Mixing
Ruler and compass construction of the unit-distance petersen graph embedding
$begingroup$
The Petersen graph is a unit distance graph, and this embedding is shown below, where each edge of the graph is one unit in length.
Is there a ruler and compass construction for this embedding? If so, what is it?
It seems like the kind of thing that you can construct with a ruler and compass, but my brief attempt and google search failed.
I tried by constructing the outer pentagon, and then drawing unit radius arcs inside it, centered at each vertex. The inner star should have it's vertices on those arcs.
I did come fairly close by eyeballing the "radius" of the star and drawing the smaller circle, but I suspect this was cheating :p
Image credit: http://en.wikipedia.org/wiki/File:Petersen_graph,_unit_distance.svg
graph-theory geometric-construction
asked Nov 1 '13 at 20:47
AlejandroAlejandro
1636
$endgroup$
add a comment |
$begingroup$
The Petersen graph is a unit distance graph, and this embedding is shown below, where each edge of the graph is one unit in length.
Is there a ruler and compass construction for this embedding? If so, what is it?
It seems like the kind of thing that you can construct with a ruler and compass, but my brief attempt and google search failed.
I tried by constructing the outer pentagon, and then drawing unit radius arcs inside it, centered at each vertex. The inner star should have it's vertices on those arcs.
I did come fairly close by eyeballing the "radius" of the star and drawing the smaller circle, but I suspect this was cheating :p
Image credit: http://en.wikipedia.org/wiki/File:Petersen_graph,_unit_distance.svg
graph-theory geometric-construction
asked Nov 1 '13 at 20:47
AlejandroAlejandro
1636
$endgroup$
add a comment |
$begingroup$
The Petersen graph is a unit distance graph, and this embedding is shown below, where each edge of the graph is one unit in length.
Is there a ruler and compass construction for this embedding? If so, what is it?
It seems like the kind of thing that you can construct with a ruler and compass, but my brief attempt and google search failed.
I tried by constructing the outer pentagon, and then drawing unit radius arcs inside it, centered at each vertex. The inner star should have it's vertices on those arcs.
I did come fairly close by eyeballing the "radius" of the star and drawing the smaller circle, but I suspect this was cheating :p
Image credit: http://en.wikipedia.org/wiki/File:Petersen_graph,_unit_distance.svg
graph-theory geometric-construction
asked Nov 1 '13 at 20:47
AlejandroAlejandro
1636
$endgroup$
The Petersen graph is a unit distance graph, and this embedding is shown below, where each edge of the graph is one unit in length.
Is there a ruler and compass construction for this embedding? If so, what is it?
It seems like the kind of thing that you can construct with a ruler and compass, but my brief attempt and google search failed.
I tried by constructing the outer pentagon, and then drawing unit radius arcs inside it, centered at each vertex. The inner star should have it's vertices on those arcs.
I did come fairly close by eyeballing the "radius" of the star and drawing the smaller circle, but I suspect this was cheating :p
Image credit: http://en.wikipedia.org/wiki/File:Petersen_graph,_unit_distance.svg
graph-theory geometric-construction
graph-theory geometric-construction
asked Nov 1 '13 at 20:47
AlejandroAlejandro
1636
asked Nov 1 '13 at 20:47
AlejandroAlejandro
1636
asked Nov 1 '13 at 20:47
AlejandroAlejandro
1636
asked Nov 1 '13 at 20:47
AlejandroAlejandro
1636
asked Nov 1 '13 at 20:47
AlejandroAlejandro
1636
1636
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
If the inner star's side are $1$ the radius of its circumflexing circle is $frac1{2r}=sinfrac25pi=sqrt{frac{5+sqrt5}8}$ which leads that $$r=sqrt{frac{5-sqrt5}{10}}$$
So you can build $r$ with ruller and compass.
Update
Actually it might become simpler: First build your unit-sized pentagon $A_1B_1C_1D_1E_1$:
Then draw perpendiculars to each side at each vertex.
These perpendiculars will cut each-other at points $A_2,D_3,B_2,E_3,C_2,A_3,D_2,B_3,E_3,C_3$ (subindexes $3$ are not shown in the drawing). You dont need them to find them all, just one or two of them (v.g. $A_2$ and $C_2$).
The figure $A_2C_2E_2B_2D_2$ is a pentagram congruent to the one in the unit-distance Petersen graph, and they both share the same circumcircle (red in the figure)
So, take $A_1A_2$ and $C_1C_2$ they cut at the center $O$ of $A_1B_1C_1D_1E_1$. Draw the circle centered in $O$ with radius $OA_2$, and I supose you can now finish the problem.
answered Nov 1 '13 at 21:11
Carlos Eugenio Thompson PinzónCarlos Eugenio Thompson Pinzón
2,526823
$endgroup$
1
$begingroup$
Can you explain your first identity?
