Proposed Problem
See also: Complete Problem 286, Collection of Geometry Problems
Level: High School, SAT Prep, College geometry
Monday, April 27, 2009
Problem 286: Square, Midpoints, Octagon
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Proposed Problem
See also: Complete Problem 286, Collection of Geometry Problems
Level: 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/12
ReplyDeleteAjit: ajitathle@gmail.com
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!
ReplyDeletebut it is approximative an equilateral octagon.
sincerely
gwengler
It is an equilateral octagon because all the vertices lie on a line of symmetry for the given square. Using the horizontal symmetry line and vertical symmetry line only and facts of rotational symmetry, one can prove that we have 8 pairs of congruent triangles via one of the four commonly used triangle congruences. Via CPCTC, we can show that all sides of the octagon are congruent thus proving the figure equilateral. A similar series of arguments could made using the diagonal symmetry lines as starting points. Equilateral doesn't necessarily mean regular here.
ReplyDeleteFor the first part see congruent forms
ReplyDeletehttps://photos.app.goo.gl/JDrBZ1sPdG864iBE6
Let EG and HF meet at O. Now EG goes thro' 2 vertices of the octagon, name them U and V
ReplyDeleteOV = VG since Tr.s CVG & HOG are congruent. And so it can easily be proved that all the lines which connect O to the vertices of the octagon are of equal length x.
Now < HOD = < COD since Tr.s HOD & COD are congruent & each and angle = 45
Therefore we have the octagon broken down into 8 congruent Tr.s SAS and all the sides are = x and the octagon is therefore a regular octagon which is the 1st required result
Now HD = a/2 = AH, VG = a/4 from the midpoint theorem and so VH = x + 2x = 3x from similar triangles
Hence CH = V5a/2 = 6x from which x = V5a/12 which is the 2nd required result
Sumith Peiris
Moratuwa
Sri Lanka
Another way to prove that the octagon is equilateral is to see that each vertice Pn has a predecessor Pn-1 and a successor Pn+1 that are such that Pn-1 and Pn+1 are the symetric of one another accross either AC, BD, EG or FH. Therefore the length of each segment Pn-1_Pn is equal to that of segment Pn_Pn+1.
ReplyDeleteThe previous comment should be improved as follows to make it completely accurate:
DeleteAnother way to prove that the octagon is equilateral is to see that each vertice Pn has a predecessor Pn-1 and a successor Pn+1 that are such that Pn-1 and Pn+1 are the symetric of one another accross either AC, BD, EG or FH to which Pn belongs. Therefore the length of each segment Pn-1_Pn is equal to that of segment Pn_Pn+1.