Mixing
Variation of the sum of distances
$begingroup$
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
62
$endgroup$
add a comment |
$begingroup$
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
62
$endgroup$
add a comment |
$begingroup$
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
62
$endgroup$
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
62
asked Dec 30 '18 at 17:01
TrumpTrump
62
edited Jan 6 at 14:51
Trump
asked Dec 30 '18 at 17:01
TrumpTrump
62
asked Dec 30 '18 at 17:01
TrumpTrump
62
asked Dec 30 '18 at 17:01
TrumpTrump
62
62
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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
10.6k31332
$endgroup$
1
$begingroup$
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?
$endgroup$
– Trump
Dec 30 '18 at 17:38
add a comment |
$begingroup$
Your inequality is false: see diagram below for a counterexample.
answered Jan 6 at 17:43
AretinoAretino
22.9k21443
$endgroup$
$begingroup$
Maybe $M$ and $M'$ should be on the same side of $P$.
$endgroup$
– BPP
Jan 16 at 17:09
$begingroup$
Maybe... But the OP, apparently, doesn't care to explain his thoughts.
$endgroup$
– Aretino
Jan 16 at 19:33
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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oldest
votes
$begingroup$
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
10.6k31332
$endgroup$
1
$begingroup$
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?
$endgroup$
– Trump
Dec 30 '18 at 17:38
add a comment |
$begingroup$
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
10.6k31332
$endgroup$
1
$begingroup$
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?
$endgroup$
– Trump
Dec 30 '18 at 17:38
add a comment |
$begingroup$
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
10.6k31332
$endgroup$
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
10.6k31332
edited Dec 30 '18 at 17:36
answered Dec 30 '18 at 17:18
David G. StorkDavid G. Stork
10.6k31332
answered Dec 30 '18 at 17:18
David G. StorkDavid G. Stork
10.6k31332
answered Dec 30 '18 at 17:18
David G. StorkDavid G. Stork
10.6k31332
10.6k31332
1
$begingroup$
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?
$endgroup$
– Trump
Dec 30 '18 at 17:38
add a comment |
1
$begingroup$
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?
$endgroup$
– Trump
Dec 30 '18 at 17:38
1
1
$begingroup$
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?
$endgroup$
– Trump
Dec 30 '18 at 17:38
$begingroup$
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?
$endgroup$
– Trump
Dec 30 '18 at 17:38
add a comment |
$begingroup$
Your inequality is false: see diagram below for a counterexample.
answered Jan 6 at 17:43
AretinoAretino
22.9k21443
$endgroup$
$begingroup$
Maybe $M$ and $M'$ should be on the same side of $P$.
$endgroup$
– BPP
Jan 16 at 17:09
$begingroup$
Maybe... But the OP, apparently, doesn't care to explain his thoughts.
$endgroup$
– Aretino
Jan 16 at 19:33
add a comment |
$begingroup$
Your inequality is false: see diagram below for a counterexample.
answered Jan 6 at 17:43
AretinoAretino
22.9k21443
$endgroup$
$begingroup$
Maybe $M$ and $M'$ should be on the same side of $P$.
$endgroup$
– BPP
Jan 16 at 17:09
$begingroup$
Maybe... But the OP, apparently, doesn't care to explain his thoughts.
$endgroup$
– Aretino
Jan 16 at 19:33
add a comment |
$begingroup$
Your inequality is false: see diagram below for a counterexample.
answered Jan 6 at 17:43
AretinoAretino
22.9k21443
$endgroup$
Your inequality is false: see diagram below for a counterexample.
answered Jan 6 at 17:43
AretinoAretino
22.9k21443
edited Jan 7 at 19:01
answered Jan 6 at 17:43
AretinoAretino
22.9k21443
answered Jan 6 at 17:43
AretinoAretino
22.9k21443
answered Jan 6 at 17:43
AretinoAretino
22.9k21443
22.9k21443
$begingroup$
Maybe $M$ and $M'$ should be on the same side of $P$.
$endgroup$
– BPP
Jan 16 at 17:09
$begingroup$
Maybe... But the OP, apparently, doesn't care to explain his thoughts.
$endgroup$
– Aretino
Jan 16 at 19:33
add a comment |
$begingroup$
Maybe $M$ and $M'$ should be on the same side of $P$.
$endgroup$
– BPP
Jan 16 at 17:09
$begingroup$
Maybe... But the OP, apparently, doesn't care to explain his thoughts.
$endgroup$
– Aretino
Jan 16 at 19:33
$begingroup$
Maybe $M$ and $M'$ should be on the same side of $P$.
$endgroup$
– BPP
Jan 16 at 17:09
$begingroup$
Maybe $M$ and $M'$ should be on the same side of $P$.
$endgroup$
– BPP
Jan 16 at 17:09
$begingroup$
Maybe... But the OP, apparently, doesn't care to explain his thoughts.
$endgroup$
– Aretino
Jan 16 at 19:33
$begingroup$
Maybe... But the OP, apparently, doesn't care to explain his thoughts.
$endgroup$
– Aretino
Jan 16 at 19:33
add a comment |
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