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In the diagram below, there is a circumference with its [#permalink]
 16 Oct 2003, 03:45
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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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GMAT Instructor
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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!!!
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993
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Re: PS: FOUR CIRCLES (MY TRIBUTE TO AKAMAIBRAH) [#permalink]
16 Oct 2003, 05:27
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, 05:53, edited 2 times in total.
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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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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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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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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993
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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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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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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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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?
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