$endgroup$
– Samuel
Nov 1 '13 at 21:17
$begingroup$
sin(2pi / 5) = opposite / hypotenus = (1/2) / r. Draw a star in a circle, and radii to its points and you'll probably see this. I still don't know how to construct r, so more elaboration is welcome.
$endgroup$
– Alejandro
Nov 1 '13 at 21:37
$begingroup$
also, I suspect the construction of r will not fit on the page, since I imagine I'll need a line segment that is at least 5 units long to start.
$endgroup$
– Alejandro
Nov 1 '13 at 22:25
add a comment |
$begingroup$
I first drew this back in (I think) 1999. It's a very nice construction, and if you base it on inscribing the outer pentagon into a circle, you can do it at full size covering most of a piece of paper, without ever running off that paper, and with barely any more real work than constructing a regular pentagon.
It's also my avatar, everywhere online I have one.
The construction, as I do it:
- First, build a large circle and a pair of perpendicular diameters. Let the center be $O$, one of the endpoints of a diameter be $A$, and one of the endpoints of the other diameter be the first vertex $V_0$.
- Find the midpoint of $OA$, and call it B.
- Find the bisector of angle $OAV_0$. Let the intersection of this bisector with $OV_0$ be $C$.
- Reflect $O$ across $C$ to get $D$, and draw the perpendicular bisector of $OD$ through $C$. This perpendicular intersects the outer circle at two vertices $V_1$ and $V_4$ of the outer pentagon we want.
- Draw the circle centered at $O$ with $D$ on it. This circle intersects $OA$ at a vertex $V_5$ of the inner pentagram.
- We have now done all the hard parts; the rest is marking off the same standard distance to find all the vertices.
From $V_1$, set your compass to length $V_0V_1$ and mark off $V_2$ on the outer circle (the other intersection) and $V_6$ on the inner circle (same angular direction as $V_5$ relative to $V_0$; it should be just inside the segment $V_0V_5$).
From $V_4$, mark off $V_3$ on the outer circle and $V_9$ on the inner.
From $V_5$, mark off $V_7$ and $V_8$ on the inner circle.
Draw in the lines, and we're done.
The key facts that make this construction as nice as it is:
- The ratio of the radii of the two circles the pentagons are inscribed in is the golden ratio $frac{sqrt{5}-1}{2}$, which we need anyway to construct a pentagon in the first place. The first step in the construction that isn't part of the construction for the outer pentagon is to draw the smaller circle.
- While that part was fairly obvious and well-known (the ratio we need is the same as the ratio between the side and diagonal of a regular pentagon), we still need to know how much to twist the inner pentagram by so that the crossing edges are the same length as the pentagon and the pentagram. Law of cosines - we need $$1^2+left(frac{sqrt{5}-1}{2}right)^2 - 2cdot 1cdot frac{sqrt{5}-1}{2}cos(theta) = (2sin(36^circ))^2$$
$$1+frac{3-sqrt{5}}{2} -(sqrt{5}-1)costheta = 2-2cos(72^circ) = 2-frac{sqrt{5}-1}{2}$$
$$frac{5-sqrt{5}}{2}-(sqrt{5}-1)costheta = frac{5-sqrt{5}}{2}$$
Huh, $costheta=0$? Exactly a $90^circ$ rotation? That very nice coincidence is what inspired me to draw it in the first place.
Now, on actually drawing it: Because the crossing lines come pretty close to the next vertices over on the inner pentagram, this diagram is visually sensitive to small errors. Draw carefully, because an error of a few millimeters out of a full-page drawing will be visible.
answered Nov 25 '18 at 20:42
jmerryjmerry
12.3k1628
$endgroup$
$begingroup$
So, looking back on my first post, the one that brought me here (initially posting as a guest) - it could have been better. So I added a picture to make it better. The diagram is color-coded; intermediate steps in blue, transient construction marks in green, labels in red, and the final graph in black. It was generated in Asymptote, directly following the construction as described.
$endgroup$
– jmerry
Jan 20 at 20:37
add a comment |
$begingroup$
with compass and ruler. -use whatever method you are comfortable with to construct a pentagon. -draw lines from each corner, for a total of 5 lines,
to the middle of the edge line connecting the next 2 angles of the
pentagon. -set your compass to the length of the edge lines
(incidentally this was the same size as the circles I used in one
particular method to construct my pentagon originally), and mark
the end of each of the most recent 5 lines. -draw lines from
where each compass intersected the interior lines to form your
pentacle/5 pointed star. if you measure each line they are the
same length
answered Oct 5 '15 at 22:20
finbarfinbar
1
$endgroup$
$begingroup$
That certainly is a simple construction, but can you justify it as being true?
