See complete Problem 150

Quadrilateral, Area, Trisection of Sides. Level: High School, SAT Prep, College geometry

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## Sunday, July 27, 2008

### Elearn Geometry Problem 150

Labels:
area,
Problem 150,
quadrilateral,
trisection

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[BAH]=[BHG]=[BGD] & [DBE]=[DEF]=[DFC] so [BGDE]=S/3,

ReplyDeleteS1=[EHGF]=[EHG]+[GFE]=[EGD]+[GEB]=[BGDE]=S/3

We can use vector production - DG=a, BE=b, AB=c and S[BGDE]=(a+c+b)*(-a)/2+(2a+c+2b)*(-b)/2 = S[ABCD]/3=(c*(-3a)/2+(-3b)*(3a+3b+c)/2)/3, here a*a=0 a*b=-b*a

ReplyDeleteIs easy to see that altitudes of AHB, HGE and GDF are in arithmetic progression, then also its areas. The same happens for triangles BHE, EGF and FDC.

ReplyDeleteThen quadrilateral areas ABEH, HEFG and GFCD also are in arithmetic progression, S1 = 1/3 S and S0 + S2 = 2/3 S.

The result is trivially extended for all the symmetrically distributed pairs of the n quadrilaterals S_1, S_2, ... S_n in which S can be divided by joining the points that divide the opposite sides into n equal segments.

S_1 + S_n = S_2 + S_{n-1} + ... = 2/n S

Then, if n is odd, S_{(n+1)/2} = 1/n S