## 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

1. ninhi5 Group Title

diagram?

2. FoolForMath Group Title

That would spoil the problem to some extent ;)

3. ninhi5 Group Title

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

4. FoolForMath Group Title

There is only one stamp.

5. ninhi5 Group Title

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

6. ninhi5 Group Title

|dw:1338096917684:dw|

7. ninhi5 Group Title

cant really draw to scale

8. ninhi5 Group Title

is that how it should be?

9. FoolForMath Group Title

No.

10. FoolForMath Group Title

@experimentX got it right in the first time itself.

11. ninhi5 Group Title

i'm confused

12. FoolForMath Group Title

me too :|

13. binary3i Group Title

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

14. FoolForMath Group Title

No.

15. SmoothMath Group Title

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

16. FoolForMath Group Title

Yes.

17. SmoothMath Group Title

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

18. binary3i Group Title

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

19. Callisto Group Title

4pi/3 - 2sqrt3?

20. FoolForMath Group Title

I believe Callisto did it :D

21. Callisto Group Title

It's actually very easy and interesting :)

22. FoolForMath Group Title

I never post hard problem(s) ;)

23. SmoothMath Group Title

That's kind of you.

24. Callisto Group Title

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

25. experimentX Group Title

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

26. ninhi5 Group Title

27. Callisto Group Title

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 Group Title

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

29. SmoothMath Group Title

Anytime =)

30. AccessDenied Group Title

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

31. SmoothMath Group Title

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

32. Callisto Group Title

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

33. Callisto Group Title

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

34. SmoothMath Group Title

Ah, quite nice.

35. inkyvoyd Group Title

Guys, you just spoiled the problem for me, ahah

36. Callisto Group Title

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

37. binary3i Group Title

i made a mistake while integrating.

38. SmoothMath Group Title

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

39. FoolForMath Group Title

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 Group Title

someone should give @SmoothMath a medal

41. SmoothMath Group Title

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

42. experimentX Group Title

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

43. FoolForMath Group Title

Keep looking ;)

44. apoorvk Group Title

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

45. Callisto Group Title

@FoolForMath 's solution is easier to understand :)

46. apoorvk Group Title

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

47. FoolForMath Group Title

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

48. apoorvk Group Title

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

49. FoolForMath Group Title

that*<- Redundant