Wikipedia:Reference desk/Archives/Mathematics/2014 September 14

= September 14 =

Star angle
Imagine pointed stars made to match the notation used at Schläfli symbol.

For example the pentagram is {5,2}.

How do I determine the angle within the points from this pair of numbers? How do I determine the exterior angle where lines from two points meet?

Thanks -- SGBailey (talk) 19:53, 14 September 2014 (UTC)
 * I think the notation for the pentagram is actually (5/2). The angle for the star polygon with symbol (n/k) is π(1-2k/n) radians. So for example the angle for the pentagram is π(1-4/5)=π/5=36°. --RDBury (talk) 00:53, 15 September 2014 (UTC)


 * Both the pentagram and the Schlafli article use {} not . -- SGBailey (talk) 06:40, 15 September 2014 (UTC)
 * Yes, but it contains one rational number $$\frac{5}{2}$$, not two integers separated by a comma. —Tamfang (talk) 06:53, 15 September 2014 (UTC)
 * Oh yes, so it does. -- SGBailey (talk) 14:41, 16 September 2014 (UTC)


 * And the sum of internal angles for a star polygon with symbol {n/k} is 2 &pi; (n - 1). From this and the convex internal angles it is easy to see that the concave internal angles are &pi; (1 + 2 (k-1)/n), or equivalently, the corresponding external angles are &pi; (1 - 2 (k-1)/n). Icek (talk) 01:09, 15 September 2014 (UTC)


 * So for {5,2) in degrees we have 180 * ( 1 + 2 * (2-1) / 5) = 180 * 1.4. That has to be wrong. I'm currently thinking 180 * ( n - 2k ) / n -- SGBailey (talk) 06:40, 15 September 2014 (UTC)
 * No, it isn't wrong for the internal concave angles of the pentagram. Measure it. The regular pentagon has internal angles of 108 degrees. Think of the pentagram as a pentagon with 5 triangles attached; then from RDBury's formula for the outer angle of such a triangle (36 degrees), you can calculate that the sum of the other 2 angles of such a triangle must be 144 degrees, therefore one such angle is 72 degrees. The internal concave angle of the pentagram can be thought of as the sum of the internal angle of the pentagon and 2 angles of 72 degrees each from the adjacent triangles. That sums to 252 degrees, the same answer as from my formula. Icek (talk) 11:20, 15 September 2014 (UTC)


 * Your equation is wrong. The answer for the {5,2} pentagram is 36° (or equivalent in radians). Your equation doesn't give 36°. It may be the correct formula for some other aspect of the shape, but not for the angle of the points of the star. -- SGBailey (talk) 15:26, 15 September 2014 (UTC)


 * RDBury's formula is perfectly adequate for that angle. I only added the formula for the concave angles, and its complement, and the latter would be your "exterior angle where lines from two points meet" (but only if the two points are adjacent): &pi; (1 - 2 (k-1)/n). Icek (talk) 01:30, 16 September 2014 (UTC)


 * I don't know where Icek's formula comes from. The sum of internal angles of a convex polygon is (n-2)π, as you can see by decomposing it into triangles, not 2(n-1)π.
 * The vertex of a regular {x} polygon, whether x is integer or not (convex or star polygon respectively), represents a turn of 2π/x, and the internal angle is π minus this, whence RDBury's formula π(1-2/x).
 * Another way to look at it: consider the triangle formed by the center and one edge of the star. The angle at the center is 2π/x, so the sum of the other two angles – each of which is half of the interior angle at a vertex – is π-2π/x. —Tamfang (talk) 07:05, 15 September 2014 (UTC)


 * The sum of the internal angles of a star-shaped polygon with n corners is (n-2)&pi; as well: You can consider the triangles formed by the center and one edge of the star, and there are n of them. The sum of angles of these triangles is n &pi;. Subtract the angles at the center (which sum to 2&pi;), and you are left with the sum of the internal angles of the star-shaped polygon.
 * But the star-shaped polygon with symbol {n/k} should have 2n corners, as long as k &gt; 1 and the fraction n/k is irreducible, hence my formula valid only for these special cases: 2&pi; (n-1).
 * The rest follows from subtracting the n pointed angles and then dividing the rest by n, because there are n concave angles. Icek (talk) 11:20, 15 September 2014 (UTC)