[RESOLVED] Pentagon Geometry Question

I have a board that is W inches wide, and I wish to cut from it a regular pentagon (sides of equal length) that has an altitude of W inches (the distance from a point to an opposite flat side.

Measured in terms of the width, W:
(1) How long (L) should the board be to cut the regular pentagon from it?
(2) What will be the length of each side, S?
(3) Assuming I can increase L, is this the largest regular pentagon that I can cut from a rectangular board of width W? My gut feeling is yes, but please advise.

Cutting the pentagon up into 5 triangles makes the relations follow pretty quickly. Each is isosceles with large angle 360/5 = 72 degrees. Note that your side length S is related to the length R of the legs of these triangles with a relation you can find by making two right triangles out of one of your isosceles triangles. Also, W = A+R where A is the length of the apothem. As for L, that's just the length of the chord from one vertex to an opposite vertex, which can be found using basically the same trick. Perhaps someone else will fill in the details; if not I will later.

For your final question, if you could increase L to be infinitely large, does another rotation of the pentagon yield a larger pentagon? No; you've already got the best rotation. The arguments I can come up with rely on symmetry and continuity and aren't terribly nice to write down. Perhaps someone can give a better one.

As a sketch of mine, consider two opposite sides of the pentagon, which is inscribed in a rectangle with the initial orientation implied by width W = A+R, with one edge parallel to the base. Imagine extending the two edge segments in both directions, and keep track of the point of intersection. Initially the intersection lies on the rectangle's edge; rotate the pentagon until the intersection point is midway between the edges of the rectangle. The "height" of the rectangle is at a maximum or minimum here, since continuing to rotate will be the same as rotating back. This orientation is a maximum since the maximum distance from any point on the pentagon to any other point is that from one vertex to an opposite vertex, and the height is exactly that distance. The initial orientation similarly has to be a minimum; no max or min can occur somewhere during the rotation (provable using more technical analysis of the extreme points of the two sides in question), so you have the best orientation already, width wise.

Quote:

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Originally Posted by jemidiah

Editing. I divided wrong.

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Don't give up, J; I need your help. I'm going to be cutting a bunch of these on the table saw. ;)

I need to make them all the same size, as close as my skills allow. :)

J, wait a minute. I have the height (width of board) fixed as W and that is the altitude. I need to know the length of the rectangle (distance across a diagonal), L, based on the width, W.

That way I can cut a whole bunch of identical rectangles from a long board first and then cut the identical regular pentagons from the rectangles. I believe that would minimize stock waste and maximize the size of the pentagon.

Got it? :ehh:

(1) What is L in terms of W (the ratio of W to L or vice versa)? And, (2) what is S in terms of W?

I also just thought of something. I believe I can minimize waste even further by cutting trapezoids rather that rectangles:
http://i156.photobucket.com/albums/t.../Pentagons.gif

A good mathematician could tell me the % saved by doing this rather than starting with rectangles alone. However, I'm a lousy mathematician. :sick:

Cutting the pentagon into 5 triangles as I mentioned above, I derived that W = S*1/2*(cot(36 degrees)+csc(36 degrees)) ~= 1.53884177*S. That looks about right visually.

For L, I'm assuming you want the length of a board given your interlocking pentagon tiling? Or do you want it for a single pentagon?


Gah, the forums are excruciatingly slow. I wanted to mention that if you can reliably draw the setup you have in mind, you could measure the proportions directly. That is, because of scale independence, the ratio of L to W and of W to S are both just constant numbers, which you could measure given a sufficiently accurate model, without even using analytic methods.

Cutting the pentagon into 5 triangles as I mentioned above, I derived that W = S*1/2*(cot(36 degrees)+csc(36 degrees)) ~= 1.53884177*S. That looks about right visually.

For L, I'm assuming you want the length of a board given your interlocking pentagon tiling? Or do you want it for a single pentagon?

Quote:

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Originally Posted by jemidiah

Cutting the pentagon into 5 triangles as I mentioned above, I derived that W = S*1/2*(cot(36 degrees)+csc(36 degrees)) ~= 1.53884177*S. That looks about right visually.

For L, I'm assuming you want the length of a board given your interlocking pentagon tiling? Or do you want it for a single pentagon?

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Single board. However, that produces more waste than if the board is cut using the illlustration shown with the orientations of the pentagons flipped as you move along. It also requires less cutting.

I believe the diagonals of a pentagon are 1.05 * W and S = 0.65 * W, and that's close enough for this OP.

In the interest of serendipity, do you see the identical "waste" isosceles triangles in my illustration? Take any ten of those and join them pie style. Suddenly you have a regular decagon. The sharp angle is 36 degrees.

Gasp!

S = 0.65*W agrees with W = 1.54*S from my previous post, cool. Computing the length of the interlocking pattern is a bit more complex. Number vertices on each pentagon from 1 to 5 starting at the unique vertex touching the top or bottom, going clockwise.

Let D be the distance from pentagon 1, vertex 4 to pentagon 2, vertex 4; D = W / sin(72 deg).
Let q be the length of the overlap of two touching sides of adjacent pentagons--that is, the distance from pentagon 1, vertex 5 to pentagon 2, vertex 5; 2*S-q = D so q = 2*S-D.
Let x be the horizontal distance from pentagon 1, vertex 5 to pentagon 2, vertex 5; x = q*cos(72 deg).
Let m be the distance from vertex 2 to vertex 4; from the law of cosines, m=S*Sqrt(2*(1-cos(108 deg))).
Let n be the number of pentagons in your tiling.

Start with one pentagon; it will have length m. Add a pentagon; it will add (m-x) length. Inductively, we then have L = (n-1)(m-x) + m.

Back substitute to find L in terms of W and S if you wish :).

Thanks, J. I followed your recommendations and cut 10 regular pentagons from one board using my table saw. The sides are exactly 5" long, and I ripped the width of the board to exactly 7.7" wide before I started. Looks very good. All of these pieces are identical in size.

I am very proud of your advice and proud of my workmanship. :thumb:

Glad it worked out :)