Existing of non-intersecting rays
$begingroup$
Given $n$ points on a plane, it seems intuitive that it’s possible to draw a ray (half-line) from each point s. t. the $n$ rays do not intersect.
But how to prove this?
geometry
asked 2 hours ago
athosathos
98611340
$endgroup$
add a comment |
$begingroup$
Given $n$ points on a plane, it seems intuitive that it’s possible to draw a ray (half-line) from each point s. t. the $n$ rays do not intersect.
But how to prove this?
geometry
asked 2 hours ago
athosathos
98611340
$endgroup$
add a comment |
$begingroup$
Given $n$ points on a plane, it seems intuitive that it’s possible to draw a ray (half-line) from each point s. t. the $n$ rays do not intersect.
But how to prove this?
geometry
asked 2 hours ago
athosathos
98611340
$endgroup$
Given $n$ points on a plane, it seems intuitive that it’s possible to draw a ray (half-line) from each point s. t. the $n$ rays do not intersect.
But how to prove this?
geometry
geometry
asked 2 hours ago
athosathos
98611340
asked 2 hours ago
athosathos
98611340
asked 2 hours ago
athosathos
98611340
asked 2 hours ago
athosathos
98611340
asked 2 hours ago
athosathos
98611340
98611340
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Pick any point $P$ in the plane that is not on a line containing two or more of the given $n$ points. At each point, draw the ray in the direction away from $P$.
One can in fact do better: It is possible to draw lines through all $n$ points that do not intersect. Choose an orientation that is not parallel to any of the lines between any two of the given points, and draw parallel lines in that orientation through each point.
answered 2 hours ago
FredHFredH
2,6041021
$endgroup$
$begingroup$
+1 for being slightly faster than me :)
$endgroup$
– Severin Schraven
2 hours ago
$begingroup$
Thx ! This is a “oh of course “ moment of me
$endgroup$
– athos
2 hours ago
$begingroup$
It is almost trivial but you need to prove the existence of a point $P$ not collinear to the rest, for example consider the case of points all lying on the same line where this property fails. Luckily the orientation method works in that case too and the existence of such an orientation is guaranteed since picking an arbitary axis the angle of the line between any two points form a finite set.
$endgroup$
– Μάρκος Καραμέρης
1 hour ago
$begingroup$
@ΜάρκοςΚαραμέρης $P$ is not one of the given points. Since a finite set of lines does not exhaust the plane, there are plenty of possible choices.
$endgroup$
– FredH
1 hour ago
$begingroup$
@FredH Ah ok makes perfect sense now!
$endgroup$
– Μάρκος Καραμέρης
1 hour ago
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%2f3160029%2fexisting-of-non-intersecting-rays%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Pick any point $P$ in the plane that is not on a line containing two or more of the given $n$ points. At each point, draw the ray in the direction away from $P$.
One can in fact do better: It is possible to draw lines through all $n$ points that do not intersect. Choose an orientation that is not parallel to any of the lines between any two of the given points, and draw parallel lines in that orientation through each point.
answered 2 hours ago
FredHFredH
2,6041021
$endgroup$
$begingroup$
+1 for being slightly faster than me :)
$endgroup$
– Severin Schraven
2 hours ago
$begingroup$
Thx ! This is a “oh of course “ moment of me
$endgroup$
– athos
2 hours ago
$begingroup$
It is almost trivial but you need to prove the existence of a point $P$ not collinear to the rest, for example consider the case of points all lying on the same line where this property fails. Luckily the orientation method works in that case too and the existence of such an orientation is guaranteed since picking an arbitary axis the angle of the line between any two points form a finite set.
$endgroup$
– Μάρκος Καραμέρης
1 hour ago
$begingroup$
@ΜάρκοςΚαραμέρης $P$ is not one of the given points. Since a finite set of lines does not exhaust the plane, there are plenty of possible choices.
$endgroup$
– FredH
1 hour ago
$begingroup$
@FredH Ah ok makes perfect sense now!
$endgroup$
– Μάρκος Καραμέρης
1 hour ago
add a comment |
$begingroup$
Pick any point $P$ in the plane that is not on a line containing two or more of the given $n$ points. At each point, draw the ray in the direction away from $P$.
One can in fact do better: It is possible to draw lines through all $n$ points that do not intersect. Choose an orientation that is not parallel to any of the lines between any two of the given points, and draw parallel lines in that orientation through each point.
answered 2 hours ago
FredHFredH
2,6041021
$endgroup$
$begingroup$
+1 for being slightly faster than me :)
$endgroup$
– Severin Schraven
2 hours ago
$begingroup$
Thx ! This is a “oh of course “ moment of me
$endgroup$
– athos
2 hours ago
$begingroup$
It is almost trivial but you need to prove the existence of a point $P$ not collinear to the rest, for example consider the case of points all lying on the same line where this property fails. Luckily the orientation method works in that case too and the existence of such an orientation is guaranteed since picking an arbitary axis the angle of the line between any two points form a finite set.
$endgroup$
– Μάρκος Καραμέρης
1 hour ago
$begingroup$
@ΜάρκοςΚαραμέρης $P$ is not one of the given points. Since a finite set of lines does not exhaust the plane, there are plenty of possible choices.
