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From: Cezanne123 on 11 Sep 2006 16:23
I am totally stuck on this problem:

Suppose k >= 3, x, y belongs to R^k, |x - y| = d > 0, and r > 0. Prove:

(a) If 2r > d, there are infinitely many z belonging to R^k such that |z - x| = |z - y| = r.
(b) If 2r = d, there is exactly one such z.
(c) If 2r < d, there is no such z.

How must these statements be modified if k is 2 or 1?

Note: x, y, z are vectors with R^k as the vector space.
From: William Elliot on 12 Sep 2006 06:22
From: Cezanne123 <Cezanne123(a)aol.com>

> Suppose k >= 3, x, y belongs to R^k, |x - y| = d > 0, and r > 0.
> Prove:

> (a) If 2r > d, there are infinitely many z belonging to R^k
> such that |z - x| = |z - y| = r.

> (b) If 2r = d, there is exactly one such z.

> (c) If 2r < d, there is no such z.

Shift problem from y to orgin and solve for y = 0.
Hence
|x| = d > 0
r > 0
|z - x| = |z| = r

Thus (c)
d = |x| = |x - z + z| <= |x - z| + |z| = 2r

and (b)
d = |x| = |x - z + z| <= |x - z| + |z| = d
|z| = |z - x| = d/2
z^2 = z^2 - 2z*x + x^2
d^2 = 2z*x = 2(d/2)d cos(x,z)
cos(x,z) = 1

> How must these statements be modified if k is 2 or 1?

Likely for (a) only.

> Note: x, y, z are vectors with R^k as the vector space.

How are they modified for k = 0? ;-)

----

From: The World Wide Wade on 12 Sep 2006 14:19
In article
<18320796.1158020622344.JavaMail.jakarta(a)nitrogen.mathforum.org>,
Cezanne123 <Cezanne123(a)aol.com> wrote:

> I am totally stuck on this problem:
>
> Suppose k >= 3, x, y belongs to R^k, |x - y| = d > 0, and r > 0. Prove:
>
> (a) If 2r > d, there are infinitely many z belonging to R^k such that |z - x|
> = |z - y| = r.
> (b) If 2r = d, there is exactly one such z.
> (c) If 2r < d, there is no such z.
>
> How must these statements be modified if k is 2 or 1?
>
> Note: x, y, z are vectors with R^k as the vector space.

Hard to believe you are totally stuck. Draw a picture. Right in
front of your face you should see the circle of z's when k = 3.
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