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FoolForMath

An intuitive geometry problem, The ticket for the Chess world cup final is in the shape of a regular hexagon, with a side of 6 units. A circular stamp, of radius 2 units, is meant to be used to stamp the ticket in the center. By, mistake the stamp get misaligned and stamps the ticket in such a way that the edge of the stamp exactly touches the vertex of the hexagon, while the center of the stamp lies on the line joining that vertex to the opposite vertex. Find the portion of the stamp which falls outside the ticket.

  • one year ago
  • one year ago

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  1. ninhi5
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    diagram?

    • one year ago
  2. FoolForMath
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    That would spoil the problem to some extent ;)

    • one year ago
  3. ninhi5
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    would there be 2 circles, one inside and one outside?

    • one year ago
  4. FoolForMath
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    There is only one stamp.

    • one year ago
  5. ninhi5
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    well if the edge is outside then the circle must be inscribed inside

    • one year ago
  6. ninhi5
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    |dw:1338096917684:dw|

    • one year ago
  7. ninhi5
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    cant really draw to scale

    • one year ago
  8. ninhi5
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    is that how it should be?

    • one year ago
  9. FoolForMath
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    No.

    • one year ago
  10. FoolForMath
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    @experimentX got it right in the first time itself.

    • one year ago
  11. ninhi5
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    i'm confused

    • one year ago
  12. FoolForMath
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    me too :|

    • one year ago
  13. binary3i
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    is that \[(\pi -3\sqrt{3})/6\]

    • one year ago
  14. FoolForMath
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    No.

    • one year ago
  15. SmoothMath
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    |dw:1338097283187:dw| Not perfect, but I believe the picture should look something like this.

    • one year ago
  16. FoolForMath
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    Yes.

    • one year ago
  17. SmoothMath
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    And the area we are trying to find is: |dw:1338097419327:dw|

    • one year ago
  18. binary3i
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    \[2\int\limits_{0}^{1/2}(\sqrt{2x-x^2} - \sqrt{3}x) dx\]

    • one year ago
  19. Callisto
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    4pi/3 - 2sqrt3?

    • one year ago
  20. FoolForMath
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    I believe Callisto did it :D

    • one year ago
  21. Callisto
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    It's actually very easy and interesting :)

    • one year ago
  22. FoolForMath
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    I never post hard problem(s) ;)

    • one year ago
  23. SmoothMath
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    That's kind of you.

    • one year ago
  24. Callisto
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    No.... you always post difficult problems :|

    • one year ago
  25. experimentX
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    ... I'm still waiting for someone to post answer to your perpendicular tangent question

    • one year ago
  26. ninhi5
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    explanation please

    • one year ago
  27. Callisto
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    First, I must admit that my way to solve it is very stupid.... radius of the circle is 2 and the centre of the circle lies on the line joining 2 opposite vertex => 2 equi. triangles with sides 2 cm are formed Area of sector = pi(2)^2 x 120/360 = 4pi/3 So, area required = 4pi/3 - 2sqrt3

    • one year ago
  28. Callisto
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    @SmoothMath Hope you don't mind me borrowing your diagram :| |dw:1338097936559:dw|

    • one year ago
  29. SmoothMath
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    Anytime =)

    • one year ago
  30. AccessDenied
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    Ehh, that's how I would have done it too. :P

    • one year ago
  31. SmoothMath
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    I got a bit caught up on how to justify the third side being 2.

    • one year ago
  32. Callisto
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    PS: When I drew the diagram, I found it very interesting that I even forgot I needed to solve the problem :P

    • one year ago
  33. Callisto
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    Each angle of hexagon = 180 (6-2) /6 = 120 divided by 2 = 60 Base angle isos triangle => all are 60 => equi. triangle

    • one year ago
  34. SmoothMath
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    Ah, quite nice.

    • one year ago
  35. inkyvoyd
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    Guys, you just spoiled the problem for me, ahah

    • one year ago
  36. Callisto
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    I only know this basic math :( @FoolForMath How do you solve it?

    • one year ago
  37. binary3i
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    i made a mistake while integrating.

    • one year ago
  38. SmoothMath
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    That's a great way to solve it, Callisto. Better to use basic math whenever you can.

    • one year ago
  39. FoolForMath
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    I did it in the same way. The key realization is the fact that there is a equilateral triangle of length 2 unit is formed. After that, the hexagon is not needed. \[2\times r^2\left(\frac {\pi \theta}{360^\circ}-\frac {\sin \theta }{2} \right) \] Here \(r=2\) and \( \theta\) is \( 60^\circ \)

    • one year ago
  40. ninhi5
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    someone should give @SmoothMath a medal

    • one year ago
  41. SmoothMath
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    Well, aren't you kind. I'm not too worried about medals though, and Callisto clearly has the "best answer." =)

    • one year ago
  42. experimentX
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    i was looking for the solution of this question ... http://openstudy.com/users/foolformath#/updates/4fa421cde4b029e9dc34a843

    • one year ago
  43. FoolForMath
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    Keep looking ;)

    • one year ago
  44. apoorvk
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    Gotcha, this was a sitter - @Callisto did it in the easiest way possible. Kudos!

    • one year ago
  45. Callisto
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    @FoolForMath 's solution is easier to understand :)

    • one year ago
  46. apoorvk
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    Yea maybe - but before the 'key realisation' - we need the 'key visualisation'. :P

    • one year ago
  47. FoolForMath
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    lol, leave it @apoovk she won't accept that her amazing solution.

    • one year ago
  48. apoorvk
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    yeah she won't, right never does. :P

    • one year ago
  49. FoolForMath
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    that*<- Redundant

    • one year ago
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