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## Main Question or Discussion Point

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 [tex]k\geq 3, |\mathbf{x}-\mathbf{y}|=d>0,\text{ and }r>0[/tex]. Prove:

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

I have:

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

I'm not sure this is a valid proof, because it would hold in [tex]\mathbb{R}^2[/tex]. 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 [tex]\mathbf{x}[/tex] and [tex]\mathbf{y}[/tex], 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: [tex]\left|\mathbf{z}-\mathbf{x}\right|=r,\left|\mathbf{z}-\mathbf{y}\right|=r[/tex] and apply the definition of norm, you get a system of two equations and two variables ([tex]z_1,z_2[/tex]). 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 ([tex]z_1,z_2,z_3[/tex]). 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 [tex]\mathbf{z}[/tex].

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

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

(c) If [tex]2r<d[/tex], there is no such [tex]\mathbf{z}[/tex].

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 [tex]\left|\mathbf{z}-\mathbf{x}\right|=\left|\mathbf{z}-\mathbf{y}\right|=r[/tex] and [tex]\left|\mathbf{z}-\mathbf{x}\right|=\left|\mathbf{x}-\mathbf{z}\right|[/tex], then [tex]2r=\left|\mathbf{x}-\mathbf{z}\right|+\left|\mathbf{z}-\mathbf{y}\right|[/tex]. It's given that [tex]\left|\mathbf{x}-\mathbf{y}\right|=d[/tex], so if [tex]2r<d[/tex], then [tex]\left|\mathbf{x}-\mathbf{z}\right|+\left|\mathbf{z}-\mathbf{y}\right|<\left|\mathbf{x}-\mathbf{y}\right|[/tex]. But this contradicts Rudin's Theorem 1.37(f), which says that [tex]\left|\mathbf{x}-\mathbf{z}\right|+\left|\mathbf{z}-\mathbf{y}\right|\geq\left|\mathbf{x}-\mathbf{y}\right|[/tex]. Therefore, no [tex]\mathbf{z}[/tex] exists.

Any input would be appreciated.

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 [tex]k\geq 3, |\mathbf{x}-\mathbf{y}|=d>0,\text{ and }r>0[/tex]. Prove:

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

I have:

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

I'm not sure this is a valid proof, because it would hold in [tex]\mathbb{R}^2[/tex]. 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 [tex]\mathbf{x}[/tex] and [tex]\mathbf{y}[/tex], 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: [tex]\left|\mathbf{z}-\mathbf{x}\right|=r,\left|\mathbf{z}-\mathbf{y}\right|=r[/tex] and apply the definition of norm, you get a system of two equations and two variables ([tex]z_1,z_2[/tex]). 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 ([tex]z_1,z_2,z_3[/tex]). 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 [tex]\mathbf{z}[/tex].

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

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

(c) If [tex]2r<d[/tex], there is no such [tex]\mathbf{z}[/tex].

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 [tex]\left|\mathbf{z}-\mathbf{x}\right|=\left|\mathbf{z}-\mathbf{y}\right|=r[/tex] and [tex]\left|\mathbf{z}-\mathbf{x}\right|=\left|\mathbf{x}-\mathbf{z}\right|[/tex], then [tex]2r=\left|\mathbf{x}-\mathbf{z}\right|+\left|\mathbf{z}-\mathbf{y}\right|[/tex]. It's given that [tex]\left|\mathbf{x}-\mathbf{y}\right|=d[/tex], so if [tex]2r<d[/tex], then [tex]\left|\mathbf{x}-\mathbf{z}\right|+\left|\mathbf{z}-\mathbf{y}\right|<\left|\mathbf{x}-\mathbf{y}\right|[/tex]. But this contradicts Rudin's Theorem 1.37(f), which says that [tex]\left|\mathbf{x}-\mathbf{z}\right|+\left|\mathbf{z}-\mathbf{y}\right|\geq\left|\mathbf{x}-\mathbf{y}\right|[/tex]. Therefore, no [tex]\mathbf{z}[/tex] exists.

Any input would be appreciated.