intersecting lines in the projective plane
Consider three points $A, B$ and $C$ in the projective plane [the white points in the picture below] not all on one line. Next, choose a point $O$ [the blue point] and draw the lines that connect $O$ with $A, B$ and $C$ respectively. We will refer to these lines as the "blue lines". Then, we repeat this procedure with another point $O'$ in the plane [the red point in the picture]; the corresponding lines will be called the "red lines".
We then define the points of intersection of the blue line through A and the red line through B, the blue line through B and the red line through C, the blue line through C and the red line through A, and connect these intersection points [the purple points] via lines with $C, A$ and $B$ respectively. Prove that the latter three lines [the white lines in the picture] are concurrent.
I would like to obtain a proof of the statement above by constructing suitable (degenerate) cubics and/or conics, and applying classical theorems.
Thanks
geometry intersection-theory
asked Nov 24 at 14:06
pdm
1264
|
show 1 more comment
Consider three points $A, B$ and $C$ in the projective plane [the white points in the picture below] not all on one line. Next, choose a point $O$ [the blue point] and draw the lines that connect $O$ with $A, B$ and $C$ respectively. We will refer to these lines as the "blue lines". Then, we repeat this procedure with another point $O'$ in the plane [the red point in the picture]; the corresponding lines will be called the "red lines".
We then define the points of intersection of the blue line through A and the red line through B, the blue line through B and the red line through C, the blue line through C and the red line through A, and connect these intersection points [the purple points] via lines with $C, A$ and $B$ respectively. Prove that the latter three lines [the white lines in the picture] are concurrent.
I would like to obtain a proof of the statement above by constructing suitable (degenerate) cubics and/or conics, and applying classical theorems.
Thanks
geometry intersection-theory
asked Nov 24 at 14:06
pdm
1264
Could you add a picture clarifying the two plans you refer to and the various points?
– Moti
Nov 24 at 15:32
Where is O? In what plan is O'? If intersected all are in same plan - right?
– Moti
Nov 24 at 15:33
1
Seems like a variant of Desargues’ theorem.
– amd
Nov 24 at 22:40
@Moti What is not clear about my question? The points O and O' are arbitrarily chosen in the plane. I will add a picture later of a possible configuration.
– pdm
Nov 24 at 23:30
1
@amd: Pappos, not Desargues. Desargues is a $(10_3,10_3)$ configuration: 10 points each incident with 3 lines, and 10 lines each incident with 3 points. Pappos is $(9_3,9_3)$. Since these two are kind of the fundamental theorems in projective incidence geometry, most things boil down to one or the other, and counting helps distinguishing them. As an interesting fact aside, note that Pappos implies Desargues but not vice versa. Not relevant here.
– MvG
Nov 25 at 16:25
|
show 1 more comment
Consider three points $A, B$ and $C$ in the projective plane [the white points in the picture below] not all on one line. Next, choose a point $O$ [the blue point] and draw the lines that connect $O$ with $A, B$ and $C$ respectively. We will refer to these lines as the "blue lines". Then, we repeat this procedure with another point $O'$ in the plane [the red point in the picture]; the corresponding lines will be called the "red lines".
We then define the points of intersection of the blue line through A and the red line through B, the blue line through B and the red line through C, the blue line through C and the red line through A, and connect these intersection points [the purple points] via lines with $C, A$ and $B$ respectively. Prove that the latter three lines [the white lines in the picture] are concurrent.
I would like to obtain a proof of the statement above by constructing suitable (degenerate) cubics and/or conics, and applying classical theorems.
Thanks
geometry intersection-theory
asked Nov 24 at 14:06
pdm
1264
Consider three points $A, B$ and $C$ in the projective plane [the white points in the picture below] not all on one line. Next, choose a point $O$ [the blue point] and draw the lines that connect $O$ with $A, B$ and $C$ respectively. We will refer to these lines as the "blue lines". Then, we repeat this procedure with another point $O'$ in the plane [the red point in the picture]; the corresponding lines will be called the "red lines".
We then define the points of intersection of the blue line through A and the red line through B, the blue line through B and the red line through C, the blue line through C and the red line through A, and connect these intersection points [the purple points] via lines with $C, A$ and $B$ respectively. Prove that the latter three lines [the white lines in the picture] are concurrent.
I would like to obtain a proof of the statement above by constructing suitable (degenerate) cubics and/or conics, and applying classical theorems.
Thanks
geometry intersection-theory
geometry intersection-theory
asked Nov 24 at 14:06
pdm
1264
asked Nov 24 at 14:06
pdm
1264
edited Nov 24 at 23:33
asked Nov 24 at 14:06
pdm
1264
asked Nov 24 at 14:06
pdm
1264
asked Nov 24 at 14:06
pdm
1264
1264
Could you add a picture clarifying the two plans you refer to and the various points?
– Moti
Nov 24 at 15:32
Where is O? In what plan is O'? If intersected all are in same plan - right?
