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Home » Mathematics » High School Intermediate Topics (Ages 13+)
 
Circle Problem
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helijet
P versus NP
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#1
Circle Problem
A large circle is tangent to each of the sides of a square. A small circle is tangent to the large circle and two sides of the square as shown. If the sides of the square have length 1, then how would you find the radius of the smaller circle?
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PostPosted: Sat Nov 07, 2009 1:51 pm  Back to top 
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ckck
Yang-Mills Theory
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#2
Re: Circle Problem
helijet wrote:
A large circle is tangent to each of the sides of a square. A small circle is tangent to the large circle and two sides of the square as shown. If the sides of the square have length 1, then how would you find the radius of the smaller circle?

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From the center of the larger circle draw the radius that runs through the smaller circle. If you notice, this length is half the length of the diagonal of the square.
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PostPosted: Sat Nov 07, 2009 5:23 pm  Back to top 
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helijet
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#3
Re: Circle Problem
ckck wrote:
helijet wrote:
A large circle is tangent to each of the sides of a square. A small circle is tangent to the large circle and two sides of the square as shown. If the sides of the square have length 1, then how would you find the radius of the smaller circle?

Click to reveal hidden content
From the center of the larger circle draw the radius that runs through the smaller circle. If you notice, this length is half the length of the diagonal of the square.


I have done what you have suggested before and I know that it is easy to find the length of the diagonal of the square just by drawing in the right triangle and find its hypotenuse. Notice that there is a little space in between the corner of the square and the smaller circle. The problem is how to find the distance between the corner of the square and the smaller circle. Here is a bigger picture so it is easier to spot the point.
Circle Problem.jpg
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PostPosted: Sat Nov 07, 2009 5:36 pm  Back to top 
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ckck
Yang-Mills Theory
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#4
Re: Circle Problem
helijet wrote:
ckck wrote:
helijet wrote:
A large circle is tangent to each of the sides of a square. A small circle is tangent to the large circle and two sides of the square as shown. If the sides of the square have length 1, then how would you find the radius of the smaller circle?

Click to reveal hidden content
From the center of the larger circle draw the radius that runs through the smaller circle. If you notice, this length is half the length of the diagonal of the square.


I have done what you have suggested before and I know that it is easy to find the length of the diagonal of the square just by drawing in the right triangle and find its hypotenuse. Notice that there is a little space in between the corner of the square and the smaller circle. The problem is how to find the distance between the corner of the square and the smaller circle. Here is a bigger picture so it is easier to spot the point.

Click to reveal hidden content
From the center of that smaller circle, draw the line from there to the vertex. As you said, there is a small bit of space. However it can be solved. Let r denote the radius of the smaller circle. Now from the center of the smaller circle, draw two lines, one which connects the center to the top side of the square, and one which connects the center to the left side of the square. Now you should have a square which contains that length. That length is r\sqrt{2} (hope you can see why). I think you know how to solve the rest of this. Smile

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PostPosted: Sat Nov 07, 2009 5:42 pm  Back to top 
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helijet
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#5
Would this be the answer to this problem because this is as far as I can get:
0.5\sqrt{2} -0.5 -r\sqrt{2} + r / 2

Unfortunately none of the choices match this answer. Sad

PostPosted: Sat Nov 07, 2009 6:39 pm  Back to top 
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ckck
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#6
Well, the length of that diagonal of the smaller square would be r\sqrt{2}. Thus the entire length of the diagonal in terms of the radii of the two circles is r\sqrt{2}+r+\frac{1}{2}. Also the length of this diagonal is \frac{\sqrt{2}}{2}. Setting these two equal and solve for r.
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PostPosted: Sat Nov 07, 2009 7:37 pm  Back to top 
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kanakmangal
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#7
3/2 - sqrt(2)

PostPosted: Wed Nov 11, 2009 9:09 am  Back to top 
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castigioni
Poincare Conjecture
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#8
The commom tangent to the circles determine an isosceles triangle whose minor circle is the incircle.The inradius is r=A/s
whose value is r=3/2-2^.5

PostPosted: Wed Nov 11, 2009 11:22 am  Back to top 
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