Mixing
Drawing approximated regular shapes on square grid
$begingroup$
I find myself often fooling around with pen and paper, preferably squared paper. So I began looking for ways to sketch geometric figures as precisely as possible without using compass and/or ruler. In particular I'm thinking of regular polygons and circles. I'm attaching a picture of my best findings, intentionally left handmade to show the kind of result I'm pursuing (following an answer I introduced a couple of digital adjustments). Explanations follow below.
Let me start with regular polygons. The rule is simple: the vertices must lie at the grid crossings, that is their coordinates must be expressible as integer numbers. Apart from the square, it is not possible to get regular polygons with this restriction, so the intent is to obtain the closest approximations keeping the figures "small". I've explored different ways to quantitatively assess the "precision" of an approximation, starting from length uniformity of sides (satisfactory only for triangles) later combined with uniformity of angles, to the mean squared distance of vertices from the ideal ones not lying on the grid. I put no requirements about the center; should it happen to fall on or close to a crossing, even better. Moreover I tested different size factors to take into account the fact that getting better shape approximations with bigger figures is somewhat obvious and also less useful for sketching. Anyway I've found substantial agreement among different strategies. In fact, I didn't really find head-to-head confrontations whose "winner" depended on the scoring details, and even if I did I would have happily declared a tie and kept all the alternatives. So, the sheet of paper in the top-left corner of the figure collects my best results of this family.
Next I'm relaxing the restriction: I'll let the vertices fall also on the middle points of the sides of paper squares. These can be located with satisfactory precision by eye, instead I'm not going to allow center points of paper squares. This new rule is equivalent to halving all the coordinates of a valid previous solution, provided that no point had both coordinates odd (one can horizontally and/or vertically translate the original figure by $1$ before attempting the division). I might also add that I don't like to see a line running parallel to the grid cutting squares in halves, but it's just a matter of taste. Obviously, at comparable dimensions, better approximations can be found with this additional freedom, and my best findings are represented in the top-right corner.
I shall proceed to circles. They are different, in that they do not have a limited number of vertices and one can look for support points –that now can belong to the circle and lie exactly at grid intersections at the same time– with no a priori restrictions of position and amount. So the goal becomes to find as many of them, and as evenly distributed, as possible. Again I'm trying to keep the circles small, with similar considerations as before about loss of interest as size grows. With this in mind, my best circles are depicted in the bottom-left corner.
Finally, again I soften the rules: support points on the grid can be approximations of exact circle points and/or they can be placed on middle points of square sides. My best findings of this kind are those in the bottom-right corner.
I couldn't find any serious dissertation on this (indeed trivial) argument, the most closely related argument being this: Circle Lattice Points.
Is anyone aware of others? Or is anybody willing to find themselves other similar (or better, under any respect) figures?
EDIT: I don't want to seem ungrateful to those who took care of my question, but I still haven't received the answer I was hoping for. The fact is that I'm looking for a way to discover whether I have "missed" some interesting figures in the described context. I exploited my (limited) programming skills to perform some automated searches, but I'm not sure that my algorithms guarantee not to miss any good solutions. I can describe and discuss these too, but I'd prefer to leave everyone free to use their original approach. Another chance would be to find some reference covering the topic, but I'm afraid that this is really too "recreational" to be dealt with in literature. Maybe something on the Internet? Someone who has done anything similar before me? I'm starting a bounty to find out! Thanks in advance!
geometry circle recreational-mathematics polygons diophantine-approximation
asked Apr 25 '17 at 15:12
lesath82lesath82
1,921226
$endgroup$
add a comment |
$begingroup$
I find myself often fooling around with pen and paper, preferably squared paper. So I began looking for ways to sketch geometric figures as precisely as possible without using compass and/or ruler. In particular I'm thinking of regular polygons and circles. I'm attaching a picture of my best findings, intentionally left handmade to show the kind of result I'm pursuing (following an answer I introduced a couple of digital adjustments). Explanations follow below.
Let me start with regular polygons. The rule is simple: the vertices must lie at the grid crossings, that is their coordinates must be expressible as integer numbers. Apart from the square, it is not possible to get regular polygons with this restriction, so the intent is to obtain the closest approximations keeping the figures "small". I've explored different ways to quantitatively assess the "precision" of an approximation, starting from length uniformity of sides (satisfactory only for triangles) later combined with uniformity of angles, to the mean squared distance of vertices from the ideal ones not lying on the grid. I put no requirements about the center; should it happen to fall on or close to a crossing, even better. Moreover I tested different size factors to take into account the fact that getting better shape approximations with bigger figures is somewhat obvious and also less useful for sketching. Anyway I've found substantial agreement among different strategies. In fact, I didn't really find head-to-head confrontations whose "winner" depended on the scoring details, and even if I did I would have happily declared a tie and kept all the alternatives. So, the sheet of paper in the top-left corner of the figure collects my best results of this family.
Next I'm relaxing the restriction: I'll let the vertices fall also on the middle points of the sides of paper squares. These can be located with satisfactory precision by eye, instead I'm not going to allow center points of paper squares. This new rule is equivalent to halving all the coordinates of a valid previous solution, provided that no point had both coordinates odd (one can horizontally and/or vertically translate the original figure by $1$ before attempting the division). I might also add that I don't like to see a line running parallel to the grid cutting squares in halves, but it's just a matter of taste. Obviously, at comparable dimensions, better approximations can be found with this additional freedom, and my best findings are represented in the top-right corner.
I shall proceed to circles. They are different, in that they do not have a limited number of vertices and one can look for support points –that now can belong to the circle and lie exactly at grid intersections at the same time– with no a priori restrictions of position and amount. So the goal becomes to find as many of them, and as evenly distributed, as possible. Again I'm trying to keep the circles small, with similar considerations as before about loss of interest as size grows. With this in mind, my best circles are depicted in the bottom-left corner.
Finally, again I soften the rules: support points on the grid can be approximations of exact circle points and/or they can be placed on middle points of square sides. My best findings of this kind are those in the bottom-right corner.
I couldn't find any serious dissertation on this (indeed trivial) argument, the most closely related argument being this: Circle Lattice Points.
Is anyone aware of others? Or is anybody willing to find themselves other similar (or better, under any respect) figures?
EDIT: I don't want to seem ungrateful to those who took care of my question, but I still haven't received the answer I was hoping for. The fact is that I'm looking for a way to discover whether I have "missed" some interesting figures in the described context. I exploited my (limited) programming skills to perform some automated searches, but I'm not sure that my algorithms guarantee not to miss any good solutions. I can describe and discuss these too, but I'd prefer to leave everyone free to use their original approach. Another chance would be to find some reference covering the topic, but I'm afraid that this is really too "recreational" to be dealt with in literature. Maybe something on the Internet? Someone who has done anything similar before me? I'm starting a bounty to find out! Thanks in advance!
geometry circle recreational-mathematics polygons diophantine-approximation
asked Apr 25 '17 at 15:12
lesath82lesath82
1,921226
$endgroup$
$begingroup$
en.wikipedia.org/wiki/Diophantine_equation
$endgroup$
– dantopa
Apr 25 '17 at 15:55
$begingroup$
This is quite similar to math.stackexchange.com/questions/64897
$endgroup$
– Mark S.
