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How does this proof of the regular dodecahedron's existence fail?
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On Tim Gowers' webpage he has an example "proof" of the regular dodecahedron's existence which he claims contains a flaw.
He writes
Of course, I have not written the above proof in a totally formal way. My question is, where would the difficulty arise if I tried to do so?
which suggests to me that he believes the proof contains a serious flaw that can't be fixed by simply adding more detail.
However I can't detect any such error. So how does the proof fail?
geometry euclidean-geometry fake-proofs polyhedra platonic-solids
asked Jul 28 at 15:08
Oscar Cunningham
9,40922560
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up vote
10
down vote
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On Tim Gowers' webpage he has an example "proof" of the regular dodecahedron's existence which he claims contains a flaw.
He writes
Of course, I have not written the above proof in a totally formal way. My question is, where would the difficulty arise if I tried to do so?
which suggests to me that he believes the proof contains a serious flaw that can't be fixed by simply adding more detail.
However I can't detect any such error. So how does the proof fail?
geometry euclidean-geometry fake-proofs polyhedra platonic-solids
asked Jul 28 at 15:08
Oscar Cunningham
9,40922560
I haven't checked the "proof". Are you sure Gowers is suggesting that the proof fails? All he's asking the reader to do is find the place where making it rigorous might be harder than you think.
– Ethan Bolker
Jul 28 at 15:11
@EthanBolker Well I can't see any particular step which needs lots of effort to formalise either, so I'd be just as interested in finding that out. Of course the regular dodecahedron does in fact exist, so any hole will be patchable eventually.
– Oscar Cunningham
Jul 28 at 15:15
1
@EthanBolker The web page starts out, "What is wrong with the following argument for the existence of a regular dodecahedron...," so it seems there's really an error.
– saulspatz
Jul 28 at 15:18
1
I'm beginning to think that this professor has a personal bugaboo that he obsesses about that others don't consider important. There isn't anything really wrong with the proof that I can. It assumes you can fold up pentagons to a bowl. But you can because the angle of a pentagon is 108. It assumes the edges line up in only one "fit". (that was my guess at the "error") But that's true because the the edges rotated make cones that intersect at a line. That the zigzag is 108 was demonostrated. And that the additional pentagons fit was demonstrated by orientation... So all is good.
– fleablood
Jul 28 at 20:49
1
I think questions should include all the details (for example, by quoting the full proof) to be self-contained, to prevent link rot.
– user202729
Jul 29 at 0:54
 |Â
show 2 more comments
up vote
10
down vote
favorite
up vote
10
down vote
favorite
On Tim Gowers' webpage he has an example "proof" of the regular dodecahedron's existence which he claims contains a flaw.
He writes
Of course, I have not written the above proof in a totally formal way. My question is, where would the difficulty arise if I tried to do so?
which suggests to me that he believes the proof contains a serious flaw that can't be fixed by simply adding more detail.
However I can't detect any such error. So how does the proof fail?
geometry euclidean-geometry fake-proofs polyhedra platonic-solids
asked Jul 28 at 15:08
Oscar Cunningham
9,40922560
On Tim Gowers' webpage he has an example "proof" of the regular dodecahedron's existence which he claims contains a flaw.
He writes
Of course, I have not written the above proof in a totally formal way. My question is, where would the difficulty arise if I tried to do so?
which suggests to me that he believes the proof contains a serious flaw that can't be fixed by simply adding more detail.
However I can't detect any such error. So how does the proof fail?
geometry euclidean-geometry fake-proofs polyhedra platonic-solids
asked Jul 28 at 15:08
Oscar Cunningham
9,40922560
edited Jul 28 at 16:03
asked Jul 28 at 15:08
Oscar Cunningham
9,40922560
asked Jul 28 at 15:08
Oscar Cunningham
9,40922560
asked Jul 28 at 15:08
Oscar Cunningham
9,40922560
9,40922560
I haven't checked the "proof". Are you sure Gowers is suggesting that the proof fails? All he's asking the reader to do is find the place where making it rigorous might be harder than you think.
– Ethan Bolker
Jul 28 at 15:11
@EthanBolker Well I can't see any particular step which needs lots of effort to formalise either, so I'd be just as interested in finding that out. Of course the regular dodecahedron does in fact exist, so any hole will be patchable eventually.
– Oscar Cunningham
Jul 28 at 15:15
1
@EthanBolker The web page starts out, "What is wrong with the following argument for the existence of a regular dodecahedron...," so it seems there's really an error.
– saulspatz
Jul 28 at 15:18
1
I'm beginning to think that this professor has a personal bugaboo that he obsesses about that others don't consider important. There isn't anything really wrong with the proof that I can. It assumes you can fold up pentagons to a bowl. But you can because the angle of a pentagon is 108. It assumes the edges line up in only one "fit". (that was my guess at the "error") But that's true because the the edges rotated make cones that intersect at a line. That the zigzag is 108 was demonostrated. And that the additional pentagons fit was demonstrated by orientation... So all is good.
– fleablood
Jul 28 at 20:49
1
I think questions should include all the details (for example, by quoting the full proof) to be self-contained, to prevent link rot.
– user202729
Jul 29 at 0:54
 |Â
show 2 more comments
I haven't checked the "proof". Are you sure Gowers is suggesting that the proof fails? All he's asking the reader to do is find the place where making it rigorous might be harder than you think.
– Ethan Bolker
Jul 28 at 15:11
@EthanBolker Well I can't see any particular step which needs lots of effort to formalise either, so I'd be just as interested in finding that out. Of course the regular dodecahedron does in fact exist, so any hole will be patchable eventually.
– Oscar Cunningham
Jul 28 at 15:15
1
@EthanBolker The web page starts out, "What is wrong with the following argument for the existence of a regular dodecahedron...," so it seems there's really an error.
– saulspatz
Jul 28 at 15:18
1
I'm beginning to think that this professor has a personal bugaboo that he obsesses about that others don't consider important. There isn't anything really wrong with the proof that I can. It assumes you can fold up pentagons to a bowl. But you can because the angle of a pentagon is 108. It assumes the edges line up in only one "fit". (that was my guess at the "error") But that's true because the the edges rotated make cones that intersect at a line. That the zigzag is 108 was demonostrated. And that the additional pentagons fit was demonstrated by orientation... So all is good.
– fleablood
Jul 28 at 20:49
1
I think questions should include all the details (for example, by quoting the full proof) to be self-contained, to prevent link rot.
– user202729
Jul 29 at 0:54
I haven't checked the "proof". Are you sure Gowers is suggesting that the proof fails? All he's asking the reader to do is find the place where making it rigorous might be harder than you think.
