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Find the smallest sum distance among three points on three circles
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There are three separate circles on a plane. How does one find three points, $a$, $b$ and $c$, one on each circle, such that the sum distance $barab + barac + barbc$ is minimal?
geometry
edited Aug 3 at 11:11
David G. Stork
7,3202728
asked Aug 3 at 10:58
user126498
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There are three separate circles on a plane. How does one find three points, $a$, $b$ and $c$, one on each circle, such that the sum distance $barab + barac + barbc$ is minimal?
geometry
edited Aug 3 at 11:11
David G. Stork
7,3202728
asked Aug 3 at 10:58
user126498
61
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Aug 3 at 11:04
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
There are three separate circles on a plane. How does one find three points, $a$, $b$ and $c$, one on each circle, such that the sum distance $barab + barac + barbc$ is minimal?
geometry
edited Aug 3 at 11:11
David G. Stork
7,3202728
asked Aug 3 at 10:58
user126498
61
There are three separate circles on a plane. How does one find three points, $a$, $b$ and $c$, one on each circle, such that the sum distance $barab + barac + barbc$ is minimal?
geometry
edited Aug 3 at 11:11
David G. Stork
7,3202728
asked Aug 3 at 10:58
user126498
61
edited Aug 3 at 11:11
David G. Stork
7,3202728
edited Aug 3 at 11:11
David G. Stork
7,3202728
edited Aug 3 at 11:11
David G. Stork
7,3202728
7,3202728
asked Aug 3 at 10:58
user126498
61
asked Aug 3 at 10:58
user126498
61
asked Aug 3 at 10:58
user126498
61
61
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Aug 3 at 11:04
add a comment |Â
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Aug 3 at 11:04
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Aug 3 at 11:04
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Aug 3 at 11:04
add a comment |Â
2 Answers
2
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oldest
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up vote
0
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Here's how to solve this (draw a figure!):
Without loss of generality, circle $C_1$ (with radius $r_1$), can be centered at the origin; circle $C_2$ (with radius $r_2$) can be placed on the $x$-axis a distance $l_2x$ from the origin; circle $C_3$ (with radius $r_3$) will be centered at coordinates $(l_3x, l_3y)$.
The position of point $a$ on $C_1$ can be specified by the single angle variable $theta_1$, and likewise for $b$ and $theta_2$ and $c$ and $theta_3$. Thus your problem reduces to finding the three angles that minimize the sum of the inter-point distances.
The distance $barab$ is
$$sqrt(r_1 cos theta_1 - (l_2x + r_2 cos theta_2))^2 + (r_1 sin theta_1 - r_2 sin theta_2)^2$$
I'll let you fill in the two other distances.
Then express the sum $L_total =barab + barac + barbc$, which will depend upon the three angles.
For $L_total$ to be minimal, the three derivatives must be zero:
- $partial L_total over partial theta_1 = 0$
- $partial L_total over partial theta_2 = 0$
- $partial L_total over partial theta_3 = 0$
This will give you a set of simultaneous equations, whose solution minimizes $L_total$.
Actually, the above will give you the maximum and the minimum distance $L_total$, so you choose the minimum by simple comparison. (You could also impose conditions on the second derivative, but that is too much work for such a simple problem.)
In pathological cases (e.g., all circle centers are co-linear), there will be multiple minimizing solutions.
answered Aug 3 at 11:24
David G. Stork
7,3202728
add a comment |Â
up vote
0
down vote
This cannot be solved by straightedge and compass since it leads to a cubic equation, but assuming that $Gamma_A,Gamma_B,Gamma_C$ are three disjoint circles through $A,B,C$, there is a simple iterative approach for finding an approximate solution:
- Move $A$ on $Gamma_A$ until the ellipse through $A$ with foci at $B,C$ is tangent to $Gamma_A$;
- Move $B$ on $Gamma_B$ until the ellipse through $B$ with foci at $A,C$ is tangent to $Gamma_B$;
- Move $C$ on $Gamma_C$ until the ellipse through $C$ with foci at $A,B$ is tangent to $Gamma_C$;
- Iterate until the required accuracy is reached.
answered Aug 3 at 14:16
Jack D'Aurizio♦
268k30261624
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Here's how to solve this (draw a figure!):
Without loss of generality, circle $C_1$ (with radius $r_1$), can be centered at the origin; circle $C_2$ (with radius $r_2$) can be placed on the $x$-axis a distance $l_2x$ from the origin; circle $C_3$ (with radius $r_3$) will be centered at coordinates $(l_3x, l_3y)$.
The position of point $a$ on $C_1$ can be specified by the single angle variable $theta_1$, and likewise for $b$ and $theta_2$ and $c$ and $theta_3$. Thus your problem reduces to finding the three angles that minimize the sum of the inter-point distances.
