but it isn't really is.

The idea is that we can break any polygon A1A2A3...AnA_1A_2A_3...A_n into n2n-2 triangles, these triangles will be of the form A1AiAi+1,iN(1,n)\triangle A_1A_{i}A_{i+1},i\in\mathbb{N}\cap (1,n)

We can then find square PiQiRiSiP_iQ_iR_iS_i such that ar(A1AiAi+1)=ar(PiQiRiSi),iN(1,n)ar(\triangle A_1A_iA_{i+1})=ar(P_iQ_iR_iS_i),i\in\mathbb{N}\cap (1,n)

And then we can find a square PQRSPQRS such that ar(PQRS)=i=2n1ar(PiQiRiSi)ar(PQRS)=\sum_{i=2}^{n-1}ar(P_iQ_iR_iS_i)

And if we are able two add at least 2 square then we can find the sum.

So we are first going to break the problem in two parts :

  • Squaring A Triangle

  • Adding Two Squares


Squaring A Triangle\large{\underline{\blue{Squaring\space A \space Triangle}}} Given ABC we need to find square PQRS such thatGiven \space \triangle ABC \space we \space need \space to \space find \space square \space PQRS \space such \space that ar(ABC)=ar(PQRS)ar(\triangle ABC)=ar(PQRS) Steps of constructionSteps \space of \space construction 1) Draw ADBC such that D lies either on BC or extended BC1) \space Draw \space \overline{AD}\perp\overline{BC} \space such \space that \space D \space lies \space either \space on \space \red{\overline{BC}} \space or \space \blue{extended \space \overline{BC}} 2) Draw a rectangle EFGH such that EF=AD,FG=BC22) \space Draw \space a \space rectangle \space EFGH \space such \space that \space \overline{EF}=\overline{AD},\overline{FG}=\dfrac{\overline{BC}}{2} 3) Draw a square PQRS such that ar(PQRS)=ar(EFGH)3) \space Draw \space a \space square \space PQRS \space such \space that \space ar(PQRS)=ar(EFGH)


Adding Two Squares\large{\underline{\red{Adding\space Two\space Squares}}} Given square ABCD and EFGH we need to find square PQRS such thatGiven \space square \space ABCD \space and \space EFGH \space we \space need \space to \space find \space square \space PQRS \space such \space that ar(PQRS)=ar(ABCD)+ar(EFGH)ar(PQRS)=ar(ABCD)+ar(EFGH) Steps of contructionSteps \space of \space contruction 1) Draw  KLM such that KLM=90°,KL=AB,LM=EF1) \space Draw \space \triangle \space KLM \space such \space that \space \angle KLM=90\degree,\overline{KL}=\overline{AB},\overline{LM}=\overline{EF} 2) Draw square PQRS such that PQ=KM2) \space Draw \space square \space PQRS \space such \space that \space \overline{PQ}=\overline{KM}

So we are done!


NOTE\large{\underline{NOTE}}

  • ar(X)ar(X) denotes the area of planar figure XX

  • If anybody need justification of any of the construction he or she may ask. (Although they are much easier to be justified)

  • The number of steps for squaring an nsidedn-sided polygon shall be increasing linearly if we used these steps, if anybody can do this in less amount of steps then please comment.

  • If anybody can prepare an animation of this either in Desmos or in .gif format etc, please share it here.

Note by Zakir Husain
3 months, 1 week ago

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Talking about a convex nn-gon, we can start successively reducing the number of sides one at a time, keeping the area unchanged, until we come up with a triangle with the same area as the initial polygon. The rest has already been described by @Zakir Husain.

Here is the course of action (referring to the figure):

Form An{{A}_{n}} draw a line parallel to A1An1{{A}_{1}}{{A}_{n-1}} wich intersects An1An2 {{A}_{n-1}}{{A}_{n-2}} at point An1{{{A}'}_{n-1}}. A1An1An1\triangle {{A}_{1}}{{{A}'}_{n-1}}{{A}_{n-1}} and A1AnAn1\triangle {{A}_{1}}{{A}_{n}}{{A}_{n-1}} have the same base A1An1A_{1} A_{n-1} and the same corresponding height, therefore the have the same area. Thus, polygon A1A2An2An1An{{A}_{1}}{{A}_{2}}\ldots {{A}_{n-2}}{{A}_{n-1}}{{A}_{n}} and A1A2An2An1{{A}_{1}}{{A}_{2}}\ldots {{A}_{n-2}}{{{A}'}_{n-1}} also have the same area, but the new polygon has n1n-1 sides. By repeating this procedure, we can transform the nn-gon to a triangle with the same area.

Thanos Petropoulos - 1 month, 2 weeks ago

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Great Idea

Zakir Husain - 1 month, 2 weeks ago

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@Mahdi Raza Can you make an animation on this?

Zakir Husain - 3 months, 1 week ago

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To square a triangle, how do you move from step 2 to step 3?

Mahdi Raza - 3 months, 1 week ago

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What did you mean by move ?

Zakir Husain - 3 months, 1 week ago

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@Zakir Husain I meant, how to go from step 2 of construction to step 3. In step 2, we have a rectangle and in step 3 we have a square with the same area. So how do you go from step 2 to step 3?

Mahdi Raza - 3 months, 1 week ago

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@Mahdi Raza See You basically made a rectangle ABCD in step 2. Now there is only one problem, we need to find a square with that area (area of triangle) not a rectangle. Now from my previous note we know that we can just square the rectangle. Then we will find a square PQRS with the same area as the rectangle, but since the area of the rectangle ABCD is equals to the area of the triangle. Therefore the areas of the triangle, the rectangle, the square are the same.

Therefore in summary we are able to find the square PQRS with the same area as the triangle.

Note :

  • The construction of the rectangle here is really just a transition period which will be useful only for some time, after we find the square there is no need of the rectangle so we can just remove it from the paper it will not make any changes.

Zakir Husain - 3 months, 1 week ago

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@Zakir Husain Oh, I see, I didn't know you had posted a note before this one. I'll try for an animation. Look interesting

Mahdi Raza - 3 months, 1 week ago

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My final question, sir, is how do I square a circle?

Jeff Giff - 3 months, 1 week ago

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Well it have been proved that we can't square a circle.

In 1837, Pierre Wantzel showed that any number xx that can be constructed using straightedge and compass have to be an algebraic number.

In 1882 Ferdinand von Lindemann proved the transcendence of ππ hence theoretically we can never construct π units\pi\space units, But rather we can construct some of it's approximations

Now your question gave me a new Theorem

"Suppose S is the set of all constructible approximations of π,f:SR,f(x):=πx  x1S  x2S:0<f(x2)<f(x1)"\boxed{"Suppose \space S \space is \space the \space set \space of \space all \space constructible \space approximations \space of \space \pi,f:S\to\mathbb{R},f(x):=|\pi-x|\space \Rightarrow\space \forall x_1\in{S} \space \exists\space x_2\in S : \red{0<f(x_2)<f(x_1)}"}

Zakir Husain - 3 months, 1 week ago

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Or how to construct a line that is equal to π\sqrt{\pi}?

Jeff Giff - 3 months, 1 week ago

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