$endgroup$
– Rory Daulton
Oct 5 '15 at 23:02
add a comment |
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3 Answers
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3 Answers
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$begingroup$
If the inner star's side are $1$ the radius of its circumflexing circle is $frac1{2r}=sinfrac25pi=sqrt{frac{5+sqrt5}8}$ which leads that $$r=sqrt{frac{5-sqrt5}{10}}$$
So you can build $r$ with ruller and compass.
Update
Actually it might become simpler: First build your unit-sized pentagon $A_1B_1C_1D_1E_1$:
Then draw perpendiculars to each side at each vertex.
These perpendiculars will cut each-other at points $A_2,D_3,B_2,E_3,C_2,A_3,D_2,B_3,E_3,C_3$ (subindexes $3$ are not shown in the drawing). You dont need them to find them all, just one or two of them (v.g. $A_2$ and $C_2$).
The figure $A_2C_2E_2B_2D_2$ is a pentagram congruent to the one in the unit-distance Petersen graph, and they both share the same circumcircle (red in the figure)
So, take $A_1A_2$ and $C_1C_2$ they cut at the center $O$ of $A_1B_1C_1D_1E_1$. Draw the circle centered in $O$ with radius $OA_2$, and I supose you can now finish the problem.
answered Nov 1 '13 at 21:11
Carlos Eugenio Thompson PinzónCarlos Eugenio Thompson Pinzón
2,526823
$endgroup$
1
$begingroup$
Can you explain your first identity?
$endgroup$
– Samuel
Nov 1 '13 at 21:17
$begingroup$
sin(2pi / 5) = opposite / hypotenus = (1/2) / r. Draw a star in a circle, and radii to its points and you'll probably see this. I still don't know how to construct r, so more elaboration is welcome.
$endgroup$
– Alejandro
Nov 1 '13 at 21:37
$begingroup$
also, I suspect the construction of r will not fit on the page, since I imagine I'll need a line segment that is at least 5 units long to start.
$endgroup$
– Alejandro
Nov 1 '13 at 22:25
add a comment |
$begingroup$
If the inner star's side are $1$ the radius of its circumflexing circle is $frac1{2r}=sinfrac25pi=sqrt{frac{5+sqrt5}8}$ which leads that $$r=sqrt{frac{5-sqrt5}{10}}$$
So you can build $r$ with ruller and compass.
Update
Actually it might become simpler: First build your unit-sized pentagon $A_1B_1C_1D_1E_1$:
Then draw perpendiculars to each side at each vertex.
These perpendiculars will cut each-other at points $A_2,D_3,B_2,E_3,C_2,A_3,D_2,B_3,E_3,C_3$ (subindexes $3$ are not shown in the drawing). You dont need them to find them all, just one or two of them (v.g. $A_2$ and $C_2$).
The figure $A_2C_2E_2B_2D_2$ is a pentagram congruent to the one in the unit-distance Petersen graph, and they both share the same circumcircle (red in the figure)
So, take $A_1A_2$ and $C_1C_2$ they cut at the center $O$ of $A_1B_1C_1D_1E_1$. Draw the circle centered in $O$ with radius $OA_2$, and I supose you can now finish the problem.
answered Nov 1 '13 at 21:11
Carlos Eugenio Thompson PinzónCarlos Eugenio Thompson Pinzón
2,526823
$endgroup$
1
$begingroup$
Can you explain your first identity?
$endgroup$
– Samuel
Nov 1 '13 at 21:17
$begingroup$
sin(2pi / 5) = opposite / hypotenus = (1/2) / r. Draw a star in a circle, and radii to its points and you'll probably see this. I still don't know how to construct r, so more elaboration is welcome.
$endgroup$
– Alejandro
Nov 1 '13 at 21:37
$begingroup$
also, I suspect the construction of r will not fit on the page, since I imagine I'll need a line segment that is at least 5 units long to start.
$endgroup$
– Alejandro
Nov 1 '13 at 22:25
add a comment |
$begingroup$
If the inner star's side are $1$ the radius of its circumflexing circle is $frac1{2r}=sinfrac25pi=sqrt{frac{5+sqrt5}8}$ which leads that $$r=sqrt{frac{5-sqrt5}{10}}$$
So you can build $r$ with ruller and compass.
Update
Actually it might become simpler: First build your unit-sized pentagon $A_1B_1C_1D_1E_1$:
Then draw perpendiculars to each side at each vertex.
