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In the diagram below, there is a circumference with its

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In the diagram below, there is a circumference with its [#permalink] New post 16 Oct 2003, 02:45
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In the diagram below, there is a circumference with its radius being equal to 1. Three smaller equal curcumferences are inscribed into the bigger one. Find the radius of the smaller circumferences.
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 [#permalink] New post 16 Oct 2003, 03:50
Thanks Stolyar!

Now this is the kind of problem I like!

I hope a lot of people try to solve this one. It is fairly puzzling, yet delightfully straightforward if one uses a little imagination until the "light goes on." A perfect challenge problem.

Good job!!!

:cool
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Re: PS: FOUR CIRCLES (MY TRIBUTE TO AKAMAIBRAH) [#permalink] New post 16 Oct 2003, 04:27
stolyar wrote:
In the diagram below, there is a circumference with its radius being equal to 1. Three smaller equal curcumferences are inscribed into the bigger one. Find the radius of the smaller circumferences.


Let me try

Stolyar needs to cut 3 equal circles out of a Circular sheet of radius 1.
Whats the maximum radius of each circle he can cut.

let radius of each of the smaller circles be r.

3 * PI * r^2 < PI * 1

r < 1/sqrt(3)

i would say... 0 < r < 1/ sqrt(3)

EDIT : On second thoughts, i think the total area of the three circles cannot equal the area of the BIGGER circle. so i would just have to settle for any value among the choices that is between 0 and 1/sqrt(3).

I wish i could give a definite answer.

Thanks
Praetorian

Last edited by Praetorian on 16 Oct 2003, 04:53, edited 2 times in total.
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 [#permalink] New post 16 Oct 2003, 04:55
I would guess it to be less than 0.5, somewhere around 0.48. I have to run for the moment so can't explain the quick observation but for the concrete answer, I'll have to do some detailed calculation. Good question!
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 [#permalink] New post 16 Oct 2003, 05:12
Approximately sqrt(3)/4?

How I did it:

It's easier if you picture this happening or draw it, which ever is easier...
C = Big circle's center
R = C's radius
r = smaller circle's radius

Connect the centers of all three smaller (inscribed) circles to make a triangle. This triangle is equaliteral. Then you draw bisectors of all three angle in the triangle to C. You get three identical 30-120-30 triangles. You just work with one of them now. Connect the midpoint of base of the triangle from C. Now you have R as hypotnus of a smaller 30-60-90 triangle. r is the base of the small triangle. The side opposite of 30 angle is R/4 = 1/4, now you can find the base (or r) which is sqrt(3/16) = sqrt(3)/4.

---
I take back what I said in the previous post. It's even bit less than 0.48.
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 [#permalink] New post 16 Oct 2003, 05:36
wonder_gmat wrote:
Approximately sqrt(3)/4?

I take back what I said in the previous post. It's even bit less than 0.48.


That is not the correct answer.

It is possible to get an EXACT answer without to much work if you see the correct relationship!
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 [#permalink] New post 16 Oct 2003, 07:16
i will give a shot.
The triangle connecting inner circles is an equilateral triangle (because all sides are of same length - 2r) where r - small circles radius.

Now drop a perpendicular from any vertex to the opposite side.
Length of this perpendicular = r*sqrt(3) (30-60-90 triangle with hypotenuse of 2r).

The center of bigger circle is at the midpoint of this perpendicular (the incenter of equilateral triangle)

Thus, r*sqrt(3)/2 + r = 1(this relation is clear after drawing the figure)
=> r = 2/[2+sqrt(3)].
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 [#permalink] New post 16 Oct 2003, 07:55
Nope again; however, you move in the right direction. Triangles, triangles, and triangles again. Frankly speaking, the question is not as difficult as it seems to be. Give it another try.
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 [#permalink] New post 18 Oct 2003, 04:13
So... nobody?

Two methods:

that of Akamaibrah) look at the diagram below. If the center of the small circle is X, the radius of small circle is r and radius of big circle is 1, then segment from center of big circle to X is 1-r. Construct a 30-60-90 triangle as shown with 1-r being the hypotenuese. Note now that the long leg has length r. Since the ratio of the hypotenuese to the long side is 1 to Sqrt(3)/2 we have the relationship:

(1-r)sqrt(3)/2 = r.

Solve for r. r=(sqrt3)/(2+sqrt3)

that of mine) consider two triangles: one with vertices on the smaller circles' centers and the other with vertices on the bigger circle. The triangles are equilateral and similar. The smaller has a side of 2r; the bigger--sqrt3. The answer is the same.
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 [#permalink] New post 18 Oct 2003, 04:33
stolyar wrote:
So... nobody?

Two methods:

that of Akamaibrah) look at the diagram below. If the center of the small circle is X, the radius of small circle is r and radius of big circle is 1, then segment from center of big circle to X is 1-r. Construct a 30-60-90 triangle as shown with 1-r being the hypotenuese. Note now that the long leg has length r. Since the ratio of the hypotenuese to the long side is 1 to Sqrt(3)/2 we have the relationship:

(1-r)sqrt(3)/2 = r.

Solve for r. r=(sqrt3)/(2+sqrt3)

that of mine) consider two triangles: one with vertices on the smaller circles' centers and the other with vertices on the bigger circle. The triangles are equilateral and similar. The smaller has a side of 2r; the bigger--sqrt3. The answer is the same.


stolyar,

could you explain your solution with a diagram?
  [#permalink] 18 Oct 2003, 04:33
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