Christmas Geometry

[mmcc79]

So I was going to string up some lights outside and came up with an interesting Geometry problem that I was sure would be posted somewhere, but I could not find the answer. So my mind wandered for an answer and would like to see if others have thought of this...

I would like to suspend a string of lights between two trees. I'm going to make the shape of a star. So two strings of equal length would make the top and bottom sides of a rectangle. Then two strings going from top string to bottom string would make the sides of the rectangle.

Now if my star was a perfect star (all sides the same length) there should be a relationship between the X and Y of the rectangle. (see picture - mind you my drawing is not perfect)

http://i18.photobucket.com/albums/b110/mmcc79/star.jpg

Now alot can be said of a perfect star.
It fits in a perfect pentagram.
It contains a perfect pentagram.
The value of Phi helps alot!
Lets call the width of the rectangle X.
Lets call the height of the rectangle Y.
Lets call the length of the inner pentagram side b.
Lets call the length from a star point to the inner pentagram a.

The lenght of the top (and bottom) of the rectangle is a breeze. Its the same as the length two points in the star. Its simply X = a + b + a.

The tricky part is finding the height of the rectangle. Y = ....

any help?

[OfficeShredder]

Two ways immediately that I see. One is to find the length of the outside edges of the pentagon (I don't see how to do that off the top of my head, but I bet a bit of geometry yields the answer). Then draw a line from the midpoint of the top of the rectangle to the midpoint of the bottom. That line, along with the two lines of the star going from top to bottom, form right triangles (one on each side of your dividing line). So you have the length a+b+a of the hypotenuse, and (somehow) found the length of one of the sides (half the base of the large pentagon), so the height is yielded through Pythagoras.

The other way is trig. The angle between those two lines going from the midpoint of the top of the rectangle to the base is easy to find (since you can get the angles in the pentagon, then the angles on the base of the little triangle formed), then you draw that same line from the midpoint, and use a trig function to calculate Y

[Epsilon=One]

...if my star was a perfect star (all sides the same length) there should be a relationship between the X and Y of the rectangle. (see picture - mind you my drawing is not perfect)

...alot can be said of a perfect star.
It fits in a perfect pentagram.
It contains a perfect pentagram.

...The lenght of the top (and bottom) of the rectangle is a breeze. Its the same as the length two points in the star. Its simply X = a + b + a.

The tricky part is finding the height of the rectangle. Y = ....

any help?For some hints and a lot more information concerning Pentagrams and Phi (the Golden Ratio), click this link. (http://www.contracosta.edu/math/pentagrm.htm)

[Epsilon=One]

For some hints and a lot more information concerning Pentagrams and Phi (the Golden Ratio), click this link. (http://www.contracosta.edu/math/pentagrm.htm)I expected a solution by now that didn't require trigonometry.

How ‘bout:When the side of a regular pentagram is one, the diameter (distance between points), the “arm” of the star, is Phi, "Φ"; the proof of this is per above link.Pythagoras gives: Y = Square Root (Φ² - ¼);
which = Square Root (Φ + ¾);
and X:Y = Φ:Square Root (Φ + ¾);
Y = X(Square Root (Φ + ¾)/Φ;
and, Φ = (Square Root (5) + 1)/2
Therefore, to six decimals: Y = 0.951057X

[thechuck]

guys this is soooo simple

ki5o4ik5o454q5

[HallsofIvy]

"guys this is soooo simple"

he says without contributing anything at all.

[mmcc79]

Thanks to all your input and some muddling myself I came up with the following solution (which is simple enough for me to understand)...

http://i18.photobucket.com/albums/b110/mmcc79/star-1.jpg

[Epsilon=One]

http://i18.photobucket.com/albums/b110/mmcc79/star-1.jpgYour "rounded" number answer is correct. However:Why have you selected the values b = a/Φ (0.618a); and,
X = aΦ² (2.618a) in the upper left figure?

What is your proof that a and b = a+b in the upper right figure?

[mmcc79]

Why have you selected the values b = a/Φ (0.618a); and,
X = aΦ² (2.618a) in the upper left figure?

What is your proof that a and b = a+b in the upper right figure?

I'm not a math scholar but I read a few math proofs on perfect stars inscribed in perfect pentagons. I learned a bit about the Golden Ratio and the value of PHI. I'm sorry that I don't have those links onhand, but that is where I got those formulas from.

I don't think my formulas are inaccurate (other then rounding), and I know that you are not disagreeing or not understanding my logic, but rather you are challenging me to prove my answers more fully.

I appreciate the challenge but I feel that in the scope of my question I didn't need to prove those formulas, but rather trust in someone elses (which I did follow their proof). If someone else would like a crack at explaining the three assumptions that I made above, please feel free to do so.

[Epsilon=One]

I'm sorry that I don't have those links onhand, but that is where I got those formulas from.Go back to the hint that I gave in Post #3 for such a link.

...I know that you are not disagreeing or not understanding my logic, but rather you are challenging me to prove my answers more fully.Yes. I thought you still knew where they came from. I wanted others to know how you made the assumptions.

If someone else would like a crack at explaining the three assumptions that I made above, please feel free to do so.Assume an edge of the pentagram is one (your a + b). Then, because of equal angles (you might need a moment or a step or two for this) your "a" and "b" also equals one. Then, with a some thought, set up a ratio of similar sides, such that: 1/x = x/(x+1). Rewrite as: x² = x + 1 (or x² – x = 1) = Φ; thus, when an edge is one x = Φ = 1.618034...

See if you can take it from here.