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random7 · October 31, 2010

THE CHALLENGE

Construct a perfect golden-rectangle using only a straightedge and compass. This is a puzzle on technique, not artistic ability; the procedure used is what we are focusing on.

INFORMATION

A golden rectangle is a rectangle whose side lengths are in the golden ratio φ:1, where φ=(1+√5)/2.

Though not required, using a picture together with your explanation will help others understand the procedure you used. Please do not forget to use spoilers! The thing that I have used below. Thank-you =)

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Once you have created a procedure for constructing a perfect golden-rectangle using only a straightedge and compass, can you think of an alternate procedure? Which of the two procedures is more efficient? That is, which creates the least error when physically constructed, and thus a more perfect golden-rectangle?

It is possible to construct the vertices of a perfect golden-rectangle with a compass alone. Of course, after finding the vertices you would have to use a straightedge to draw the actual edges of the perfect golden-rectangle. Finding the vertices involves replacing any use of the straightedge with an equivalent procedure that only uses the compass, while still achieving the same result. Since I have not done this challenge myself, I will not be able to post a solution to this particular challenge; God speed =)

October 31, 2010

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Construct a simple square
Draw a line from the midpoint of one side of the square to an opposite corner
Use that line as the radius to draw an arc that defines the height of the rectangle

October 31, 2010

One simple solution

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We draw a straight line starting from the magnetic north pole to as far as we can go for more accuracy (ignoring the Earth's curvature). Now the compass is pointing north along the line we just drew. at a right angle from the line, we head west or east drawing a perpendicular line on our original line. We keep on walking and drawing until the compass has moved a certain degree. We then have a golden L shape which is half our desired rectangle so the rest will simply be completing the rectangle.

The key here is to calculate this specific angle at where we stop. Actually we have two choices and one is more accurate than the other. The better choice is making the second line the longer one. To calculate the angle, we use simple trigonometry. Theta= tan-1(Phi) where Phi is the golden ratio=1.618....

I hope I was clear without any drawings

pdqkemp · October 31, 2010

On 10/31/2010 at 11:20 AM, Moze said:
One simple solution

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We draw a straight line starting from the magnetic north pole to as far as we can go for more accuracy (ignoring the Earth's curvature). Now the compass is pointing north along the line we just drew. at a right angle from the line, we head west or east drawing a perpendicular line on our original line. We keep on walking and drawing until the compass has moved a certain degree. We then have a golden L shape which is half our desired rectangle so the rest will simply be completing the rectangle.

The key here is to calculate this specific angle at where we stop. Actually we have two choices and one is more accurate than the other. The better choice is making the second line the longer one. To calculate the angle, we use simple trigonometry. Theta= tan-1(Phi) where Phi is the golden ratio=1.618....

I hope I was clear without any drawings

This puzzle is talking about a GEOMETRIC COMPASS. Not a compass that finds North/South/East/West.

One of these: http://www.dailyclipart.net/wp-content/uploads/medium/Math1.jpg

(Great puzzle. Well beyond the compass of my ability!)

Edited October 31, 2010 by pdqkemp

October 31, 2010

On 10/31/2010 at 7:59 AM, random7 said:
THE CHALLENGE
Construct a perfect golden-rectangle using only a straightedge and compass. This is a puzzle on technique, not artistic ability; the procedure used is what we are focusing on.

INFORMATION
A golden rectangle is a rectangle whose side lengths are in the golden ratio φ:1, where φ=(1+√5)/2.

I hope I will be getting straight line of length '1' and an straight edge ( with no length measurements ) and a compass ..

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I will construct two rectangles
1 ) rectangle of [ 1 X sqrt(5) /2 ] .. so the diagonal of this will rectangle will b 3/2 .
By getting the middle point of line with length '1' ( by using two arcs from end points ) and by extending it by 1/2 u get length 3/2 .
we could extend the line of length to 1 to length 2 . draw a perpendicular to it ( by using two arcs from end points ) . intersect length 3/2 to the perpendicular line , we got 3 rd vertex of the rectangle.
2 ) rectangle [ 1 X 1/2 ] ...
it could be constructed in the same way as above ...

