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.

- anonymous

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- anonymous

diagram?

- anonymous

That would spoil the problem to some extent ;)

- anonymous

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

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## More answers

- anonymous

There is only one stamp.

- anonymous

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

- anonymous

|dw:1338096917684:dw|

- anonymous

cant really draw to scale

- anonymous

is that how it should be?

- anonymous

No.

- anonymous

@experimentX got it right in the first time itself.

- anonymous

i'm confused

- anonymous

me too :|

- binary3i

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

- anonymous

No.

- anonymous

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

- anonymous

Yes.

- anonymous

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

- binary3i

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

- Callisto

4pi/3 - 2sqrt3?

- anonymous

I believe Callisto did it :D

- Callisto

It's actually very easy and interesting :)

- anonymous

I never post hard problem(s) ;)

- anonymous

That's kind of you.

- Callisto

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

- experimentX

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

- anonymous

explanation please

- 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

- Callisto

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

- anonymous

Anytime =)

- AccessDenied

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

- anonymous

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

- Callisto

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

- Callisto

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

- anonymous

Ah, quite nice.

- inkyvoyd

Guys, you just spoiled the problem for me, ahah

- Callisto

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

- binary3i

i made a mistake while integrating.

- anonymous

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

- anonymous

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 \( 60^\circ \)

- anonymous

someone should give @SmoothMath a medal

- anonymous

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

- experimentX

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

- anonymous

Keep looking ;)

- apoorvk

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

- Callisto

@FoolForMath 's solution is easier to understand :)

- apoorvk

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

- anonymous

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

- apoorvk

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

- anonymous

that*<- Redundant

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