– Ethan Bolker
Jul 28 at 15:11
I haven't checked the "proof". Are you sure Gowers is suggesting that the proof fails? All he's asking the reader to do is find the place where making it rigorous might be harder than you think.
– Ethan Bolker
Jul 28 at 15:11
@EthanBolker Well I can't see any particular step which needs lots of effort to formalise either, so I'd be just as interested in finding that out. Of course the regular dodecahedron does in fact exist, so any hole will be patchable eventually.
– Oscar Cunningham
Jul 28 at 15:15
@EthanBolker Well I can't see any particular step which needs lots of effort to formalise either, so I'd be just as interested in finding that out. Of course the regular dodecahedron does in fact exist, so any hole will be patchable eventually.
– Oscar Cunningham
Jul 28 at 15:15
1
1
@EthanBolker The web page starts out, "What is wrong with the following argument for the existence of a regular dodecahedron...," so it seems there's really an error.
– saulspatz
Jul 28 at 15:18
@EthanBolker The web page starts out, "What is wrong with the following argument for the existence of a regular dodecahedron...," so it seems there's really an error.
– saulspatz
Jul 28 at 15:18
1
1
I'm beginning to think that this professor has a personal bugaboo that he obsesses about that others don't consider important. There isn't anything really wrong with the proof that I can. It assumes you can fold up pentagons to a bowl. But you can because the angle of a pentagon is 108. It assumes the edges line up in only one "fit". (that was my guess at the "error") But that's true because the the edges rotated make cones that intersect at a line. That the zigzag is 108 was demonostrated. And that the additional pentagons fit was demonstrated by orientation... So all is good.
– fleablood
Jul 28 at 20:49
I'm beginning to think that this professor has a personal bugaboo that he obsesses about that others don't consider important. There isn't anything really wrong with the proof that I can. It assumes you can fold up pentagons to a bowl. But you can because the angle of a pentagon is 108. It assumes the edges line up in only one "fit". (that was my guess at the "error") But that's true because the the edges rotated make cones that intersect at a line. That the zigzag is 108 was demonostrated. And that the additional pentagons fit was demonstrated by orientation... So all is good.
– fleablood
Jul 28 at 20:49
1
1
I think questions should include all the details (for example, by quoting the full proof) to be self-contained, to prevent link rot.
– user202729
Jul 29 at 0:54
I think questions should include all the details (for example, by quoting the full proof) to be self-contained, to prevent link rot.
– user202729
Jul 29 at 0:54
 |Â
show 2 more comments
4 Answers
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To find the flaw, tile the plane with regular hexagons and apply the “proof†to the tiling. There's nothing specific to pentagons in it, so it should go through with hexagons, too – but of course it doesn't, as there's no Platonic solid with hexagonal facets.
Everything works out up to the point “if you reflect in $P$, then $G$ maps to $A$â€; but the following statement “$H$ maps to a pentagon that shares an edge with $A$ and $C$†is false for hexagons, and there's no justification why it should be true for pentagons.
answered Jul 28 at 16:30
joriki
164k10179328
" no justification why it should be true for pentagons." The angle of a pentagon is is 108 < 120 = 360/3. That is a justification as to why it would be true for pentagons.
– fleablood
Jul 28 at 16:45
@fleablood: No, that's a justification why you wouldn't get an infinitely repeating tiling for pentagons like you do for hexagons, but it's not a justification for the statement “$H$ maps to a pentagon that shares an edge with $A$ and $C$â€Â, and the inference drawn from it that the pentagons match up.
– joriki
Jul 28 at 16:49
Ah, I think I see your point and I think I agree with it. But the argument "you can't do this with hexagons" fails because hexagons tile the plane and can't "fold up". Pentagons have angles too acute to tile the plane and therefore do fold up. But once they fold up, my argument (and I think yours???) is that we can't know that the edges will line up perfectly exactly once.... well, we can, but that was not argued int the proof ...
– fleablood
Jul 28 at 17:06
3
Specifically, $H$ clearly does map to a pentagon which shares an edge with $C$ since $C$ is preserved under the reflection (and in the hexagonal case, $H$ maps to a hexagon which shares an edge with $C$ also), but the statement that the image of $H$ shares an edge with $A$ is unjustified.
– B. Mehta
Jul 28 at 17:09
2
I wonder if the gap can be fixed by noting that the shared edge of $D$ and $H$ lies on $P$ (and it doesn't for hexagons), so the image of $H$ shares an edge with $C$ and $D$, and thus must be $D$.
– B. Mehta
Jul 28 at 17:12
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It's hard to pin down a specific error, since the construction, as sketched, is a valid one. One line of attack is to ask: which step of the construction is valid for pentagons, but would no longer work if one tried to construct a platonic solid out of, say, heptagons? And that is the very first step: one can always glue $n$ copies of an $n$-gon to a starting $n$-gon, but it is not always possible to fold those copies upwards so that neighbors meet flush at their edges.
The other pentagon-specific argument is the precise accounting of how many pentagons you get at each "tier" of the construction. If you try the construction for hexagons, each step still works (including the reflection argument) but you never stop placing hexagons. That's not an error in the pentagon construction argument, though.
answered Jul 28 at 16:33
user7530
33.3k558109
Why is that not an error in the pentagon construction argument? If the construction doesn't work for hexagons and no reason is provided why it should work for pentagons, isn't than an error in the argument? That is, if we take the argument seriously as the "argument for the existence of a regular dodecahedron" that it claims to be, and not just a construction manual that happens to work. (Not that anyone who's ever bought IKEA furniture would lack appreciation for a construction manual that happens to work...)
– joriki
Jul 28 at 16:36
I don't that that is a valid argument against. "Fleshing" out that the angle of a pentagon is 108 so three is less than 360 can be done extremely easily.
– fleablood
Jul 28 at 16:43
@joriki The part of the construction that breaks down for hexagons, though, is that you no longer get five hexagons in the second step, and one in the last: but you can still perform the reflections?
– user7530
Jul 28 at 16:44
@fleablood: See my response to your comment under my answer.
– joriki
Jul 28 at 16:51
@user7530: Yes, you can perform the reflections -- but I'm not sure how that's relevant to the question. The question was to find the flaw in the argument, and you found it -- I don't see how the fact that you can keep performing reflections mitigates the fact that the purported proof that the reflected pentagons join up is flawed.
– joriki
Jul 28 at 16:53
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This might not be the error but when you fold up the two pentagons B and C (both sharing sides with pentagon A) there is one and only one alignment where the sides of B and C meet. For each angle of the fold in B and each angle in the fold in C there will be a unique orientation of the edges of B and the edges of C. But how do we know any pair of angles will result in the two edges lining up perfectly? Why not some angles a point of mulitple points meet but none where all points meet? Or multiple pairs of angles where all points meet?
