A more "explicit" solution to problem 151. Let S5 be the area of the central quadrilateral. As proved in problem 150, S(GHPQ) = S/3 and S(EFMN) = S/3. Then S(GHPQ) = S1 + S3 + S5 = S/3 and S(EFMN) = S2 + S4 + S5. Hence S1 + S3 = S2 + S4 = S/3 – S5.
Los escaques están en una doble progresión aritmética, por lo que los simétricamente dispuestos suman 2/n^2 y el central, si lo hay, 1/n^2. Applet de #GeoGebra: http://bit.ly/2JuTtNe Twitter: https://twitter.com/ilarrosac/status/1005973237505937409
proved from Proposed Problem 150, EFMN=S/3=PQGH, so S1+S3=S2+S4
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ReplyDeleteA more "explicit" solution to problem 151.
ReplyDeleteLet S5 be the area of the central quadrilateral. As proved in problem 150,
S(GHPQ) = S/3 and S(EFMN) = S/3.
Then S(GHPQ) = S1 + S3 + S5 = S/3 and
S(EFMN) = S2 + S4 + S5.
Hence S1 + S3 = S2 + S4 = S/3 – S5.
Solution posted on Twitter:
ReplyDeleteCan't upload a picture here...
https://twitter.com/panlepan/status/894134252383719424
Your solution on Twitter has been embedded at http://www.gogeometry.com/problem/p151_quadrilateral_area_trisection.htm
DeleteThanks Vincent
Los escaques están en una doble progresión aritmética, por lo que los simétricamente dispuestos suman 2/n^2 y el central, si lo hay, 1/n^2.
ReplyDeleteApplet de #GeoGebra: http://bit.ly/2JuTtNe
Twitter: https://twitter.com/ilarrosac/status/1005973237505937409