Variation of the sum of distances
Let $l$ be a line and $A$ and $B$ two points on the same side of $l$. To find the point $P$ for which $AP+PB$ is minimum we take the intersection of $l$ and the line joining $B$ and the symmetric $A'$ of $A$ with respect to $l$. For any point $M$ of $l$ other than $P$ we have $AM+MB=A'M+MB>A'B=AP+PB$ so $P$ is the desired point.
My question is $color{red}{text{how to prove that $AM+MB$ increases with $PM$?}}$.
My attempt: If $M'$ is another point of $l$ such that $PM'>PM$ then $AM<AM'+MM'$ and $BM<BM'+MM'$. I want to prove that $AM+MB<AM'+M'B$ using only the triangle inequality.
geometry euclidean-geometry triangle reflection
asked Dec 30 '18 at 17:01
TrumpTrump
42
add a comment |
Let $l$ be a line and $A$ and $B$ two points on the same side of $l$. To find the point $P$ for which $AP+PB$ is minimum we take the intersection of $l$ and the line joining $B$ and the symmetric $A'$ of $A$ with respect to $l$. For any point $M$ of $l$ other than $P$ we have $AM+MB=A'M+MB>A'B=AP+PB$ so $P$ is the desired point.
My question is $color{red}{text{how to prove that $AM+MB$ increases with $PM$?}}$.
My attempt: If $M'$ is another point of $l$ such that $PM'>PM$ then $AM<AM'+MM'$ and $BM<BM'+MM'$. I want to prove that $AM+MB<AM'+M'B$ using only the triangle inequality.
geometry euclidean-geometry triangle reflection
asked Dec 30 '18 at 17:01
TrumpTrump
42
add a comment |
Let $l$ be a line and $A$ and $B$ two points on the same side of $l$. To find the point $P$ for which $AP+PB$ is minimum we take the intersection of $l$ and the line joining $B$ and the symmetric $A'$ of $A$ with respect to $l$. For any point $M$ of $l$ other than $P$ we have $AM+MB=A'M+MB>A'B=AP+PB$ so $P$ is the desired point.
My question is $color{red}{text{how to prove that $AM+MB$ increases with $PM$?}}$.
My attempt: If $M'$ is another point of $l$ such that $PM'>PM$ then $AM<AM'+MM'$ and $BM<BM'+MM'$. I want to prove that $AM+MB<AM'+M'B$ using only the triangle inequality.
geometry euclidean-geometry triangle reflection
asked Dec 30 '18 at 17:01
TrumpTrump
42
Let $l$ be a line and $A$ and $B$ two points on the same side of $l$. To find the point $P$ for which $AP+PB$ is minimum we take the intersection of $l$ and the line joining $B$ and the symmetric $A'$ of $A$ with respect to $l$. For any point $M$ of $l$ other than $P$ we have $AM+MB=A'M+MB>A'B=AP+PB$ so $P$ is the desired point.
My question is $color{red}{text{how to prove that $AM+MB$ increases with $PM$?}}$.
My attempt: If $M'$ is another point of $l$ such that $PM'>PM$ then $AM<AM'+MM'$ and $BM<BM'+MM'$. I want to prove that $AM+MB<AM'+M'B$ using only the triangle inequality.
geometry euclidean-geometry triangle reflection
geometry euclidean-geometry triangle reflection
asked Dec 30 '18 at 17:01
TrumpTrump
42
asked Dec 30 '18 at 17:01
TrumpTrump
42
edited Jan 6 at 14:51
Trump
asked Dec 30 '18 at 17:01
TrumpTrump
42
asked Dec 30 '18 at 17:01
TrumpTrump
42
asked Dec 30 '18 at 17:01
TrumpTrump
42
42
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Sometimes a figure is worth a thousand words:
This distance from $A$ to $B$ via the line is the same as the distance from $A$ to $B'$... the shortest of which is a straight line. Using the elementary fact from Euclidean geometry that the shortest distance between two points is a straight line we see:
$color{red}{text{all other such paths (see green) must be longer.}}$ Proof completed.
answered Dec 30 '18 at 17:18
David G. StorkDavid G. Stork
10.1k21332
1
Using your image Let $M'$ be another point on the black line. Draw the path $AM'B'$ in blue. How to prove that if $PM'>PM$ then the blue path is longer then the green path?
