Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

FoolForMath Group Title

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.

  • 2 years ago
  • 2 years ago

  • This Question is Closed
  1. ninhi5 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    diagram?

    • 2 years ago
  2. FoolForMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    That would spoil the problem to some extent ;)

    • 2 years ago
  3. ninhi5 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    would there be 2 circles, one inside and one outside?

    • 2 years ago
  4. FoolForMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    There is only one stamp.

    • 2 years ago
  5. ninhi5 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    well if the edge is outside then the circle must be inscribed inside

    • 2 years ago
  6. ninhi5 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1338096917684:dw|

    • 2 years ago
  7. ninhi5 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    cant really draw to scale

    • 2 years ago
  8. ninhi5 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    is that how it should be?

    • 2 years ago
  9. FoolForMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    No.

    • 2 years ago
  10. FoolForMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    @experimentX got it right in the first time itself.

    • 2 years ago
  11. ninhi5 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    i'm confused

    • 2 years ago
  12. FoolForMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    me too :|

    • 2 years ago
  13. binary3i Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    is that \[(\pi -3\sqrt{3})/6\]

    • 2 years ago
  14. FoolForMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    No.

    • 2 years ago
  15. SmoothMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1338097283187:dw| Not perfect, but I believe the picture should look something like this.

    • 2 years ago
  16. FoolForMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    Yes.

    • 2 years ago
  17. SmoothMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    And the area we are trying to find is: |dw:1338097419327:dw|

    • 2 years ago
  18. binary3i Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    \[2\int\limits_{0}^{1/2}(\sqrt{2x-x^2} - \sqrt{3}x) dx\]

    • 2 years ago
  19. Callisto Group Title
    Best Response
    You've already chosen the best response.
    Medals 7

    4pi/3 - 2sqrt3?

    • 2 years ago
  20. FoolForMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    I believe Callisto did it :D

    • 2 years ago
  21. Callisto Group Title
    Best Response
    You've already chosen the best response.
    Medals 7

    It's actually very easy and interesting :)

    • 2 years ago
  22. FoolForMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    I never post hard problem(s) ;)

    • 2 years ago
  23. SmoothMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    That's kind of you.

    • 2 years ago
  24. Callisto Group Title
    Best Response
    You've already chosen the best response.
    Medals 7

    No.... you always post difficult problems :|

    • 2 years ago
  25. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    ... I'm still waiting for someone to post answer to your perpendicular tangent question

    • 2 years ago
  26. ninhi5 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    explanation please

    • 2 years ago
  27. Callisto Group Title
    Best Response
    You've already chosen the best response.
    Medals 7

    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

    • 2 years ago
  28. Callisto Group Title
    Best Response
    You've already chosen the best response.
    Medals 7

    @SmoothMath Hope you don't mind me borrowing your diagram :| |dw:1338097936559:dw|

    • 2 years ago
  29. SmoothMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Anytime =)

    • 2 years ago
  30. AccessDenied Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Ehh, that's how I would have done it too. :P

    • 2 years ago
  31. SmoothMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    I got a bit caught up on how to justify the third side being 2.

    • 2 years ago
  32. Callisto Group Title
    Best Response
    You've already chosen the best response.
    Medals 7

    PS: When I drew the diagram, I found it very interesting that I even forgot I needed to solve the problem :P

    • 2 years ago
  33. Callisto Group Title
    Best Response
    You've already chosen the best response.
    Medals 7

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

    • 2 years ago
  34. SmoothMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Ah, quite nice.

    • 2 years ago
  35. inkyvoyd Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Guys, you just spoiled the problem for me, ahah

    • 2 years ago
  36. Callisto Group Title
    Best Response
    You've already chosen the best response.
    Medals 7

    I only know this basic math :( @FoolForMath How do you solve it?

    • 2 years ago
  37. binary3i Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    i made a mistake while integrating.

    • 2 years ago
  38. SmoothMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    That's a great way to solve it, Callisto. Better to use basic math whenever you can.

    • 2 years ago
  39. FoolForMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    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 \( 60^\circ \)

    • 2 years ago
  40. ninhi5 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    someone should give @SmoothMath a medal

    • 2 years ago
  41. SmoothMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • 2 years ago
  42. experimentX Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    i was looking for the solution of this question ... http://openstudy.com/users/foolformath#/updates/4fa421cde4b029e9dc34a843

    • 2 years ago
  43. FoolForMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    Keep looking ;)

    • 2 years ago
  44. apoorvk Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Gotcha, this was a sitter - @Callisto did it in the easiest way possible. Kudos!

    • 2 years ago
  45. Callisto Group Title
    Best Response
    You've already chosen the best response.
    Medals 7

    @FoolForMath 's solution is easier to understand :)

    • 2 years ago
  46. apoorvk Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Yea maybe - but before the 'key realisation' - we need the 'key visualisation'. :P

    • 2 years ago
  47. FoolForMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    lol, leave it @apoovk she won't accept that her amazing solution.

    • 2 years ago
  48. apoorvk Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah she won't, right never does. :P

    • 2 years ago
  49. FoolForMath Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    that*<- Redundant

    • 2 years ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.