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Pentagon
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- Lemon Quarter
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Pentagon
Constructing a hexagon is easy. As soon as we were old enough to play with the sharp, pointy thing in the metal-cased geometry set we had for Christmas, it probably didn't take any of us long to insert a wooden pencil into the instrument, secure said pencil by tightening the thin knurled nut, draw a circle with it, and discover how easy it was to divide that circle into six equal parts.
Using just a pencil, compasses and a straight edge (and a piece of plain paper), how can you construct a regular pentagon?
Julian F. G. W.
Using just a pencil, compasses and a straight edge (and a piece of plain paper), how can you construct a regular pentagon?
Julian F. G. W.
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- Lemon Slice
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Re: Pentagon
Not your standard mathematical straight edge/compasses construction, but you can make a decent regular pentagon by tying a knot in a length of ribbon.
Cinelli
Cinelli
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Re: Pentagon
cinelli wrote:Not your standard mathematical straight edge/compasses construction, but you can make a decent regular pentagon by tying a knot in a length of ribbon.
Or a strip of plain paper.
That's not the answer though.
How many of you knew that a diagonal of a regular pentagon is equal to the length of a side times (1 + sqrt 5)/2 ?
Julian F. G. W.
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- Lemon Quarter
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Re: Pentagon
For a good approximation:
Draw a line with your straight edge. Measure the line, call it 2 units. Separate the points of your compass by 3.4 units. Place the pointy end of compass such that the pencil touches both ends of your line. Continue drawing your circle until the pencil now touches the point on straight edge (origin on one end of your original line) on which you’ve marked 2 units. Repeat until you come full circle.
Based on sin 36 = 0.5878.
Draw a line with your straight edge. Measure the line, call it 2 units. Separate the points of your compass by 3.4 units. Place the pointy end of compass such that the pencil touches both ends of your line. Continue drawing your circle until the pencil now touches the point on straight edge (origin on one end of your original line) on which you’ve marked 2 units. Repeat until you come full circle.
Based on sin 36 = 0.5878.
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- Lemon Pip
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Re: Pentagon
I imagine that this is likely to be some sort of Euclidian or similar construction from antiquity. Notwithstanding what my kids tell me, I'm not that old.
So I have a geometric solution using only a compass and straight edge from 1983 tech drawing class in New Zealand. Thanks Mr Thompson...
Greg
Spoiler (first time I ever used it, did I get the colour right?):
Draw a straight line of length A. Use the compass to define the third vertex of an equilateral triangle, x1. Draw a circle of radius A from that point. [This defines a circle containing a hexagon of side length A]
Next step is to find the centre of a circle that exactly circumscribes a square of length A. Use the compass to bisect A. Draw the perpendicular line, define point y1 where it intersects A. From y1, use the compass to define point y2 at a distance A. Use the compass to bisect the line y1-y2. This point z will be the centre of a circle that circumscribes the square whose base is also the base of the hexagon we started with.
Then the magic happens. We have created two points which define the centres for circles that circumscribe a square and a hexagon. The centre point for a pentagon's circle will be halfway between them.
So use the compass to bisect the line between x1 and z, call this point c. Draw a circle of radius A from point c. Use the compass to mark this circle with chords of length A, starting from either end of the original line. Join the marks using the straight edge and, hey presto, a pentagon of side length A should appear.
So I have a geometric solution using only a compass and straight edge from 1983 tech drawing class in New Zealand. Thanks Mr Thompson...
Greg
Spoiler (first time I ever used it, did I get the colour right?):
Draw a straight line of length A. Use the compass to define the third vertex of an equilateral triangle, x1. Draw a circle of radius A from that point. [This defines a circle containing a hexagon of side length A]
Next step is to find the centre of a circle that exactly circumscribes a square of length A. Use the compass to bisect A. Draw the perpendicular line, define point y1 where it intersects A. From y1, use the compass to define point y2 at a distance A. Use the compass to bisect the line y1-y2. This point z will be the centre of a circle that circumscribes the square whose base is also the base of the hexagon we started with.
Then the magic happens. We have created two points which define the centres for circles that circumscribe a square and a hexagon. The centre point for a pentagon's circle will be halfway between them.
So use the compass to bisect the line between x1 and z, call this point c. Draw a circle of radius A from point c. Use the compass to mark this circle with chords of length A, starting from either end of the original line. Join the marks using the straight edge and, hey presto, a pentagon of side length A should appear.