I don't think that the "hexagon lie flat" is an error. It's easy to see that the angle of a pentagon is $108 < frac 3603$. Thus we can fold three pentagons up. We can do the same for squares: $90 < frac 3603$ and we can do it for triangles with $3,4$ or even $5$ at a vertex as $60 < frac 3603,frac 3604, frac 3605$.
But we can't do it for hexagons or polygons with more sides and $(textangle of an n-gon; n ge 6)ge frac360k ge 3$. (And to have a solid you need at least three polygons meeting at a vertex.)
answered Jul 28 at 17:01
fleablood
60.3k22575
On the first paragraph: If two lines coincide in two points, they coincide fully. The "proof" is right in arguing that if you tilt all five pentagons simultaneously, there must come a point where their edges touch. They already share a point at the base; as soon as they share another one, they fully coincide. Regarding the second paragraph, I think we've now agreed on where the error lies in the comments under my answer. I didn't claim that "lying flat" had anything to do with it; I also didn't understand user7530 to be saying that, but I may have misunderstood their argument.
– joriki
Jul 28 at 17:24
"If two lines coincide in two points" Well, now that was a poorly thought ought statement on my part. D'oh! " tilt all five pentagons simultaneously, there must come a point where their edges touch". Well my argument is that wasn't adressed that they do ever meet. "as soon as they share another one, they fully coincide." and that they might meet in more than one place. But I think a simple argument that two cones (the paths of the edges form cones) with a common vertex intersect at two lines (reflexive across a plane). I'm not sure that there is any error.
– fleablood
Jul 28 at 20:43
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I think @jorki is right to single out "if you reflect in P, then G maps to A, and H maps to a pentagon that shares an edge with A and C,". I think that conclusion that the image of $H$ shares an edge with $A$ is based on circular reasoning. Let $H'$ be the reflection of $H$ in plane $P$. The justification that $H'$ shares an edge with $A$ seems to be that $A$ is the reflection of $G$ and that $H$ shares an edge with $G$. But wait. He hasn't yet shown that $H$ and $G$ share an edge. In fact that's the very thing that he concludes in the next sentence on the basis that $H'$ (i.e., $D$) shares an edge with $G'$ (i.e., $A$).
answered Jul 30 at 1:21
Theodore Norvell
39119
See @BMeta's amendment under jorki's answer. I think it fixes the problem and thus the proof. In the end you get a neat construction with a proof that is easy for someone like me, who never took Geometry past high school, to follow.
– Theodore Norvell
Jul 30 at 10:35
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
To find the flaw, tile the plane with regular hexagons and apply the “proof†to the tiling. There's nothing specific to pentagons in it, so it should go through with hexagons, too – but of course it doesn't, as there's no Platonic solid with hexagonal facets.
Everything works out up to the point “if you reflect in $P$, then $G$ maps to $A$â€; but the following statement “$H$ maps to a pentagon that shares an edge with $A$ and $C$†is false for hexagons, and there's no justification why it should be true for pentagons.
answered Jul 28 at 16:30
joriki
164k10179328
" no justification why it should be true for pentagons." The angle of a pentagon is is 108 < 120 = 360/3. That is a justification as to why it would be true for pentagons.
– fleablood
Jul 28 at 16:45
@fleablood: No, that's a justification why you wouldn't get an infinitely repeating tiling for pentagons like you do for hexagons, but it's not a justification for the statement “$H$ maps to a pentagon that shares an edge with $A$ and $C$â€Â, and the inference drawn from it that the pentagons match up.
– joriki
Jul 28 at 16:49
Ah, I think I see your point and I think I agree with it. But the argument "you can't do this with hexagons" fails because hexagons tile the plane and can't "fold up". Pentagons have angles too acute to tile the plane and therefore do fold up. But once they fold up, my argument (and I think yours???) is that we can't know that the edges will line up perfectly exactly once.... well, we can, but that was not argued int the proof ...
– fleablood
Jul 28 at 17:06
3
Specifically, $H$ clearly does map to a pentagon which shares an edge with $C$ since $C$ is preserved under the reflection (and in the hexagonal case, $H$ maps to a hexagon which shares an edge with $C$ also), but the statement that the image of $H$ shares an edge with $A$ is unjustified.
– B. Mehta
Jul 28 at 17:09
2
I wonder if the gap can be fixed by noting that the shared edge of $D$ and $H$ lies on $P$ (and it doesn't for hexagons), so the image of $H$ shares an edge with $C$ and $D$, and thus must be $D$.
– B. Mehta
Jul 28 at 17:12
 |Â
show 1 more comment
up vote
5
down vote
accepted
To find the flaw, tile the plane with regular hexagons and apply the “proof†to the tiling. There's nothing specific to pentagons in it, so it should go through with hexagons, too – but of course it doesn't, as there's no Platonic solid with hexagonal facets.
Everything works out up to the point “if you reflect in $P$, then $G$ maps to $A$â€; but the following statement “$H$ maps to a pentagon that shares an edge with $A$ and $C$†is false for hexagons, and there's no justification why it should be true for pentagons.
answered Jul 28 at 16:30
joriki
164k10179328
" no justification why it should be true for pentagons." The angle of a pentagon is is 108 < 120 = 360/3. That is a justification as to why it would be true for pentagons.
– fleablood
Jul 28 at 16:45
@fleablood: No, that's a justification why you wouldn't get an infinitely repeating tiling for pentagons like you do for hexagons, but it's not a justification for the statement “$H$ maps to a pentagon that shares an edge with $A$ and $C$â€Â, and the inference drawn from it that the pentagons match up.
– joriki
Jul 28 at 16:49
Ah, I think I see your point and I think I agree with it. But the argument "you can't do this with hexagons" fails because hexagons tile the plane and can't "fold up". Pentagons have angles too acute to tile the plane and therefore do fold up. But once they fold up, my argument (and I think yours???) is that we can't know that the edges will line up perfectly exactly once.... well, we can, but that was not argued int the proof ...
– fleablood
Jul 28 at 17:06
3
Specifically, $H$ clearly does map to a pentagon which shares an edge with $C$ since $C$ is preserved under the reflection (and in the hexagonal case, $H$ maps to a hexagon which shares an edge with $C$ also), but the statement that the image of $H$ shares an edge with $A$ is unjustified.
– B. Mehta
Jul 28 at 17:09
2
I wonder if the gap can be fixed by noting that the shared edge of $D$ and $H$ lies on $P$ (and it doesn't for hexagons), so the image of $H$ shares an edge with $C$ and $D$, and thus must be $D$.
