Wikipedia:Reference desk/Archives/Mathematics/2007 June 23

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June 23

Euclidean spaces

Hello. Suppose $k\geq 3$, x,y are in R^k, |x-y| = d > 0 and r > 0. Prove that if 2r > d, there are infinitely many z in R^k such that |z-x| = |z-y| =r. If 2r = d, there is exactly one such z and if 2r < d there is no such z.

All I can think of is to take a sphere centered at x with radius r and then somehow taking a cone with vertex y and base containing that circle on the sphere for which |z-y| = r. But I have no idea how to prove anything analytically. Any help will be appreciated. --Shahab 10:18, 23 June 2007 (UTC)

For the "no such" part, think triangle inequality, which for R^k is a consequence of the Cauchy–Schwarz inequality. The case of strict equality can be used to handle "exactly one such", although it is not hard to do that from first principles. For 2r > d, visualize two equally large spheres with radius r centred at, respectively, x and y. One sphere is the locus of ||z−x|| = r, the other of ||z−y|| = r. Where do you have that ||z−x|| = ||z−y|| = r? It should not be hard to come up with a formula or construction for some vector z_c parametrized by a value c such that ||z_c-x|| = ||z_c-y|| = r for all c, where different choices of c give you infinitely many different vectors z_c.  --Lambiam^Talk 10:53, 23 June 2007 (UTC)

Thankx for the reply. However I still have trouble in constructing z_c. In the case k = 3 actual spheres intersect and so I can use the result that their intersection is a circle. But I don't know how to do that in general. I think I can handle the other cases. Can you explain the construction of z_c?--Shahab 16:47, 23 June 2007 (UTC)

A fact that may be helpful is that the set of points equidistant from two given points, the perpendicular bisector, is flat. In 2D it's a line; in 3D, a plane; and so on. The set of points at a fixed distance from one given point is a "sphere", in a generalized sense. In 2D it's a circle; in 3D, a spherical surface; and so on. Now intersect. --KSmrq^T 19:23, 23 June 2007 (UTC)

WLOG we can assume that $\mathbf {x}$ lies at the origin and $\mathbf {y} =d\mathbf {e} _{1}+0\mathbf {e} _{2}+0\mathbf {e} _{3}+\cdots +0\mathbf {e} _{k}$. Let $\mathbf {z} (\varphi )=(d/2)\mathbf {e} _{1}+{\sqrt {r^{2}-(d/2)^{2}}}(\mathbf {e_{2}} \sin \varphi +\mathbf {e_{3}} \cos \varphi )$, $0\leq \varphi <2\pi$. These form a circle around $(d/2)\mathbf {e} _{1}$, with the axis along $\mathbf {e} _{1}$ and with radius ${\sqrt {r^{2}-(d/2)^{2}}}$. We verify

${\begin{aligned}||\mathbf {z} (\varphi )-\mathbf {x} ||&={\sqrt {(d/2)^{2}+({\sqrt {r^{2}-(d/2)^{2}}}\sin \varphi )^{2}+({\sqrt {r^{2}-(d/2)^{2}}}\cos \varphi )^{2}}}\\&={\sqrt {(d/2)^{2}+r^{2}-(d/2)^{2}}}=r\end{aligned}}$

and

${\begin{aligned}||\mathbf {z} (\varphi )-\mathbf {y} ||&={\sqrt {(d-d/2)^{2}+({\sqrt {r^{2}-(d/2)^{2}}}\sin \varphi )^{2}+({\sqrt {r^{2}-(d/2)^{2}}}\cos \varphi )^{2}}}\\&={\sqrt {(d/2)^{2}+r^{2}-(d/2)^{2}}}=r\end{aligned}}$.

Did I just do someone's homework? ;-) You could complete it by proving that the number of $\mathbf {z} (\varphi )$, $0\leq \varphi <2\pi$, are infinite. You could also do better than me and find all $\mathbf {z}$ that satisfies the criterion. I only give one circle of solutions, no matter how high the dimension $k$ is. (And find my mistakes. No intentional ones, I promise!) Bonne chance. —Bromskloss 20:31, 23 June 2007 (UTC)