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November 6th, 2017, 12:42 AM   #1
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circumscribed circle

How I can prove that the only rectangle that circumscribes a circle (I mean the rectangle is around the circle) is a square?
[I think that is 3 ways: One is geometric, one is algebraic and one is analytic.]
But I don't know how to do and it is also a theorem so it is hard to prove.
So help me to do it.
Thank you...

Last edited by skipjack; November 6th, 2017 at 07:18 AM.
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November 6th, 2017, 07:26 AM   #2
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From each vertex of the rectangle, there are two tangent lines that just reach the circle and each is half the length of a side of the rectangle. A theorem (easily proved if you haven't prior knowledge of it) tells you that the two tangents have equal length, so the sides of the rectangle have equal length (which means it's a square).
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November 6th, 2017, 11:58 AM   #3
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How I can prove it analytically....

If I use a rectangle that start from big rectangle and slowly slowly go into a square so it the minimum rectangle that cover the circle. Can I prove it in this way?

Last edited by skipjack; November 6th, 2017 at 06:54 PM.
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November 6th, 2017, 02:07 PM   #4
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Start from what "big rectangle"? Any rectangle that contains the circle? What mechanism would you use for "going slowly into a square"? Are you going to treat some property of the rectangle as a parameter and change that parameter?

I suppose what you could do is this: choose any point on the circle and construct a tangent to the circle at that point. Also draw the diameter of the circle at that point. At the other end of that diameter, construct the tangent to the circle. You can prove that those two tangents are parallel. The distance between them is the length of the diameter of the circle, 2r where r is the radius of the circle. Choose a point on one side of the first tangent such that a perpendicular to the tangent does not cross the circle. Call the distance from that point to the point of tangency "x". x will be greater than r. Do the same on the other side of the tangent, calling the distance from the point of tangency to the perpendicular "y". Those two tangents, together with the two perpendiculars, form a rectangle that contains the circle. It has two sides of length 2r and two sides of length x+ y. So the rectangle has perimeter 4r+ 2r+ 2y and area 2r(x+ y). Now, "move" those two perpendiculars toward the circle, reducing x and y (those are the "parameters" I mentioned before). They will make contact with the circle, so forming the "smallest" rectangle that contains the circle, when x= y= r. That rectangle has two sides of length 2r, as before, and two sides of length x+ y= 2r also. That is a square with perimeter 4r and area $\displaystyle 4r^2$.

That seems a heck of a lot more complicated than what skipjack suggested!

Last edited by Country Boy; November 6th, 2017 at 02:21 PM.
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