– B. Mehta
Jul 28 at 17:12
 |Â
show 1 more comment
up vote
5
down vote
accepted
up vote
5
down vote
accepted
To find the flaw, tile the plane with regular hexagons and apply the “proof†to the tiling. There's nothing specific to pentagons in it, so it should go through with hexagons, too – but of course it doesn't, as there's no Platonic solid with hexagonal facets.
Everything works out up to the point “if you reflect in $P$, then $G$ maps to $A$â€; but the following statement “$H$ maps to a pentagon that shares an edge with $A$ and $C$†is false for hexagons, and there's no justification why it should be true for pentagons.
answered Jul 28 at 16:30
joriki
164k10179328
To find the flaw, tile the plane with regular hexagons and apply the “proof†to the tiling. There's nothing specific to pentagons in it, so it should go through with hexagons, too – but of course it doesn't, as there's no Platonic solid with hexagonal facets.
Everything works out up to the point “if you reflect in $P$, then $G$ maps to $A$â€; but the following statement “$H$ maps to a pentagon that shares an edge with $A$ and $C$†is false for hexagons, and there's no justification why it should be true for pentagons.
answered Jul 28 at 16:30
joriki
164k10179328
answered Jul 28 at 16:30
joriki
164k10179328
answered Jul 28 at 16:30
joriki
164k10179328
answered Jul 28 at 16:30
joriki
164k10179328
164k10179328
" no justification why it should be true for pentagons." The angle of a pentagon is is 108 < 120 = 360/3. That is a justification as to why it would be true for pentagons.
– fleablood
Jul 28 at 16:45
@fleablood: No, that's a justification why you wouldn't get an infinitely repeating tiling for pentagons like you do for hexagons, but it's not a justification for the statement “$H$ maps to a pentagon that shares an edge with $A$ and $C$â€Â, and the inference drawn from it that the pentagons match up.
– joriki
Jul 28 at 16:49
Ah, I think I see your point and I think I agree with it. But the argument "you can't do this with hexagons" fails because hexagons tile the plane and can't "fold up". Pentagons have angles too acute to tile the plane and therefore do fold up. But once they fold up, my argument (and I think yours???) is that we can't know that the edges will line up perfectly exactly once.... well, we can, but that was not argued int the proof ...
– fleablood
Jul 28 at 17:06
3
Specifically, $H$ clearly does map to a pentagon which shares an edge with $C$ since $C$ is preserved under the reflection (and in the hexagonal case, $H$ maps to a hexagon which shares an edge with $C$ also), but the statement that the image of $H$ shares an edge with $A$ is unjustified.
– B. Mehta
Jul 28 at 17:09
2
I wonder if the gap can be fixed by noting that the shared edge of $D$ and $H$ lies on $P$ (and it doesn't for hexagons), so the image of $H$ shares an edge with $C$ and $D$, and thus must be $D$.
– B. Mehta
Jul 28 at 17:12
 |Â
show 1 more comment
" no justification why it should be true for pentagons." The angle of a pentagon is is 108 < 120 = 360/3. That is a justification as to why it would be true for pentagons.
– fleablood
Jul 28 at 16:45
@fleablood: No, that's a justification why you wouldn't get an infinitely repeating tiling for pentagons like you do for hexagons, but it's not a justification for the statement “$H$ maps to a pentagon that shares an edge with $A$ and $C$â€Â, and the inference drawn from it that the pentagons match up.
– joriki
Jul 28 at 16:49
Ah, I think I see your point and I think I agree with it. But the argument "you can't do this with hexagons" fails because hexagons tile the plane and can't "fold up". Pentagons have angles too acute to tile the plane and therefore do fold up. But once they fold up, my argument (and I think yours???) is that we can't know that the edges will line up perfectly exactly once.... well, we can, but that was not argued int the proof ...
– fleablood
Jul 28 at 17:06
3
Specifically, $H$ clearly does map to a pentagon which shares an edge with $C$ since $C$ is preserved under the reflection (and in the hexagonal case, $H$ maps to a hexagon which shares an edge with $C$ also), but the statement that the image of $H$ shares an edge with $A$ is unjustified.
– B. Mehta
Jul 28 at 17:09
2
I wonder if the gap can be fixed by noting that the shared edge of $D$ and $H$ lies on $P$ (and it doesn't for hexagons), so the image of $H$ shares an edge with $C$ and $D$, and thus must be $D$.
– B. Mehta
Jul 28 at 17:12
" no justification why it should be true for pentagons." The angle of a pentagon is is 108 < 120 = 360/3. That is a justification as to why it would be true for pentagons.
– fleablood
Jul 28 at 16:45
" no justification why it should be true for pentagons." The angle of a pentagon is is 108 < 120 = 360/3. That is a justification as to why it would be true for pentagons.
– fleablood
Jul 28 at 16:45
@fleablood: No, that's a justification why you wouldn't get an infinitely repeating tiling for pentagons like you do for hexagons, but it's not a justification for the statement “$H$ maps to a pentagon that shares an edge with $A$ and $C$â€Â, and the inference drawn from it that the pentagons match up.
– joriki
Jul 28 at 16:49
@fleablood: No, that's a justification why you wouldn't get an infinitely repeating tiling for pentagons like you do for hexagons, but it's not a justification for the statement “$H$ maps to a pentagon that shares an edge with $A$ and $C$â€Â, and the inference drawn from it that the pentagons match up.
– joriki
Jul 28 at 16:49
Ah, I think I see your point and I think I agree with it. But the argument "you can't do this with hexagons" fails because hexagons tile the plane and can't "fold up". Pentagons have angles too acute to tile the plane and therefore do fold up. But once they fold up, my argument (and I think yours???) is that we can't know that the edges will line up perfectly exactly once.... well, we can, but that was not argued int the proof ...
– fleablood
Jul 28 at 17:06
Ah, I think I see your point and I think I agree with it. But the argument "you can't do this with hexagons" fails because hexagons tile the plane and can't "fold up". Pentagons have angles too acute to tile the plane and therefore do fold up. But once they fold up, my argument (and I think yours???) is that we can't know that the edges will line up perfectly exactly once.... well, we can, but that was not argued int the proof ...
– fleablood
Jul 28 at 17:06
3
3
Specifically, $H$ clearly does map to a pentagon which shares an edge with $C$ since $C$ is preserved under the reflection (and in the hexagonal case, $H$ maps to a hexagon which shares an edge with $C$ also), but the statement that the image of $H$ shares an edge with $A$ is unjustified.
– B. Mehta
Jul 28 at 17:09
Specifically, $H$ clearly does map to a pentagon which shares an edge with $C$ since $C$ is preserved under the reflection (and in the hexagonal case, $H$ maps to a hexagon which shares an edge with $C$ also), but the statement that the image of $H$ shares an edge with $A$ is unjustified.