May 13 '17 at 13:20
$begingroup$
You are perfectly right, @Mark , I missed that! There is also an interesting keyword there by Robert Israel: "simultaneous diophantine approximation", I'll check if this leads me to something useful!
$endgroup$
– lesath82
May 13 '17 at 13:47
add a comment |
$begingroup$
I find myself often fooling around with pen and paper, preferably squared paper. So I began looking for ways to sketch geometric figures as precisely as possible without using compass and/or ruler. In particular I'm thinking of regular polygons and circles. I'm attaching a picture of my best findings, intentionally left handmade to show the kind of result I'm pursuing (following an answer I introduced a couple of digital adjustments). Explanations follow below.
Let me start with regular polygons. The rule is simple: the vertices must lie at the grid crossings, that is their coordinates must be expressible as integer numbers. Apart from the square, it is not possible to get regular polygons with this restriction, so the intent is to obtain the closest approximations keeping the figures "small". I've explored different ways to quantitatively assess the "precision" of an approximation, starting from length uniformity of sides (satisfactory only for triangles) later combined with uniformity of angles, to the mean squared distance of vertices from the ideal ones not lying on the grid. I put no requirements about the center; should it happen to fall on or close to a crossing, even better. Moreover I tested different size factors to take into account the fact that getting better shape approximations with bigger figures is somewhat obvious and also less useful for sketching. Anyway I've found substantial agreement among different strategies. In fact, I didn't really find head-to-head confrontations whose "winner" depended on the scoring details, and even if I did I would have happily declared a tie and kept all the alternatives. So, the sheet of paper in the top-left corner of the figure collects my best results of this family.
Next I'm relaxing the restriction: I'll let the vertices fall also on the middle points of the sides of paper squares. These can be located with satisfactory precision by eye, instead I'm not going to allow center points of paper squares. This new rule is equivalent to halving all the coordinates of a valid previous solution, provided that no point had both coordinates odd (one can horizontally and/or vertically translate the original figure by $1$ before attempting the division). I might also add that I don't like to see a line running parallel to the grid cutting squares in halves, but it's just a matter of taste. Obviously, at comparable dimensions, better approximations can be found with this additional freedom, and my best findings are represented in the top-right corner.
I shall proceed to circles. They are different, in that they do not have a limited number of vertices and one can look for support points –that now can belong to the circle and lie exactly at grid intersections at the same time– with no a priori restrictions of position and amount. So the goal becomes to find as many of them, and as evenly distributed, as possible. Again I'm trying to keep the circles small, with similar considerations as before about loss of interest as size grows. With this in mind, my best circles are depicted in the bottom-left corner.
Finally, again I soften the rules: support points on the grid can be approximations of exact circle points and/or they can be placed on middle points of square sides. My best findings of this kind are those in the bottom-right corner.
I couldn't find any serious dissertation on this (indeed trivial) argument, the most closely related argument being this: Circle Lattice Points.
Is anyone aware of others? Or is anybody willing to find themselves other similar (or better, under any respect) figures?
EDIT: I don't want to seem ungrateful to those who took care of my question, but I still haven't received the answer I was hoping for. The fact is that I'm looking for a way to discover whether I have "missed" some interesting figures in the described context. I exploited my (limited) programming skills to perform some automated searches, but I'm not sure that my algorithms guarantee not to miss any good solutions. I can describe and discuss these too, but I'd prefer to leave everyone free to use their original approach. Another chance would be to find some reference covering the topic, but I'm afraid that this is really too "recreational" to be dealt with in literature. Maybe something on the Internet? Someone who has done anything similar before me? I'm starting a bounty to find out! Thanks in advance!
geometry circle recreational-mathematics polygons diophantine-approximation
asked Apr 25 '17 at 15:12
lesath82lesath82
1,921226
$endgroup$
I find myself often fooling around with pen and paper, preferably squared paper. So I began looking for ways to sketch geometric figures as precisely as possible without using compass and/or ruler. In particular I'm thinking of regular polygons and circles. I'm attaching a picture of my best findings, intentionally left handmade to show the kind of result I'm pursuing (following an answer I introduced a couple of digital adjustments). Explanations follow below.
Let me start with regular polygons. The rule is simple: the vertices must lie at the grid crossings, that is their coordinates must be expressible as integer numbers. Apart from the square, it is not possible to get regular polygons with this restriction, so the intent is to obtain the closest approximations keeping the figures "small". I've explored different ways to quantitatively assess the "precision" of an approximation, starting from length uniformity of sides (satisfactory only for triangles) later combined with uniformity of angles, to the mean squared distance of vertices from the ideal ones not lying on the grid. I put no requirements about the center; should it happen to fall on or close to a crossing, even better. Moreover I tested different size factors to take into account the fact that getting better shape approximations with bigger figures is somewhat obvious and also less useful for sketching. Anyway I've found substantial agreement among different strategies. In fact, I didn't really find head-to-head confrontations whose "winner" depended on the scoring details, and even if I did I would have happily declared a tie and kept all the alternatives. So, the sheet of paper in the top-left corner of the figure collects my best results of this family.
Next I'm relaxing the restriction: I'll let the vertices fall also on the middle points of the sides of paper squares. These can be located with satisfactory precision by eye, instead I'm not going to allow center points of paper squares. This new rule is equivalent to halving all the coordinates of a valid previous solution, provided that no point had both coordinates odd (one can horizontally and/or vertically translate the original figure by $1$ before attempting the division). I might also add that I don't like to see a line running parallel to the grid cutting squares in halves, but it's just a matter of taste. Obviously, at comparable dimensions, better approximations can be found with this additional freedom, and my best findings are represented in the top-right corner.
I shall proceed to circles. They are different, in that they do not have a limited number of vertices and one can look for support points –that now can belong to the circle and lie exactly at grid intersections at the same time– with no a priori restrictions of position and amount. So the goal becomes to find as many of them, and as evenly distributed, as possible. Again I'm trying to keep the circles small, with similar considerations as before about loss of interest as size grows. With this in mind, my best circles are depicted in the bottom-left corner.
Finally, again I soften the rules: support points on the grid can be approximations of exact circle points and/or they can be placed on middle points of square sides. My best findings of this kind are those in the bottom-right corner.
I couldn't find any serious dissertation on this (indeed trivial) argument, the most closely related argument being this: Circle Lattice Points.
Is anyone aware of others? Or is anybody willing to find themselves other similar (or better, under any respect) figures?
EDIT: I don't want to seem ungrateful to those who took care of my question, but I still haven't received the answer I was hoping for. The fact is that I'm looking for a way to discover whether I have "missed" some interesting figures in the described context. I exploited my (limited) programming skills to perform some automated searches, but I'm not sure that my algorithms guarantee not to miss any good solutions. I can describe and discuss these too, but I'd prefer to leave everyone free to use their original approach. Another chance would be to find some reference covering the topic, but I'm afraid that this is really too "recreational" to be dealt with in literature. Maybe something on the Internet? Someone who has done anything similar before me? I'm starting a bounty to find out! Thanks in advance!
geometry circle recreational-mathematics polygons diophantine-approximation
geometry circle recreational-mathematics polygons diophantine-approximation
asked Apr 25 '17 at 15:12
lesath82lesath82
1,921226
asked Apr 25 '17 at 15:12
lesath82lesath82
1,921226
edited May 20 '17 at 21:59
lesath82
asked Apr 25 '17 at 15:12
lesath82lesath82
1,921226
asked Apr 25 '17 at 15:12
lesath82lesath82
1,921226
asked Apr 25 '17 at 15:12
lesath82lesath82
1,921226
1,921226
$begingroup$
en.wikipedia.org/wiki/Diophantine_equation
$endgroup$
– dantopa
Apr 25 '17 at 15:55
$begingroup$
This is quite similar to math.stackexchange.com/questions/64897
$endgroup$
– Mark S.
