A lovely one
If , inside a big circle , exacty 36 small circles , each of radius 5 , can be drawn in such a way that each small circle touches the big circle and also touches both its adjacent small circles , then the radius of the big circle is a
Find [ a ]
Clarification -
-
The image is just shown to understand the given situation , here all the small circles are of same radius and each one touches its adjacent one and the circle in which it is inscribed.
-
[.] - Greatest integer function
Note - If you can give a general formula for the radius of the big circle such the there are more than or equal to 3 small circles inscribed in it , you will be appreciated
The answer is 62.
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4 solutions
As for a general formula, with n adjacent circles each of radius r the radius R of the big circle is
R = 1 − cos ( n 3 6 0 ∘ ) 2 ∗ r + r .
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or the general formula can be written as R = r ( 1 + c o s e c n π )
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Yes, that's a much better expression.
1 − cos ( n 2 π ) = 2 ∗ sin 2 ( n π ) , transforming my formula to
R = 2 sin 2 ( n π ) 2 ∗ r + r = r ( 1 + csc ( n π ) )
as you have found.
THEGENERAL FORMULA FOR RADIUS OF BIG CIRCLE (R) TO ACCOMODATE n NO SMALL CIRCLE (RADIUS r ) IS R = r(1+Cosec pi/n) , for n greater than equal to 3.
Give a proof for that formula
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this question is an absolutely fantastic problem!!!the problem maker is a genius!!!thanks for posting this wonderful problem!
In response to Mark Vincent Mamigo, I am up loading two images and two ways of solving.
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I don't know how to put a drawing.. Just imagine this: inside the big circle, there are 36 small circles with radius 5.. If we tend to imagine, connecting the radii will form a polygon with 72 sides.. Dividing one side of the polygon (which is the radius of the smaller circle) will yield 2.5 units.. We now have a right triangle.. The angle opposite the 2.5 units will be 360/72=2.5 degrees.. Getting the value of the hypotenuse, we have 2.5/sine (2.5 degrees)=57.31396407.. Adding 5 (the radius of the smaller circle) will yield the radius of the bigger circle which is 57.31396407+5=62.31396407 units.. Getting its G.I.F., we have 62..
I hope someone can upload an image to my solution for the benefit of everybody.. I really don't know how to upload an image for my solution..
I think it is 36-gon and not 72-gon. Since I can not up load an image directly here I am putting in my solution.
Let centre and radius of bigger circle be C and R respectively and centre of 2 consecutive small circles be O 1 and O 2 .
Join O C 1 C 2 .
Now, ∠ O 1 C O 2 = 3 6 3 6 0 = 1 0 ° O 1 O 2 = 1 0 , C O 1 = C O 2 = R − 5
Using cosine rule, one can get R easily which would be 62.368, so our answer will be 6 2