Edited October 31, 2010 by Rajesh MS

random7 · October 31, 2010

Could sridharang please put his/her response in a spoiler? And copy and pasting straight from another website is not appreciated; any member could find that quote easily, if they wanted to, but someone wanting to solve this puzzle themselves would have it spoiled by that quote.

To pdqkemp: Love the pun =D

On 10/31/2010 at 1:54 PM, Rajesh MS said:
I hope I will be getting straight line of length '1' and an straight edge ( with no length measurements ) and a compass ..

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I will construct two rectangles
1 ) rectangle of [ 1 X sqrt(5) /2 ] .. so the diagonal of this will rectangle will b 3/2 .
By getting the middle point of line with length '1' ( by using two arcs from end points ) and by extending it by 1/2 u get length 3/2 .
we could extend the line of length to 1 to length 2 . draw a perpendicular to it ( by using two arcs from end points ) . intersect length 3/2 to the perpendicular line , we got 3 rd vertex of the rectangle.
2 ) rectangle [ 1 X 1/2 ] ...
it could be constructed in the same way as above ...

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Although vague, I realise what you mean. So far you are the first with a correct procedure.
Now, if you are wanting to go further: how efficient is your method? That is, how many times did you independently use either the compass or the straightedge? How many times did you have to change the radius of the compass? Each of these adds a certain degree of error when actually constructing it on paper.

random7 · November 2, 2010

Ok looks like it is time for a hint, hopefully to help some more people in getting started.

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You can use a compass to find exact surd lengths, using Pythagoras' theorem and some special right-angle triangles. Here is an example:

The question now is, how can you find √5/2 exactly using a compass, Pythagoras' theorem, and a right-angle triangle? Once you have found this, it is simple to just add 1/2 to √5/2 to get the length φ.

I will give another hint tomorrow.

random7 · November 3, 2010

Ok, this topic isn't going so well, so this will be the last hint before posting the solutions tomorrow.

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Say you have two points, and these two points are joined with a line using the straightedge. By constructing two circles about these two points you will get another two points where the circles intersect. By using the straight edge to join these new points, you create a perpendicular to the original line.
In an addition to this, if both circles have the same radius, the perpendicular will also bisect the original line; that is, it will cut it in half. This can be seen in the example below:

Good luck to those working on this problem; I will post my solutions, a picture of Rajesh MS's solution, and a discussion on the efficiency of each, around this time tomorrow.

random7 · November 4, 2010

On 10/31/2010 at 1:54 PM, Rajesh MS said:
I hope I will be getting straight line of length '1' and an straight edge ( with no length measurements ) and a compass ..

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I will construct two rectangles
1 ) rectangle of [ 1 X sqrt(5) /2 ] .. so the diagonal of this will rectangle will b 3/2 .
By getting the middle point of line with length '1' ( by using two arcs from end points ) and by extending it by 1/2 u get length 3/2 .
we could extend the line of length to 1 to length 2 . draw a perpendicular to it ( by using two arcs from end points ) . intersect length 3/2 to the perpendicular line , we got 3 rd vertex of the rectangle.
2 ) rectangle [ 1 X 1/2 ] ...
it could be constructed in the same way as above ...

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The overall idea I used to construct a perfect golden-rectangle is:

Construct a unit-square.
Place the compass at the midpoint of one of the sides and draw an arc from one of the opposite corners to the extension of that side; this creates the height of the perfect golden-rectangle.
Complete the perfect golden-rectangle.

As can be seen in the diagram, using Pythagoras’ theorem to calculate the length of the hypotenuse of the right-angle triangle:

(½)² + 1² = h²
∴ h = √5 /2

We now have a total length of 1/2 + √5 /2 = φ and a width of 1, therefore we have the dimensions of a perfect golden-rectangle.

