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Q: Isoceles Triangle / Pentagram Measurements ( Answered ,   1 Comment )
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 Subject: Isoceles Triangle / Pentagram Measurements Category: Science > Math Asked by: wholzem-ga 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 33-feet long. I want to form it into a 5-pointed star, only along the parimeter of the star. So each side of the star will be 1/10 of 33-feet, 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 72-72-36, 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?``` Request for Question Clarification by mathtalk-ga on 03 Aug 2004 08:14 PDT ```Hi, wholzem-ga: I understand the lengths that you are talking about, which are chords of a circle passing through the five "outer" points of the star (pentacle). However, have you considered that the five pieces of lumber will tend to intersect one another? If they all lay flat in a plane, then they would intersect, just as the chords of a circle must. And, because there are an odd number of lengths here, we cannot simply separate them into "top" and "bottom" lines (as one might with a six-pointed star, ie. two overlaid equilateral triangles). You might arrange them in a "slanting" overlay, in which each board is "top" at one end and "bottom" at the other end. Such a slant will reduce slightly the effective length of the boards, probably not enough to be concerned about here if the thickness of the lumber is 2 inches (and as you already know, their lengths are in excess of 6 feet). regards, mathtalk-ga```
 ```Hi, wholzem-ga: As you found, the exterior segments of a "regular" (equal-sided) pentacle (five-pointed 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 one-half 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 point-to-point 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 straight-edge. See here: [The Pentagon] http://hept.port5.com/pentagon.htm regards, mathtalk-ga```
 wholzem-ga rated this answer: `Thanks!`
 ```It would be a bit tricky to do unless you have a (table-mounted) 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, mathtalk-ga```