Variation of the sum of distances
Let $l$ be a line and $A$ and $B$ two points not on $l$. 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
Trump
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Let $l$ be a line and $A$ and $B$ two points not on $l$. 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
Trump
62
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Trump is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Let $l$ be a line and $A$ and $B$ two points not on $l$. 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
Trump
62
New contributor
Trump is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Let $l$ be a line and $A$ and $B$ two points not on $l$. 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
Trump
62
New contributor
Trump is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked Dec 30 '18 at 17:01
Trump
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edited yesterday
asked Dec 30 '18 at 17:01
Trump
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asked Dec 30 '18 at 17:01
Trump
62
asked Dec 30 '18 at 17:01
Trump
62
62
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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. Stork
9,90521232
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
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1 Answer
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1 Answer
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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. Stork
9,90521232
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. Stork
9,90521232
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. Stork
9,90521232
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. Stork
9,90521232
edited Dec 30 '18 at 17:36
answered Dec 30 '18 at 17:18
David G. Stork
9,90521232
answered Dec 30 '18 at 17:18
David G. Stork
9,90521232
answered Dec 30 '18 at 17:18
David G. Stork
9,90521232
9,90521232
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 |
Trump is a new contributor. Be nice, and check out our Code of Conduct.
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