FoolForMath
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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ninhi5
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diagram?
FoolForMath
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2
That would spoil the problem to some extent ;)
ninhi5
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would there be 2 circles, one inside and one outside?
FoolForMath
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There is only one stamp.
ninhi5
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well if the edge is outside then the circle must be inscribed inside
ninhi5
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|dw:1338096917684:dw|
ninhi5
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cant really draw to scale
ninhi5
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is that how it should be?
FoolForMath
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2
No.
FoolForMath
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@experimentX got it right in the first time itself.
ninhi5
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i'm confused
FoolForMath
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me too :|
binary3i
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is that \[(\pi -3\sqrt{3})/6\]
FoolForMath
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No.
SmoothMath
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|dw:1338097283187:dw|
Not perfect, but I believe the picture should look something like this.
FoolForMath
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2
Yes.
SmoothMath
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And the area we are trying to find is:
|dw:1338097419327:dw|
binary3i
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\[2\int\limits_{0}^{1/2}(\sqrt{2x-x^2} - \sqrt{3}x) dx\]
Callisto
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7
4pi/3 - 2sqrt3?
FoolForMath
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I believe Callisto did it :D
Callisto
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It's actually very easy and interesting :)
FoolForMath
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I never post hard problem(s) ;)
SmoothMath
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That's kind of you.
Callisto
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No.... you always post difficult problems :|
experimentX
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... I'm still waiting for someone to post answer to your perpendicular tangent question
ninhi5
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explanation please
Callisto
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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
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@SmoothMath Hope you don't mind me borrowing your diagram :|
|dw:1338097936559:dw|
SmoothMath
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Anytime =)
AccessDenied
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Ehh, that's how I would have done it too. :P
SmoothMath
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I got a bit caught up on how to justify the third side being 2.
Callisto
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PS: When I drew the diagram, I found it very interesting that I even forgot I needed to solve the problem :P
Callisto
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Each angle of hexagon = 180 (6-2) /6 = 120
divided by 2 = 60
Base angle isos triangle => all are 60 => equi. triangle
SmoothMath
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Ah, quite nice.
inkyvoyd
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Guys, you just spoiled the problem for me, ahah
Callisto
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I only know this basic math :(
@FoolForMath How do you solve it?
binary3i
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i made a mistake while integrating.
SmoothMath
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That's a great way to solve it, Callisto. Better to use basic math whenever you can.
FoolForMath
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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 \( 60^\circ \)
ninhi5
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someone should give @SmoothMath a medal
SmoothMath
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Well, aren't you kind. I'm not too worried about medals though, and Callisto clearly has the "best answer." =)
FoolForMath
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Keep looking ;)
apoorvk
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Gotcha, this was a sitter - @Callisto did it in the easiest way possible. Kudos!
Callisto
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@FoolForMath 's solution is easier to understand :)
apoorvk
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Yea maybe - but before the 'key realisation' - we need the 'key visualisation'. :P
FoolForMath
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lol, leave it @apoovk she won't accept that her amazing solution.
apoorvk
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yeah she won't, right never does. :P
FoolForMath
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that*<- Redundant