1. H. T. & E. hte
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    31 Aug '07 18:11
    ABCDE is an irregular pentagon such that each of the triangles ABC, BCD, CDE, DEF and EFA has equal area, say x. The puzzle is to find the area of the pentagon in terms of x.
    Any takers for this puzzle?
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    31 Aug '07 20:361 edit
    Originally posted by ranjan sinha
    ABCDE is an irregular pentagon such that each of the triangles ABC, BCD, CDE, DEF and EFA has equal area, say x. The puzzle is to find the area of the pentagon in terms of x.
    Any takers for this puzzle?
    "F" ?

    edit: 4x?
  3. Dixie
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    01 Sep '07 00:26
    Originally posted by ranjan sinha
    ABCDE is an irregular pentagon such that each of the triangles ABC, BCD, CDE, DEF and EFA has equal area, say x. The puzzle is to find the area of the pentagon in terms of x.
    Any takers for this puzzle?
    I'm with Mephisto... where is point F in a pentagon, irregular or not?
  4. SubscriberAThousandYoung
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    01 Sep '07 01:251 edit
    Originally posted by ranjan sinha
    ABCDE is an irregular pentagon such that each of the triangles ABC, BCD, CDE, DEF and EFA has equal area, say x. The puzzle is to find the area of the pentagon in terms of x.
    Any takers for this puzzle?
    5x, assuming "DEF" is really "DEA" and "EFA" is really "EAB".

    Oh, no, I'm wrong.

    Is F the center?
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    01 Sep '07 07:18
    Originally posted by AThousandYoung
    5x, assuming "DEF" is really "DEA" and "EFA" is really "EAB".

    Oh, no, I'm wrong.

    Is F the center?
    My guess is that "F" is where the the irregular pentagon (one side from E to A) would be come a regular (or at least symmetrical) hexagon.
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    02 Sep '07 09:092 edits
    Is the pentagon convex?
  7. H. T. & E. hte
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    09 Sep '07 11:31
    Originally posted by Mephisto2
    "F" ?

    edit: 4x?
    Nope..
  8. H. T. & E. hte
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    09 Sep '07 13:39
    Originally posted by jebrydzagin
    I'm with Mephisto... where is point F in a pentagon, irregular or not?
    Both of you -you and Mephisto are right....
    Apologies for the inadvertant error....
    The puzzle should actually read as follows:-
    " ABCDE is an irregular convex pentagon such that each of the five triangles ABC, BCD, CDE, DEA and EAB have equal area say 'x'.
    Find the area of the pentagon in terms of 'x'."

    However, in spite of the inadvertant error, the spirit and the substance of the puzzle was quite clear...

    I stand corrected...Plz solve the puzzle now...
  9. H. T. & E. hte
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    09 Sep '07 13:411 edit
    Originally posted by jebrydzagin
    I'm with Mephisto... where is point F in a pentagon, irregular or not?
    That was an inadvertant error..Apologies..it has been clarified in the preceding post..
  10. H. T. & E. hte
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    09 Sep '07 13:431 edit
    Originally posted by AThousandYoung
    5x, assuming "DEF" is really "DEA" and "EFA" is really "EAB".

    Oh, no, I'm wrong.

    Is F the center?
    No, it is not 5x. it involves irrational ratios...The challenge is to derive it just from the given conditions /...
  11. H. T. & E. hte
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    09 Sep '07 13:44
    Originally posted by David113
    Is the pentagon convex?
    Yes it has to be....
  12. H. T. & E. hte
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    10 Sep '07 15:03
    Originally posted by AThousandYoung
    5x, assuming "DEF" is really "DEA" and "EFA" is really "EAB".
    Yes you are right..about DEF really being DEA and EFA really being EAB.
    But the answer is not 5x.
  13. at the centre
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    14 Sep '07 14:281 edit
    Originally posted by ranjan sinha
    Both of you -you and Mephisto are right....
    Apologies for the inadvertant error....
    The puzzle should actually read as follows:-
    " ABCDE is an irregular convex pentagon such that each of the five triangles ABC, BCD, CDE, DEA and EAB have equal area say 'x'.
    Find the area of the pentagon in terms of 'x'."

    However, in spite of the inadvertan ...[text shortened]... nce of the puzzle was quite clear...

    I stand corrected...Plz solve the puzzle now...
    Actually...under the given condition [of all the peripheral triangles ABC, BCD, CDE, DEA and EAB being equal in area], the pentagon has to be convex. This is because, equal area of each of the peripheral triangles will imply that the line joining AD is parallel to BC.
    Similarly the line joining BE is parallel to CD; the line joining CA is parallel to DE. And so on... for each of the five sides...

    Hence the pentagon ABCDE , under such conditions, cannot be concave. It cannot but be convex. Thus stating the additional condition of convexity is redundant in fact.
  14. SubscriberAThousandYoung
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    15 Sep '07 10:47
    Originally posted by howzzat
    Actually...under the given condition [of all the peripheral triangles ABC, BCD, CDE, DEA and EAB being equal in area], the pentagon has to be convex. This is because, equal area of each of the peripheral triangles will imply that the line joining AD is parallel to BC.
    Similarly the line joining BE is parallel to CD; the line joining CA is parallel ...[text shortened]... t cannot but be convex. Thus stating the additional condition of convexity is redundant in fact.
    How does the area imply that?
  15. SubscriberAThousandYoung
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    16 Sep '07 08:38
    This problem is too damn hard.
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