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FoolForMath

  • 2 years ago

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.

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  1. ninhi5
    • 2 years ago
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    diagram?

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

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

  4. FoolForMath
    • 2 years ago
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    There is only one stamp.

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

  6. ninhi5
    • 2 years ago
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    |dw:1338096917684:dw|

  7. ninhi5
    • 2 years ago
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    cant really draw to scale

  8. ninhi5
    • 2 years ago
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    is that how it should be?

  9. FoolForMath
    • 2 years ago
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    No.

  10. FoolForMath
    • 2 years ago
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    @experimentX got it right in the first time itself.

  11. ninhi5
    • 2 years ago
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    i'm confused

  12. FoolForMath
    • 2 years ago
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    me too :|

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

  14. FoolForMath
    • 2 years ago
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    No.

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

  16. FoolForMath
    • 2 years ago
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    Yes.

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

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

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

  20. FoolForMath
    • 2 years ago
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    I believe Callisto did it :D

  21. Callisto
    • 2 years ago
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    It's actually very easy and interesting :)

  22. FoolForMath
    • 2 years ago
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    I never post hard problem(s) ;)

  23. SmoothMath
    • 2 years ago
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    That's kind of you.

  24. Callisto
    • 2 years ago
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    No.... you always post difficult problems :|

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

  26. ninhi5
    • 2 years ago
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    explanation please

  27. Callisto
    • 2 years ago
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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

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

  29. SmoothMath
    • 2 years ago
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    Anytime =)

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

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

  32. Callisto
    • 2 years ago
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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

  33. Callisto
    • 2 years ago
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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

  34. SmoothMath
    • 2 years ago
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    Ah, quite nice.

  35. inkyvoyd
    • 2 years ago
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    Guys, you just spoiled the problem for me, ahah

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

  37. binary3i
    • 2 years ago
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    i made a mistake while integrating.

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

  39. FoolForMath
    • 2 years ago
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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 \( 60^\circ \)

  40. ninhi5
    • 2 years ago
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    someone should give @SmoothMath a medal

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

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

  43. FoolForMath
    • 2 years ago
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    Keep looking ;)

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

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

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

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

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

  49. FoolForMath
    • 2 years ago
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    that*<- Redundant

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