May 13 '17 at 13:20
$begingroup$
You are perfectly right, @Mark , I missed that! There is also an interesting keyword there by Robert Israel: "simultaneous diophantine approximation", I'll check if this leads me to something useful!
$endgroup$
– lesath82
May 13 '17 at 13:47
add a comment |
$begingroup$
en.wikipedia.org/wiki/Diophantine_equation
$endgroup$
– dantopa
Apr 25 '17 at 15:55
$begingroup$
This is quite similar to math.stackexchange.com/questions/64897
$endgroup$
– Mark S.
May 13 '17 at 13:20
$begingroup$
You are perfectly right, @Mark , I missed that! There is also an interesting keyword there by Robert Israel: "simultaneous diophantine approximation", I'll check if this leads me to something useful!
$endgroup$
– lesath82
May 13 '17 at 13:47
$begingroup$
en.wikipedia.org/wiki/Diophantine_equation
$endgroup$
– dantopa
Apr 25 '17 at 15:55
$begingroup$
en.wikipedia.org/wiki/Diophantine_equation
$endgroup$
– dantopa
Apr 25 '17 at 15:55
$begingroup$
This is quite similar to math.stackexchange.com/questions/64897
$endgroup$
– Mark S.
May 13 '17 at 13:20
$begingroup$
This is quite similar to math.stackexchange.com/questions/64897
$endgroup$
– Mark S.
May 13 '17 at 13:20
$begingroup$
You are perfectly right, @Mark , I missed that! There is also an interesting keyword there by Robert Israel: "simultaneous diophantine approximation", I'll check if this leads me to something useful!
$endgroup$
– lesath82
May 13 '17 at 13:47
$begingroup$
You are perfectly right, @Mark , I missed that! There is also an interesting keyword there by Robert Israel: "simultaneous diophantine approximation", I'll check if this leads me to something useful!
$endgroup$
– lesath82
May 13 '17 at 13:47
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Historically, a similar problem came to people trying to use gear ratios to approximate irrational proportionality, as there is always an natural number of teeth on any particular gear. Short story very short: The convergents of continued fractions are the "best" approximation of the true value of an irrational number in the sense that there is no other rational number that is closer to the true value with a smaller denominator. (Actually the guarantees about approximation are stronger than this, which you can see at wikipedia's entry on continued fractions.)
The convergents of continued fractions of various irrationals will tell you what kind of scaling you would need to achieve your approximations. In practice this would mean that you have something like an exact expression for placement of the points, which you then convert to a rational approximation with an error less than what you consider the finest acceptable deviation (perhaps the thickness of your line, maybe 1/20 of a grid spacing).
Consider the problem of having a point at $sqrt{2}$ in some axis. The square root of 2 is approximately 1. Well, not 1. More like $1 + frac{1}{2}$. Well, not quite 2, more like $2 + frac{1}{2}$. Actually the continued fraction of
$$sqrt{2} = 1 + frac{1}{2 + frac{1}{2 + cdots}} = [1~ 2~ 2~ 2~ 2~ 2~ ldots ] = [1~ overline{2}]$$
And its convergents are $1,~ frac{3}{2},~ frac{7}{5},~ frac{17}{12},~ frac{41}{29},~ ldots$ which are respectively less than, greater than, less than... the true value. If we mentally rescale our sheet so that 12 units equals "1", then 17 units is approximately $sqrt{2}$.
answered Apr 25 '17 at 15:45
law-of-fiveslaw-of-fives
1,812313
$endgroup$
$begingroup$
You made a good point, even if I believe it's not really helpful finding well-approximated small figures. Furthermore, for circles, I'm not having exact expression for designated points, since any of the uncountable points of the circumference is candidate to be used as a support point.
$endgroup$
– lesath82
Apr 25 '17 at 21:44
$begingroup$
@lesath82 There are many algebraic points on the circle, the most common being what we learn in early trig at 30, 45, and 60 degrees, expressible with $sqrt{2}$ and $sqrt{3}$. Other algebraic points are not necessarily very nice, but just these (and zero degrees) would give you four points per quadrant.
$endgroup$
– law-of-fives
Apr 25 '17 at 22:07
add a comment |
$begingroup$
This is not really an answer, it used to be three long comments, now put together to be easier to read.
If you look at your hexagon (near the pentagon) you could consider each side to be the hypotenuse of a right triangle. If each square (from the squared paper) has side length $1$, then that hexagon has four sides of length
$sqrt{1^2+4^2}=sqrt{17}$ and two sides of length $sqrt{3^2+3^2}=sqrt{18}$.
Your other hexagon doesn't quite have vertices on the grid, has a couple of vertices at midpoints of "grid unit segments", so one has to double all sides to fit it on the grid. If that is done, then it has two horizontal sides of length $8=sqrt{64}$, and four sides of length $sqrt{4^2+7^2}=sqrt{65}$.
So, a step (of course incomplete) would be to look for (small, integer)
$a,b,p,q$ with $a^2+b^2approx p^2+q^2$, for example
$a^2+b^2=p^2+q^2pm1$, where $a,b,p>0$ and $q≥0$.
This could be done with a computer.
Then, it would be a next step (which I didn't try to solve)
once you find such $a,b,p,q$, try to find a right triangle with sides
$a,b$ and another right triangle with sides $p,q$,
and try to put them together so as to use the hypotenuses as sides of your polygon.
More generally look for $a_1^2+b_1^2approx a_2^2+b_2^2approx...approx a_n^2+b_n^2$ for a suitable $n$ (for example your septagon has $n=4$).
Your septagon has one point which is a midpoints of a "grid unit segment", so multiplying all sides by $2$, I get lengths (from bottom to top, squares of length of hypotenuses, rather than lengths themselves):
$10^2+0^2=100=6^2+8^2approx104=10^2+2^2approx97=4^2+9^2$.
This seems related to the link that you provide in the statement of your question, as well as to the link in the comment by @dantopa (about Diophantine equations), as well as to (not exactly) Pythagorean triples (but we do not look for $a^2+b^2=c^2$, only for
$a^2+b^2=$ integer $approx$ another integer $=p^2+q^2$).
answered May 13 '17 at 0:17
MirkoMirko
7,778930
$endgroup$
$begingroup$
Thank you for your inputs! I have a couple of considerations about them. If I can find $a,b,p,q$ so that the sum of their pairwise squares is $=$ and not $approx$ it would be even better! Also I'm not ruling out $a^2+b^2=c^2$ (or at least $approx$), since suche a $c$ would represent a side aligned with the grid. The problem is that I have to collect many copious groups with these features, since I'll have to find a collection that I can arrange respecting angle requirements too! I'm afraid this is not going to be an efficient way, but I'll give it a try!
$endgroup$
– lesath82
May 13 '17 at 22:56
$begingroup$
It may not be efficient, but if only small numbers are interesting and if you don't find many of these, then you might be able to rule out such cases. But I admit I do not know what to think about it, before I see what a computer search for such numbers might show. Thank you !
