I was given an extra credit problem* about Regular Pentagons. I need to know how to construct one using only a compass and a straight-edge. I would really appreciate it if someone can please explain it to me, or point towards a website that can. I’ve looked but apparently search engines are smarter than I am.
There’s also a second part of it. Using a regular pentagon prove that phi/1=phi. I don’t know anything about math, so I really have no idea about how to do this.
*[sub]This is not cheating. The Professor told us to ask people for help and to investigate online if we aren’t “math types”. The course is “Math in Art” so there are many of us who aren’t “math types”. [/sub]
A google search for pentagon +construct +compass yielded this, which should help you with the first part. It doesn’t explain how to find a perpendicular bisector, but all you need to do is set the point of your compass at an endpoint and open it to a width greater than half the length of the line and make a circle. Without changing the width, set the point at the other endpoint of the line and make another circle. Draw a line between the intersections of those circles, and hey, presto! Perpendicular bisector. As for proving the identity property of multiplication, I didn’t know that was possible using a pentagon. In fact, I thought that was axiomatic. Good luck with it, though.
Prove that phi/1 = phi using a regular pentagon, huh? Nope, sorry, no clue. It’s not axiomatic, but it sure requires something more sophisticated than geometry. If you want the best rough proof I know, either ask here or e-mail me.
There’s an applet off this page that will let you walk through constructing a pentagon (and other constructions). It’s a really neat page, and you might want to look at it.
You know what? You may have copied the second part of the problem incorrectly. Phi satisfies the equation 1/phi = 1 - phi, which you might be able to prove using a pentagon.
Sorry, make that 1/phi = phi - 1.
Ok, maybe I typed it wrong. I really don’t know a lot about it. He gave us this picture
http://www.geocities.com/pepperlandgirl4/pentagon.html
This page has the regular pentagon/golden ratio relationship:
http://www.cut-the-knot.com/pythagoras/pentagon.html
It’s in the paragraph next to figure 2. (Here they use g=phi).
By the way, here’s another interesting way to construct a regular pentagon. Take a strip of paper and make the “standard knot” in it (I don’t know what the knot is called (overhand, maybe?)–make a loop with the paper and stick an end through it, you know what I mean). Tighten up the knot, flatten it out, and you’ll get a regular pentagon.
Cabbage wrote:
Hey! I’d forgotten all about that! I wonder if it could possibly help in trying to fold a business card into a pentagon (using nothing but business cards - no rulers that can’t be made from B.C.s, definitely no scissors, tape, etc.)
I’ve figured out squares (easy enough) and equilateral triangles (a little bit tougher)… Both with “tabs” in the right places in order to lock mutliple business cards together to make four out of the five regular polyhedra (tetrahedron, cube, octahedron, and icosahedron).
All I’m missing is the dodecahedron, which I gave up on after having failed to find a way to fold a B.C. into a pentagon. But then later I realized that even if I could figure it out, they would fail to lock together to make a dodecahedron because the five sides would forbid a regular over-under-over-under “weave” between tabs and card edges. They’d have to be an over-over or an under-under, meaning one corner of each pentagon would be flapping about, and the thing would collapse.
Well, one might possibly do it if one could fold a B.C. into two pentagons which shared a side. Then there would be eight edges, hopefully only four with tabs. Six of these B.C.s might be coerced into forming a dodecahedron, but I’ve misplaced my D&D dice, and I’m having trouble wrapping my head around the problem now.
Oh, sorry for the hijack.
DaveW, you may find this page interesting. I didn’t read through it closely enough to know if it’s exactly what you’re looking for, but if not, at least it may be a start. (Doing a quick search for this, I ran into the name Valerie Vann several times, who wrote this article. If this page doesn’t have what you’re looking for, maybe searching on her name could turn something up):
I learned to construct a regular polygon this way. Take n as the number of sides(in this case five) Subtract 2, and multiply that figure by 180. Divide THIS figure by n. The resulting number is the angle that the regular polygon will have. So for a pentagon it’s {(5-2)*180}/5=108 So if you start with a line of any length, measure a 108 degree angle, another line, another 108 and so on, you will have a regular pentagon.
If it were a regular triangle it would be {(3-2)*108}/3=60 So the angles on a regular triangle are 60 degrees.
That’s interesting. It can’t work for every n, though; I believe it’s been shown impossible to construct a regular heptagon (7-sided figure, in case I named it incorrectly) with just a compass and a straightedge.
Actually, Baker’s method will work for every n. The problem is, it’s not a “compass and straight edge” construction. For Baker’s method, you need something to measure angles with.
A similar example of that type of problem is the trisection of an angle. A well known theorem is that you can’t trisect an angle with a compass and straight edge. Every once in a while, someone comes along claiming to have done it. Generally, what has happened in these cases is that they’ve discovered the fact that you can trisect an angle with a compass and a ruler (a straight edge with one unit lenth measured off on it).
I see your point about the heptagon when I work with it. Is is possible to have a fractional angle(don’t know what else to call it), rather than a whole angle? If it was the angle would be 128 4/7 degrees.
Wow, post a reply THEN realize I’d screwed up on a device to measure angles. You are so right Cabbage. Duh. Well, at least if I had a protractor the method would work.
I remember writing a basic program for X-cred to determine the same thing, back in the dark ages…
Wow, I got some great answers! Thanks so much. This is the main reason I love the SDMB, ask any question any time and within hours get some wonderful responses!
Thank you all again for your time. I really appreciate it.
Jeez, how’d I miss that? It was late at night, I guess.
For anyone who’s curious why we insist on using a straightedge and not a ruler…well, basically it’s Plato’s fault. He decreed that geometric constructions could only be done with a straightedge and compass, and it took till sometime in the 1800’s for somebody to consider constructions done with other instruments. FYI.
Cabbage wrote:
My god! Thank you very much for this and the link, Cabbage. I honestly never even thought to search the Web for “American Business Card Oragami.” Of course, I never thought of the stuff I’d done as anything more than trivia, but apparently some people have being working with business cards as high art. I am awed.
But no, that particular construction of a dodecahedron is not what I was thinking about. In every other figure I’ve made except the cube (er, ‘Mosley cube’), each pair of faces is made from one B.C., so the construction by Vann looks “ugly” to me (since each face appears to be two B.C.s), and my models don’t have any sort of internal structure supporting the model. Highly interesting, and I might have to just copy it for my little collection, but not quite what I was looking for.
I do now feel a need to figure out how to make a tetrahedron from one B.C. though. Mine takes two.
Thanks again.