# Baby Rudin problem driving me crazy

1. Feb 15, 2010

### slipperypete

I started my way through baby Rudin recently, been able to answer every question so far, but I just can't get #16 from Chapter 1.

I (think) understand the geometry of it perfectly. I just can't prove it rigorously/analytically. I haven't been able to do anything for 3 days now, except think about this problem. So any help would be appreciated.

Suppose $$k\geq 3, |\mathbf{x}-\mathbf{y}|=d>0,\text{ and }r>0$$. Prove:

(a) If 2r>d, there are infinitely many $$\mathbf{z}\in\mathbb{R}^k$$ such that $$\left|\mathbf{z}-\mathbf{x}\right|=\left|\mathbf{z}-\mathbf{y}\right|=r$$.

I have:

Since $$\left|\mathbf{z}-\mathbf{x}\right|=\left|\mathbf{z}-\mathbf{y}\right|=r$$, then $$2r=\left|\mathbf{z}-\mathbf{x}\right|+\left|\mathbf{z}-\mathbf{y}\right|$$. Also, so $$\left|\mathbf{z}-\mathbf{x}\right|=\left|\mathbf{x}-\mathbf{z}\right|$$, so $$2r=\left|\mathbf{z}-\mathbf{x}\right|+\left|\mathbf{z}-\mathbf{y}\right|=\left|\mathbf{x}-\mathbf{z}\right|+\left|\mathbf{z}-\mathbf{y}\right|$$. By Rudin's Proposition 1.37(f), $$\left|\mathbf{x}-\mathbf{y}\right|\leq\left|\mathbf{x}-\mathbf{z}\right|+\left|\mathbf{z}-\mathbf{y}\right|$$ for any $$\mathbf{x},\mathbf{y},\mathbf{z}\in\mathbb{R}^{k}$$. Since $$\left|\mathbf{x}-\mathbf{y}\right|=d$$, this proposition makes it is clear that, given $$\mathbf{x},\mathbf{y}\in\mathbb{R}^{k}$$, there exists infinitely many $$\mathbf{z}\in\mathbb{R}^{k}$$ satisfying the equations $$2r>d,\left|\mathbf{z}-\mathbf{x}\right|=\left|\mathbf{z}-\mathbf{y}\right|=r$$.

I'm not sure this is a valid proof, because it would hold in $$\mathbb{R}^2$$. But if I'm imagining the geometry correctly, there should be exactly 2 solutions in 2-space (i.e., the 2 points at which the circles of radius r, drawn about $$\mathbf{x}$$ and $$\mathbf{y}$$, intersect), not infinitely many. So I'm sure I'm missing something, but I don't see any holes in the proof.

I've also observed that under 2-space, if you square these equations: $$\left|\mathbf{z}-\mathbf{x}\right|=r,\left|\mathbf{z}-\mathbf{y}\right|=r$$ and apply the definition of norm, you get a system of two equations and two variables ($$z_1,z_2$$). I know there's a solution there, which meshes perfectly with my understanding of the geometry behind this problem. Moreover, if you move on to 3-space and proceed in the same way, you get a system of two equations and three variables ($$z_1,z_2,z_3$$). I know that this system will have infinitely many solutions. But I don't know how to prove rigorously that the 2-variable system is solvable or that the 3-variable system is unsolvable. Also, the way I understand the problem, r is fixed. But it is subject to some constraint (e.g., 2r>d). I don't know how to incorporate this constraint into the proof.

(b) If 2r=d, there is exactly one such $$\mathbf{z}$$.

I can't even figure out where to begin. I've noticed that if you set $$\mathbf{z}=\tfrac{1}{2}\left(\mathbf{x}+\mathbf{y}\right)$$, then $$\left|\mathbf{z}-\mathbf{x}\right|=\left|\mathbf{z}-\mathbf{y}\right|$$. Also, this makes $$2r=2\left|\mathbf{z}-\mathbf{x}\right|=2\left|\tfrac{\mathbf{x}+\mathbf{y}}{2}-\mathbf{x}\right|=\left|\mathbf{y}-\mathbf{x}\right|=d$$.

I can derive this "solution" for $$\mathbf{z}$$ by squaring $$\left|\mathbf{z}-\mathbf{x}\right|=\left|\mathbf{z}-\mathbf{y}\right|$$ and going from there. But no relationship between 2r and d is needed to figure this out, so it would seem that $$\mathbf{z}=\tfrac{1}{2}\left(\mathbf{x}+\mathbf{y}\right)$$ should be true in the cases of 2r>d and 2r<d. So I'm stuck on this one.

(c) If $$2r<d$$, there is no such $$\mathbf{z}$$.

This is the one I think I've made the most progress on. I think Rudin's Theorem 1.37(f) should apply. Let me know if this proof works/doesn't work:

Assume a solution does exist. Since $$\left|\mathbf{z}-\mathbf{x}\right|=\left|\mathbf{z}-\mathbf{y}\right|=r$$ and $$\left|\mathbf{z}-\mathbf{x}\right|=\left|\mathbf{x}-\mathbf{z}\right|$$, then $$2r=\left|\mathbf{x}-\mathbf{z}\right|+\left|\mathbf{z}-\mathbf{y}\right|$$. It's given that $$\left|\mathbf{x}-\mathbf{y}\right|=d$$, so if $$2r<d$$, then $$\left|\mathbf{x}-\mathbf{z}\right|+\left|\mathbf{z}-\mathbf{y}\right|<\left|\mathbf{x}-\mathbf{y}\right|$$. But this contradicts Rudin's Theorem 1.37(f), which says that $$\left|\mathbf{x}-\mathbf{z}\right|+\left|\mathbf{z}-\mathbf{y}\right|\geq\left|\mathbf{x}-\mathbf{y}\right|$$. Therefore, no $$\mathbf{z}$$ exists.

Any input would be appreciated.

2. May 17, 2012