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Subdivide a circle according to a subdivided segment
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Given a segment $AB$, subdivided in $m$ sub-segments, I have to draw a circumference $C$, subdivided in $m$ arcs, in such a way that the ratios between the lengths of the sub-segments of $AB$ are the same as the ratios between the lengths of the arcs subdividing $C$.
Here's an (illustrative) example with $m=3$.
Is there a simple geometric construction to solve this problem?
I apologize in case this is an obvious question. Thanks for your suggestions!
geometry euclidean-geometry geometric-construction
asked Jul 26 at 16:33
Andrea Prunotto
569115
 |Â
show 4 more comments
up vote
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Given a segment $AB$, subdivided in $m$ sub-segments, I have to draw a circumference $C$, subdivided in $m$ arcs, in such a way that the ratios between the lengths of the sub-segments of $AB$ are the same as the ratios between the lengths of the arcs subdividing $C$.
Here's an (illustrative) example with $m=3$.
Is there a simple geometric construction to solve this problem?
I apologize in case this is an obvious question. Thanks for your suggestions!
geometry euclidean-geometry geometric-construction
asked Jul 26 at 16:33
Andrea Prunotto
569115
1
No, not for a general number of equal segments (and not more generally). Any line segment can be equally subdivided into $mge 1$ equal segments, but in dividing a circle into equal arcs only special numbers of them can be constructed geometrically (a famous result by Gauss tells which numbers).
– hardmath
Jul 26 at 16:39
@hardmath Thanks for your comment! Any chance with $m=3$?
– Andrea Prunotto
Jul 26 at 16:41
@hardmath Sorry, I check your link first. Thanks again for your observation!
– Andrea Prunotto
Jul 26 at 16:44
As you illustrate, the case of equal subsegments and $m=3$ can be done, but if you were to allow unequal subsegments, already the case $m=2$ would be intractable by ruler and compass constructions. Perhaps you have in mind a more general geometric "construction"?
– hardmath
Jul 26 at 16:44
1
It is feasible when $c$ is one of Gauss's special numbers, which have a prime factorization related to Fermat primes. I'll see if there is a previous Question which lays out the details, and if not I'll write something up for you.
– hardmath
Jul 26 at 16:53
 |Â
show 4 more comments
up vote
0
down vote
favorite
1
up vote
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down vote
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1
Given a segment $AB$, subdivided in $m$ sub-segments, I have to draw a circumference $C$, subdivided in $m$ arcs, in such a way that the ratios between the lengths of the sub-segments of $AB$ are the same as the ratios between the lengths of the arcs subdividing $C$.
Here's an (illustrative) example with $m=3$.
Is there a simple geometric construction to solve this problem?
I apologize in case this is an obvious question. Thanks for your suggestions!
geometry euclidean-geometry geometric-construction
asked Jul 26 at 16:33
Andrea Prunotto
569115
Given a segment $AB$, subdivided in $m$ sub-segments, I have to draw a circumference $C$, subdivided in $m$ arcs, in such a way that the ratios between the lengths of the sub-segments of $AB$ are the same as the ratios between the lengths of the arcs subdividing $C$.
Here's an (illustrative) example with $m=3$.
Is there a simple geometric construction to solve this problem?
I apologize in case this is an obvious question. Thanks for your suggestions!
geometry euclidean-geometry geometric-construction
asked Jul 26 at 16:33
Andrea Prunotto
569115
asked Jul 26 at 16:33
Andrea Prunotto
569115
asked Jul 26 at 16:33
Andrea Prunotto
569115
asked Jul 26 at 16:33
Andrea Prunotto
569115
569115
1
No, not for a general number of equal segments (and not more generally). Any line segment can be equally subdivided into $mge 1$ equal segments, but in dividing a circle into equal arcs only special numbers of them can be constructed geometrically (a famous result by Gauss tells which numbers).
– hardmath
Jul 26 at 16:39
@hardmath Thanks for your comment! Any chance with $m=3$?
– Andrea Prunotto
Jul 26 at 16:41
@hardmath Sorry, I check your link first. Thanks again for your observation!
– Andrea Prunotto
Jul 26 at 16:44
As you illustrate, the case of equal subsegments and $m=3$ can be done, but if you were to allow unequal subsegments, already the case $m=2$ would be intractable by ruler and compass constructions. Perhaps you have in mind a more general geometric "construction"?