The distance $barab$ is
$$sqrt(r_1 cos theta_1 - (l_2x + r_2 cos theta_2))^2 + (r_1 sin theta_1 - r_2 sin theta_2)^2$$
I'll let you fill in the two other distances.
Then express the sum $L_total =barab + barac + barbc$, which will depend upon the three angles.
For $L_total$ to be minimal, the three derivatives must be zero:
- $partial L_total over partial theta_1 = 0$
- $partial L_total over partial theta_2 = 0$
- $partial L_total over partial theta_3 = 0$
This will give you a set of simultaneous equations, whose solution minimizes $L_total$.
Actually, the above will give you the maximum and the minimum distance $L_total$, so you choose the minimum by simple comparison. (You could also impose conditions on the second derivative, but that is too much work for such a simple problem.)
In pathological cases (e.g., all circle centers are co-linear), there will be multiple minimizing solutions.
answered Aug 3 at 11:24
David G. Stork
7,3202728
add a comment |Â
up vote
0
down vote
Here's how to solve this (draw a figure!):
Without loss of generality, circle $C_1$ (with radius $r_1$), can be centered at the origin; circle $C_2$ (with radius $r_2$) can be placed on the $x$-axis a distance $l_2x$ from the origin; circle $C_3$ (with radius $r_3$) will be centered at coordinates $(l_3x, l_3y)$.
The position of point $a$ on $C_1$ can be specified by the single angle variable $theta_1$, and likewise for $b$ and $theta_2$ and $c$ and $theta_3$. Thus your problem reduces to finding the three angles that minimize the sum of the inter-point distances.
The distance $barab$ is
$$sqrt(r_1 cos theta_1 - (l_2x + r_2 cos theta_2))^2 + (r_1 sin theta_1 - r_2 sin theta_2)^2$$
I'll let you fill in the two other distances.
Then express the sum $L_total =barab + barac + barbc$, which will depend upon the three angles.
For $L_total$ to be minimal, the three derivatives must be zero:
- $partial L_total over partial theta_1 = 0$
- $partial L_total over partial theta_2 = 0$
- $partial L_total over partial theta_3 = 0$
This will give you a set of simultaneous equations, whose solution minimizes $L_total$.
Actually, the above will give you the maximum and the minimum distance $L_total$, so you choose the minimum by simple comparison. (You could also impose conditions on the second derivative, but that is too much work for such a simple problem.)
In pathological cases (e.g., all circle centers are co-linear), there will be multiple minimizing solutions.
answered Aug 3 at 11:24
David G. Stork
7,3202728
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Here's how to solve this (draw a figure!):
Without loss of generality, circle $C_1$ (with radius $r_1$), can be centered at the origin; circle $C_2$ (with radius $r_2$) can be placed on the $x$-axis a distance $l_2x$ from the origin; circle $C_3$ (with radius $r_3$) will be centered at coordinates $(l_3x, l_3y)$.
The position of point $a$ on $C_1$ can be specified by the single angle variable $theta_1$, and likewise for $b$ and $theta_2$ and $c$ and $theta_3$. Thus your problem reduces to finding the three angles that minimize the sum of the inter-point distances.
The distance $barab$ is
$$sqrt(r_1 cos theta_1 - (l_2x + r_2 cos theta_2))^2 + (r_1 sin theta_1 - r_2 sin theta_2)^2$$
I'll let you fill in the two other distances.
Then express the sum $L_total =barab + barac + barbc$, which will depend upon the three angles.
For $L_total$ to be minimal, the three derivatives must be zero:
- $partial L_total over partial theta_1 = 0$
- $partial L_total over partial theta_2 = 0$
- $partial L_total over partial theta_3 = 0$
This will give you a set of simultaneous equations, whose solution minimizes $L_total$.
Actually, the above will give you the maximum and the minimum distance $L_total$, so you choose the minimum by simple comparison. (You could also impose conditions on the second derivative, but that is too much work for such a simple problem.)
In pathological cases (e.g., all circle centers are co-linear), there will be multiple minimizing solutions.
answered Aug 3 at 11:24
David G. Stork
7,3202728
Here's how to solve this (draw a figure!):
Without loss of generality, circle $C_1$ (with radius $r_1$), can be centered at the origin; circle $C_2$ (with radius $r_2$) can be placed on the $x$-axis a distance $l_2x$ from the origin; circle $C_3$ (with radius $r_3$) will be centered at coordinates $(l_3x, l_3y)$.
The position of point $a$ on $C_1$ can be specified by the single angle variable $theta_1$, and likewise for $b$ and $theta_2$ and $c$ and $theta_3$. Thus your problem reduces to finding the three angles that minimize the sum of the inter-point distances.