– Moti
Nov 24 at 15:33
1
Seems like a variant of Desargues’ theorem.
– amd
Nov 24 at 22:40
@Moti What is not clear about my question? The points O and O' are arbitrarily chosen in the plane. I will add a picture later of a possible configuration.
– pdm
Nov 24 at 23:30
1
@amd: Pappos, not Desargues. Desargues is a $(10_3,10_3)$ configuration: 10 points each incident with 3 lines, and 10 lines each incident with 3 points. Pappos is $(9_3,9_3)$. Since these two are kind of the fundamental theorems in projective incidence geometry, most things boil down to one or the other, and counting helps distinguishing them. As an interesting fact aside, note that Pappos implies Desargues but not vice versa. Not relevant here.
– MvG
Nov 25 at 16:25
|
show 1 more comment
Could you add a picture clarifying the two plans you refer to and the various points?
– Moti
Nov 24 at 15:32
Where is O? In what plan is O'? If intersected all are in same plan - right?
– Moti
Nov 24 at 15:33
1
Seems like a variant of Desargues’ theorem.
– amd
Nov 24 at 22:40
@Moti What is not clear about my question? The points O and O' are arbitrarily chosen in the plane. I will add a picture later of a possible configuration.
– pdm
Nov 24 at 23:30
1
@amd: Pappos, not Desargues. Desargues is a $(10_3,10_3)$ configuration: 10 points each incident with 3 lines, and 10 lines each incident with 3 points. Pappos is $(9_3,9_3)$. Since these two are kind of the fundamental theorems in projective incidence geometry, most things boil down to one or the other, and counting helps distinguishing them. As an interesting fact aside, note that Pappos implies Desargues but not vice versa. Not relevant here.
– MvG
Nov 25 at 16:25
Could you add a picture clarifying the two plans you refer to and the various points?
– Moti
Nov 24 at 15:32
Could you add a picture clarifying the two plans you refer to and the various points?
– Moti
Nov 24 at 15:32
Where is O? In what plan is O'? If intersected all are in same plan - right?
– Moti
Nov 24 at 15:33
Where is O? In what plan is O'? If intersected all are in same plan - right?
– Moti
Nov 24 at 15:33
1
1
Seems like a variant of Desargues’ theorem.
– amd
Nov 24 at 22:40
Seems like a variant of Desargues’ theorem.
– amd
Nov 24 at 22:40
@Moti What is not clear about my question? The points O and O' are arbitrarily chosen in the plane. I will add a picture later of a possible configuration.
– pdm
Nov 24 at 23:30
@Moti What is not clear about my question? The points O and O' are arbitrarily chosen in the plane. I will add a picture later of a possible configuration.
– pdm
Nov 24 at 23:30
1
1
@amd: Pappos, not Desargues. Desargues is a $(10_3,10_3)$ configuration: 10 points each incident with 3 lines, and 10 lines each incident with 3 points. Pappos is $(9_3,9_3)$. Since these two are kind of the fundamental theorems in projective incidence geometry, most things boil down to one or the other, and counting helps distinguishing them. As an interesting fact aside, note that Pappos implies Desargues but not vice versa. Not relevant here.
– MvG
Nov 25 at 16:25
@amd: Pappos, not Desargues. Desargues is a $(10_3,10_3)$ configuration: 10 points each incident with 3 lines, and 10 lines each incident with 3 points. Pappos is $(9_3,9_3)$. Since these two are kind of the fundamental theorems in projective incidence geometry, most things boil down to one or the other, and counting helps distinguishing them. As an interesting fact aside, note that Pappos implies Desargues but not vice versa. Not relevant here.
– MvG
Nov 25 at 16:25
|
show 1 more comment
2 Answers
2
active
oldest
votes
Inspect the combinatorics of your configuration. You have $9$ points, each of which is incident with $3$ lines of the configuration. Conversely you have $9$ lines, each incident with $3$ points. So the whole configuration is a $(9_3,9_3)$ configuration. There is just one such configuration which corresponds to a projective incidence theorem, namely Pappos' theorem (or Pappus', if you prefer that transliteration). Take a close look and you should be able to identify that configuration.
If you want to, you can view Pappos' theorem as a special case of Pascal's theorem, where the conic degenerates to a pair of lines. Or you might consider triplets of lines as a degenerate cubic and then treat Pappos' theorem as a special case of the Cayley-Bacharach theorem.
answered Nov 25 at 8:33
MvG
30.7k449101
add a comment |
The first step is to use the projective feature of the plane in the following way -
Let me know if this get you closer to the solution you looked for
answered Nov 26 at 2:56
Moti
1,335712
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3011591%2fintersecting-lines-in-the-projective-plane%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Inspect the combinatorics of your configuration. You have $9$ points, each of which is incident with $3$ lines of the configuration. Conversely you have $9$ lines, each incident with $3$ points. So the whole configuration is a $(9_3,9_3)$ configuration. There is just one such configuration which corresponds to a projective incidence theorem, namely Pappos' theorem (or Pappus', if you prefer that transliteration). Take a close look and you should be able to identify that configuration.