– B. Mehta
Jul 28 at 17:09
2
2
I wonder if the gap can be fixed by noting that the shared edge of $D$ and $H$ lies on $P$ (and it doesn't for hexagons), so the image of $H$ shares an edge with $C$ and $D$, and thus must be $D$.
– B. Mehta
Jul 28 at 17:12
I wonder if the gap can be fixed by noting that the shared edge of $D$ and $H$ lies on $P$ (and it doesn't for hexagons), so the image of $H$ shares an edge with $C$ and $D$, and thus must be $D$.
– B. Mehta
Jul 28 at 17:12
 |Â
show 1 more comment
up vote
4
down vote
It's hard to pin down a specific error, since the construction, as sketched, is a valid one. One line of attack is to ask: which step of the construction is valid for pentagons, but would no longer work if one tried to construct a platonic solid out of, say, heptagons? And that is the very first step: one can always glue $n$ copies of an $n$-gon to a starting $n$-gon, but it is not always possible to fold those copies upwards so that neighbors meet flush at their edges.
The other pentagon-specific argument is the precise accounting of how many pentagons you get at each "tier" of the construction. If you try the construction for hexagons, each step still works (including the reflection argument) but you never stop placing hexagons. That's not an error in the pentagon construction argument, though.
answered Jul 28 at 16:33
user7530
33.3k558109
Why is that not an error in the pentagon construction argument? If the construction doesn't work for hexagons and no reason is provided why it should work for pentagons, isn't than an error in the argument? That is, if we take the argument seriously as the "argument for the existence of a regular dodecahedron" that it claims to be, and not just a construction manual that happens to work. (Not that anyone who's ever bought IKEA furniture would lack appreciation for a construction manual that happens to work...)
– joriki
Jul 28 at 16:36
I don't that that is a valid argument against. "Fleshing" out that the angle of a pentagon is 108 so three is less than 360 can be done extremely easily.
– fleablood
Jul 28 at 16:43
@joriki The part of the construction that breaks down for hexagons, though, is that you no longer get five hexagons in the second step, and one in the last: but you can still perform the reflections?
– user7530
Jul 28 at 16:44
@fleablood: See my response to your comment under my answer.
– joriki
Jul 28 at 16:51
@user7530: Yes, you can perform the reflections -- but I'm not sure how that's relevant to the question. The question was to find the flaw in the argument, and you found it -- I don't see how the fact that you can keep performing reflections mitigates the fact that the purported proof that the reflected pentagons join up is flawed.
– joriki
Jul 28 at 16:53
add a comment |Â
up vote
4
down vote
It's hard to pin down a specific error, since the construction, as sketched, is a valid one. One line of attack is to ask: which step of the construction is valid for pentagons, but would no longer work if one tried to construct a platonic solid out of, say, heptagons? And that is the very first step: one can always glue $n$ copies of an $n$-gon to a starting $n$-gon, but it is not always possible to fold those copies upwards so that neighbors meet flush at their edges.
The other pentagon-specific argument is the precise accounting of how many pentagons you get at each "tier" of the construction. If you try the construction for hexagons, each step still works (including the reflection argument) but you never stop placing hexagons. That's not an error in the pentagon construction argument, though.
answered Jul 28 at 16:33
user7530
33.3k558109
Why is that not an error in the pentagon construction argument? If the construction doesn't work for hexagons and no reason is provided why it should work for pentagons, isn't than an error in the argument? That is, if we take the argument seriously as the "argument for the existence of a regular dodecahedron" that it claims to be, and not just a construction manual that happens to work. (Not that anyone who's ever bought IKEA furniture would lack appreciation for a construction manual that happens to work...)
– joriki
Jul 28 at 16:36
I don't that that is a valid argument against. "Fleshing" out that the angle of a pentagon is 108 so three is less than 360 can be done extremely easily.
– fleablood
Jul 28 at 16:43
@joriki The part of the construction that breaks down for hexagons, though, is that you no longer get five hexagons in the second step, and one in the last: but you can still perform the reflections?
– user7530
Jul 28 at 16:44
@fleablood: See my response to your comment under my answer.
– joriki
Jul 28 at 16:51
@user7530: Yes, you can perform the reflections -- but I'm not sure how that's relevant to the question. The question was to find the flaw in the argument, and you found it -- I don't see how the fact that you can keep performing reflections mitigates the fact that the purported proof that the reflected pentagons join up is flawed.
– joriki
Jul 28 at 16:53
add a comment |Â
up vote
4
down vote
up vote
4
down vote
It's hard to pin down a specific error, since the construction, as sketched, is a valid one. One line of attack is to ask: which step of the construction is valid for pentagons, but would no longer work if one tried to construct a platonic solid out of, say, heptagons? And that is the very first step: one can always glue $n$ copies of an $n$-gon to a starting $n$-gon, but it is not always possible to fold those copies upwards so that neighbors meet flush at their edges.
The other pentagon-specific argument is the precise accounting of how many pentagons you get at each "tier" of the construction. If you try the construction for hexagons, each step still works (including the reflection argument) but you never stop placing hexagons. That's not an error in the pentagon construction argument, though.
answered Jul 28 at 16:33
user7530
33.3k558109
It's hard to pin down a specific error, since the construction, as sketched, is a valid one. One line of attack is to ask: which step of the construction is valid for pentagons, but would no longer work if one tried to construct a platonic solid out of, say, heptagons? And that is the very first step: one can always glue $n$ copies of an $n$-gon to a starting $n$-gon, but it is not always possible to fold those copies upwards so that neighbors meet flush at their edges.
The other pentagon-specific argument is the precise accounting of how many pentagons you get at each "tier" of the construction. If you try the construction for hexagons, each step still works (including the reflection argument) but you never stop placing hexagons. That's not an error in the pentagon construction argument, though.
answered Jul 28 at 16:33
user7530
33.3k558109
answered Jul 28 at 16:33
user7530
33.3k558109
answered Jul 28 at 16:33
user7530
33.3k558109
answered Jul 28 at 16:33
user7530
33.3k558109
33.3k558109
Why is that not an error in the pentagon construction argument? If the construction doesn't work for hexagons and no reason is provided why it should work for pentagons, isn't than an error in the argument? That is, if we take the argument seriously as the "argument for the existence of a regular dodecahedron" that it claims to be, and not just a construction manual that happens to work. (Not that anyone who's ever bought IKEA furniture would lack appreciation for a construction manual that happens to work...)
– joriki
Jul 28 at 16:36
I don't that that is a valid argument against. "Fleshing" out that the angle of a pentagon is 108 so three is less than 360 can be done extremely easily.
– fleablood
Jul 28 at 16:43
@joriki The part of the construction that breaks down for hexagons, though, is that you no longer get five hexagons in the second step, and one in the last: but you can still perform the reflections?