$endgroup$
– Mirko
May 15 '17 at 19:12
add a comment |
$begingroup$
At first: idea is nice for design.
My first attempt is hard and primitive, but the idea can get more successful implementation.
And the second attempt can be more interesting, because this, as if, is the only non-convex polygon with pronounced symmetry axes, all "convex" vertices of which lie on one circle. And which can be paved with a plane.
In addition, in a vertical arrangement, it is very similar to the reward for the positive, which is present in the topic.
answered May 13 '17 at 11:01
Yuri NegometyanovYuri Negometyanov
11.4k1728
$endgroup$
$begingroup$
I'm not sure what you mean talking about "the reward for the positive". Yet your picture turned out to be extremely useful: at the beginning I discarded circles defined by 8 points as "too easy", but afterwards I decided to include one (not even smaller) with 6... this made no sense, so I corrected my picture. Thank you!
$endgroup$
– lesath82
May 13 '17 at 22:22
add a comment |
$begingroup$
For a hexagon, using the continued fraction for $sqrt{3}$, [1; 1, 2, 1, 2, 1, 2, 1... might work really well. I haven't figured out a proof but I think that for a pentagon, the highest possible accuracy approaches a perfect pentagon at a much slower rate with respect to size limit. I think a better way to figure out how to draw a more accurate pentagon is to drop the criterion of the verticies being at exact intersections on the graph paper and instead do the math and figure for some exact pentagon which exact intersections are on which side of each of its edges and then draw lines that are on the same side of each of those points as the edges of the exact pentagon. The end point of an edge is not necessarily going to be really close to an intersection on the grid but some part of the edge might be much closer to one. Don't bother using your eyes to verify the accuracy of the shapes you drew. It will not teach you how to draw more accurately because as described in my answer to Visually deceptive "proofs" which are mathematically wrong, the eyes are prone to optical illusions even of mathematically impossible images. The math will more reliably tell you whether there exists a pentagon that contains all of a certain set of intersection points on the grid and only those intersection points. There is a way to mathematically define a certain pentagon and figure out mathematically for each intersection point on the grid which side of each edge of the pentagon extended to infinity it is on.
answered Jan 9 at 5:50
TimothyTimothy
301212
$endgroup$
add a comment |
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4 Answers
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4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Historically, a similar problem came to people trying to use gear ratios to approximate irrational proportionality, as there is always an natural number of teeth on any particular gear. Short story very short: The convergents of continued fractions are the "best" approximation of the true value of an irrational number in the sense that there is no other rational number that is closer to the true value with a smaller denominator. (Actually the guarantees about approximation are stronger than this, which you can see at wikipedia's entry on continued fractions.)
The convergents of continued fractions of various irrationals will tell you what kind of scaling you would need to achieve your approximations. In practice this would mean that you have something like an exact expression for placement of the points, which you then convert to a rational approximation with an error less than what you consider the finest acceptable deviation (perhaps the thickness of your line, maybe 1/20 of a grid spacing).
Consider the problem of having a point at $sqrt{2}$ in some axis. The square root of 2 is approximately 1. Well, not 1. More like $1 + frac{1}{2}$. Well, not quite 2, more like $2 + frac{1}{2}$. Actually the continued fraction of
$$sqrt{2} = 1 + frac{1}{2 + frac{1}{2 + cdots}} = [1~ 2~ 2~ 2~ 2~ 2~ ldots ] = [1~ overline{2}]$$
And its convergents are $1,~ frac{3}{2},~ frac{7}{5},~ frac{17}{12},~ frac{41}{29},~ ldots$ which are respectively less than, greater than, less than... the true value. If we mentally rescale our sheet so that 12 units equals "1", then 17 units is approximately $sqrt{2}$.
answered Apr 25 '17 at 15:45
law-of-fiveslaw-of-fives
1,812313
$endgroup$
$begingroup$
You made a good point, even if I believe it's not really helpful finding well-approximated small figures. Furthermore, for circles, I'm not having exact expression for designated points, since any of the uncountable points of the circumference is candidate to be used as a support point.
$endgroup$
– lesath82
Apr 25 '17 at 21:44
$begingroup$
@lesath82 There are many algebraic points on the circle, the most common being what we learn in early trig at 30, 45, and 60 degrees, expressible with $sqrt{2}$ and $sqrt{3}$. Other algebraic points are not necessarily very nice, but just these (and zero degrees) would give you four points per quadrant.
$endgroup$
– law-of-fives
Apr 25 '17 at 22:07
add a comment |
$begingroup$
Historically, a similar problem came to people trying to use gear ratios to approximate irrational proportionality, as there is always an natural number of teeth on any particular gear. Short story very short: The convergents of continued fractions are the "best" approximation of the true value of an irrational number in the sense that there is no other rational number that is closer to the true value with a smaller denominator. (Actually the guarantees about approximation are stronger than this, which you can see at wikipedia's entry on continued fractions.)
The convergents of continued fractions of various irrationals will tell you what kind of scaling you would need to achieve your approximations. In practice this would mean that you have something like an exact expression for placement of the points, which you then convert to a rational approximation with an error less than what you consider the finest acceptable deviation (perhaps the thickness of your line, maybe 1/20 of a grid spacing).
Consider the problem of having a point at $sqrt{2}$ in some axis. The square root of 2 is approximately 1. Well, not 1. More like $1 + frac{1}{2}$. Well, not quite 2, more like $2 + frac{1}{2}$. Actually the continued fraction of
$$sqrt{2} = 1 + frac{1}{2 + frac{1}{2 + cdots}} = [1~ 2~ 2~ 2~ 2~ 2~ ldots ] = [1~ overline{2}]$$
And its convergents are $1,~ frac{3}{2},~ frac{7}{5},~ frac{17}{12},~ frac{41}{29},~ ldots$ which are respectively less than, greater than, less than... the true value. If we mentally rescale our sheet so that 12 units equals "1", then 17 units is approximately $sqrt{2}$.
answered Apr 25 '17 at 15:45
law-of-fiveslaw-of-fives
1,812313
$endgroup$
$begingroup$
You made a good point, even if I believe it's not really helpful finding well-approximated small figures. Furthermore, for circles, I'm not having exact expression for designated points, since any of the uncountable points of the circumference is candidate to be used as a support point.
$endgroup$
– lesath82
Apr 25 '17 at 21:44
$begingroup$
@lesath82 There are many algebraic points on the circle, the most common being what we learn in early trig at 30, 45, and 60 degrees, expressible with $sqrt{2}$ and $sqrt{3}$. Other algebraic points are not necessarily very nice, but just these (and zero degrees) would give you four points per quadrant.
$endgroup$
– law-of-fives
Apr 25 '17 at 22:07
add a comment |
$begingroup$
Historically, a similar problem came to people trying to use gear ratios to approximate irrational proportionality, as there is always an natural number of teeth on any particular gear. Short story very short: The convergents of continued fractions are the "best" approximation of the true value of an irrational number in the sense that there is no other rational number that is closer to the true value with a smaller denominator. (Actually the guarantees about approximation are stronger than this, which you can see at wikipedia's entry on continued fractions.)
The convergents of continued fractions of various irrationals will tell you what kind of scaling you would need to achieve your approximations. In practice this would mean that you have something like an exact expression for placement of the points, which you then convert to a rational approximation with an error less than what you consider the finest acceptable deviation (perhaps the thickness of your line, maybe 1/20 of a grid spacing).