These perpendiculars will cut each-other at points $A_2,D_3,B_2,E_3,C_2,A_3,D_2,B_3,E_3,C_3$ (subindexes $3$ are not shown in the drawing). You dont need them to find them all, just one or two of them (v.g. $A_2$ and $C_2$).
The figure $A_2C_2E_2B_2D_2$ is a pentagram congruent to the one in the unit-distance Petersen graph, and they both share the same circumcircle (red in the figure)
So, take $A_1A_2$ and $C_1C_2$ they cut at the center $O$ of $A_1B_1C_1D_1E_1$. Draw the circle centered in $O$ with radius $OA_2$, and I supose you can now finish the problem.
answered Nov 1 '13 at 21:11
Carlos Eugenio Thompson PinzónCarlos Eugenio Thompson Pinzón
2,526823
$endgroup$
If the inner star's side are $1$ the radius of its circumflexing circle is $frac1{2r}=sinfrac25pi=sqrt{frac{5+sqrt5}8}$ which leads that $$r=sqrt{frac{5-sqrt5}{10}}$$
So you can build $r$ with ruller and compass.
Update
Actually it might become simpler: First build your unit-sized pentagon $A_1B_1C_1D_1E_1$:
Then draw perpendiculars to each side at each vertex.
These perpendiculars will cut each-other at points $A_2,D_3,B_2,E_3,C_2,A_3,D_2,B_3,E_3,C_3$ (subindexes $3$ are not shown in the drawing). You dont need them to find them all, just one or two of them (v.g. $A_2$ and $C_2$).
The figure $A_2C_2E_2B_2D_2$ is a pentagram congruent to the one in the unit-distance Petersen graph, and they both share the same circumcircle (red in the figure)
So, take $A_1A_2$ and $C_1C_2$ they cut at the center $O$ of $A_1B_1C_1D_1E_1$. Draw the circle centered in $O$ with radius $OA_2$, and I supose you can now finish the problem.
answered Nov 1 '13 at 21:11
Carlos Eugenio Thompson PinzónCarlos Eugenio Thompson Pinzón
2,526823
edited Nov 2 '13 at 3:54
answered Nov 1 '13 at 21:11
Carlos Eugenio Thompson PinzónCarlos Eugenio Thompson Pinzón
2,526823
answered Nov 1 '13 at 21:11
Carlos Eugenio Thompson PinzónCarlos Eugenio Thompson Pinzón
2,526823
answered Nov 1 '13 at 21:11
Carlos Eugenio Thompson PinzónCarlos Eugenio Thompson Pinzón
2,526823
2,526823
1
$begingroup$
Can you explain your first identity?
$endgroup$
– Samuel
Nov 1 '13 at 21:17
$begingroup$
sin(2pi / 5) = opposite / hypotenus = (1/2) / r. Draw a star in a circle, and radii to its points and you'll probably see this. I still don't know how to construct r, so more elaboration is welcome.
$endgroup$
– Alejandro
Nov 1 '13 at 21:37
$begingroup$
also, I suspect the construction of r will not fit on the page, since I imagine I'll need a line segment that is at least 5 units long to start.
$endgroup$
– Alejandro
Nov 1 '13 at 22:25
add a comment |
1
$begingroup$
Can you explain your first identity?
$endgroup$
– Samuel
Nov 1 '13 at 21:17
$begingroup$
sin(2pi / 5) = opposite / hypotenus = (1/2) / r. Draw a star in a circle, and radii to its points and you'll probably see this. I still don't know how to construct r, so more elaboration is welcome.
$endgroup$
– Alejandro
Nov 1 '13 at 21:37
$begingroup$
also, I suspect the construction of r will not fit on the page, since I imagine I'll need a line segment that is at least 5 units long to start.
$endgroup$
– Alejandro
Nov 1 '13 at 22:25
1
1
$begingroup$
Can you explain your first identity?
$endgroup$
– Samuel
Nov 1 '13 at 21:17
$begingroup$
Can you explain your first identity?
$endgroup$
– Samuel
Nov 1 '13 at 21:17
$begingroup$
sin(2pi / 5) = opposite / hypotenus = (1/2) / r. Draw a star in a circle, and radii to its points and you'll probably see this. I still don't know how to construct r, so more elaboration is welcome.
$endgroup$
– Alejandro
Nov 1 '13 at 21:37
$begingroup$
sin(2pi / 5) = opposite / hypotenus = (1/2) / r. Draw a star in a circle, and radii to its points and you'll probably see this. I still don't know how to construct r, so more elaboration is welcome.
$endgroup$
– Alejandro
Nov 1 '13 at 21:37
$begingroup$
also, I suspect the construction of r will not fit on the page, since I imagine I'll need a line segment that is at least 5 units long to start.
