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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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That would spoil the problem to some extent ;)
would there be 2 circles, one inside and one outside?

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Other answers:

There is only one stamp.
well if the edge is outside then the circle must be inscribed inside
cant really draw to scale
is that how it should be?
@experimentX got it right in the first time itself.
i'm confused
me too :|
is that \[(\pi -3\sqrt{3})/6\]
|dw:1338097283187:dw| Not perfect, but I believe the picture should look something like this.
And the area we are trying to find is: |dw:1338097419327:dw|
\[2\int\limits_{0}^{1/2}(\sqrt{2x-x^2} - \sqrt{3}x) dx\]
4pi/3 - 2sqrt3?
I believe Callisto did it :D
It's actually very easy and interesting :)
I never post hard problem(s) ;)
That's kind of you.
No.... you always post difficult problems :|
... I'm still waiting for someone to post answer to your perpendicular tangent question
explanation please
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
@SmoothMath Hope you don't mind me borrowing your diagram :| |dw:1338097936559:dw|
Anytime =)
Ehh, that's how I would have done it too. :P
I got a bit caught up on how to justify the third side being 2.
PS: When I drew the diagram, I found it very interesting that I even forgot I needed to solve the problem :P
Each angle of hexagon = 180 (6-2) /6 = 120 divided by 2 = 60 Base angle isos triangle => all are 60 => equi. triangle
Ah, quite nice.
Guys, you just spoiled the problem for me, ahah
I only know this basic math :( @FoolForMath How do you solve it?
i made a mistake while integrating.
That's a great way to solve it, Callisto. Better to use basic math whenever you can.
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 \( 60^\circ \)
someone should give @SmoothMath a medal
Well, aren't you kind. I'm not too worried about medals though, and Callisto clearly has the "best answer." =)
i was looking for the solution of this question ...
Keep looking ;)
Gotcha, this was a sitter - @Callisto did it in the easiest way possible. Kudos!
@FoolForMath 's solution is easier to understand :)
Yea maybe - but before the 'key realisation' - we need the 'key visualisation'. :P
lol, leave it @apoovk she won't accept that her amazing solution.
yeah she won't, right never does. :P
that*<- Redundant

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