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Proposed ProblemSee also: Complete Problem 286, Collection of Geometry ProblemsLevel: High School, SAT Prep, College geometry
One may do this by using Menelaus's Theorem or by analytical geom. Following the latter, let A be (0,0)and name the polygon vertex near H as P and then go clockwise with the vertices as Q,R,S,T,U,V & W. Then B:(0,a),C:(a,a) and D:(a,0).Now AG is y=x/2 and DE is x/a+2y/a=1. Thus, P is (a/2,a/4). Likewise, BH is 2x/a+y/a=1. Hence, Q is (a/3,a/3) and PQ^2 =(a/2-a/3)^2+(a/4-a/3)^2 or PQ =V5*a/12. Likewise, one may show QR=RS=ST=TU=UV=VW=WP=V5*a/12. Hence PQRSTUVW is a regular polygon of side = V5*a/12Ajit: firstname.lastname@example.org
Sorry, the octagon can't be regular (equilateral) because if it would be, the points A,H,D,G,C,F,B,E of the intersections would lay on a circle!but it is approximative an equilateral octagon.sincerelygwengler