See complete Problem 161
Parallelogram, Midpoints, Octagon, Areas. Level: High School, SAT Prep, College geometry
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Sunday, August 17, 2008
Elearn Geometry Problem 161
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The octagon have vertices, clockwise from top, M,N,O,P,Q,R,S,T, with center X. by rotational symmetry, F,M,X,Q,H ; E,S,X,O,G collinear. N is the centroid of XFG, XN cut FG at I. MI//XG & OI//XF, so [XMNO]=[NFG]; XM=MF & XO=OG, so [XMNO]=[NMF]+[NOG]; get [XMNO]/[XFG]=1/3. S1/S=[XMNO]/[XFCG]=1/2*[XMNO]/[XFG]=1/6
ReplyDeletehttps://photos.app.goo.gl/rvrwkFGoy7JVYggg6
ReplyDeleteSee diagram here.
ReplyDeleteIf IJKL is the parallelogram in the center of the figure, from problem 177 [IJKL] = S/5.
The surface S1 of octagon MNPQRSTU is equal to [IJKL] minus the areas of the 4 small triangles [ITU], [JMN], [KPQ] and [LRS] which are identical.
Take ITU: it is similar to JPC (all sides parallel 2 by 2).
In ETFC, ET//FC and ET = FC/2 => TU = UF/2 => TU = TF/3 = PC/3 so [ITU] = [JPC/9].
In FPCE, FC//PE and PE = FC.3/2 => PJ = JF.3/2 => PJ = PF.3/5 => [JPC] = [CFP].3/5
But [CFP] = [CFD]/2 = S/8.
Hence, [ITU] = S.3 /(9*5*8) = S/(3*5*8) and 4.[ITU] = S/(2*3*5).
So S1 = S/5 –S/(2*3*5) = S/6.
QED