– Trump
Dec 30 '18 at 17:38
add a comment |
Your inequality is false: see diagram below for a counterexample.
answered Jan 6 at 17:43
AretinoAretino
22.8k21443
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%2f3057009%2fvariation-of-the-sum-of-distances%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
Sometimes a figure is worth a thousand words:
This distance from $A$ to $B$ via the line is the same as the distance from $A$ to $B'$... the shortest of which is a straight line. Using the elementary fact from Euclidean geometry that the shortest distance between two points is a straight line we see:
$color{red}{text{all other such paths (see green) must be longer.}}$ Proof completed.
answered Dec 30 '18 at 17:18
David G. StorkDavid G. Stork
10.1k21332
1
Using your image Let $M'$ be another point on the black line. Draw the path $AM'B'$ in blue. How to prove that if $PM'>PM$ then the blue path is longer then the green path?
– Trump
Dec 30 '18 at 17:38
add a comment |
Sometimes a figure is worth a thousand words:
This distance from $A$ to $B$ via the line is the same as the distance from $A$ to $B'$... the shortest of which is a straight line. Using the elementary fact from Euclidean geometry that the shortest distance between two points is a straight line we see:
$color{red}{text{all other such paths (see green) must be longer.}}$ Proof completed.
answered Dec 30 '18 at 17:18
David G. StorkDavid G. Stork
10.1k21332
1
Using your image Let $M'$ be another point on the black line. Draw the path $AM'B'$ in blue. How to prove that if $PM'>PM$ then the blue path is longer then the green path?
– Trump
Dec 30 '18 at 17:38
add a comment |
Sometimes a figure is worth a thousand words:
This distance from $A$ to $B$ via the line is the same as the distance from $A$ to $B'$... the shortest of which is a straight line. Using the elementary fact from Euclidean geometry that the shortest distance between two points is a straight line we see:
$color{red}{text{all other such paths (see green) must be longer.}}$ Proof completed.
answered Dec 30 '18 at 17:18
David G. StorkDavid G. Stork
10.1k21332
Sometimes a figure is worth a thousand words:
This distance from $A$ to $B$ via the line is the same as the distance from $A$ to $B'$... the shortest of which is a straight line. Using the elementary fact from Euclidean geometry that the shortest distance between two points is a straight line we see:
$color{red}{text{all other such paths (see green) must be longer.}}$ Proof completed.
answered Dec 30 '18 at 17:18
David G. StorkDavid G. Stork
10.1k21332
edited Dec 30 '18 at 17:36
answered Dec 30 '18 at 17:18
David G. StorkDavid G. Stork
10.1k21332
answered Dec 30 '18 at 17:18
David G. StorkDavid G. Stork
10.1k21332
answered Dec 30 '18 at 17:18
David G. StorkDavid G. Stork
10.1k21332
10.1k21332
1
Using your image Let $M'$ be another point on the black line. Draw the path $AM'B'$ in blue. How to prove that if $PM'>PM$ then the blue path is longer then the green path?
– Trump
Dec 30 '18 at 17:38
add a comment |
1
Using your image Let $M'$ be another point on the black line. Draw the path $AM'B'$ in blue. How to prove that if $PM'>PM$ then the blue path is longer then the green path?
– Trump
Dec 30 '18 at 17:38
1
1
Using your image Let $M'$ be another point on the black line. Draw the path $AM'B'$ in blue. How to prove that if $PM'>PM$ then the blue path is longer then the green path?
– Trump
Dec 30 '18 at 17:38
Using your image Let $M'$ be another point on the black line. Draw the path $AM'B'$ in blue. How to prove that if $PM'>PM$ then the blue path is longer then the green path?
– Trump
Dec 30 '18 at 17:38
add a comment |
Your inequality is false: see diagram below for a counterexample.
answered Jan 6 at 17:43
AretinoAretino
22.8k21443
add a comment |
Your inequality is false: see diagram below for a counterexample.
answered Jan 6 at 17:43
AretinoAretino
22.8k21443
add a comment |
Your inequality is false: see diagram below for a counterexample.
answered Jan 6 at 17:43
AretinoAretino
22.8k21443
Your inequality is false: see diagram below for a counterexample.
answered Jan 6 at 17:43
AretinoAretino
22.8k21443
edited 2 days ago
answered Jan 6 at 17:43
AretinoAretino
22.8k21443
answered Jan 6 at 17:43
AretinoAretino
22.8k21443
answered Jan 6 at 17:43
AretinoAretino
22.8k21443
22.8k21443
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%2f3057009%2fvariation-of-the-sum-of-distances%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
