Geometry Construction proof
What is the locus of the midpoint of a line segment of varying length where each end of the segment moves around a circle?
Basically speaking if you have two circles, A and B, of any radii and a point on each of the two circles C, and D then the locus of all points M the midpoint of line CD is an annulus with the center being the midpoint of the line connecting the centers. How would I prove this statement? I know the maximum and minimum distance the midpoint can be: (r1+r2)/2 and (r1-r2)/2, respectively. All I have to do is show that for any given point M in the annulus that two points C and D exist such that M is the midpoint of CD. Can you please help? Here is a link to help visualize the problem: Locus of Midpoint of Line joining Two Circles Press animate to begin the animation. Thank you for taking the time to read this and 10 points for best answer!!!
Additional Details
No the actual question is (sorry if I didn't make it clear) If you picked a point M within the annulus region how would construct point C on circle A and point D on circle B such that M is the midpoint of line CD? It has nothing to do with rates (again I am sorry if I didn't make that clear).
12 minutes ago
Re: Geometry Construction proof
I think it is clear that you don't have enough information. Certainly the rate at which each point moves around it circle has a lot to do with this, as well as the radii of the two circles and possibly the starting points. For example, if you were to start with the two points at the points of the circles closest together, the two circles having the same radii, and the points move around the circle at the same rate, one clockwise and the other counter-clockwise, the center point will move on a line segment on the perpendicular bisector of the line segment from the center of one circle to the other. And if you have the same starting position, two circles with the same radii, and the two points moving with the same rate, both clockwise, then the locus of the midpoints will be a single point- the midpoint of the line segment between centers.
Re: Geometry Construction proof
No the actual question is (sorry if I didn't make it clear) If you picked a point M within the annulus region how would construct point C on circle A and point D on circle B such that M is the midpoint of line CD? It has nothing to do with rates (again I am sorry if I didn't make that clear).
Re: Geometry Construction proof
Quote:
Originally Posted by
HallsofIvy 
I think it is clear that you don't have enough information. Certainly the rate at which each point moves around it circle has a lot to do with this, as well as the radii of the two circles and possibly the starting points. For example, if you were to start with the two points at the points of the circles closest together, the two circles having the same radii, and the points move around the circle at the same rate, one clockwise and the other counter-clockwise, the center point will move on a line segment on the perpendicular bisector of the line segment from the center of one circle to the other. And if you have the same starting position, two circles with the same radii, and the two points moving with the same rate, both clockwise, then the locus of the midpoints will be a single point- the midpoint of the line segment between centers.
No the actual question is (sorry if I didn't make it clear) If you picked a point M within the annulus region how would construct point C on circle A and point D on circle B such that M is the midpoint of line CD? It has nothing to do with rates (again I am sorry if I didn't make that clear).
1 Attachment(s)
Re: Geometry Construction proof
Locus of point m=mid-point of two points, one on circle r1, another on circle r2
http://mathhelpforum.com/attachment....1&d=1349655933
Re: Geometry Construction proof
Wait Max Jasper can you tell how me how you came to the answer please? Thank you for your solution!
Re: Geometry Construction proof
Use the following parametric formula:
![x(\theta ,\omega )=\frac{1}{2} \text{r2}* \text{Cos}[\theta ]-\frac{1}{2} \text{r1}* \text{Cos}[\omega ]](http://latex.codecogs.com/png.latex?x(\theta ,\omega )=\frac{1}{2} \text{r2}* \text{Cos}[\theta ]-\frac{1}{2} \text{r1}* \text{Cos}[\omega ])
![y(\theta ,\omega )=\frac{1}{2} \text{r2} *\text{Sin}[\theta ]-\frac{1}{2} \text{r1}* \text{Sin}[\omega ]](http://latex.codecogs.com/png.latex?y(\theta ,\omega )=\frac{1}{2} \text{r2} *\text{Sin}[\theta ]-\frac{1}{2} \text{r1}* \text{Sin}[\omega ])
r2>r1
Re: Geometry Construction proof
Quote:
Originally Posted by
MaxJasper
Wait Max Jasper can you tell how me how you came to the answer please? Thank you for your solution!
Re: Geometry Construction proof
Quote:
Originally Posted by
MaxJasper
But how does this show that the region is an annulus? I'm kind of slow at parameters but I get that it really is just a difference of the x and y values. I mean we get a picture of the annulus but how does this prive. Sorry if this may seem obvious to you but thanks again you have already helped me a great deal!
Re: Geometry Construction proof
Use vectors to define each point on circle r1 & r2....find coordinates of m at the mid-point as a vector ( all vectors from any point (x,y) can do)...then you have the locus of m as a function of parameters of the the two original vectors or the two points on r1 & r2....Simplest is to use complex variables.
Re: Geometry Construction proof
Quote:
Originally Posted by
MaxJasper Use vectors to define each point on circle r1 & r2....find coordinates of m at the mid-point as a vector ( all vectors from any point (x,y) can do)...then you have the locus of m as a function of parameters of the the two original vectors or the two points on r1 & r2....Simplest is to use complex variables.
I see that they are vectors but how does this prove that the region is an annulus? I mean I can clearly see what the max and the min of this region is but how do I know that this curve goes through every single point within that region? I know it is continuous but is there a way that if I pick a point in the annulus I can show that it lies on the curve? Perhaps you have already answered this but thanks again for helping already!