– hardmath
Jul 26 at 16:44
1
It is feasible when $c$ is one of Gauss's special numbers, which have a prime factorization related to Fermat primes. I'll see if there is a previous Question which lays out the details, and if not I'll write something up for you.
– hardmath
Jul 26 at 16:53
 |Â
show 4 more comments
1
No, not for a general number of equal segments (and not more generally). Any line segment can be equally subdivided into $mge 1$ equal segments, but in dividing a circle into equal arcs only special numbers of them can be constructed geometrically (a famous result by Gauss tells which numbers).
– hardmath
Jul 26 at 16:39
@hardmath Thanks for your comment! Any chance with $m=3$?
– Andrea Prunotto
Jul 26 at 16:41
@hardmath Sorry, I check your link first. Thanks again for your observation!
– Andrea Prunotto
Jul 26 at 16:44
As you illustrate, the case of equal subsegments and $m=3$ can be done, but if you were to allow unequal subsegments, already the case $m=2$ would be intractable by ruler and compass constructions. Perhaps you have in mind a more general geometric "construction"?
– hardmath
Jul 26 at 16:44
1
It is feasible when $c$ is one of Gauss's special numbers, which have a prime factorization related to Fermat primes. I'll see if there is a previous Question which lays out the details, and if not I'll write something up for you.
– hardmath
Jul 26 at 16:53
1
1
No, not for a general number of equal segments (and not more generally). Any line segment can be equally subdivided into $mge 1$ equal segments, but in dividing a circle into equal arcs only special numbers of them can be constructed geometrically (a famous result by Gauss tells which numbers).
– hardmath
Jul 26 at 16:39
No, not for a general number of equal segments (and not more generally). Any line segment can be equally subdivided into $mge 1$ equal segments, but in dividing a circle into equal arcs only special numbers of them can be constructed geometrically (a famous result by Gauss tells which numbers).
– hardmath
Jul 26 at 16:39
@hardmath Thanks for your comment! Any chance with $m=3$?
– Andrea Prunotto
Jul 26 at 16:41
@hardmath Thanks for your comment! Any chance with $m=3$?
– Andrea Prunotto
Jul 26 at 16:41
@hardmath Sorry, I check your link first. Thanks again for your observation!
– Andrea Prunotto
Jul 26 at 16:44
@hardmath Sorry, I check your link first. Thanks again for your observation!
– Andrea Prunotto
Jul 26 at 16:44
As you illustrate, the case of equal subsegments and $m=3$ can be done, but if you were to allow unequal subsegments, already the case $m=2$ would be intractable by ruler and compass constructions. Perhaps you have in mind a more general geometric "construction"?
– hardmath
Jul 26 at 16:44
As you illustrate, the case of equal subsegments and $m=3$ can be done, but if you were to allow unequal subsegments, already the case $m=2$ would be intractable by ruler and compass constructions. Perhaps you have in mind a more general geometric "construction"?
– hardmath
Jul 26 at 16:44
1
1
It is feasible when $c$ is one of Gauss's special numbers, which have a prime factorization related to Fermat primes. I'll see if there is a previous Question which lays out the details, and if not I'll write something up for you.
– hardmath
Jul 26 at 16:53
It is feasible when $c$ is one of Gauss's special numbers, which have a prime factorization related to Fermat primes. I'll see if there is a previous Question which lays out the details, and if not I'll write something up for you.
– hardmath
Jul 26 at 16:53
 |Â
show 4 more comments
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1
No, not for a general number of equal segments (and not more generally). Any line segment can be equally subdivided into $mge 1$ equal segments, but in dividing a circle into equal arcs only special numbers of them can be constructed geometrically (a famous result by Gauss tells which numbers).
– hardmath
Jul 26 at 16:39
@hardmath Thanks for your comment! Any chance with $m=3$?
– Andrea Prunotto
Jul 26 at 16:41
@hardmath Sorry, I check your link first. Thanks again for your observation!
– Andrea Prunotto
Jul 26 at 16:44
As you illustrate, the case of equal subsegments and $m=3$ can be done, but if you were to allow unequal subsegments, already the case $m=2$ would be intractable by ruler and compass constructions. Perhaps you have in mind a more general geometric "construction"?
– hardmath
Jul 26 at 16:44
1
It is feasible when $c$ is one of Gauss's special numbers, which have a prime factorization related to Fermat primes. I'll see if there is a previous Question which lays out the details, and if not I'll write something up for you.
– hardmath
Jul 26 at 16:53