$endgroup$
– Alejandro
Nov 1 '13 at 22:25
$begingroup$
also, I suspect the construction of r will not fit on the page, since I imagine I'll need a line segment that is at least 5 units long to start.
$endgroup$
– Alejandro
Nov 1 '13 at 22:25
add a comment |
$begingroup$
I first drew this back in (I think) 1999. It's a very nice construction, and if you base it on inscribing the outer pentagon into a circle, you can do it at full size covering most of a piece of paper, without ever running off that paper, and with barely any more real work than constructing a regular pentagon.
It's also my avatar, everywhere online I have one.
The construction, as I do it:
- First, build a large circle and a pair of perpendicular diameters. Let the center be $O$, one of the endpoints of a diameter be $A$, and one of the endpoints of the other diameter be the first vertex $V_0$.
- Find the midpoint of $OA$, and call it B.
- Find the bisector of angle $OAV_0$. Let the intersection of this bisector with $OV_0$ be $C$.
- Reflect $O$ across $C$ to get $D$, and draw the perpendicular bisector of $OD$ through $C$. This perpendicular intersects the outer circle at two vertices $V_1$ and $V_4$ of the outer pentagon we want.
- Draw the circle centered at $O$ with $D$ on it. This circle intersects $OA$ at a vertex $V_5$ of the inner pentagram.
- We have now done all the hard parts; the rest is marking off the same standard distance to find all the vertices.
From $V_1$, set your compass to length $V_0V_1$ and mark off $V_2$ on the outer circle (the other intersection) and $V_6$ on the inner circle (same angular direction as $V_5$ relative to $V_0$; it should be just inside the segment $V_0V_5$).
From $V_4$, mark off $V_3$ on the outer circle and $V_9$ on the inner.
From $V_5$, mark off $V_7$ and $V_8$ on the inner circle.
Draw in the lines, and we're done.
The key facts that make this construction as nice as it is:
- The ratio of the radii of the two circles the pentagons are inscribed in is the golden ratio $frac{sqrt{5}-1}{2}$, which we need anyway to construct a pentagon in the first place. The first step in the construction that isn't part of the construction for the outer pentagon is to draw the smaller circle.
- While that part was fairly obvious and well-known (the ratio we need is the same as the ratio between the side and diagonal of a regular pentagon), we still need to know how much to twist the inner pentagram by so that the crossing edges are the same length as the pentagon and the pentagram. Law of cosines - we need $$1^2+left(frac{sqrt{5}-1}{2}right)^2 - 2cdot 1cdot frac{sqrt{5}-1}{2}cos(theta) = (2sin(36^circ))^2$$
$$1+frac{3-sqrt{5}}{2} -(sqrt{5}-1)costheta = 2-2cos(72^circ) = 2-frac{sqrt{5}-1}{2}$$
$$frac{5-sqrt{5}}{2}-(sqrt{5}-1)costheta = frac{5-sqrt{5}}{2}$$
Huh, $costheta=0$? Exactly a $90^circ$ rotation? That very nice coincidence is what inspired me to draw it in the first place.
Now, on actually drawing it: Because the crossing lines come pretty close to the next vertices over on the inner pentagram, this diagram is visually sensitive to small errors. Draw carefully, because an error of a few millimeters out of a full-page drawing will be visible.
answered Nov 25 '18 at 20:42
jmerryjmerry
12.3k1628
$endgroup$
$begingroup$
So, looking back on my first post, the one that brought me here (initially posting as a guest) - it could have been better. So I added a picture to make it better. The diagram is color-coded; intermediate steps in blue, transient construction marks in green, labels in red, and the final graph in black. It was generated in Asymptote, directly following the construction as described.
$endgroup$
– jmerry
Jan 20 at 20:37
add a comment |
$begingroup$
I first drew this back in (I think) 1999. It's a very nice construction, and if you base it on inscribing the outer pentagon into a circle, you can do it at full size covering most of a piece of paper, without ever running off that paper, and with barely any more real work than constructing a regular pentagon.
It's also my avatar, everywhere online I have one.
The construction, as I do it:
- First, build a large circle and a pair of perpendicular diameters. Let the center be $O$, one of the endpoints of a diameter be $A$, and one of the endpoints of the other diameter be the first vertex $V_0$.
- Find the midpoint of $OA$, and call it B.
- Find the bisector of angle $OAV_0$. Let the intersection of this bisector with $OV_0$ be $C$.
- Reflect $O$ across $C$ to get $D$, and draw the perpendicular bisector of $OD$ through $C$. This perpendicular intersects the outer circle at two vertices $V_1$ and $V_4$ of the outer pentagon we want.