The steps I took to construct a perfect golden-rectangle, using only a compass and straightedge, are:

Draw a horizon with the straightedge.
Set the compass at a fixed distance; this will define 2units in length.
Mark a point A on the horizon and draw a circle with centre at point A using the compass, that is, draw a circle about point A; mark one of the intersections of the horizon as point B.
From point B, draw arcs to intersect the circle about point A, at point C and point D.
Draw a line connecting point C and point D; this line is a perpendicular bisector of the interval AB.
Mark the intersection of CD and AB as point E and draw a circle about this point.
Mark the closer intersection, to point A, of this circle with the horizon as point F.
From point F, draw arcs to intersect the circle about point E, at point G and point H.
Draw a line which connects point G and point H; extend this line past point G to intersect the circle about point A, at point I; this line is a perpendicular bisector of the interval EF.
From point I, draw arcs to intersect the circle about point A, at point J and point K.
Draw a line connecting point J and point K; this line is a perpendicular bisector of the interval AI; we now have a unit-square.
Mark the intersection of JK and AI as point L.
From point L, draw arcs to intersect the circle about point A, at point M and point N.
Draw a line connecting point M and point N; this line is a perpendicular bisector of the interval AL.
Mark the intersection of MN and AL as point O, and the intersection of MN and CD as point P; these two points mark the midpoints of two sides of the unit-square.
Set the compass to the distance between point P and point L, and draw an arc from point P to the line CD; the intersection nearest the endpoint; mark this point Q.
Draw an arc from point O to the line GH; the intersection nearest the endpoint; mark this point R.
Draw a line from point Q to point R, thus completing the construction of the perfect golden-rectangle, which has point A, point E, point Q and point R as its vertices.

After the development of this first method, I found another way to construct a unit-square. Following this new direction, the following procedure was developed:

Now, Rajesh MS has proposed another procedure, which also uses a completely different overall idea to mine. Instead of using the diagonal from a midpoint on a unit-square, Rajesh MS uses a diagonal length of 3/2 for a rectangle of height √5/2 and length 1. These following pictures show three different adaptations of Rajesh MS’s method:

This last adaptation is just a modification of the second, since I realised that one of the circles is redundant.

EFFICIENCY
Now I want to talk about the efficiency of each method. Every time you need to locate a point exactly on a piece of paper there is a chance of error, since we do not have the ability to locate a point exactly. Whenever we need to change the radius of the compass exactly there is a chance of error, since we do not have the ability to exactly change the radius of a compass, let alone the ability to set it to the distance between two exact points.
Some example situations of attempting to locate a point exactly are: placing the compass at a point; lining-up the straightedge with two points so you may draw a straight line between them; setting the compass to the distance between two points. It should be noted that the horizon is a line that you draw ambiguously, and is thus free from any error; also, the first radius that the compass is set-to only affects the scale of the resulting golden rectangle, and is thus ambiguously set, and thus free from any error.
Using these concepts, the count for potential error for each procedure is:
Procedure Number 1 = 19
Procedure Number 2 = 21
Procedure Number 3a = 17
Procedure Number 3b = 17
Procedure Number 3c = 16

Which shows that all three adaptations of Rajesh MS’s solution beat my two solutions at being more efficient, and that Procedure Number 3c is the most efficient of them all; well done! = )
One more thing about efficiency: if you want to join two points with a straightedge, a more precise line can be drawn if the two points are very far apart; the same is true with using large circles. Therefore, if a method uses larger circles and distances, it will generally be slightly more efficient. Thus, we can see that 3b is more efficient than 3a, even though they have the same count for potential error (as seen above). Also, since Procedure Number 1 uses the largest circles and the longest lines, it utilises this concept best out of all of the others.
All of the still-diagrams construct a golden rectangle of the same size. This shows that if space is an issue, Procedure Number 2 is best used; since it constructs the same sized golden rectangle as the others, while needing the least space to construct.
All of these points are interesting when looking at efficiency, since they show that sometimes there is no simple answer of which procedure is the most efficient.

CHALLENGE NUMBER 2
If you were to figure-out how to do this challenge, I would suggest using Procedure Number 3a, since it almost finds all of the vertices of the golden rectangle, with the compass alone.

Now that you can construct a golden-rectangle, you will be able to construct an approximate golden-spiral at your leisure! Using a compass and straightedge, of course.

= ) Enjoy