Existing of non-intersecting raysConstructing a circle through a point in the interior of an angleHow many...
What was the exact wording from Ivanhoe of this advice on how to free yourself from slavery?
Biological Blimps: Propulsion
Why did the EU agree to delay the Brexit deadline?
Longest common substring in linear time
What are the purposes of autoencoders?
A social experiment. What is the worst that can happen?
Drawing ramified coverings with tikz
Is it safe to use olive oil to clean the ear wax?
Redundant comparison & "if" before assignment
What is this cable/device?
It grows, but water kills it
Store Credit Card Information in Password Manager?
What does routing an IP address mean?
The IT department bottlenecks progress. How should I handle this?
Travelling outside the UK without a passport
why `nmap 192.168.1.97` returns less services than `nmap 127.0.0.1`?
Why should universal income be universal?
Multiplicative persistence
If a character has darkvision, can they see through an area of nonmagical darkness filled with lightly obscuring gas?
Count the occurrence of each unique word in the file
Should I stop contributing to retirement accounts?
Yosemite Fire Rings - What to Expect?
Are the IPv6 address space and IPv4 address space completely disjoint?
How much character growth crosses the line into breaking the character
Existing of non-intersecting rays
Constructing a circle through a point in the interior of an angleHow many rays can made from $4$ collinear points?Angle between different rays (3d line segments) and computing their angular relationshipsIntersecting three rays and a sphere of known radiusDesigning a distance function between raysLines between point and sphere surface intersecting a planeIntersecting planes stereometry problemIs a single line, line segment, or ray a valid angle?Coxeter, Introduction to Geometry, ordered geometry, parallelism of rays and linesNon-congruent angle of an isosceles triangle
$begingroup$
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 14 mins ago
athosathos
98411340
$endgroup$
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 14 mins ago
athosathos
98411340
$endgroup$
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 14 mins ago
athosathos
98411340
$endgroup$
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 14 mins ago
athosathos
98411340
asked 14 mins ago
athosathos
98411340
asked 14 mins ago
athosathos
98411340
asked 14 mins ago
athosathos
98411340
asked 14 mins ago
athosathos
98411340
98411340
add a comment |
add a comment |
2 Answers
2
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 $P$.
answered 3 mins ago
FredHFredH
2,5891021
$endgroup$
$begingroup$
+1 for being slightly faster than me :)
$endgroup$
– Severin Schraven
2 mins ago
$begingroup$
Thx ! This is a “oh of course “ moment of me
$endgroup$
– athos
1 min ago
add a comment |
$begingroup$
I would do it by induction. Say you already have $n$ points, each with a ray, and all rays are non-intersecting. You then place a new point in the plane. Because all of the rays already in the plane are non-intersecting, it must be that this point is not completely enclosed by rays. Because it is not completely enclosed, it must be possible for it to move continually away from its current position on some straight line path, forever, without hitting a ray. This path itself is a ray from the point which intersects no others! The base case of the induction is obviously true, so the proof is complete.
answered 1 min ago
CyborgOctopusCyborgOctopus
1407
$endgroup$
$begingroup$
No . the new point might fall on an existing line
$endgroup$
– athos
1 min 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
2 Answers
2
active
oldest
votes
2 Answers
2
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 $P$.
answered 3 mins ago
FredHFredH
2,5891021
$endgroup$
$begingroup$
+1 for being slightly faster than me :)
$endgroup$
– Severin Schraven
2 mins ago
$begingroup$
Thx ! This is a “oh of course “ moment of me
$endgroup$
– athos
1 min 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 $P$.
answered 3 mins ago
FredHFredH
2,5891021
$endgroup$
$begingroup$
+1 for being slightly faster than me :)
$endgroup$
– Severin Schraven
2 mins ago
$begingroup$
Thx ! This is a “oh of course “ moment of me
$endgroup$
– athos
1 min 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 $P$.
answered 3 mins ago
FredHFredH
2,5891021
$endgroup$
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 $P$.