- Draw the circle centered at $O$ with $D$ on it. This circle intersects $OA$ at a vertex $V_5$ of the inner pentagram.
- We have now done all the hard parts; the rest is marking off the same standard distance to find all the vertices.
From $V_1$, set your compass to length $V_0V_1$ and mark off $V_2$ on the outer circle (the other intersection) and $V_6$ on the inner circle (same angular direction as $V_5$ relative to $V_0$; it should be just inside the segment $V_0V_5$).
From $V_4$, mark off $V_3$ on the outer circle and $V_9$ on the inner.
From $V_5$, mark off $V_7$ and $V_8$ on the inner circle.
Draw in the lines, and we're done.
The key facts that make this construction as nice as it is:
- The ratio of the radii of the two circles the pentagons are inscribed in is the golden ratio $frac{sqrt{5}-1}{2}$, which we need anyway to construct a pentagon in the first place. The first step in the construction that isn't part of the construction for the outer pentagon is to draw the smaller circle.
- While that part was fairly obvious and well-known (the ratio we need is the same as the ratio between the side and diagonal of a regular pentagon), we still need to know how much to twist the inner pentagram by so that the crossing edges are the same length as the pentagon and the pentagram. Law of cosines - we need $$1^2+left(frac{sqrt{5}-1}{2}right)^2 - 2cdot 1cdot frac{sqrt{5}-1}{2}cos(theta) = (2sin(36^circ))^2$$
$$1+frac{3-sqrt{5}}{2} -(sqrt{5}-1)costheta = 2-2cos(72^circ) = 2-frac{sqrt{5}-1}{2}$$
$$frac{5-sqrt{5}}{2}-(sqrt{5}-1)costheta = frac{5-sqrt{5}}{2}$$
Huh, $costheta=0$? Exactly a $90^circ$ rotation? That very nice coincidence is what inspired me to draw it in the first place.
Now, on actually drawing it: Because the crossing lines come pretty close to the next vertices over on the inner pentagram, this diagram is visually sensitive to small errors. Draw carefully, because an error of a few millimeters out of a full-page drawing will be visible.
answered Nov 25 '18 at 20:42
jmerryjmerry
12.3k1628
$endgroup$
$begingroup$
So, looking back on my first post, the one that brought me here (initially posting as a guest) - it could have been better. So I added a picture to make it better. The diagram is color-coded; intermediate steps in blue, transient construction marks in green, labels in red, and the final graph in black. It was generated in Asymptote, directly following the construction as described.
$endgroup$
– jmerry
Jan 20 at 20:37
add a comment |
$begingroup$
I first drew this back in (I think) 1999. It's a very nice construction, and if you base it on inscribing the outer pentagon into a circle, you can do it at full size covering most of a piece of paper, without ever running off that paper, and with barely any more real work than constructing a regular pentagon.
It's also my avatar, everywhere online I have one.
The construction, as I do it:
- First, build a large circle and a pair of perpendicular diameters. Let the center be $O$, one of the endpoints of a diameter be $A$, and one of the endpoints of the other diameter be the first vertex $V_0$.
- Find the midpoint of $OA$, and call it B.
- Find the bisector of angle $OAV_0$. Let the intersection of this bisector with $OV_0$ be $C$.
- Reflect $O$ across $C$ to get $D$, and draw the perpendicular bisector of $OD$ through $C$. This perpendicular intersects the outer circle at two vertices $V_1$ and $V_4$ of the outer pentagon we want.
- Draw the circle centered at $O$ with $D$ on it. This circle intersects $OA$ at a vertex $V_5$ of the inner pentagram.
- We have now done all the hard parts; the rest is marking off the same standard distance to find all the vertices.
From $V_1$, set your compass to length $V_0V_1$ and mark off $V_2$ on the outer circle (the other intersection) and $V_6$ on the inner circle (same angular direction as $V_5$ relative to $V_0$; it should be just inside the segment $V_0V_5$).
From $V_4$, mark off $V_3$ on the outer circle and $V_9$ on the inner.
From $V_5$, mark off $V_7$ and $V_8$ on the inner circle.
Draw in the lines, and we're done.
The key facts that make this construction as nice as it is:
- The ratio of the radii of the two circles the pentagons are inscribed in is the golden ratio $frac{sqrt{5}-1}{2}$, which we need anyway to construct a pentagon in the first place. The first step in the construction that isn't part of the construction for the outer pentagon is to draw the smaller circle.