The distance $barab$ is
$$sqrt(r_1 cos theta_1 - (l_2x + r_2 cos theta_2))^2 + (r_1 sin theta_1 - r_2 sin theta_2)^2$$
I'll let you fill in the two other distances.
Then express the sum $L_total =barab + barac + barbc$, which will depend upon the three angles.
For $L_total$ to be minimal, the three derivatives must be zero:
- $partial L_total over partial theta_1 = 0$
- $partial L_total over partial theta_2 = 0$
- $partial L_total over partial theta_3 = 0$
This will give you a set of simultaneous equations, whose solution minimizes $L_total$.
Actually, the above will give you the maximum and the minimum distance $L_total$, so you choose the minimum by simple comparison. (You could also impose conditions on the second derivative, but that is too much work for such a simple problem.)
In pathological cases (e.g., all circle centers are co-linear), there will be multiple minimizing solutions.
answered Aug 3 at 11:24
David G. Stork
7,3202728
edited Aug 3 at 11:37
answered Aug 3 at 11:24
David G. Stork
7,3202728
answered Aug 3 at 11:24
David G. Stork
7,3202728
answered Aug 3 at 11:24
David G. Stork
7,3202728
7,3202728
add a comment |Â
add a comment |Â
up vote
0
down vote
This cannot be solved by straightedge and compass since it leads to a cubic equation, but assuming that $Gamma_A,Gamma_B,Gamma_C$ are three disjoint circles through $A,B,C$, there is a simple iterative approach for finding an approximate solution:
- Move $A$ on $Gamma_A$ until the ellipse through $A$ with foci at $B,C$ is tangent to $Gamma_A$;
- Move $B$ on $Gamma_B$ until the ellipse through $B$ with foci at $A,C$ is tangent to $Gamma_B$;
- Move $C$ on $Gamma_C$ until the ellipse through $C$ with foci at $A,B$ is tangent to $Gamma_C$;
- Iterate until the required accuracy is reached.
answered Aug 3 at 14:16
Jack D'Aurizio♦
268k30261624
add a comment |Â
up vote
0
down vote
This cannot be solved by straightedge and compass since it leads to a cubic equation, but assuming that $Gamma_A,Gamma_B,Gamma_C$ are three disjoint circles through $A,B,C$, there is a simple iterative approach for finding an approximate solution:
- Move $A$ on $Gamma_A$ until the ellipse through $A$ with foci at $B,C$ is tangent to $Gamma_A$;
- Move $B$ on $Gamma_B$ until the ellipse through $B$ with foci at $A,C$ is tangent to $Gamma_B$;
- Move $C$ on $Gamma_C$ until the ellipse through $C$ with foci at $A,B$ is tangent to $Gamma_C$;
- Iterate until the required accuracy is reached.
answered Aug 3 at 14:16
Jack D'Aurizio♦
268k30261624
add a comment |Â
up vote
0
down vote
up vote
0
down vote
This cannot be solved by straightedge and compass since it leads to a cubic equation, but assuming that $Gamma_A,Gamma_B,Gamma_C$ are three disjoint circles through $A,B,C$, there is a simple iterative approach for finding an approximate solution:
- Move $A$ on $Gamma_A$ until the ellipse through $A$ with foci at $B,C$ is tangent to $Gamma_A$;
- Move $B$ on $Gamma_B$ until the ellipse through $B$ with foci at $A,C$ is tangent to $Gamma_B$;
- Move $C$ on $Gamma_C$ until the ellipse through $C$ with foci at $A,B$ is tangent to $Gamma_C$;
- Iterate until the required accuracy is reached.
answered Aug 3 at 14:16
Jack D'Aurizio♦
268k30261624
This cannot be solved by straightedge and compass since it leads to a cubic equation, but assuming that $Gamma_A,Gamma_B,Gamma_C$ are three disjoint circles through $A,B,C$, there is a simple iterative approach for finding an approximate solution:
- Move $A$ on $Gamma_A$ until the ellipse through $A$ with foci at $B,C$ is tangent to $Gamma_A$;
- Move $B$ on $Gamma_B$ until the ellipse through $B$ with foci at $A,C$ is tangent to $Gamma_B$;
- Move $C$ on $Gamma_C$ until the ellipse through $C$ with foci at $A,B$ is tangent to $Gamma_C$;
- Iterate until the required accuracy is reached.
answered Aug 3 at 14:16
Jack D'Aurizio♦
268k30261624
answered Aug 3 at 14:16
Jack D'Aurizio♦
268k30261624
answered Aug 3 at 14:16
Jack D'Aurizio♦
268k30261624
answered Aug 3 at 14:16
Jack D'Aurizio♦
268k30261624
268k30261624
add a comment |Â
add a comment |Â
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Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Aug 3 at 11:04