answered 3 mins ago
FredHFredH
2,5891021
answered 3 mins ago
FredHFredH
2,5891021
answered 3 mins ago
FredHFredH
2,5891021
answered 3 mins ago
FredHFredH
2,5891021
2,5891021
$begingroup$
+1 for being slightly faster than me :)
$endgroup$
– Severin Schraven
2 mins ago
$begingroup$
Thx ! This is a “oh of course “ moment of me
$endgroup$
– athos
1 min ago
add a comment |
$begingroup$
+1 for being slightly faster than me :)
$endgroup$
– Severin Schraven
2 mins ago
$begingroup$
Thx ! This is a “oh of course “ moment of me
$endgroup$
– athos
1 min ago
$begingroup$
+1 for being slightly faster than me :)
$endgroup$
– Severin Schraven
2 mins ago
$begingroup$
+1 for being slightly faster than me :)
$endgroup$
– Severin Schraven
2 mins ago
$begingroup$
Thx ! This is a “oh of course “ moment of me
$endgroup$
– athos
1 min ago
$begingroup$
Thx ! This is a “oh of course “ moment of me
$endgroup$
– athos
1 min ago
add a comment |
$begingroup$
I would do it by induction. Say you already have $n$ points, each with a ray, and all rays are non-intersecting. You then place a new point in the plane. Because all of the rays already in the plane are non-intersecting, it must be that this point is not completely enclosed by rays. Because it is not completely enclosed, it must be possible for it to move continually away from its current position on some straight line path, forever, without hitting a ray. This path itself is a ray from the point which intersects no others! The base case of the induction is obviously true, so the proof is complete.
answered 1 min ago
CyborgOctopusCyborgOctopus
1407
$endgroup$
$begingroup$
No . the new point might fall on an existing line
$endgroup$
– athos
1 min ago
add a comment |
$begingroup$
I would do it by induction. Say you already have $n$ points, each with a ray, and all rays are non-intersecting. You then place a new point in the plane. Because all of the rays already in the plane are non-intersecting, it must be that this point is not completely enclosed by rays. Because it is not completely enclosed, it must be possible for it to move continually away from its current position on some straight line path, forever, without hitting a ray. This path itself is a ray from the point which intersects no others! The base case of the induction is obviously true, so the proof is complete.
answered 1 min ago
CyborgOctopusCyborgOctopus
1407
$endgroup$
$begingroup$
No . the new point might fall on an existing line
$endgroup$
– athos
1 min ago
add a comment |
$begingroup$
I would do it by induction. Say you already have $n$ points, each with a ray, and all rays are non-intersecting. You then place a new point in the plane. Because all of the rays already in the plane are non-intersecting, it must be that this point is not completely enclosed by rays. Because it is not completely enclosed, it must be possible for it to move continually away from its current position on some straight line path, forever, without hitting a ray. This path itself is a ray from the point which intersects no others! The base case of the induction is obviously true, so the proof is complete.
answered 1 min ago
CyborgOctopusCyborgOctopus
1407
$endgroup$
I would do it by induction. Say you already have $n$ points, each with a ray, and all rays are non-intersecting. You then place a new point in the plane. Because all of the rays already in the plane are non-intersecting, it must be that this point is not completely enclosed by rays. Because it is not completely enclosed, it must be possible for it to move continually away from its current position on some straight line path, forever, without hitting a ray. This path itself is a ray from the point which intersects no others! The base case of the induction is obviously true, so the proof is complete.
answered 1 min ago
CyborgOctopusCyborgOctopus
1407
answered 1 min ago
CyborgOctopusCyborgOctopus
1407
answered 1 min ago
CyborgOctopusCyborgOctopus
1407
answered 1 min ago
CyborgOctopusCyborgOctopus
1407
1407
$begingroup$
No . the new point might fall on an existing line
$endgroup$
– athos
1 min ago
add a comment |
$begingroup$
No . the new point might fall on an existing line
$endgroup$
– athos
1 min ago
$begingroup$
No . the new point might fall on an existing line
$endgroup$
– athos
1 min ago
$begingroup$
No . the new point might fall on an existing line
$endgroup$
– athos
1 min 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
