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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
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

This Question is Closed

FoolForMath Group TitleBest ResponseYou've already chosen the best response.2
That would spoil the problem to some extent ;)
 2 years ago

ninhi5 Group TitleBest ResponseYou've already chosen the best response.0
would there be 2 circles, one inside and one outside?
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.2
There is only one stamp.
 2 years ago

ninhi5 Group TitleBest ResponseYou've already chosen the best response.0
well if the edge is outside then the circle must be inscribed inside
 2 years ago

ninhi5 Group TitleBest ResponseYou've already chosen the best response.0
dw:1338096917684:dw
 2 years ago

ninhi5 Group TitleBest ResponseYou've already chosen the best response.0
cant really draw to scale
 2 years ago

ninhi5 Group TitleBest ResponseYou've already chosen the best response.0
is that how it should be?
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.2
@experimentX got it right in the first time itself.
 2 years ago

ninhi5 Group TitleBest ResponseYou've already chosen the best response.0
i'm confused
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.2
me too :
 2 years ago

binary3i Group TitleBest ResponseYou've already chosen the best response.0
is that \[(\pi 3\sqrt{3})/6\]
 2 years ago

SmoothMath Group TitleBest ResponseYou've already chosen the best response.0
dw:1338097283187:dw Not perfect, but I believe the picture should look something like this.
 2 years ago

SmoothMath Group TitleBest ResponseYou've already chosen the best response.0
And the area we are trying to find is: dw:1338097419327:dw
 2 years ago

binary3i Group TitleBest ResponseYou've already chosen the best response.0
\[2\int\limits_{0}^{1/2}(\sqrt{2xx^2}  \sqrt{3}x) dx\]
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.7
4pi/3  2sqrt3?
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.2
I believe Callisto did it :D
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.7
It's actually very easy and interesting :)
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.2
I never post hard problem(s) ;)
 2 years ago

SmoothMath Group TitleBest ResponseYou've already chosen the best response.0
That's kind of you.
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.7
No.... you always post difficult problems :
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
... I'm still waiting for someone to post answer to your perpendicular tangent question
 2 years ago

ninhi5 Group TitleBest ResponseYou've already chosen the best response.0
explanation please
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.7
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
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.7
@SmoothMath Hope you don't mind me borrowing your diagram : dw:1338097936559:dw
 2 years ago

SmoothMath Group TitleBest ResponseYou've already chosen the best response.0
Anytime =)
 2 years ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.0
Ehh, that's how I would have done it too. :P
 2 years ago

SmoothMath Group TitleBest ResponseYou've already chosen the best response.0
I got a bit caught up on how to justify the third side being 2.
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.7
PS: When I drew the diagram, I found it very interesting that I even forgot I needed to solve the problem :P
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.7
Each angle of hexagon = 180 (62) /6 = 120 divided by 2 = 60 Base angle isos triangle => all are 60 => equi. triangle
 2 years ago

SmoothMath Group TitleBest ResponseYou've already chosen the best response.0
Ah, quite nice.
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
Guys, you just spoiled the problem for me, ahah
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.7
I only know this basic math :( @FoolForMath How do you solve it?
 2 years ago

binary3i Group TitleBest ResponseYou've already chosen the best response.0
i made a mistake while integrating.
 2 years ago

SmoothMath Group TitleBest ResponseYou've already chosen the best response.0
That's a great way to solve it, Callisto. Better to use basic math whenever you can.
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.2
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 \( 60^\circ \)
 2 years ago

ninhi5 Group TitleBest ResponseYou've already chosen the best response.0
someone should give @SmoothMath a medal
 2 years ago

SmoothMath Group TitleBest ResponseYou've already chosen the best response.0
Well, aren't you kind. I'm not too worried about medals though, and Callisto clearly has the "best answer." =)
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
i was looking for the solution of this question ... http://openstudy.com/users/foolformath#/updates/4fa421cde4b029e9dc34a843
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.2
Keep looking ;)
 2 years ago

apoorvk Group TitleBest ResponseYou've already chosen the best response.0
Gotcha, this was a sitter  @Callisto did it in the easiest way possible. Kudos!
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.7
@FoolForMath 's solution is easier to understand :)
 2 years ago

apoorvk Group TitleBest ResponseYou've already chosen the best response.0
Yea maybe  but before the 'key realisation'  we need the 'key visualisation'. :P
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.2
lol, leave it @apoovk she won't accept that her amazing solution.
 2 years ago

apoorvk Group TitleBest ResponseYou've already chosen the best response.0
yeah she won't, right never does. :P
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.2
that*< Redundant
 2 years ago
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