Consider the problem of having a point at $sqrt{2}$ in some axis. The square root of 2 is approximately 1. Well, not 1. More like $1 + frac{1}{2}$. Well, not quite 2, more like $2 + frac{1}{2}$. Actually the continued fraction of
$$sqrt{2} = 1 + frac{1}{2 + frac{1}{2 + cdots}} = [1~ 2~ 2~ 2~ 2~ 2~ ldots ] = [1~ overline{2}]$$
And its convergents are $1,~ frac{3}{2},~ frac{7}{5},~ frac{17}{12},~ frac{41}{29},~ ldots$ which are respectively less than, greater than, less than... the true value. If we mentally rescale our sheet so that 12 units equals "1", then 17 units is approximately $sqrt{2}$.
answered Apr 25 '17 at 15:45
law-of-fiveslaw-of-fives
1,812313
$endgroup$
Historically, a similar problem came to people trying to use gear ratios to approximate irrational proportionality, as there is always an natural number of teeth on any particular gear. Short story very short: The convergents of continued fractions are the "best" approximation of the true value of an irrational number in the sense that there is no other rational number that is closer to the true value with a smaller denominator. (Actually the guarantees about approximation are stronger than this, which you can see at wikipedia's entry on continued fractions.)
The convergents of continued fractions of various irrationals will tell you what kind of scaling you would need to achieve your approximations. In practice this would mean that you have something like an exact expression for placement of the points, which you then convert to a rational approximation with an error less than what you consider the finest acceptable deviation (perhaps the thickness of your line, maybe 1/20 of a grid spacing).
Consider the problem of having a point at $sqrt{2}$ in some axis. The square root of 2 is approximately 1. Well, not 1. More like $1 + frac{1}{2}$. Well, not quite 2, more like $2 + frac{1}{2}$. Actually the continued fraction of
$$sqrt{2} = 1 + frac{1}{2 + frac{1}{2 + cdots}} = [1~ 2~ 2~ 2~ 2~ 2~ ldots ] = [1~ overline{2}]$$
And its convergents are $1,~ frac{3}{2},~ frac{7}{5},~ frac{17}{12},~ frac{41}{29},~ ldots$ which are respectively less than, greater than, less than... the true value. If we mentally rescale our sheet so that 12 units equals "1", then 17 units is approximately $sqrt{2}$.
answered Apr 25 '17 at 15:45
law-of-fiveslaw-of-fives
1,812313
answered Apr 25 '17 at 15:45
law-of-fiveslaw-of-fives
1,812313
answered Apr 25 '17 at 15:45
law-of-fiveslaw-of-fives
1,812313
answered Apr 25 '17 at 15:45
law-of-fiveslaw-of-fives
1,812313
1,812313
$begingroup$
You made a good point, even if I believe it's not really helpful finding well-approximated small figures. Furthermore, for circles, I'm not having exact expression for designated points, since any of the uncountable points of the circumference is candidate to be used as a support point.
$endgroup$
– lesath82
Apr 25 '17 at 21:44
$begingroup$
@lesath82 There are many algebraic points on the circle, the most common being what we learn in early trig at 30, 45, and 60 degrees, expressible with $sqrt{2}$ and $sqrt{3}$. Other algebraic points are not necessarily very nice, but just these (and zero degrees) would give you four points per quadrant.
$endgroup$
– law-of-fives
Apr 25 '17 at 22:07
add a comment |
$begingroup$
You made a good point, even if I believe it's not really helpful finding well-approximated small figures. Furthermore, for circles, I'm not having exact expression for designated points, since any of the uncountable points of the circumference is candidate to be used as a support point.
$endgroup$
– lesath82
Apr 25 '17 at 21:44
$begingroup$
@lesath82 There are many algebraic points on the circle, the most common being what we learn in early trig at 30, 45, and 60 degrees, expressible with $sqrt{2}$ and $sqrt{3}$. Other algebraic points are not necessarily very nice, but just these (and zero degrees) would give you four points per quadrant.
$endgroup$
– law-of-fives
Apr 25 '17 at 22:07
$begingroup$
You made a good point, even if I believe it's not really helpful finding well-approximated small figures. Furthermore, for circles, I'm not having exact expression for designated points, since any of the uncountable points of the circumference is candidate to be used as a support point.
$endgroup$
– lesath82
Apr 25 '17 at 21:44
$begingroup$
You made a good point, even if I believe it's not really helpful finding well-approximated small figures. Furthermore, for circles, I'm not having exact expression for designated points, since any of the uncountable points of the circumference is candidate to be used as a support point.
$endgroup$
– lesath82
Apr 25 '17 at 21:44
$begingroup$
@lesath82 There are many algebraic points on the circle, the most common being what we learn in early trig at 30, 45, and 60 degrees, expressible with $sqrt{2}$ and $sqrt{3}$. Other algebraic points are not necessarily very nice, but just these (and zero degrees) would give you four points per quadrant.
$endgroup$
– law-of-fives
Apr 25 '17 at 22:07
$begingroup$
@lesath82 There are many algebraic points on the circle, the most common being what we learn in early trig at 30, 45, and 60 degrees, expressible with $sqrt{2}$ and $sqrt{3}$. Other algebraic points are not necessarily very nice, but just these (and zero degrees) would give you four points per quadrant.
$endgroup$
– law-of-fives
Apr 25 '17 at 22:07
add a comment |
$begingroup$
This is not really an answer, it used to be three long comments, now put together to be easier to read.
If you look at your hexagon (near the pentagon) you could consider each side to be the hypotenuse of a right triangle. If each square (from the squared paper) has side length $1$, then that hexagon has four sides of length
$sqrt{1^2+4^2}=sqrt{17}$ and two sides of length $sqrt{3^2+3^2}=sqrt{18}$.
Your other hexagon doesn't quite have vertices on the grid, has a couple of vertices at midpoints of "grid unit segments", so one has to double all sides to fit it on the grid. If that is done, then it has two horizontal sides of length $8=sqrt{64}$, and four sides of length $sqrt{4^2+7^2}=sqrt{65}$.
So, a step (of course incomplete) would be to look for (small, integer)
$a,b,p,q$ with $a^2+b^2approx p^2+q^2$, for example
$a^2+b^2=p^2+q^2pm1$, where $a,b,p>0$ and $q≥0$.
This could be done with a computer.
Then, it would be a next step (which I didn't try to solve)
once you find such $a,b,p,q$, try to find a right triangle with sides
$a,b$ and another right triangle with sides $p,q$,
and try to put them together so as to use the hypotenuses as sides of your polygon.
More generally look for $a_1^2+b_1^2approx a_2^2+b_2^2approx...approx a_n^2+b_n^2$ for a suitable $n$ (for example your septagon has $n=4$).
Your septagon has one point which is a midpoints of a "grid unit segment", so multiplying all sides by $2$, I get lengths (from bottom to top, squares of length of hypotenuses, rather than lengths themselves):
$10^2+0^2=100=6^2+8^2approx104=10^2+2^2approx97=4^2+9^2$.