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- Lemon Quarter
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Re: Pentagon
vandefrosty wrote: We have created two points which define the centres for circles that circumscribe a square and a hexagon. The centre point for a pentagon's circle will be halfway between them.
I suggest that you check this for yourself. It is actually very close and no doubt good enough for technical drawing but it is not exact. I suspect that a similar method could be used to construct an approximation to a regular heptagon, for which a perfect solution does not exist.
As to the original question, I have already given you a clue.
Julian F. G. W.
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- Lemon Pip
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Re: Pentagon
jfgw wrote:
I suggest that you check this for yourself. It is actually very close and no doubt good enough for technical drawing but it is not exact. I suspect that a similar method could be used to construct an approximation to a regular heptagon, for which a perfect solution does not exist.
As to the original question, I have already given you a clue.
Julian F. G. W.
Yes, I was subsequently interested to check out Wikipedia and saw that this isn't one of the rigorous methods. Oh well, no future in the field of pure maths for me!
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- Lemon Quarter
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Re: Pentagon
jfgw wrote:vandefrosty wrote: We have created two points which define the centres for circles that circumscribe a square and a hexagon. The centre point for a pentagon's circle will be halfway between them.
I suggest that you check this for yourself. It is actually very close and no doubt good enough for technical drawing but it is not exact. I suspect that a similar method could be used to construct an approximation to a regular heptagon, for which a perfect solution does not exist.
As to the original question, I have already given you a clue.
Julian F. G. W.
0. Set an arbitrary unit length.
1. Construct sqrt(5) as the hypotenuse of a 1,2,sqrt(5) triangle.
2. Add 1 and bisect to get length of (1+sqrt(5))/2
3. Construct a 1,(1+sqrt(5)/2),(1+sqrt(5)/2) isosceles triangle.
4. This gives you three corners of a pentagram.
5. Trisect the two longer sides, and joint up the appropriate points to complete a regular pentagram.
6. Read this post backward.
7. When the deamon appears order him to draw a regular pentagon.
Okay there may be easier alternatives to steps 5-7, but where's the fun in that? If you wanted easy you'd use a protractor. And this construction also works for nonagons.
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- Lemon Quarter
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Re: Pentagon
9873210 wrote:jfgw wrote:vandefrosty wrote: We have created two points which define the centres for circles that circumscribe a square and a hexagon. The centre point for a pentagon's circle will be halfway between them.
I suggest that you check this for yourself. It is actually very close and no doubt good enough for technical drawing but it is not exact. I suspect that a similar method could be used to construct an approximation to a regular heptagon, for which a perfect solution does not exist.
As to the original question, I have already given you a clue.
Julian F. G. W.
0. Set an arbitrary unit length.
1. Construct sqrt(5) as the hypotenuse of a 1,2,sqrt(5) triangle.
2. Add 1 and bisect to get length of (1+sqrt(5))/2
3. Construct a 1,(1+sqrt(5)/2),(1+sqrt(5)/2) isosceles triangle.
4. This gives you three corners of a pentagram.
5. Trisect the two longer sides, and joint up the appropriate points to complete a regular pentagram.
6. Read this post backward.
7. When the deamon appears order him to draw a regular pentagon.
Okay there may be easier alternatives to steps 5-7, but where's the fun in that? If you wanted easy you'd use a protractor. And this construction also works for nonagons.
Oops was hasty at step 4. Instead of trisecting take a unit length from each end.
Must remember: coffee before necromancy.
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- Lemon Quarter
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Re: Pentagon
With the amendment, that is essentially the method I had in mind. There are others.
Posted previously on LL,
https://i.kym-cdn.com/photos/images/ori ... 50/3ce.png
Julian F. G. W.
9873210 wrote:6. Read this post backward.
7. When the deamon appears order him to draw a regular pentagon.
Posted previously on LL,
https://i.kym-cdn.com/photos/images/ori ... 50/3ce.png
Julian F. G. W.
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- Lemon Slice
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Re: Pentagon
If you wondered why the diagonal of a regular pentagon has the value (1 + sqrt(5))/2, which equals phi, the golden ratio of classical architecture, I came upon this beautiful animation:
http://www.matematicasvisuales.com/engl ... gonal.html
Cinelli
http://www.matematicasvisuales.com/engl ... gonal.html
Cinelli
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