$endgroup$
– FredH
1 hour ago
$begingroup$
@FredH Ah ok makes perfect sense now!
$endgroup$
– Μάρκος Καραμέρης
1 hour ago
add a comment |
$begingroup$
Pick any point $P$ in the plane that is not on a line containing two or more of the given $n$ points. At each point, draw the ray in the direction away from $P$.
One can in fact do better: It is possible to draw lines through all $n$ points that do not intersect. Choose an orientation that is not parallel to any of the lines between any two of the given points, and draw parallel lines in that orientation through each point.
answered 2 hours ago
FredHFredH
2,6041021
$endgroup$
Pick any point $P$ in the plane that is not on a line containing two or more of the given $n$ points. At each point, draw the ray in the direction away from $P$.
One can in fact do better: It is possible to draw lines through all $n$ points that do not intersect. Choose an orientation that is not parallel to any of the lines between any two of the given points, and draw parallel lines in that orientation through each point.
answered 2 hours ago
FredHFredH
2,6041021
edited 2 hours ago
answered 2 hours ago
FredHFredH
2,6041021
answered 2 hours ago
FredHFredH
2,6041021
answered 2 hours ago
FredHFredH
2,6041021
2,6041021
$begingroup$
+1 for being slightly faster than me :)
$endgroup$
– Severin Schraven
2 hours ago
$begingroup$
Thx ! This is a “oh of course “ moment of me
$endgroup$
– athos
2 hours ago
$begingroup$
It is almost trivial but you need to prove the existence of a point $P$ not collinear to the rest, for example consider the case of points all lying on the same line where this property fails. Luckily the orientation method works in that case too and the existence of such an orientation is guaranteed since picking an arbitary axis the angle of the line between any two points form a finite set.
$endgroup$
– Μάρκος Καραμέρης
1 hour ago
$begingroup$
@ΜάρκοςΚαραμέρης $P$ is not one of the given points. Since a finite set of lines does not exhaust the plane, there are plenty of possible choices.
$endgroup$
– FredH
1 hour ago
$begingroup$
@FredH Ah ok makes perfect sense now!
$endgroup$
– Μάρκος Καραμέρης
1 hour ago
add a comment |
$begingroup$
+1 for being slightly faster than me :)
$endgroup$
– Severin Schraven
2 hours ago
$begingroup$
Thx ! This is a “oh of course “ moment of me
$endgroup$
– athos
2 hours ago
$begingroup$
It is almost trivial but you need to prove the existence of a point $P$ not collinear to the rest, for example consider the case of points all lying on the same line where this property fails. Luckily the orientation method works in that case too and the existence of such an orientation is guaranteed since picking an arbitary axis the angle of the line between any two points form a finite set.
$endgroup$
– Μάρκος Καραμέρης
1 hour ago
$begingroup$
@ΜάρκοςΚαραμέρης $P$ is not one of the given points. Since a finite set of lines does not exhaust the plane, there are plenty of possible choices.
$endgroup$
– FredH
1 hour ago
$begingroup$
@FredH Ah ok makes perfect sense now!
$endgroup$
– Μάρκος Καραμέρης
1 hour ago
$begingroup$
+1 for being slightly faster than me :)
$endgroup$
– Severin Schraven
2 hours ago
$begingroup$
+1 for being slightly faster than me :)
$endgroup$
– Severin Schraven
2 hours ago
$begingroup$
Thx ! This is a “oh of course “ moment of me
$endgroup$
– athos
2 hours ago
$begingroup$
Thx ! This is a “oh of course “ moment of me
$endgroup$
– athos
2 hours ago
$begingroup$
It is almost trivial but you need to prove the existence of a point $P$ not collinear to the rest, for example consider the case of points all lying on the same line where this property fails. Luckily the orientation method works in that case too and the existence of such an orientation is guaranteed since picking an arbitary axis the angle of the line between any two points form a finite set.
$endgroup$
– Μάρκος Καραμέρης
1 hour ago
$begingroup$
It is almost trivial but you need to prove the existence of a point $P$ not collinear to the rest, for example consider the case of points all lying on the same line where this property fails. Luckily the orientation method works in that case too and the existence of such an orientation is guaranteed since picking an arbitary axis the angle of the line between any two points form a finite set.
$endgroup$
– Μάρκος Καραμέρης
1 hour ago
$begingroup$
@ΜάρκοςΚαραμέρης $P$ is not one of the given points. Since a finite set of lines does not exhaust the plane, there are plenty of possible choices.
$endgroup$
– FredH
1 hour ago
$begingroup$
@ΜάρκοςΚαραμέρης $P$ is not one of the given points. Since a finite set of lines does not exhaust the plane, there are plenty of possible choices.
$endgroup$
– FredH
1 hour ago
$begingroup$
@FredH Ah ok makes perfect sense now!
$endgroup$
– Μάρκος Καραμέρης
1 hour ago
$begingroup$
@FredH Ah ok makes perfect sense now!
$endgroup$
– Μάρκος Καραμέρης
1 hour ago
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.
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%2f3160029%2fexisting-of-non-intersecting-rays%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