This seems related to the link that you provide in the statement of your question, as well as to the link in the comment by @dantopa (about Diophantine equations), as well as to (not exactly) Pythagorean triples (but we do not look for $a^2+b^2=c^2$, only for
$a^2+b^2=$ integer $approx$ another integer $=p^2+q^2$).
answered May 13 '17 at 0:17
MirkoMirko
7,778930
$endgroup$
$begingroup$
Thank you for your inputs! I have a couple of considerations about them. If I can find $a,b,p,q$ so that the sum of their pairwise squares is $=$ and not $approx$ it would be even better! Also I'm not ruling out $a^2+b^2=c^2$ (or at least $approx$), since suche a $c$ would represent a side aligned with the grid. The problem is that I have to collect many copious groups with these features, since I'll have to find a collection that I can arrange respecting angle requirements too! I'm afraid this is not going to be an efficient way, but I'll give it a try!
$endgroup$
– lesath82
May 13 '17 at 22:56
$begingroup$
It may not be efficient, but if only small numbers are interesting and if you don't find many of these, then you might be able to rule out such cases. But I admit I do not know what to think about it, before I see what a computer search for such numbers might show. Thank you !
$endgroup$
– Mirko
May 15 '17 at 19:12
add a comment |
$begingroup$
This is not really an answer, it used to be three long comments, now put together to be easier to read.
If you look at your hexagon (near the pentagon) you could consider each side to be the hypotenuse of a right triangle. If each square (from the squared paper) has side length $1$, then that hexagon has four sides of length
$sqrt{1^2+4^2}=sqrt{17}$ and two sides of length $sqrt{3^2+3^2}=sqrt{18}$.
Your other hexagon doesn't quite have vertices on the grid, has a couple of vertices at midpoints of "grid unit segments", so one has to double all sides to fit it on the grid. If that is done, then it has two horizontal sides of length $8=sqrt{64}$, and four sides of length $sqrt{4^2+7^2}=sqrt{65}$.
So, a step (of course incomplete) would be to look for (small, integer)
$a,b,p,q$ with $a^2+b^2approx p^2+q^2$, for example
$a^2+b^2=p^2+q^2pm1$, where $a,b,p>0$ and $q≥0$.
This could be done with a computer.
Then, it would be a next step (which I didn't try to solve)
once you find such $a,b,p,q$, try to find a right triangle with sides
$a,b$ and another right triangle with sides $p,q$,
and try to put them together so as to use the hypotenuses as sides of your polygon.
More generally look for $a_1^2+b_1^2approx a_2^2+b_2^2approx...approx a_n^2+b_n^2$ for a suitable $n$ (for example your septagon has $n=4$).
Your septagon has one point which is a midpoints of a "grid unit segment", so multiplying all sides by $2$, I get lengths (from bottom to top, squares of length of hypotenuses, rather than lengths themselves):
$10^2+0^2=100=6^2+8^2approx104=10^2+2^2approx97=4^2+9^2$.
This seems related to the link that you provide in the statement of your question, as well as to the link in the comment by @dantopa (about Diophantine equations), as well as to (not exactly) Pythagorean triples (but we do not look for $a^2+b^2=c^2$, only for
$a^2+b^2=$ integer $approx$ another integer $=p^2+q^2$).
answered May 13 '17 at 0:17
MirkoMirko
7,778930
$endgroup$
$begingroup$
Thank you for your inputs! I have a couple of considerations about them. If I can find $a,b,p,q$ so that the sum of their pairwise squares is $=$ and not $approx$ it would be even better! Also I'm not ruling out $a^2+b^2=c^2$ (or at least $approx$), since suche a $c$ would represent a side aligned with the grid. The problem is that I have to collect many copious groups with these features, since I'll have to find a collection that I can arrange respecting angle requirements too! I'm afraid this is not going to be an efficient way, but I'll give it a try!
$endgroup$
– lesath82
May 13 '17 at 22:56
$begingroup$
It may not be efficient, but if only small numbers are interesting and if you don't find many of these, then you might be able to rule out such cases. But I admit I do not know what to think about it, before I see what a computer search for such numbers might show. Thank you !
$endgroup$
– Mirko
May 15 '17 at 19:12
add a comment |
$begingroup$
This is not really an answer, it used to be three long comments, now put together to be easier to read.
If you look at your hexagon (near the pentagon) you could consider each side to be the hypotenuse of a right triangle. If each square (from the squared paper) has side length $1$, then that hexagon has four sides of length
$sqrt{1^2+4^2}=sqrt{17}$ and two sides of length $sqrt{3^2+3^2}=sqrt{18}$.
Your other hexagon doesn't quite have vertices on the grid, has a couple of vertices at midpoints of "grid unit segments", so one has to double all sides to fit it on the grid. If that is done, then it has two horizontal sides of length $8=sqrt{64}$, and four sides of length $sqrt{4^2+7^2}=sqrt{65}$.
So, a step (of course incomplete) would be to look for (small, integer)
$a,b,p,q$ with $a^2+b^2approx p^2+q^2$, for example
$a^2+b^2=p^2+q^2pm1$, where $a,b,p>0$ and $q≥0$.
This could be done with a computer.
Then, it would be a next step (which I didn't try to solve)
once you find such $a,b,p,q$, try to find a right triangle with sides
$a,b$ and another right triangle with sides $p,q$,
and try to put them together so as to use the hypotenuses as sides of your polygon.
More generally look for $a_1^2+b_1^2approx a_2^2+b_2^2approx...approx a_n^2+b_n^2$ for a suitable $n$ (for example your septagon has $n=4$).
Your septagon has one point which is a midpoints of a "grid unit segment", so multiplying all sides by $2$, I get lengths (from bottom to top, squares of length of hypotenuses, rather than lengths themselves):
$10^2+0^2=100=6^2+8^2approx104=10^2+2^2approx97=4^2+9^2$.
This seems related to the link that you provide in the statement of your question, as well as to the link in the comment by @dantopa (about Diophantine equations), as well as to (not exactly) Pythagorean triples (but we do not look for $a^2+b^2=c^2$, only for
$a^2+b^2=$ integer $approx$ another integer $=p^2+q^2$).
answered May 13 '17 at 0:17
MirkoMirko
7,778930
$endgroup$
This is not really an answer, it used to be three long comments, now put together to be easier to read.
If you look at your hexagon (near the pentagon) you could consider each side to be the hypotenuse of a right triangle. If each square (from the squared paper) has side length $1$, then that hexagon has four sides of length
$sqrt{1^2+4^2}=sqrt{17}$ and two sides of length $sqrt{3^2+3^2}=sqrt{18}$.
Your other hexagon doesn't quite have vertices on the grid, has a couple of vertices at midpoints of "grid unit segments", so one has to double all sides to fit it on the grid. If that is done, then it has two horizontal sides of length $8=sqrt{64}$, and four sides of length $sqrt{4^2+7^2}=sqrt{65}$.
So, a step (of course incomplete) would be to look for (small, integer)
$a,b,p,q$ with $a^2+b^2approx p^2+q^2$, for example
$a^2+b^2=p^2+q^2pm1$, where $a,b,p>0$ and $q≥0$.
This could be done with a computer.
Then, it would be a next step (which I didn't try to solve)
once you find such $a,b,p,q$, try to find a right triangle with sides
$a,b$ and another right triangle with sides $p,q$,
and try to put them together so as to use the hypotenuses as sides of your polygon.
