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# Reconstructing a Pentagon

## Question

A pentagon and its mid points are drawn. The original pentagon is then erased, leaving just its mid points visible. Is it possible to reconstruct the pentagon?

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I don't know why, but I apologize if I did anything wrong.

Let me try to reword my previous answer in a more familiar context.

Instead of a pentagon, consider a hexagon. A hexagon can be subdivided into four triangles: one triangle in the "center", and three "outer" triangles*. Each of these outer triangles share one side with the center triangle and two sides with the hexagon. Basic geometry tells us that the side shared with the center triangle is exactly double in length and equal in pitch to the line formed by connecting the midpoints of the two sides shared with the hexagon. Thus, it is possible to construct the shape of the center triangle given only the midpoints - but only if the three pairs of midpoints give us lines of length and pitch that form a closed triangle.

Now, since we are actually dealing with a pentagon, we have five midpoints. However, we can gladly oblige and place an imaginary sixth midpoint in such a way that a closed triangle can be made. Then, we can recreate the shape of the center triangle. Since our triangle doesn't have a fixed position, we can put it anywhere. It doesn't matter where; any valid center triangle and six midpoints always define a unique hexagon**. Here's an idea: we can fix the triangle's position by fixing one of its vertices - say, directly on top of our imaginary midpoint. This would reduce the length of one of the hexagon's sides to zero, effectively eliminating it, which gives us a pentagon instead.

All that's left is to cross a few dots with a straightedge and - boom. We have our pentagon.

If you look at my previous solution, you can find proof of its uniqueness, a more generic explanation that extends to n-gons, and other methods of reconstruction.

*(quotations are used because these names can be misleading in the case of a non-convex hexagon)

**(in the same way that a base and two midpoints uniquely define a triangle)

Edited by gavinksong

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does this approach hold if the pentagon is not regular or convex?

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I think it is possible to reconstruct the pentagon (unless we can show that a solution is not unique), but I don't think TSLF's solution would work for all cases. Like BMAD said, what if the pentagon is not regular?

How does TSLF know the appropriate radius for the circles? Also, once one of the sides is drawn, the other four sides have already been determined. It is not necessary to use five circles.

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Notice that after choosing a vertex, we can flip it across the next midpoint to find the next vertex and so on. After flipping the point across five midpoints, the resulting point may or may not coincide with the original. If it does, then we have found a solution.

Secondly, notice that flipping across a midpoint maps the plane to itself. Namely, it flips the entire plane across the midpoint. A composition of an even number of such functions is equivalent to a shifting function, and an odd number is equivalent to another flipping function. A function that flips that plane always has one unique solution. Thus, there is always an unambiguous reconstruction of the pentagon.

That is enough to answer the OP, but I can also find an explicit method of reconstruction. Flipping across two points always results in a shift by double the vector going from the first point to the second point. A composition of two shifting functions is simply a shift by the sum of their vectors. Thus, flipping a vertex across the first four midpoints in sequence is always equal to a shift by an easily calculable vector. When the vertex is flipped across the final midpoint, it needs to return to its original position. If we draw the shifting vector such that its midpoint coincides with the fifth midpoint, then it constitutes a side of the reconstructed pentagon. The rest of the pentagon can be constructed unambiguously.

Edited by gavinksong
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Since flipping across all five midpoints in sequence is equivalent to flipping across the solution vertex, if you choose any abritrary point and flip it across the five midpoints in sequence, the solution vertex should be precisely midway between the original point and the resulting point.

Edited by gavinksong
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Interestingly, this also means that the midpoints of the sides of any quadrilateral are the vertices of a parallelogram. It also suggests that reconstruction of a triangle is always possible.

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I think it is possible to reconstruct the pentagon (unless we can show that a solution is not unique), but I don't think TSLF's solution would work for all cases. Like BMAD said, what if the pentagon is not regular?

How does TSLF know the appropriate radius for the circles? Also, once one of the sides is drawn, the other four sides have already been determined. It is not necessary to use five circles.

to answer the spoiler question, i believe the circle radius needs to be greater than (equal) 1/2 the distance between two points.

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I think it is possible to reconstruct the pentagon (unless we can show that a solution is not unique), but I don't think TSLF's solution would work for all cases. Like BMAD said, what if the pentagon is not regular?

How does TSLF know the appropriate radius for the circles? Also, once one of the sides is drawn, the other four sides have already been determined. It is not necessary to use five circles.

to answer the spoiler question, i believe the circle radius needs to be greater than (equal) 1/2 the distance between two points.

This is not true. You must mean that the sum of the radii of the two points in question needs to be larger than the distance between the points.

To know what the appropriate radii are is equivalent to knowing the length of each side. TSLF never showed how he found these lengths, yet his solution depends on knowing them.

If you don't believe me, imagine trying to reconstruct a triangle using TSLF's method. Unless you know beforehand what the side lengths are, you will probably have to do a lot of trial and error to find radii that lead to a sensible reconstruction.

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Unless you know beforehand what the side lengths are, you will probably have to do a lot of trial and error to find radii that lead to a sensible reconstruction.

if only one radius, or set of radii, leads to a sensible reconstruction, one could model the reconstruction algebraically and solve for it/them. Then do the reconstruction. Or does the OP rule out math?
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Unless you know beforehand what the side lengths are, you will probably have to do a lot of trial and error to find radii that lead to a sensible reconstruction.

if only one radius, or set of radii, leads to a sensible reconstruction, one could model the reconstruction algebraically and solve for it/them. Then do the reconstruction. Or does the OP rule out math?

I think that's what I did above, except instead of solving for a radius, I solved for a vertex. However, I use "the math" mainly as a proof of concept for a method where a specific solution can be found out using a straightedge and a compass.

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Unless you know beforehand what the side lengths are, you will probably have to do a lot of trial and error to find radii that lead to a sensible reconstruction.

if only one radius, or set of radii, leads to a sensible reconstruction, one could model the reconstruction algebraically and solve for it/them. Then do the reconstruction. Or does the OP rule out math?

math is permitted

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I agree gavinksong.  Sorry for not committing before.  Initially I found your solution to be merely an alternative to the original.  But after consideration, I think yours is better.

I have switch to gavinksong's solution as it appears to be more robust.  I am open to arguments for/against either TSLF or Gavin's or another better option if anyone has interest.

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i dont   see my drawing is sufficient for all  pentagons ..just want to answer the  op.

Gavinsong got interesting  approach

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