- While that part was fairly obvious and well-known (the ratio we need is the same as the ratio between the side and diagonal of a regular pentagon), we still need to know how much to twist the inner pentagram by so that the crossing edges are the same length as the pentagon and the pentagram. Law of cosines - we need $$1^2+left(frac{sqrt{5}-1}{2}right)^2 - 2cdot 1cdot frac{sqrt{5}-1}{2}cos(theta) = (2sin(36^circ))^2$$
$$1+frac{3-sqrt{5}}{2} -(sqrt{5}-1)costheta = 2-2cos(72^circ) = 2-frac{sqrt{5}-1}{2}$$
$$frac{5-sqrt{5}}{2}-(sqrt{5}-1)costheta = frac{5-sqrt{5}}{2}$$
Huh, $costheta=0$? Exactly a $90^circ$ rotation? That very nice coincidence is what inspired me to draw it in the first place.
Now, on actually drawing it: Because the crossing lines come pretty close to the next vertices over on the inner pentagram, this diagram is visually sensitive to small errors. Draw carefully, because an error of a few millimeters out of a full-page drawing will be visible.
answered Nov 25 '18 at 20:42
jmerryjmerry
12.3k1628
$endgroup$
I first drew this back in (I think) 1999. It's a very nice construction, and if you base it on inscribing the outer pentagon into a circle, you can do it at full size covering most of a piece of paper, without ever running off that paper, and with barely any more real work than constructing a regular pentagon.
It's also my avatar, everywhere online I have one.
The construction, as I do it:
- First, build a large circle and a pair of perpendicular diameters. Let the center be $O$, one of the endpoints of a diameter be $A$, and one of the endpoints of the other diameter be the first vertex $V_0$.
- Find the midpoint of $OA$, and call it B.
- Find the bisector of angle $OAV_0$. Let the intersection of this bisector with $OV_0$ be $C$.
- Reflect $O$ across $C$ to get $D$, and draw the perpendicular bisector of $OD$ through $C$. This perpendicular intersects the outer circle at two vertices $V_1$ and $V_4$ of the outer pentagon we want.
- Draw the circle centered at $O$ with $D$ on it. This circle intersects $OA$ at a vertex $V_5$ of the inner pentagram.
- We have now done all the hard parts; the rest is marking off the same standard distance to find all the vertices.
From $V_1$, set your compass to length $V_0V_1$ and mark off $V_2$ on the outer circle (the other intersection) and $V_6$ on the inner circle (same angular direction as $V_5$ relative to $V_0$; it should be just inside the segment $V_0V_5$).
From $V_4$, mark off $V_3$ on the outer circle and $V_9$ on the inner.
From $V_5$, mark off $V_7$ and $V_8$ on the inner circle.
Draw in the lines, and we're done.
The key facts that make this construction as nice as it is:
- The ratio of the radii of the two circles the pentagons are inscribed in is the golden ratio $frac{sqrt{5}-1}{2}$, which we need anyway to construct a pentagon in the first place. The first step in the construction that isn't part of the construction for the outer pentagon is to draw the smaller circle.
- While that part was fairly obvious and well-known (the ratio we need is the same as the ratio between the side and diagonal of a regular pentagon), we still need to know how much to twist the inner pentagram by so that the crossing edges are the same length as the pentagon and the pentagram. Law of cosines - we need $$1^2+left(frac{sqrt{5}-1}{2}right)^2 - 2cdot 1cdot frac{sqrt{5}-1}{2}cos(theta) = (2sin(36^circ))^2$$
$$1+frac{3-sqrt{5}}{2} -(sqrt{5}-1)costheta = 2-2cos(72^circ) = 2-frac{sqrt{5}-1}{2}$$
$$frac{5-sqrt{5}}{2}-(sqrt{5}-1)costheta = frac{5-sqrt{5}}{2}$$
Huh, $costheta=0$? Exactly a $90^circ$ rotation? That very nice coincidence is what inspired me to draw it in the first place.
Now, on actually drawing it: Because the crossing lines come pretty close to the next vertices over on the inner pentagram, this diagram is visually sensitive to small errors. Draw carefully, because an error of a few millimeters out of a full-page drawing will be visible.
answered Nov 25 '18 at 20:42
jmerryjmerry
12.3k1628
edited Jan 20 at 20:34
answered Nov 25 '18 at 20:42
jmerryjmerry
12.3k1628
answered Nov 25 '18 at 20:42
jmerryjmerry
12.3k1628
answered Nov 25 '18 at 20:42
jmerryjmerry
12.3k1628
12.3k1628
$begingroup$
So, looking back on my first post, the one that brought me here (initially posting as a guest) - it could have been better. So I added a picture to make it better. The diagram is color-coded; intermediate steps in blue, transient construction marks in green, labels in red, and the final graph in black. It was generated in Asymptote, directly following the construction as described.
