FoolForMath 3 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.

1. ninhi5

diagram?

2. FoolForMath

That would spoil the problem to some extent ;)

3. ninhi5

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

4. FoolForMath

There is only one stamp.

5. ninhi5

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

6. ninhi5

|dw:1338096917684:dw|

7. ninhi5

cant really draw to scale

8. ninhi5

is that how it should be?

9. FoolForMath

No.

10. FoolForMath

@experimentX got it right in the first time itself.

11. ninhi5

i'm confused

12. FoolForMath

me too :|

13. binary3i

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

14. FoolForMath

No.

15. SmoothMath

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

16. FoolForMath

Yes.

17. SmoothMath

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

18. binary3i

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

19. Callisto

4pi/3 - 2sqrt3?

20. FoolForMath

I believe Callisto did it :D

21. Callisto

It's actually very easy and interesting :)

22. FoolForMath

I never post hard problem(s) ;)

23. SmoothMath

That's kind of you.

24. Callisto

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

25. experimentX

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

26. ninhi5

27. Callisto

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

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

29. SmoothMath

Anytime =)

30. AccessDenied

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

31. SmoothMath

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

32. Callisto

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

33. Callisto

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

34. SmoothMath

Ah, quite nice.

35. inkyvoyd

Guys, you just spoiled the problem for me, ahah

36. Callisto

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

37. binary3i

i made a mistake while integrating.

38. SmoothMath

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

39. FoolForMath

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

someone should give @SmoothMath a medal

41. SmoothMath

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

42. experimentX

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

43. FoolForMath

Keep looking ;)

44. apoorvk

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

45. Callisto

@FoolForMath 's solution is easier to understand :)

46. apoorvk

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

47. FoolForMath

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

48. apoorvk

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

49. FoolForMath

that*<- Redundant