If you want to, you can view Pappos' theorem as a special case of Pascal's theorem, where the conic degenerates to a pair of lines. Or you might consider triplets of lines as a degenerate cubic and then treat Pappos' theorem as a special case of the Cayley-Bacharach theorem.
answered Nov 25 at 8:33
MvG
30.7k449101
add a comment |
Inspect the combinatorics of your configuration. You have $9$ points, each of which is incident with $3$ lines of the configuration. Conversely you have $9$ lines, each incident with $3$ points. So the whole configuration is a $(9_3,9_3)$ configuration. There is just one such configuration which corresponds to a projective incidence theorem, namely Pappos' theorem (or Pappus', if you prefer that transliteration). Take a close look and you should be able to identify that configuration.
If you want to, you can view Pappos' theorem as a special case of Pascal's theorem, where the conic degenerates to a pair of lines. Or you might consider triplets of lines as a degenerate cubic and then treat Pappos' theorem as a special case of the Cayley-Bacharach theorem.
answered Nov 25 at 8:33
MvG
30.7k449101
add a comment |
Inspect the combinatorics of your configuration. You have $9$ points, each of which is incident with $3$ lines of the configuration. Conversely you have $9$ lines, each incident with $3$ points. So the whole configuration is a $(9_3,9_3)$ configuration. There is just one such configuration which corresponds to a projective incidence theorem, namely Pappos' theorem (or Pappus', if you prefer that transliteration). Take a close look and you should be able to identify that configuration.
If you want to, you can view Pappos' theorem as a special case of Pascal's theorem, where the conic degenerates to a pair of lines. Or you might consider triplets of lines as a degenerate cubic and then treat Pappos' theorem as a special case of the Cayley-Bacharach theorem.
answered Nov 25 at 8:33
MvG
30.7k449101
Inspect the combinatorics of your configuration. You have $9$ points, each of which is incident with $3$ lines of the configuration. Conversely you have $9$ lines, each incident with $3$ points. So the whole configuration is a $(9_3,9_3)$ configuration. There is just one such configuration which corresponds to a projective incidence theorem, namely Pappos' theorem (or Pappus', if you prefer that transliteration). Take a close look and you should be able to identify that configuration.
If you want to, you can view Pappos' theorem as a special case of Pascal's theorem, where the conic degenerates to a pair of lines. Or you might consider triplets of lines as a degenerate cubic and then treat Pappos' theorem as a special case of the Cayley-Bacharach theorem.
answered Nov 25 at 8:33
MvG
30.7k449101
edited Nov 25 at 16:31
answered Nov 25 at 8:33
MvG
30.7k449101
answered Nov 25 at 8:33
MvG
30.7k449101
answered Nov 25 at 8:33
MvG
30.7k449101
30.7k449101
add a comment |
add a comment |
The first step is to use the projective feature of the plane in the following way -
Let me know if this get you closer to the solution you looked for
answered Nov 26 at 2:56
Moti
1,335712
add a comment |
The first step is to use the projective feature of the plane in the following way -
Let me know if this get you closer to the solution you looked for
answered Nov 26 at 2:56
Moti
1,335712
add a comment |
The first step is to use the projective feature of the plane in the following way -
Let me know if this get you closer to the solution you looked for
answered Nov 26 at 2:56
Moti
1,335712
The first step is to use the projective feature of the plane in the following way -
Let me know if this get you closer to the solution you looked for
answered Nov 26 at 2:56
Moti
1,335712
answered Nov 26 at 2:56
Moti
1,335712
answered Nov 26 at 2:56
Moti
1,335712
answered Nov 26 at 2:56
Moti
1,335712
1,335712
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3011591%2fintersecting-lines-in-the-projective-plane%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Could you add a picture clarifying the two plans you refer to and the various points?
– Moti
Nov 24 at 15:32
Where is O? In what plan is O'? If intersected all are in same plan - right?
– Moti
Nov 24 at 15:33
1
Seems like a variant of Desargues’ theorem.
– amd
Nov 24 at 22:40
@Moti What is not clear about my question? The points O and O' are arbitrarily chosen in the plane. I will add a picture later of a possible configuration.
– pdm
Nov 24 at 23:30
1
@amd: Pappos, not Desargues. Desargues is a $(10_3,10_3)$ configuration: 10 points each incident with 3 lines, and 10 lines each incident with 3 points. Pappos is $(9_3,9_3)$. Since these two are kind of the fundamental theorems in projective incidence geometry, most things boil down to one or the other, and counting helps distinguishing them. As an interesting fact aside, note that Pappos implies Desargues but not vice versa. Not relevant here.
– MvG
Nov 25 at 16:25