Three lines moving at same velocity form a triangle. What is the formula for their lengths at time t?
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Three infinite lines on a plane forming a triangle where they cross. The lines are travelling perpendicular to their length at a constant speed $v=1$ such as the triangle is getting smaller.
The lengths of the sides of the triangle at time $t=0$ are $(x_0,y_0,z_0)$. Find the lengths of the sides of $t$.
This is a question I invented while thinking about physics. Extra marks for an intuitive solution without using vectors.
Considering the space-time diagram at some time $t$ all three lines will meet a point forming a triangular pyramid in space-time. Maybe this gives a clue to solving it.
geometry dynamical-systems
asked Nov 18 at 2:41
zooby
946616
add a comment |
up vote
5
down vote
favorite
Three infinite lines on a plane forming a triangle where they cross. The lines are travelling perpendicular to their length at a constant speed $v=1$ such as the triangle is getting smaller.
The lengths of the sides of the triangle at time $t=0$ are $(x_0,y_0,z_0)$. Find the lengths of the sides of $t$.
This is a question I invented while thinking about physics. Extra marks for an intuitive solution without using vectors.
Considering the space-time diagram at some time $t$ all three lines will meet a point forming a triangular pyramid in space-time. Maybe this gives a clue to solving it.
geometry dynamical-systems
asked Nov 18 at 2:41
zooby
946616
add a comment |
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Three infinite lines on a plane forming a triangle where they cross. The lines are travelling perpendicular to their length at a constant speed $v=1$ such as the triangle is getting smaller.
The lengths of the sides of the triangle at time $t=0$ are $(x_0,y_0,z_0)$. Find the lengths of the sides of $t$.
This is a question I invented while thinking about physics. Extra marks for an intuitive solution without using vectors.
Considering the space-time diagram at some time $t$ all three lines will meet a point forming a triangular pyramid in space-time. Maybe this gives a clue to solving it.
geometry dynamical-systems
asked Nov 18 at 2:41
zooby
946616
Three infinite lines on a plane forming a triangle where they cross. The lines are travelling perpendicular to their length at a constant speed $v=1$ such as the triangle is getting smaller.
The lengths of the sides of the triangle at time $t=0$ are $(x_0,y_0,z_0)$. Find the lengths of the sides of $t$.
This is a question I invented while thinking about physics. Extra marks for an intuitive solution without using vectors.
Considering the space-time diagram at some time $t$ all three lines will meet a point forming a triangular pyramid in space-time. Maybe this gives a clue to solving it.
geometry dynamical-systems
geometry dynamical-systems
asked Nov 18 at 2:41
zooby
946616
asked Nov 18 at 2:41
zooby
946616
asked Nov 18 at 2:41
zooby
946616
asked Nov 18 at 2:41
zooby
946616
asked Nov 18 at 2:41
zooby
946616
946616
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
Put the incentre at the origin, the result is just homothety $r(t)=r(0)-t$, for $t<r(0)=r$ (where $r(t)$ is the inradius at time $t$) and hence also on the side lengths $x(t)=x(0)(1-frac{t}{r(0)})$, etc.
answered Nov 18 at 3:02
user10354138
6,324623
Excellent! Answer very quick!
– zooby
Nov 18 at 3:09
add a comment |
up vote
1
down vote
A little more definitive answer..........
$$x_t = x (1 - frac{2 Vcdot t}{(x+z-y)tan(frac{A}{2})})$$
$$text{where} zge yge x text{and} (x+z-y)tan(frac{A}{2})ge Vcdot t$$
Substitute $y$ and $y_t$, or $z$ and $z_t$ for $x$ and $x_t$ to get $y_t$ or $z_t$ in the above formula.
answered Nov 18 at 5:29
Phil H
3,8782312
Yes, but you don't need $tan(A/2)$ because you can calculate that from the sides of the triangle. $L^2 = (x+y-z)(x-y+z)(z-x+y)/(x+y+z)$ give or take a factor of 2.
– zooby
Nov 18 at 15:10
Yes, that works.
– Phil H
Nov 18 at 18:06
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Put the incentre at the origin, the result is just homothety $r(t)=r(0)-t$, for $t<r(0)=r$ (where $r(t)$ is the inradius at time $t$) and hence also on the side lengths $x(t)=x(0)(1-frac{t}{r(0)})$, etc.
answered Nov 18 at 3:02
user10354138
6,324623
Excellent! Answer very quick!
– zooby
Nov 18 at 3:09
add a comment |
up vote
3
down vote
accepted
Put the incentre at the origin, the result is just homothety $r(t)=r(0)-t$, for $t<r(0)=r$ (where $r(t)$ is the inradius at time $t$) and hence also on the side lengths $x(t)=x(0)(1-frac{t}{r(0)})$, etc.
answered Nov 18 at 3:02
user10354138
6,324623
Excellent! Answer very quick!
