

Subject:
Isoceles Triangle / Pentagram Measurements
Category: Science > Math Asked by: wholzemga List Price: $2.00 
Posted:
03 Aug 2004 00:09 PDT
Expires: 02 Sep 2004 00:09 PDT Question ID: 382812 
I have a string of Christmas lights that is 33feet long. I want to form it into a 5pointed star, only along the parimeter of the star. So each side of the star will be 1/10 of 33feet, or about 39 1/2 inches. But I want to build the star with just 5 pieces of lumber. So these pieces of lumber need to be 79 inches, plus the distance between two ajoining internal points of the star. Or the same as the parimeter of an isosceles triangle with angles of 727236, and the two equal sides of 39 1/2 inches. What is the missing measurement? How long does each piece of lumber need to be?  


Subject:
Re: Isoceles Triangle / Pentagram Measurements
Answered By: mathtalkga on 03 Aug 2004 10:12 PDT Rated: 
Hi, wholzemga: As you found, the exterior segments of a "regular" (equalsided) pentacle (fivepointed star) will each be one tenth of its perimeter, so with a string of lights of length: 33 feet = 396 inches each of those ten lengths would be 39.6 inches. Now the "bases" of the five isoceles triangles that form the points can be found by trigonometry. As you already know, these triangles have two equal base angles of 72 degrees and a third acute "apex" angle of 36 degrees. Dropping a perpendicular bisector from the apex (which is also an angle bisector) divides the isoceles triangle into two congruent right triangles. The ratio of the base of one of these new right triangles (half the base of the isoceles triangle) to its hypotenuse (one of the "long" sides of the isoceles triangle) is: sin(18 deg) = cos(72 deg) So the length of the base of an original isoceles triangle is: 2 * cos(18 deg) * 39.6 = 24.474... inches or roughly 2 feet and onehalf inch. Adding 2 * 39.6 = 79.2 inches to this would give 103.674... inches, or very nearly: 8 feet 7 & 5/8 inches for the pointtopoint chords of a regular pentacle inscribed in a circle. There are several ways to construct a regular pentagon (equiv. a regular pentacle) with compass and straightedge. See here: [The Pentagon] http://hept.port5.com/pentagon.htm regards, mathtalkga 
wholzemga
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Thanks! 

Subject:
Re: Isoceles Triangle / Pentagram Measurements
From: mathtalkga on 03 Aug 2004 11:24 PDT 
It would be a bit tricky to do unless you have a (tablemounted) router, but two notches could be cut halfway into the lumber (on alternating sides) at the places where the boards must cross at the interior angles. These notches would make 72 degree angles with the edge of the wood, and their width (measured perpendicular to the edge of the cuts) would be that of the "other" piece of lumber with the matching notch (assuming equal widths, the notches would be rhombus shaped). Where precisely to locate these notches depends on whether the "outer" point are supposed to line up on the centerlines of the boards or along an edge (probably an outer edge, so that the points of the star are most distinct). regards, mathtalkga 
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