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![[Post New]](/templates/default/images/icon_minipost_new.gif) 23 May 2007 09:58:11 IST
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Two regular hexagons are inscribed in a circle of unit radius then what is the minimum value of their common area ????
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Ken
From: UNITED STATES, Green Bay, Wisconsin
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23 May 2007 10:45:05 IST
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is the answer 6+3root3 / 4 ?
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GBXDHNXDFNHBHRSDRGWEASGSEDH |
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23 May 2007 17:51:18 IST
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well ,
u see for a given amount of volumr a circle is supposed to have minimum area.
now as we will go on increasing the no. of sides in a polygon
then it will tend to a circle,reducing the area...
in this question u have two hexagon,as such when it is put inside a circle
then if it forms a dodecagon(12 sided figure)
it would have the minimum area
now the other half has the application of MAXIMA n MINIMA
(one of the part of calculas),this however is not known by me.
plss tell me if my logic is crt...
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IIT- Imposible Is This(atleast fr meeeeeeeee) |
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23 May 2007 19:43:27 IST
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Nick you are going right and Prathima are you 100 % sure the answer is the one you have mentioned??
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Ken
From: UNITED STATES, Green Bay, Wisconsin
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23 May 2007 20:17:31 IST
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is the ans 3
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23 May 2007 20:26:34 IST
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Are you sure partha....
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Ken
From: UNITED STATES, Green Bay, Wisconsin
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24 May 2007 09:07:01 IST
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nick do u mean all the vertices of dodecogon lies on the circle?
i m getting the answer as 6+3root3 / 4
plz tell me if it was right mr.iitian007
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GBXDHNXDFNHBHRSDRGWEASGSEDH |
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24 May 2007 10:04:09 IST
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they have to PRATHIMA
as it has to be "inscribed" in the circle....
also angle at each vertice will be 150......RIGHT????
using formula ( n-2)180/n
joining to vetices at the centre we have 12 triangles....
angle subtended at the centre by each triangle will be 30
as it bisects each vertices angle....
using trigo. area of this triangle will be
1/2 ab sinC
where a=b=r=1
C=30
= 1/4
no. of triangles=12
therefore minimum area= 12*1/4=3....
which i think should be the anssssssssss
crt me if wronggggggg
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IIT- Imposible Is This(atleast fr meeeeeeeee) |
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24 May 2007 10:12:14 IST
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u r finding the common area of two hexagons whose vertices r on circle
i think the dodecagon vertices do not present on the circle
i dont know whether i m right r wrong
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GBXDHNXDFNHBHRSDRGWEASGSEDH |
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24 May 2007 10:18:47 IST
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GBXDHNXDFNHBHRSDRGWEASGSEDH |
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24 May 2007 10:45:23 IST
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No No No No No No No..............
Prathima.The hexagons are inscribed in the circle so the vertices have to lie on the circle.Thanks for trying prathima but I don't think your answer is correct.i agree with nick's technique But nick, Are you sure your method is right???Well I agree with you.
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Ken
From: UNITED STATES, Green Bay, Wisconsin
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24 May 2007 10:47:12 IST
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ok thank u
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GBXDHNXDFNHBHRSDRGWEASGSEDH |
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24 May 2007 10:58:40 IST
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But WOW WOW WOW WOW WOW WOW WOW WOW WOW WOW WOW WOW ................................................INFINITY
WHERE are the experts ???????? I hope that I am right about this
space bieng an expert panel.
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Ken
From: UNITED STATES, Green Bay, Wisconsin
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24 May 2007 11:04:38 IST
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well ya dude,
i think this is the only...............becoZ
u caant increase or decrease the angles of a polygon which is regular by any means as they are constant
alsoooooooooo u CANNOT BREAK THE SIDES OF THE HEX.
to give u various com. of other sides
annnd finally as i mentioned there might be diff. approach to the Q
through CALCULUS,...............
CHEERSSSS
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IIT- Imposible Is This(atleast fr meeeeeeeee) |
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24 May 2007 11:13:14 IST
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