1. H. T. & E. hte
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    16 Sep '07 09:59
    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.
    You are quite right.
  2. Standard memberAThousandYoung
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    16 Sep '07 10:01
    If that's true, then doesn't that mean that this has to be a symmetric pentagon?
  3. at the centre
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    16 Sep '07 10:063 edits
    Originally posted by AThousandYoung
    How does the area imply that?
    That's coz both the triangles ABC and DBC can be viewed as having the common base BC. Hence their vertices A and D must be at equal heights from the base BC. This would mean that AD is parallel to BC. Similarly for the other sides CD, DE etc...
  4. Backside of desert
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    16 Sep '07 18:19
    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.
    I believe that a pentagon with all peripherary triangles being equal must be a regular pentagon; but I am still working on the proof.
  5. Standard memberAThousandYoung
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    16 Sep '07 18:34
    Originally posted by preachingforjesus
    I believe that a pentagon with all peripherary triangles being equal must be a regular pentagon; but I am still working on the proof.
    That came from someone named preachingforjesus? Well, let's see if you can manage a proof. I certainly can't.
  6. Standard memberAThousandYoung
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    16 Sep '07 18:42
    Originally posted by howzzat
    That's coz both the triangles ABC and DBC can be viewed as having the common base BC. Hence their vertices A and D must be at equal heights from the base BC. This would mean that AD is parallel to BC. Similarly for the other sides CD, DE etc...
    This would mean that AD is parallel to BC.

    I'm not sure I buy this part of your analysis. Can you prove it?
  7. Joined
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    16 Sep '07 18:59
    Originally posted by ranjan sinha
    " 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'."
    Is the answer simply (2 + phi)x?

    [where phi = golden ratio]
  8. Account suspended
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    16 Sep '07 20:23
    Originally posted by LemonJello
    Is the answer simply (2 + phi)x?

    [where phi = golden ratio]
    Given: pentagon made up of five triangles
    Given: all triangles have equal areas, x
    fact: area of pentagon = sum of areas of inscribed triangles

    therefore, area of pentagon equals x + x + x + x +x equals 5 x
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    16 Sep '07 20:351 edit
    Originally posted by rubberjaw30
    Given: pentagon made up of five triangles
    Given: all triangles have equal areas, x
    fact: area of pentagon = sum of areas of inscribed triangles

    therefore, area of pentagon equals x + x + x + x +x equals 5 x
    No. Just sketch an irregular pentagon ABCDE and then draw the triangles specified in the problem, and you'll see why you are mistaken.
  10. Standard memberAThousandYoung
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    16 Sep '07 23:06
    Originally posted by rubberjaw30
    Given: pentagon made up of five triangles
    Given: all triangles have equal areas, x
    fact: area of pentagon = sum of areas of inscribed triangles

    therefore, area of pentagon equals x + x + x + x +x equals 5 x
    The triangles you're referring to are not the ones mentioned in the problem. You're referring to triangles with one vertex in the center.
  11. at the centre
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    17 Sep '07 12:21
    Originally posted by AThousandYoung
    If that's true, then doesn't that mean that this has to be a symmetric pentagon?
    Why? How can you say that ?...
  12. at the centre
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    17 Sep '07 12:231 edit
    Originally posted by LemonJello
    Is the answer simply (2 + phi)x?

    [where phi = golden ratio]
    Is that a conjecture? Or is there any proof? I too have a feeling that the area of the pentagon should be a linear function of x.
  13. at the centre
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    17 Sep '07 12:332 edits
    Originally posted by AThousandYoung
    [b]This would mean that AD is parallel to BC.

    I'm not sure I buy this part of your analysis. Can you prove it?[/b]
    The triangles ABC and BCD have common base BC. Right?
    Area of a triangle =(1/2)*Base*Height. Right?
    If base is common, the heights of the two triangles must be equal. Hence if perpendiculars are drawn from A and D up to the line BC, these perpendiculars will have equal length. Under the conditions of the problem it seems clear that both the points A and D are on the same side of the line BC. Hence the line joining AD will have to be parallel to BC.

    P.S. - Of course this argument will fail if A and D lye on opposite sides of the line BC.
  14. H. T. & E. hte
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    17 Sep '07 13:141 edit
    Originally posted by preachingforjesus
    I believe that a pentagon with all peripherary triangles being equal must be a regular pentagon; but I am still working on the proof.
    I don't think you are right in believing so.

    Take for example the irregular pentagon having angles
    angle/_A=115 deg; angle/_B=125 deg; angle/_C=95 deg;
    angle/_D=130deg & angle/_E=105 degrees.

    I have constructed this example by working backwards using:-

    /_CAD=25deg; /_DBE=40deg; /_ECA=30deg; /_ADB=45deg; /_BEC=40deg.

    This pentagon satisfies the conditions of the problem. But obviously it is not a regular pentagon.
  15. H. T. & E. hte
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    17 Sep '07 13:25
    Originally posted by AThousandYoung
    This problem is too damn hard.
    Not really... It can be solved using just High School level skills in mathematics and geometry.
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