– user7530
Jul 28 at 16:44
@fleablood: See my response to your comment under my answer.
– joriki
Jul 28 at 16:51
@user7530: Yes, you can perform the reflections -- but I'm not sure how that's relevant to the question. The question was to find the flaw in the argument, and you found it -- I don't see how the fact that you can keep performing reflections mitigates the fact that the purported proof that the reflected pentagons join up is flawed.
– joriki
Jul 28 at 16:53
add a comment |Â
Why is that not an error in the pentagon construction argument? If the construction doesn't work for hexagons and no reason is provided why it should work for pentagons, isn't than an error in the argument? That is, if we take the argument seriously as the "argument for the existence of a regular dodecahedron" that it claims to be, and not just a construction manual that happens to work. (Not that anyone who's ever bought IKEA furniture would lack appreciation for a construction manual that happens to work...)
– joriki
Jul 28 at 16:36
I don't that that is a valid argument against. "Fleshing" out that the angle of a pentagon is 108 so three is less than 360 can be done extremely easily.
– fleablood
Jul 28 at 16:43
@joriki The part of the construction that breaks down for hexagons, though, is that you no longer get five hexagons in the second step, and one in the last: but you can still perform the reflections?
– user7530
Jul 28 at 16:44
@fleablood: See my response to your comment under my answer.
– joriki
Jul 28 at 16:51
@user7530: Yes, you can perform the reflections -- but I'm not sure how that's relevant to the question. The question was to find the flaw in the argument, and you found it -- I don't see how the fact that you can keep performing reflections mitigates the fact that the purported proof that the reflected pentagons join up is flawed.
– joriki
Jul 28 at 16:53
Why is that not an error in the pentagon construction argument? If the construction doesn't work for hexagons and no reason is provided why it should work for pentagons, isn't than an error in the argument? That is, if we take the argument seriously as the "argument for the existence of a regular dodecahedron" that it claims to be, and not just a construction manual that happens to work. (Not that anyone who's ever bought IKEA furniture would lack appreciation for a construction manual that happens to work...)
– joriki
Jul 28 at 16:36
Why is that not an error in the pentagon construction argument? If the construction doesn't work for hexagons and no reason is provided why it should work for pentagons, isn't than an error in the argument? That is, if we take the argument seriously as the "argument for the existence of a regular dodecahedron" that it claims to be, and not just a construction manual that happens to work. (Not that anyone who's ever bought IKEA furniture would lack appreciation for a construction manual that happens to work...)
– joriki
Jul 28 at 16:36
I don't that that is a valid argument against. "Fleshing" out that the angle of a pentagon is 108 so three is less than 360 can be done extremely easily.
– fleablood
Jul 28 at 16:43
I don't that that is a valid argument against. "Fleshing" out that the angle of a pentagon is 108 so three is less than 360 can be done extremely easily.
– fleablood
Jul 28 at 16:43
@joriki The part of the construction that breaks down for hexagons, though, is that you no longer get five hexagons in the second step, and one in the last: but you can still perform the reflections?
– user7530
Jul 28 at 16:44
@joriki The part of the construction that breaks down for hexagons, though, is that you no longer get five hexagons in the second step, and one in the last: but you can still perform the reflections?
– user7530
Jul 28 at 16:44
@fleablood: See my response to your comment under my answer.
– joriki
Jul 28 at 16:51
@fleablood: See my response to your comment under my answer.
– joriki
Jul 28 at 16:51
@user7530: Yes, you can perform the reflections -- but I'm not sure how that's relevant to the question. The question was to find the flaw in the argument, and you found it -- I don't see how the fact that you can keep performing reflections mitigates the fact that the purported proof that the reflected pentagons join up is flawed.
– joriki
Jul 28 at 16:53
@user7530: Yes, you can perform the reflections -- but I'm not sure how that's relevant to the question. The question was to find the flaw in the argument, and you found it -- I don't see how the fact that you can keep performing reflections mitigates the fact that the purported proof that the reflected pentagons join up is flawed.
– joriki
Jul 28 at 16:53
add a comment |Â
up vote
1
down vote
This might not be the error but when you fold up the two pentagons B and C (both sharing sides with pentagon A) there is one and only one alignment where the sides of B and C meet. For each angle of the fold in B and each angle in the fold in C there will be a unique orientation of the edges of B and the edges of C. But how do we know any pair of angles will result in the two edges lining up perfectly? Why not some angles a point of mulitple points meet but none where all points meet? Or multiple pairs of angles where all points meet?
I don't think that the "hexagon lie flat" is an error. It's easy to see that the angle of a pentagon is $108 < frac 3603$. Thus we can fold three pentagons up. We can do the same for squares: $90 < frac 3603$ and we can do it for triangles with $3,4$ or even $5$ at a vertex as $60 < frac 3603,frac 3604, frac 3605$.
But we can't do it for hexagons or polygons with more sides and $(textangle of an n-gon; n ge 6)ge frac360k ge 3$. (And to have a solid you need at least three polygons meeting at a vertex.)
answered Jul 28 at 17:01
fleablood
60.3k22575
On the first paragraph: If two lines coincide in two points, they coincide fully. The "proof" is right in arguing that if you tilt all five pentagons simultaneously, there must come a point where their edges touch. They already share a point at the base; as soon as they share another one, they fully coincide. Regarding the second paragraph, I think we've now agreed on where the error lies in the comments under my answer. I didn't claim that "lying flat" had anything to do with it; I also didn't understand user7530 to be saying that, but I may have misunderstood their argument.
– joriki
Jul 28 at 17:24
"If two lines coincide in two points" Well, now that was a poorly thought ought statement on my part. D'oh! " tilt all five pentagons simultaneously, there must come a point where their edges touch". Well my argument is that wasn't adressed that they do ever meet. "as soon as they share another one, they fully coincide." and that they might meet in more than one place. But I think a simple argument that two cones (the paths of the edges form cones) with a common vertex intersect at two lines (reflexive across a plane). I'm not sure that there is any error.
– fleablood
Jul 28 at 20:43
add a comment |Â
up vote
1
down vote
This might not be the error but when you fold up the two pentagons B and C (both sharing sides with pentagon A) there is one and only one alignment where the sides of B and C meet. For each angle of the fold in B and each angle in the fold in C there will be a unique orientation of the edges of B and the edges of C. But how do we know any pair of angles will result in the two edges lining up perfectly? Why not some angles a point of mulitple points meet but none where all points meet? Or multiple pairs of angles where all points meet?
I don't think that the "hexagon lie flat" is an error. It's easy to see that the angle of a pentagon is $108 < frac 3603$. Thus we can fold three pentagons up. We can do the same for squares: $90 < frac 3603$ and we can do it for triangles with $3,4$ or even $5$ at a vertex as $60 < frac 3603,frac 3604, frac 3605$.
