March 4th, 2010, 09:40 PM  #1 
Member Joined: Jan 2010 Posts: 43 Thanks: 0  Circle Proof
Given a circle centered at point O and an arbitrary point P, consider the locus of all points Y that occur as midpoints of segments PX, where X lies on the given circle. Show that this locus is a circle with radius half that of the original circle. Locate the center of the locus. I'm completely lost with this problem, so any help is greatly appreciated. Thanks!

March 5th, 2010, 04:22 AM  #2 
Senior Member Joined: Nov 2008 Posts: 199 Thanks: 0  Re: Circle Proof
This is really more of a high school geometry question but here's a hint. First suppose your circle is defined by x^2+y^2=r^2, so in other words is centered on the origin in the cartesian plane. Suppose that P is a point on the x axis, (p,0) say. Take some arbitrary point (x,y) on the circle, draw your line between (p,0) and (x,y,) and label the midpoint (x',y'). You should be able to express x' in terms of x and p and y' in terms of y. Having done this you can get expressions for x in terms of x' and p and y in terms of y'. If you substitute these into the original circle equation, with a bit of fiddling you'll see you have another circle equation, this time for x' and y'. This is the circle described by the midpoints and you can read its center off from the equation. To drop the restriction that P lies on the x axis the method is very similar.

March 5th, 2010, 08:32 AM  #3 
Senior Member Joined: Mar 2007 Posts: 428 Thanks: 0  Re: Circle Proof
Precisely the method I used. Calling the locus point (m,n) it is only 4 or 5 steps to write the equation of the the relationship between m and n and see immediately that it is a circle. Hopefully mef1stOpheles [love the name by the way] will give it a try and get back if still having difficulty before it is done for him. I'd point out that it is not necessary to use a circle about the origin, but a more general (x  h)^2 + (y  k)^2 = R^2. The math isn't at all difficult as it need not be expanded nor should be, but involves only a simple substitution from the relation for m and n found from being the midpoint.

March 5th, 2010, 10:02 AM  #4 
Member Joined: Jan 2010 Posts: 43 Thanks: 0  Re: Circle Proof
thanks for the help, though I'm wondering if it is possible to do this proof without cartesian coordinates or circle equations. I'm in a geometry class focused on analytical proofs so my professor doesn't want us to use coordinates nor the equation of a circle. Is there another way to do this without using the above methods?

March 5th, 2010, 11:27 AM  #5 
Senior Member Joined: Nov 2008 Posts: 199 Thanks: 0  Re: Circle Proof
I think you must mean 'synthetic' rather than 'analytic' as analytic geometry is about using coordinate systems in geometry. My feeling is you should be able to use the intuition gained from the analytic proof to cook something up involving triangles if that's what's required. Since you know where the new origin is now you could try showing that the triangles created by the line connecting the point on the circle to the old origin and the line from the point perpendicular to the line from the old origin to P give rise to similar triangles involving the new origin.

March 8th, 2010, 07:33 PM  #6 
Member Joined: Jan 2010 Posts: 43 Thanks: 0  Re: Circle Proof
I'm still having trouble understanding the question and I can't even draw the picture. I do mean synthetic NOT analytic, my apologies about the error before. It doesn't make sense to me that an arbitrary point P will always produce a circle with half the radius

March 8th, 2010, 10:36 PM  #7 
Global Moderator Joined: Dec 2006 Posts: 20,372 Thanks: 2009 
For a diagram, draw two distinct circles, one twice the radius of the other. It doesn't matter whether or not they intersect. Draw a straight line through the center, O, of the larger circle, and the center of the smaller circle. By observing where this line intersects the circles, you can work out where the point P should lie on the line.

March 9th, 2010, 06:19 AM  #8 
Senior Member Joined: Mar 2007 Posts: 428 Thanks: 0  Re: Circle Proof
"A picture is worth a thousand words." For the PC, I found DeltaCad to be excellent in terms of intuitive ease of use. Presently, I am using a full Pro architectural version for the Apple that was given to me by the company since I do my tutoring to raise funds for cancer research. A bit overboard, but something like that is still a viable option. I believe that programs such as Cabri [?] are available for both PC an Apple? Screen capture is good enough for this purpose if there is no direct way to capture a drawn image. The problem is that one can not apparently upload directly from the computer to here, as is done in other similar forums. I'm not knocking this fine forum at all, but would say it might be a decided advantage to be able to do so in this advanced age of computer technology.


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