More generally look for $a_1^2+b_1^2approx a_2^2+b_2^2approx...approx a_n^2+b_n^2$ for a suitable $n$ (for example your septagon has $n=4$).
Your septagon has one point which is a midpoints of a "grid unit segment", so multiplying all sides by $2$, I get lengths (from bottom to top, squares of length of hypotenuses, rather than lengths themselves):
$10^2+0^2=100=6^2+8^2approx104=10^2+2^2approx97=4^2+9^2$.
This seems related to the link that you provide in the statement of your question, as well as to the link in the comment by @dantopa (about Diophantine equations), as well as to (not exactly) Pythagorean triples (but we do not look for $a^2+b^2=c^2$, only for
$a^2+b^2=$ integer $approx$ another integer $=p^2+q^2$).
answered May 13 '17 at 0:17
MirkoMirko
7,778930
edited May 13 '17 at 0:28
answered May 13 '17 at 0:17
MirkoMirko
7,778930
answered May 13 '17 at 0:17
MirkoMirko
7,778930
answered May 13 '17 at 0:17
MirkoMirko
7,778930
7,778930
$begingroup$
Thank you for your inputs! I have a couple of considerations about them. If I can find $a,b,p,q$ so that the sum of their pairwise squares is $=$ and not $approx$ it would be even better! Also I'm not ruling out $a^2+b^2=c^2$ (or at least $approx$), since suche a $c$ would represent a side aligned with the grid. The problem is that I have to collect many copious groups with these features, since I'll have to find a collection that I can arrange respecting angle requirements too! I'm afraid this is not going to be an efficient way, but I'll give it a try!
$endgroup$
– lesath82
May 13 '17 at 22:56
$begingroup$
It may not be efficient, but if only small numbers are interesting and if you don't find many of these, then you might be able to rule out such cases. But I admit I do not know what to think about it, before I see what a computer search for such numbers might show. Thank you !
$endgroup$
– Mirko
May 15 '17 at 19:12
add a comment |
$begingroup$
Thank you for your inputs! I have a couple of considerations about them. If I can find $a,b,p,q$ so that the sum of their pairwise squares is $=$ and not $approx$ it would be even better! Also I'm not ruling out $a^2+b^2=c^2$ (or at least $approx$), since suche a $c$ would represent a side aligned with the grid. The problem is that I have to collect many copious groups with these features, since I'll have to find a collection that I can arrange respecting angle requirements too! I'm afraid this is not going to be an efficient way, but I'll give it a try!
$endgroup$
– lesath82
May 13 '17 at 22:56
$begingroup$
It may not be efficient, but if only small numbers are interesting and if you don't find many of these, then you might be able to rule out such cases. But I admit I do not know what to think about it, before I see what a computer search for such numbers might show. Thank you !
$endgroup$
– Mirko
May 15 '17 at 19:12
$begingroup$
Thank you for your inputs! I have a couple of considerations about them. If I can find $a,b,p,q$ so that the sum of their pairwise squares is $=$ and not $approx$ it would be even better! Also I'm not ruling out $a^2+b^2=c^2$ (or at least $approx$), since suche a $c$ would represent a side aligned with the grid. The problem is that I have to collect many copious groups with these features, since I'll have to find a collection that I can arrange respecting angle requirements too! I'm afraid this is not going to be an efficient way, but I'll give it a try!
$endgroup$
– lesath82
May 13 '17 at 22:56
$begingroup$
Thank you for your inputs! I have a couple of considerations about them. If I can find $a,b,p,q$ so that the sum of their pairwise squares is $=$ and not $approx$ it would be even better! Also I'm not ruling out $a^2+b^2=c^2$ (or at least $approx$), since suche a $c$ would represent a side aligned with the grid. The problem is that I have to collect many copious groups with these features, since I'll have to find a collection that I can arrange respecting angle requirements too! I'm afraid this is not going to be an efficient way, but I'll give it a try!
$endgroup$
– lesath82
May 13 '17 at 22:56
$begingroup$
It may not be efficient, but if only small numbers are interesting and if you don't find many of these, then you might be able to rule out such cases. But I admit I do not know what to think about it, before I see what a computer search for such numbers might show. Thank you !
$endgroup$
– Mirko
May 15 '17 at 19:12
$begingroup$
It may not be efficient, but if only small numbers are interesting and if you don't find many of these, then you might be able to rule out such cases. But I admit I do not know what to think about it, before I see what a computer search for such numbers might show. Thank you !
$endgroup$
– Mirko
May 15 '17 at 19:12
add a comment |
$begingroup$
At first: idea is nice for design.
My first attempt is hard and primitive, but the idea can get more successful implementation.
And the second attempt can be more interesting, because this, as if, is the only non-convex polygon with pronounced symmetry axes, all "convex" vertices of which lie on one circle. And which can be paved with a plane.
In addition, in a vertical arrangement, it is very similar to the reward for the positive, which is present in the topic.
answered May 13 '17 at 11:01
Yuri NegometyanovYuri Negometyanov
11.4k1728
$endgroup$
$begingroup$
I'm not sure what you mean talking about "the reward for the positive". Yet your picture turned out to be extremely useful: at the beginning I discarded circles defined by 8 points as "too easy", but afterwards I decided to include one (not even smaller) with 6... this made no sense, so I corrected my picture. Thank you!
$endgroup$
– lesath82
May 13 '17 at 22:22
add a comment |
$begingroup$
At first: idea is nice for design.
My first attempt is hard and primitive, but the idea can get more successful implementation.
And the second attempt can be more interesting, because this, as if, is the only non-convex polygon with pronounced symmetry axes, all "convex" vertices of which lie on one circle. And which can be paved with a plane.
In addition, in a vertical arrangement, it is very similar to the reward for the positive, which is present in the topic.
answered May 13 '17 at 11:01
Yuri NegometyanovYuri Negometyanov
11.4k1728
$endgroup$
$begingroup$
I'm not sure what you mean talking about "the reward for the positive". Yet your picture turned out to be extremely useful: at the beginning I discarded circles defined by 8 points as "too easy", but afterwards I decided to include one (not even smaller) with 6... this made no sense, so I corrected my picture. Thank you!
$endgroup$
– lesath82
May 13 '17 at 22:22
add a comment |
$begingroup$
At first: idea is nice for design.
My first attempt is hard and primitive, but the idea can get more successful implementation.
And the second attempt can be more interesting, because this, as if, is the only non-convex polygon with pronounced symmetry axes, all "convex" vertices of which lie on one circle. And which can be paved with a plane.
In addition, in a vertical arrangement, it is very similar to the reward for the positive, which is present in the topic.
answered May 13 '17 at 11:01
Yuri NegometyanovYuri Negometyanov
11.4k1728
$endgroup$
At first: idea is nice for design.
My first attempt is hard and primitive, but the idea can get more successful implementation.
And the second attempt can be more interesting, because this, as if, is the only non-convex polygon with pronounced symmetry axes, all "convex" vertices of which lie on one circle. And which can be paved with a plane.