But we can't do it for hexagons or polygons with more sides and $(textangle of an n-gon; n ge 6)ge frac360k ge 3$. (And to have a solid you need at least three polygons meeting at a vertex.)
answered Jul 28 at 17:01
fleablood
60.3k22575
On the first paragraph: If two lines coincide in two points, they coincide fully. The "proof" is right in arguing that if you tilt all five pentagons simultaneously, there must come a point where their edges touch. They already share a point at the base; as soon as they share another one, they fully coincide. Regarding the second paragraph, I think we've now agreed on where the error lies in the comments under my answer. I didn't claim that "lying flat" had anything to do with it; I also didn't understand user7530 to be saying that, but I may have misunderstood their argument.
– joriki
Jul 28 at 17:24
"If two lines coincide in two points" Well, now that was a poorly thought ought statement on my part. D'oh! " tilt all five pentagons simultaneously, there must come a point where their edges touch". Well my argument is that wasn't adressed that they do ever meet. "as soon as they share another one, they fully coincide." and that they might meet in more than one place. But I think a simple argument that two cones (the paths of the edges form cones) with a common vertex intersect at two lines (reflexive across a plane). I'm not sure that there is any error.
– fleablood
Jul 28 at 20:43
add a comment |Â
up vote
1
down vote
up vote
1
down vote
This might not be the error but when you fold up the two pentagons B and C (both sharing sides with pentagon A) there is one and only one alignment where the sides of B and C meet. For each angle of the fold in B and each angle in the fold in C there will be a unique orientation of the edges of B and the edges of C. But how do we know any pair of angles will result in the two edges lining up perfectly? Why not some angles a point of mulitple points meet but none where all points meet? Or multiple pairs of angles where all points meet?
I don't think that the "hexagon lie flat" is an error. It's easy to see that the angle of a pentagon is $108 < frac 3603$. Thus we can fold three pentagons up. We can do the same for squares: $90 < frac 3603$ and we can do it for triangles with $3,4$ or even $5$ at a vertex as $60 < frac 3603,frac 3604, frac 3605$.
But we can't do it for hexagons or polygons with more sides and $(textangle of an n-gon; n ge 6)ge frac360k ge 3$. (And to have a solid you need at least three polygons meeting at a vertex.)
answered Jul 28 at 17:01
fleablood
60.3k22575
This might not be the error but when you fold up the two pentagons B and C (both sharing sides with pentagon A) there is one and only one alignment where the sides of B and C meet. For each angle of the fold in B and each angle in the fold in C there will be a unique orientation of the edges of B and the edges of C. But how do we know any pair of angles will result in the two edges lining up perfectly? Why not some angles a point of mulitple points meet but none where all points meet? Or multiple pairs of angles where all points meet?
I don't think that the "hexagon lie flat" is an error. It's easy to see that the angle of a pentagon is $108 < frac 3603$. Thus we can fold three pentagons up. We can do the same for squares: $90 < frac 3603$ and we can do it for triangles with $3,4$ or even $5$ at a vertex as $60 < frac 3603,frac 3604, frac 3605$.
But we can't do it for hexagons or polygons with more sides and $(textangle of an n-gon; n ge 6)ge frac360k ge 3$. (And to have a solid you need at least three polygons meeting at a vertex.)
answered Jul 28 at 17:01
fleablood
60.3k22575
edited Jul 28 at 17:11
answered Jul 28 at 17:01
fleablood
60.3k22575
answered Jul 28 at 17:01
fleablood
60.3k22575
answered Jul 28 at 17:01
fleablood
60.3k22575
60.3k22575
On the first paragraph: If two lines coincide in two points, they coincide fully. The "proof" is right in arguing that if you tilt all five pentagons simultaneously, there must come a point where their edges touch. They already share a point at the base; as soon as they share another one, they fully coincide. Regarding the second paragraph, I think we've now agreed on where the error lies in the comments under my answer. I didn't claim that "lying flat" had anything to do with it; I also didn't understand user7530 to be saying that, but I may have misunderstood their argument.
– joriki
Jul 28 at 17:24
"If two lines coincide in two points" Well, now that was a poorly thought ought statement on my part. D'oh! " tilt all five pentagons simultaneously, there must come a point where their edges touch". Well my argument is that wasn't adressed that they do ever meet. "as soon as they share another one, they fully coincide." and that they might meet in more than one place. But I think a simple argument that two cones (the paths of the edges form cones) with a common vertex intersect at two lines (reflexive across a plane). I'm not sure that there is any error.
– fleablood
Jul 28 at 20:43
add a comment |Â
On the first paragraph: If two lines coincide in two points, they coincide fully. The "proof" is right in arguing that if you tilt all five pentagons simultaneously, there must come a point where their edges touch. They already share a point at the base; as soon as they share another one, they fully coincide. Regarding the second paragraph, I think we've now agreed on where the error lies in the comments under my answer. I didn't claim that "lying flat" had anything to do with it; I also didn't understand user7530 to be saying that, but I may have misunderstood their argument.
– joriki
Jul 28 at 17:24
"If two lines coincide in two points" Well, now that was a poorly thought ought statement on my part. D'oh! " tilt all five pentagons simultaneously, there must come a point where their edges touch". Well my argument is that wasn't adressed that they do ever meet. "as soon as they share another one, they fully coincide." and that they might meet in more than one place. But I think a simple argument that two cones (the paths of the edges form cones) with a common vertex intersect at two lines (reflexive across a plane). I'm not sure that there is any error.
– fleablood
Jul 28 at 20:43
On the first paragraph: If two lines coincide in two points, they coincide fully. The "proof" is right in arguing that if you tilt all five pentagons simultaneously, there must come a point where their edges touch. They already share a point at the base; as soon as they share another one, they fully coincide. Regarding the second paragraph, I think we've now agreed on where the error lies in the comments under my answer. I didn't claim that "lying flat" had anything to do with it; I also didn't understand user7530 to be saying that, but I may have misunderstood their argument.
– joriki
Jul 28 at 17:24
On the first paragraph: If two lines coincide in two points, they coincide fully. The "proof" is right in arguing that if you tilt all five pentagons simultaneously, there must come a point where their edges touch. They already share a point at the base; as soon as they share another one, they fully coincide. Regarding the second paragraph, I think we've now agreed on where the error lies in the comments under my answer. I didn't claim that "lying flat" had anything to do with it; I also didn't understand user7530 to be saying that, but I may have misunderstood their argument.