– zooby
Nov 18 at 3:09
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Put the incentre at the origin, the result is just homothety $r(t)=r(0)-t$, for $t<r(0)=r$ (where $r(t)$ is the inradius at time $t$) and hence also on the side lengths $x(t)=x(0)(1-frac{t}{r(0)})$, etc.
answered Nov 18 at 3:02
user10354138
6,324623
Put the incentre at the origin, the result is just homothety $r(t)=r(0)-t$, for $t<r(0)=r$ (where $r(t)$ is the inradius at time $t$) and hence also on the side lengths $x(t)=x(0)(1-frac{t}{r(0)})$, etc.
answered Nov 18 at 3:02
user10354138
6,324623
edited Nov 18 at 13:15
answered Nov 18 at 3:02
user10354138
6,324623
answered Nov 18 at 3:02
user10354138
6,324623
answered Nov 18 at 3:02
user10354138
6,324623
6,324623
Excellent! Answer very quick!
– zooby
Nov 18 at 3:09
add a comment |
Excellent! Answer very quick!
– zooby
Nov 18 at 3:09
Excellent! Answer very quick!
– zooby
Nov 18 at 3:09
Excellent! Answer very quick!
– zooby
Nov 18 at 3:09
add a comment |
up vote
1
down vote
A little more definitive answer..........
$$x_t = x (1 - frac{2 Vcdot t}{(x+z-y)tan(frac{A}{2})})$$
$$text{where} zge yge x text{and} (x+z-y)tan(frac{A}{2})ge Vcdot t$$
Substitute $y$ and $y_t$, or $z$ and $z_t$ for $x$ and $x_t$ to get $y_t$ or $z_t$ in the above formula.
answered Nov 18 at 5:29
Phil H
3,8782312
Yes, but you don't need $tan(A/2)$ because you can calculate that from the sides of the triangle. $L^2 = (x+y-z)(x-y+z)(z-x+y)/(x+y+z)$ give or take a factor of 2.
– zooby
Nov 18 at 15:10
Yes, that works.
– Phil H
Nov 18 at 18:06
add a comment |
up vote
1
down vote
A little more definitive answer..........
$$x_t = x (1 - frac{2 Vcdot t}{(x+z-y)tan(frac{A}{2})})$$
$$text{where} zge yge x text{and} (x+z-y)tan(frac{A}{2})ge Vcdot t$$
Substitute $y$ and $y_t$, or $z$ and $z_t$ for $x$ and $x_t$ to get $y_t$ or $z_t$ in the above formula.
answered Nov 18 at 5:29
Phil H
3,8782312
Yes, but you don't need $tan(A/2)$ because you can calculate that from the sides of the triangle. $L^2 = (x+y-z)(x-y+z)(z-x+y)/(x+y+z)$ give or take a factor of 2.
– zooby
Nov 18 at 15:10
Yes, that works.
– Phil H
Nov 18 at 18:06
add a comment |
up vote
1
down vote
up vote
1
down vote
A little more definitive answer..........
$$x_t = x (1 - frac{2 Vcdot t}{(x+z-y)tan(frac{A}{2})})$$
$$text{where} zge yge x text{and} (x+z-y)tan(frac{A}{2})ge Vcdot t$$
Substitute $y$ and $y_t$, or $z$ and $z_t$ for $x$ and $x_t$ to get $y_t$ or $z_t$ in the above formula.
answered Nov 18 at 5:29
Phil H
3,8782312
A little more definitive answer..........
$$x_t = x (1 - frac{2 Vcdot t}{(x+z-y)tan(frac{A}{2})})$$
$$text{where} zge yge x text{and} (x+z-y)tan(frac{A}{2})ge Vcdot t$$
Substitute $y$ and $y_t$, or $z$ and $z_t$ for $x$ and $x_t$ to get $y_t$ or $z_t$ in the above formula.
answered Nov 18 at 5:29
Phil H
3,8782312
edited Nov 18 at 14:03
answered Nov 18 at 5:29
Phil H
3,8782312
answered Nov 18 at 5:29
Phil H
3,8782312
answered Nov 18 at 5:29
Phil H
3,8782312
3,8782312
Yes, but you don't need $tan(A/2)$ because you can calculate that from the sides of the triangle. $L^2 = (x+y-z)(x-y+z)(z-x+y)/(x+y+z)$ give or take a factor of 2.
– zooby
Nov 18 at 15:10
Yes, that works.
– Phil H
Nov 18 at 18:06
add a comment |
Yes, but you don't need $tan(A/2)$ because you can calculate that from the sides of the triangle. $L^2 = (x+y-z)(x-y+z)(z-x+y)/(x+y+z)$ give or take a factor of 2.
– zooby
Nov 18 at 15:10
Yes, that works.
– Phil H
Nov 18 at 18:06
Yes, but you don't need $tan(A/2)$ because you can calculate that from the sides of the triangle. $L^2 = (x+y-z)(x-y+z)(z-x+y)/(x+y+z)$ give or take a factor of 2.
– zooby
Nov 18 at 15:10
Yes, but you don't need $tan(A/2)$ because you can calculate that from the sides of the triangle. $L^2 = (x+y-z)(x-y+z)(z-x+y)/(x+y+z)$ give or take a factor of 2.
– zooby
Nov 18 at 15:10
Yes, that works.
– Phil H
Nov 18 at 18:06
Yes, that works.
– Phil H
Nov 18 at 18:06
add a comment |
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