Thursday, December 10, 2009

Problem 402. Hexagon, Triangle, Centroid, Parallel, Congruence, Similarity

Proposed Problem
Click the figure below to see the complete problem 402 about Hexagon, Triangle, Centroid, Parallel, Congruence, Similarity.

Problem 402. Hexagon, Triangle, Centroid, Parallel, Congruence, Similarity.
See also:
Complete Problem 402
Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 11:49 AM
Labels: , , , , ,

4 comments:

  1. AnonymousDecember 11, 2009 at 10:47 AM

    My problems for this picture.
    Proff
    1)Area hexagon ABCDEF/ Area hexagon GHJKLM = 9/4.

    2) Area polygon ABCDE/ Area polygon GHJKL = 117/32.

    2) Area polygon ABCD/ Area polygon GHJK = 9.

    ReplyDelete
  2. AnonymousDecember 12, 2009 at 12:59 AM

    the affix of triangle ABC centroid G is
    g=(a+b+c)/3
    then
    3(h-g)= b+c+d-a-b-c= d-a
    3(k-l)= d+e+f-e+f+a= d-a
    this means that GH=LK and GH // LK
    .........

    ReplyDelete
  3. AnonymousDecember 12, 2009 at 9:14 AM

    We see that
    XH=(XB+XC+XD)/3
    XG=(XB+XC+XA)/3
    XK=(XD+XE+XF)/3
    XL=(XE+XF+XA)/3
    and HG=(XH-XG,YH-YG)=(XD-XA,YD-YA)/3
    LK=(XK-XL,YK-YL)=(XD-XA,YD-YA)/3
    etc.
    And my question for this picture - prove
    area hexagon ABCDEF/area hexagon GHJKLM = 9/4

    ReplyDelete
  4. MdeVAugust 3, 2021 at 8:17 AM

    Interesting about the areas, but the result about the formed hexagon having opposite sides equal & parallel can be generalized for 2n-gons as shown here: http://dynamicmathematicslearning.com/octagoncentroids.html

    ReplyDelete

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