– joriki
Jul 28 at 17:24
"If two lines coincide in two points" Well, now that was a poorly thought ought statement on my part. D'oh! " tilt all five pentagons simultaneously, there must come a point where their edges touch". Well my argument is that wasn't adressed that they do ever meet. "as soon as they share another one, they fully coincide." and that they might meet in more than one place. But I think a simple argument that two cones (the paths of the edges form cones) with a common vertex intersect at two lines (reflexive across a plane). I'm not sure that there is any error.
– fleablood
Jul 28 at 20:43
"If two lines coincide in two points" Well, now that was a poorly thought ought statement on my part. D'oh! " tilt all five pentagons simultaneously, there must come a point where their edges touch". Well my argument is that wasn't adressed that they do ever meet. "as soon as they share another one, they fully coincide." and that they might meet in more than one place. But I think a simple argument that two cones (the paths of the edges form cones) with a common vertex intersect at two lines (reflexive across a plane). I'm not sure that there is any error.
– fleablood
Jul 28 at 20:43
add a comment |Â
up vote
1
down vote
I think @jorki is right to single out "if you reflect in P, then G maps to A, and H maps to a pentagon that shares an edge with A and C,". I think that conclusion that the image of $H$ shares an edge with $A$ is based on circular reasoning. Let $H'$ be the reflection of $H$ in plane $P$. The justification that $H'$ shares an edge with $A$ seems to be that $A$ is the reflection of $G$ and that $H$ shares an edge with $G$. But wait. He hasn't yet shown that $H$ and $G$ share an edge. In fact that's the very thing that he concludes in the next sentence on the basis that $H'$ (i.e., $D$) shares an edge with $G'$ (i.e., $A$).
answered Jul 30 at 1:21
Theodore Norvell
39119
See @BMeta's amendment under jorki's answer. I think it fixes the problem and thus the proof. In the end you get a neat construction with a proof that is easy for someone like me, who never took Geometry past high school, to follow.
– Theodore Norvell
Jul 30 at 10:35
add a comment |Â
up vote
1
down vote
I think @jorki is right to single out "if you reflect in P, then G maps to A, and H maps to a pentagon that shares an edge with A and C,". I think that conclusion that the image of $H$ shares an edge with $A$ is based on circular reasoning. Let $H'$ be the reflection of $H$ in plane $P$. The justification that $H'$ shares an edge with $A$ seems to be that $A$ is the reflection of $G$ and that $H$ shares an edge with $G$. But wait. He hasn't yet shown that $H$ and $G$ share an edge. In fact that's the very thing that he concludes in the next sentence on the basis that $H'$ (i.e., $D$) shares an edge with $G'$ (i.e., $A$).
answered Jul 30 at 1:21
Theodore Norvell
39119
See @BMeta's amendment under jorki's answer. I think it fixes the problem and thus the proof. In the end you get a neat construction with a proof that is easy for someone like me, who never took Geometry past high school, to follow.
– Theodore Norvell
Jul 30 at 10:35
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I think @jorki is right to single out "if you reflect in P, then G maps to A, and H maps to a pentagon that shares an edge with A and C,". I think that conclusion that the image of $H$ shares an edge with $A$ is based on circular reasoning. Let $H'$ be the reflection of $H$ in plane $P$. The justification that $H'$ shares an edge with $A$ seems to be that $A$ is the reflection of $G$ and that $H$ shares an edge with $G$. But wait. He hasn't yet shown that $H$ and $G$ share an edge. In fact that's the very thing that he concludes in the next sentence on the basis that $H'$ (i.e., $D$) shares an edge with $G'$ (i.e., $A$).
answered Jul 30 at 1:21
Theodore Norvell
39119
I think @jorki is right to single out "if you reflect in P, then G maps to A, and H maps to a pentagon that shares an edge with A and C,". I think that conclusion that the image of $H$ shares an edge with $A$ is based on circular reasoning. Let $H'$ be the reflection of $H$ in plane $P$. The justification that $H'$ shares an edge with $A$ seems to be that $A$ is the reflection of $G$ and that $H$ shares an edge with $G$. But wait. He hasn't yet shown that $H$ and $G$ share an edge. In fact that's the very thing that he concludes in the next sentence on the basis that $H'$ (i.e., $D$) shares an edge with $G'$ (i.e., $A$).
answered Jul 30 at 1:21
Theodore Norvell
39119
edited Jul 30 at 1:51
answered Jul 30 at 1:21
Theodore Norvell
39119
answered Jul 30 at 1:21
Theodore Norvell
39119
answered Jul 30 at 1:21
Theodore Norvell
39119
39119
See @BMeta's amendment under jorki's answer. I think it fixes the problem and thus the proof. In the end you get a neat construction with a proof that is easy for someone like me, who never took Geometry past high school, to follow.
– Theodore Norvell
Jul 30 at 10:35
add a comment |Â
See @BMeta's amendment under jorki's answer. I think it fixes the problem and thus the proof. In the end you get a neat construction with a proof that is easy for someone like me, who never took Geometry past high school, to follow.
– Theodore Norvell
Jul 30 at 10:35
See @BMeta's amendment under jorki's answer. I think it fixes the problem and thus the proof. In the end you get a neat construction with a proof that is easy for someone like me, who never took Geometry past high school, to follow.
– Theodore Norvell
Jul 30 at 10:35
See @BMeta's amendment under jorki's answer. I think it fixes the problem and thus the proof. In the end you get a neat construction with a proof that is easy for someone like me, who never took Geometry past high school, to follow.
– Theodore Norvell
Jul 30 at 10:35
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I haven't checked the "proof". Are you sure Gowers is suggesting that the proof fails? All he's asking the reader to do is find the place where making it rigorous might be harder than you think.
– Ethan Bolker
Jul 28 at 15:11
@EthanBolker Well I can't see any particular step which needs lots of effort to formalise either, so I'd be just as interested in finding that out. Of course the regular dodecahedron does in fact exist, so any hole will be patchable eventually.
– Oscar Cunningham
Jul 28 at 15:15
1
@EthanBolker The web page starts out, "What is wrong with the following argument for the existence of a regular dodecahedron...," so it seems there's really an error.
– saulspatz
Jul 28 at 15:18
1
I'm beginning to think that this professor has a personal bugaboo that he obsesses about that others don't consider important. There isn't anything really wrong with the proof that I can. It assumes you can fold up pentagons to a bowl. But you can because the angle of a pentagon is 108. It assumes the edges line up in only one "fit". (that was my guess at the "error") But that's true because the the edges rotated make cones that intersect at a line. That the zigzag is 108 was demonostrated. And that the additional pentagons fit was demonstrated by orientation... So all is good.
– fleablood
Jul 28 at 20:49
1
I think questions should include all the details (for example, by quoting the full proof) to be self-contained, to prevent link rot.
– user202729
Jul 29 at 0:54