anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
diagram?
anonymous
  • anonymous
That would spoil the problem to some extent ;)
anonymous
  • anonymous
would there be 2 circles, one inside and one outside?

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More answers

anonymous
  • anonymous
There is only one stamp.
anonymous
  • anonymous
well if the edge is outside then the circle must be inscribed inside
anonymous
  • anonymous
|dw:1338096917684:dw|
anonymous
  • anonymous
cant really draw to scale
anonymous
  • anonymous
is that how it should be?
anonymous
  • anonymous
No.
anonymous
  • anonymous
@experimentX got it right in the first time itself.
anonymous
  • anonymous
i'm confused
anonymous
  • anonymous
me too :|
binary3i
  • binary3i
is that \[(\pi -3\sqrt{3})/6\]
anonymous
  • anonymous
No.
anonymous
  • anonymous
|dw:1338097283187:dw| Not perfect, but I believe the picture should look something like this.
anonymous
  • anonymous
Yes.
anonymous
  • anonymous
And the area we are trying to find is: |dw:1338097419327:dw|
binary3i
  • binary3i
\[2\int\limits_{0}^{1/2}(\sqrt{2x-x^2} - \sqrt{3}x) dx\]
Callisto
  • Callisto
4pi/3 - 2sqrt3?
anonymous
  • anonymous
I believe Callisto did it :D
Callisto
  • Callisto
It's actually very easy and interesting :)
anonymous
  • anonymous
I never post hard problem(s) ;)
anonymous
  • anonymous
That's kind of you.
Callisto
  • Callisto
No.... you always post difficult problems :|
experimentX
  • experimentX
... I'm still waiting for someone to post answer to your perpendicular tangent question
anonymous
  • anonymous
explanation please
Callisto
  • 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
  • Callisto
@SmoothMath Hope you don't mind me borrowing your diagram :| |dw:1338097936559:dw|
anonymous
  • anonymous
Anytime =)
AccessDenied
  • AccessDenied
Ehh, that's how I would have done it too. :P
anonymous
  • anonymous
I got a bit caught up on how to justify the third side being 2.
Callisto
  • Callisto
PS: When I drew the diagram, I found it very interesting that I even forgot I needed to solve the problem :P
Callisto
  • Callisto
Each angle of hexagon = 180 (6-2) /6 = 120 divided by 2 = 60 Base angle isos triangle => all are 60 => equi. triangle
anonymous
  • anonymous
Ah, quite nice.
inkyvoyd
  • inkyvoyd
Guys, you just spoiled the problem for me, ahah
Callisto
  • Callisto
I only know this basic math :( @FoolForMath How do you solve it?
binary3i
  • binary3i
i made a mistake while integrating.
anonymous
  • anonymous
That's a great way to solve it, Callisto. Better to use basic math whenever you can.
anonymous
  • 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
  • anonymous
someone should give @SmoothMath a medal
anonymous
  • anonymous
Well, aren't you kind. I'm not too worried about medals though, and Callisto clearly has the "best answer." =)
experimentX
  • experimentX
i was looking for the solution of this question ... http://openstudy.com/users/foolformath#/updates/4fa421cde4b029e9dc34a843
anonymous
  • anonymous
Keep looking ;)
apoorvk
  • apoorvk
Gotcha, this was a sitter - @Callisto did it in the easiest way possible. Kudos!
Callisto
  • Callisto
@FoolForMath 's solution is easier to understand :)
apoorvk
  • apoorvk
Yea maybe - but before the 'key realisation' - we need the 'key visualisation'. :P
anonymous
  • anonymous
lol, leave it @apoovk she won't accept that her amazing solution.
apoorvk
  • apoorvk
yeah she won't, right never does. :P
anonymous
  • anonymous
that*<- Redundant

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