In addition, in a vertical arrangement, it is very similar to the reward for the positive, which is present in the topic.
answered May 13 '17 at 11:01
Yuri NegometyanovYuri Negometyanov
11.4k1728
edited May 13 '17 at 11:20
answered May 13 '17 at 11:01
Yuri NegometyanovYuri Negometyanov
11.4k1728
answered May 13 '17 at 11:01
Yuri NegometyanovYuri Negometyanov
11.4k1728
answered May 13 '17 at 11:01
Yuri NegometyanovYuri Negometyanov
11.4k1728
11.4k1728
$begingroup$
I'm not sure what you mean talking about "the reward for the positive". Yet your picture turned out to be extremely useful: at the beginning I discarded circles defined by 8 points as "too easy", but afterwards I decided to include one (not even smaller) with 6... this made no sense, so I corrected my picture. Thank you!
$endgroup$
– lesath82
May 13 '17 at 22:22
add a comment |
$begingroup$
I'm not sure what you mean talking about "the reward for the positive". Yet your picture turned out to be extremely useful: at the beginning I discarded circles defined by 8 points as "too easy", but afterwards I decided to include one (not even smaller) with 6... this made no sense, so I corrected my picture. Thank you!
$endgroup$
– lesath82
May 13 '17 at 22:22
$begingroup$
I'm not sure what you mean talking about "the reward for the positive". Yet your picture turned out to be extremely useful: at the beginning I discarded circles defined by 8 points as "too easy", but afterwards I decided to include one (not even smaller) with 6... this made no sense, so I corrected my picture. Thank you!
$endgroup$
– lesath82
May 13 '17 at 22:22
$begingroup$
I'm not sure what you mean talking about "the reward for the positive". Yet your picture turned out to be extremely useful: at the beginning I discarded circles defined by 8 points as "too easy", but afterwards I decided to include one (not even smaller) with 6... this made no sense, so I corrected my picture. Thank you!
$endgroup$
– lesath82
May 13 '17 at 22:22
add a comment |
$begingroup$
For a hexagon, using the continued fraction for $sqrt{3}$, [1; 1, 2, 1, 2, 1, 2, 1... might work really well. I haven't figured out a proof but I think that for a pentagon, the highest possible accuracy approaches a perfect pentagon at a much slower rate with respect to size limit. I think a better way to figure out how to draw a more accurate pentagon is to drop the criterion of the verticies being at exact intersections on the graph paper and instead do the math and figure for some exact pentagon which exact intersections are on which side of each of its edges and then draw lines that are on the same side of each of those points as the edges of the exact pentagon. The end point of an edge is not necessarily going to be really close to an intersection on the grid but some part of the edge might be much closer to one. Don't bother using your eyes to verify the accuracy of the shapes you drew. It will not teach you how to draw more accurately because as described in my answer to Visually deceptive "proofs" which are mathematically wrong, the eyes are prone to optical illusions even of mathematically impossible images. The math will more reliably tell you whether there exists a pentagon that contains all of a certain set of intersection points on the grid and only those intersection points. There is a way to mathematically define a certain pentagon and figure out mathematically for each intersection point on the grid which side of each edge of the pentagon extended to infinity it is on.
answered Jan 9 at 5:50
TimothyTimothy
301212
$endgroup$
add a comment |
$begingroup$
For a hexagon, using the continued fraction for $sqrt{3}$, [1; 1, 2, 1, 2, 1, 2, 1... might work really well. I haven't figured out a proof but I think that for a pentagon, the highest possible accuracy approaches a perfect pentagon at a much slower rate with respect to size limit. I think a better way to figure out how to draw a more accurate pentagon is to drop the criterion of the verticies being at exact intersections on the graph paper and instead do the math and figure for some exact pentagon which exact intersections are on which side of each of its edges and then draw lines that are on the same side of each of those points as the edges of the exact pentagon. The end point of an edge is not necessarily going to be really close to an intersection on the grid but some part of the edge might be much closer to one. Don't bother using your eyes to verify the accuracy of the shapes you drew. It will not teach you how to draw more accurately because as described in my answer to Visually deceptive "proofs" which are mathematically wrong, the eyes are prone to optical illusions even of mathematically impossible images. The math will more reliably tell you whether there exists a pentagon that contains all of a certain set of intersection points on the grid and only those intersection points. There is a way to mathematically define a certain pentagon and figure out mathematically for each intersection point on the grid which side of each edge of the pentagon extended to infinity it is on.
answered Jan 9 at 5:50
TimothyTimothy
301212
$endgroup$
add a comment |
$begingroup$
For a hexagon, using the continued fraction for $sqrt{3}$, [1; 1, 2, 1, 2, 1, 2, 1... might work really well. I haven't figured out a proof but I think that for a pentagon, the highest possible accuracy approaches a perfect pentagon at a much slower rate with respect to size limit. I think a better way to figure out how to draw a more accurate pentagon is to drop the criterion of the verticies being at exact intersections on the graph paper and instead do the math and figure for some exact pentagon which exact intersections are on which side of each of its edges and then draw lines that are on the same side of each of those points as the edges of the exact pentagon. The end point of an edge is not necessarily going to be really close to an intersection on the grid but some part of the edge might be much closer to one. Don't bother using your eyes to verify the accuracy of the shapes you drew. It will not teach you how to draw more accurately because as described in my answer to Visually deceptive "proofs" which are mathematically wrong, the eyes are prone to optical illusions even of mathematically impossible images. The math will more reliably tell you whether there exists a pentagon that contains all of a certain set of intersection points on the grid and only those intersection points. There is a way to mathematically define a certain pentagon and figure out mathematically for each intersection point on the grid which side of each edge of the pentagon extended to infinity it is on.
answered Jan 9 at 5:50
TimothyTimothy
301212
$endgroup$
For a hexagon, using the continued fraction for $sqrt{3}$, [1; 1, 2, 1, 2, 1, 2, 1... might work really well. I haven't figured out a proof but I think that for a pentagon, the highest possible accuracy approaches a perfect pentagon at a much slower rate with respect to size limit. I think a better way to figure out how to draw a more accurate pentagon is to drop the criterion of the verticies being at exact intersections on the graph paper and instead do the math and figure for some exact pentagon which exact intersections are on which side of each of its edges and then draw lines that are on the same side of each of those points as the edges of the exact pentagon. The end point of an edge is not necessarily going to be really close to an intersection on the grid but some part of the edge might be much closer to one. Don't bother using your eyes to verify the accuracy of the shapes you drew. It will not teach you how to draw more accurately because as described in my answer to Visually deceptive "proofs" which are mathematically wrong, the eyes are prone to optical illusions even of mathematically impossible images. The math will more reliably tell you whether there exists a pentagon that contains all of a certain set of intersection points on the grid and only those intersection points. There is a way to mathematically define a certain pentagon and figure out mathematically for each intersection point on the grid which side of each edge of the pentagon extended to infinity it is on.
answered Jan 9 at 5:50
TimothyTimothy
301212
answered Jan 9 at 5:50
TimothyTimothy
301212
answered Jan 9 at 5:50
TimothyTimothy
301212
answered Jan 9 at 5:50
TimothyTimothy
301212
301212
add a comment |
add a comment |
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$begingroup$
en.wikipedia.org/wiki/Diophantine_equation
$endgroup$
– dantopa
Apr 25 '17 at 15:55
$begingroup$
This is quite similar to math.stackexchange.com/questions/64897
$endgroup$
– Mark S.
May 13 '17 at 13:20
$begingroup$
You are perfectly right, @Mark , I missed that! There is also an interesting keyword there by Robert Israel: "simultaneous diophantine approximation", I'll check if this leads me to something useful!
$endgroup$
– lesath82
May 13 '17 at 13:47