$endgroup$
– jmerry
Jan 20 at 20:37
add a comment |
$begingroup$
So, looking back on my first post, the one that brought me here (initially posting as a guest) - it could have been better. So I added a picture to make it better. The diagram is color-coded; intermediate steps in blue, transient construction marks in green, labels in red, and the final graph in black. It was generated in Asymptote, directly following the construction as described.
$endgroup$
– jmerry
Jan 20 at 20:37
$begingroup$
So, looking back on my first post, the one that brought me here (initially posting as a guest) - it could have been better. So I added a picture to make it better. The diagram is color-coded; intermediate steps in blue, transient construction marks in green, labels in red, and the final graph in black. It was generated in Asymptote, directly following the construction as described.
$endgroup$
– jmerry
Jan 20 at 20:37
$begingroup$
So, looking back on my first post, the one that brought me here (initially posting as a guest) - it could have been better. So I added a picture to make it better. The diagram is color-coded; intermediate steps in blue, transient construction marks in green, labels in red, and the final graph in black. It was generated in Asymptote, directly following the construction as described.
$endgroup$
– jmerry
Jan 20 at 20:37
add a comment |
$begingroup$
with compass and ruler. -use whatever method you are comfortable with to construct a pentagon. -draw lines from each corner, for a total of 5 lines,
to the middle of the edge line connecting the next 2 angles of the
pentagon. -set your compass to the length of the edge lines
(incidentally this was the same size as the circles I used in one
particular method to construct my pentagon originally), and mark
the end of each of the most recent 5 lines. -draw lines from
where each compass intersected the interior lines to form your
pentacle/5 pointed star. if you measure each line they are the
same length
answered Oct 5 '15 at 22:20
finbarfinbar
1
$endgroup$
$begingroup$
That certainly is a simple construction, but can you justify it as being true?
$endgroup$
– Rory Daulton
Oct 5 '15 at 23:02
add a comment |
$begingroup$
with compass and ruler. -use whatever method you are comfortable with to construct a pentagon. -draw lines from each corner, for a total of 5 lines,
to the middle of the edge line connecting the next 2 angles of the
pentagon. -set your compass to the length of the edge lines
(incidentally this was the same size as the circles I used in one
particular method to construct my pentagon originally), and mark
the end of each of the most recent 5 lines. -draw lines from
where each compass intersected the interior lines to form your
pentacle/5 pointed star. if you measure each line they are the
same length
answered Oct 5 '15 at 22:20
finbarfinbar
1
$endgroup$
$begingroup$
That certainly is a simple construction, but can you justify it as being true?
$endgroup$
– Rory Daulton
Oct 5 '15 at 23:02
add a comment |
$begingroup$
with compass and ruler. -use whatever method you are comfortable with to construct a pentagon. -draw lines from each corner, for a total of 5 lines,
to the middle of the edge line connecting the next 2 angles of the
pentagon. -set your compass to the length of the edge lines
(incidentally this was the same size as the circles I used in one
particular method to construct my pentagon originally), and mark
the end of each of the most recent 5 lines. -draw lines from
where each compass intersected the interior lines to form your
pentacle/5 pointed star. if you measure each line they are the
same length
answered Oct 5 '15 at 22:20
finbarfinbar
1
$endgroup$
with compass and ruler. -use whatever method you are comfortable with to construct a pentagon. -draw lines from each corner, for a total of 5 lines,
to the middle of the edge line connecting the next 2 angles of the
pentagon. -set your compass to the length of the edge lines
(incidentally this was the same size as the circles I used in one
particular method to construct my pentagon originally), and mark
the end of each of the most recent 5 lines. -draw lines from
where each compass intersected the interior lines to form your
pentacle/5 pointed star. if you measure each line they are the
same length
answered Oct 5 '15 at 22:20
finbarfinbar
1
answered Oct 5 '15 at 22:20
finbarfinbar
1
answered Oct 5 '15 at 22:20
finbarfinbar
1
answered Oct 5 '15 at 22:20
finbarfinbar
1
1
$begingroup$
That certainly is a simple construction, but can you justify it as being true?
$endgroup$
– Rory Daulton
Oct 5 '15 at 23:02
add a comment |
$begingroup$
That certainly is a simple construction, but can you justify it as being true?
$endgroup$
– Rory Daulton
Oct 5 '15 at 23:02
$begingroup$
That certainly is a simple construction, but can you justify it as being true?
$endgroup$
– Rory Daulton
Oct 5 '15 at 23:02
$begingroup$
That certainly is a simple construction, but can you justify it as being true?
$endgroup$
– Rory Daulton
Oct 5 '15 at 23:02
add a comment |
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