Online Geometry theorems, problems, solutions, and related topics.
See complete Problem 63Regular Heptagon, side and diagonals. Level: High School, SAT Prep, College geometryPost your solutions or ideas in the comments.
Let AB = CD = DE = s, AC = CE = c, AD = AE = d.Applying Ptolemy's Theorem to cyclic quadrilateral ACDE, we obtain cs + ds = cd.Then s(c + d) = cd, and so 1/s = (c + d)/cd = 1/d + 1/c.Therefore, in regular heptagon ABCDEFG, 1/s = 1/c + 1/d
Problem 63 solution. This solution was submitted by Dan Suttin from San Antonio, TX Thanks Dan.
Apply inversion on A with any inversion radius (A -> A' ; B -> B' ...etc)Then we can rewrite the statement by proving A'B'=A'C'+A'D'.The inversion image is easy : AB,AC,AD remains unchanged while the circumcircle becomes a line B'C'D' not passing through A.By the conformal properties, ∠BAC=∠B'A'C'=∠C'A'D'=∠A'B'C'=180/7Construct a point E' on A'B' such that A'E' = A'C', so we simply need to prove E'B'=A'D'.This is actually quite obvious. A'C'=B'C'∠E'B'C'=∠C'A'D'=180/7∠B'E'C'=720/7=∠B'D'A'So ΔB'C'E'=ΔA'C'D', and hence E'B'=A'DQ.E.D.