A Menelaus-Type Theorem for the Pentagram

81.34 Tangential properties of pentagrams

“…Results comparable to theorem 1 involving products and quotients of sides of a pentagram were published in [1]. Using figure 1 but without the circle we have:…”

89.49 A Ceva-type theorem for the cyclic quadrilateral

“…One of the surprising results in geometry is that you can take a product of ratios of segments traversing a non-regular polygon and get a product of 1. Ceva's theorem is the most widely acclaimed result of this nature, but there are others such as [1], [2], and [3]. In this note we give two proofs of an unknown, or not widely known, similar result.…”

“…This result is well known, and is easily proved by adding the volumes of the tetrahedra OBCD, OACD, OABD and OABC (cf. [2]). Our reason for giving a somewhat longer proof is that the notion of volume comes quite late in the scheme of formal Euclidean geomerty.…”

90.47 A cyclic property of equilateral polygons

“…Since BDI DC = rfT 1 1 ry~l = ylfi etc., where r denotes the inradius, the points D, E, F, where the incircle touches the sides of the triangle ABC, may be assigned relative areal coordinates D(0, fi, y), E(a, 0, y), F(a, P, 0). Thus the Gergonne point G(a, p, y) lies on AD, BE, DF.…”

“…-bf = ai(bi + i+m i+ i) for AA,A, + iA, + 2 and s 2 -a 2 = £,(m,+ a,_i) for AA, ; _ iAiA,• +1 for i = 1, 2, ... , 5 (modulo 5). The quotient of these can then be written ass 2 -bf = ai(b i + l + m i + l ) s 2 -a} bi(mi + a,_i)From[1] and the generalisation for all n in[6] we get the cyclic product of ratios fli a 2 a 3 a A a5 , , a t + m 2 at, + m 5 a 2 + m 3 a 5 + w. a 3 + m A = 1 and = 1 b\ b 2 &3 b A b 5 m?, + b 3 m\+ b\ m^ + b 4 m 2 + b 2 m$ + b 5 for an arbitrary pentagram (see Figure 2). A* s Ai FIGURE Therefore using the reciprocals of the second formula we have ai a 2 a 3 a 4 a 5 \/w 3 + b 3 m\ + b\ m 4 + b 4 m 2 + b 2 m$ + b 5 \ bi b 2 b-x, b 4 b 5 j\ai + m 2 a 4 + m$ a 2 + m 3 a 5 + mi a 3 + m 4 j a l (m 2 + b 2 ) a 2 (m 3 + b 3 ) a 3 (m 4 + b 4 ) a 4 (m 5 + b 5 ) a 5 (m 1 + b x ) b\ (05 + m\) b 2 (a\ + m 2 ) b 3 (a 2 + m 3 ) b 4 (a 3 + m 4 ) b$ (a 4 + W5) s -a 2 s -aj s -a 4 s -ai ~ s 2 -b\s 2 -bis 1 -bys 2 -bis 1 -bj'…”