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标题: 阿曼拼贴 [打印本页]
作者: ping88 时间: 2012-5-18 20:18
标题: 阿曼拼贴
简单几个图形，无限生长Constructions for the Golden Ratio Let's start by showing how to construct the golden section points on any line: first a line phi (0·618..) times as long as the original and then a line Phi (1·618..) times as long. John Turner has nicknamed the two points that divide a line at the golden ratio (0.618 of the way from either end) as gold points.
Constructing the internal golden section points: phi If we have a line with end-points A and B, how can we find the point which divides it at the golden section point?
We can do this using compasses for drawing circles and a set-square for drawing lines at right-angles to other lines, and we don't need a ruler at all for measuring lengths!
(In fact we can do it with just the compasses, but how to do it without the set-square is left as an exercise for you.)
We want to find a point G between A and B so that AG:AB = phi (0·61803...) by which we mean that G is phi of the way along the line. This will also mean that the smaller segment GB is 0·61803.. times the size of the longer segment AG too.
| AG | = | GB | = phi = 0·618033.. = | √5 – 1 |
| AB | AG | 2 |
Here's how to construct point G using set-square and compasses only: - First we find the mid point of AB. To do this without a ruler, put your compasses on one end, open them out to be somewhere near the other end of the line and draw a semicircle over the line AB. Repeat this at the other end of the line without altering the compass size. The two points where the semicircles cross can then be joined and this new line will cross AB at its mid point.
- Now we are going to draw a line half the length of AB at point B, but at right-angles to the original line. This is where you use the set-square (but you CAN do this just using your compasses too - how?). So first draw a line at right angles to AB at end B.
- Put your compasses on B, open them to the mid-point of AB and draw an arc to find the point on your new line which is half as long as AB. Now you have a new line at B at right angles to the AB and BC is half as long as the original line AB.
- Join the point just found to the other end of the original line (A) to make a triangle. Putting the compass point at the top point of the triangle and opening it out to point B (so it has a radius along the right-angle line) mark out a point on the diagonal which will also be half as long as the original line.
- Finally, put the compass point at point A, open it out to the new point just found on the diagonal and mark a point the same distance along the original line. This point is now divides the original line AB into two parts, where the longer part AG is phi (0·61803..) times as long as the original line AB.
Why does this work?
It works because, if we call the top point of the triangle T, then BT is half as long as AB. So suppose we say AB has length 1. Then BT will have length 1/2. We can find the length of the other side of the triangle, the diagonal AT, by using Pythagoras' Theorem: AT2 = AB2 + BT2
i.e. AT2 = 12 + (1/2)2
AT2 = 1 + 1/4 = 5/4
Now, taking the square-root of each side gives: AT = (√5)/2
Point V was drawn so that TV is the same length as TB = AB/2 = 1/2.
So AV is just AT - TV = (√5)/2 – 1/2 = phi.
The final construction is to mark a point G which is same distance (AV) along the original line (AB) which we do using the compasses. So AG is phi times as long as AB!
1·61803 39887 49894 84820 45868 34365 63811 77203 09179 80576 ..
More..
- First repeat the steps 1 and 2 above so that we have found the mid-point of AB and also have a line at right angles at point B.
- Now place the compass point on B and open them out to touch A so that you can mark a point T on the vertical line which is as long as the original line.
- Placing the compass point on the mid-point M of AB, open them out to the new point T on the vertical line and draw an arc on the original line extended past point B to a new point G.
- The line AG is now Phi times as long as the original line AB.
Why does this work?
If you followed the reasoning for why the first construction (for phi) worked, you should find it quite easy to prove that AG is Phi times the length of AB, that is, that AG = (√5/2 + 1/2) times AB.
Hint:
Let AB have length 1 again and so AM=MB=1/2. Since BT is now also 1, how long is MT? This is the same length as MG so you can now find out how long AG is since AG=AM+MG.
Hofstetter's 3 Constructions of Gold Points Using only circles Kurt Hofstetter has found a beautifully simple construction for a line and its golden section point using only compasses to draw 4 circles:
- On any straight line S, pick two points X and Y.
With each as centre draw a circle (green) through the other point labelling their points of intersection G (top) and B (bottom) and the points where they meet line S as P and Q;
- With centre X, draw a circle through Q (blue);
- With centre Y, draw a circle through P (blue); labelling the top point of intersection of the blue circles A;
G is a gold point of AB (G is at the golden section of AB).
A Simple Construction of the Golden Section, Kurt Hofstetter in Forum Geometricorum Vol 2 (2002) pages 65-66 which has the proof too.
作者: ping88 时间: 2012-5-18 20:18
Another even simpler method! Kurt Hofstetter has discovered a very simple constructions of the gold point on a line AB just using circles and one line to find the gold point(s) on a given line segment AB:
- With centre A, draw a circle through B;
- With centre B, draw a circle through A;
- Extend BA to meet the first circle (centre A) at C;
- Label the lower point on both circles as D;
- With centre C draw another circle through B;
- Where the larger circle meets the circle centred on B, label the upper point E;
Line DE crosses line AB at G.
G is a gold point of AB
The other gold point on AB can be constructed similarly. Uli Eggermann of Germany points out that we also have A as the gold point of CG!
It is quite simple to prove too using the properties of Phi and phi that phi=1/Phi and 1+phi=Phi:
- If AB is 1 then so is CA: they are both radii of the same circle
- Above we constructed G, the gold point of AB (=1), so that AG is phi
- CG = CA + AG = 1 + phi = Phi
- Since CA is 1 and CG is Phi then CA/CG = 1/Phi = phi
- so A is the gold point of CG
Kurt Hofstetter's proof was published as Another 5-step Division of a Segment in the Golden Section in Forum Geometricorum Vol 4 (2004) pages 21-22. Lemoine's Construction of the Golden Section In this construction to find the gold point of a given line segment AB rediscovered by K Hofstetter but going back to 1902, we again need to construct only one line: - With A as centre, draw the first circle through B;
- With B as centre, draw a second circle through A using labels C (top) and D (bottom) for the two points of intersection with the first circle;
- Through the top point, C, draw a third circle through A and B and use label E for its other intersection with the first circle;
- Draw the line CD and where it meets the third circle, label this point F;
- With centre E draw a circle though F; where it meets the line AB label this point G;
G is a gold point of AB.
Also, if BA is extended to meet the larger circle with centre E at G', then
A is a gold point of G'B. A 5-step Division of a Segment in the Golden Section K Hofstetter in Forum Geometricorum Vol 3 (2003) 205-206.
The earliest reference to this construction appears to be in Géométrographie ou Art des Constructions Géométriques by E. Lemoine, published by C. Naud, Paris, 1902, page 51. Phi and Pentagons There is an intimate connection between the golden section and the regular 5-sided shape called a pentagon and its variation - the pentagram - that we explore first. Pentagons and Pentagrams | The pentagram is a symmetrical 5-pointed star that fits inside a pentagon. Starting from a pentagon, by joining each vertex to the next-but-one you can draw a pentagram without taking your pen off the paper. |
The pentagram has 5 triangles on the edges of another pentagon at its centre. Let's focus on one of the triangles and the central pentagon as shown here. All the orange angles at the vertices of the pentagon are equal. They are called the external angles of the polygon. What size are they? This practical demonstration will give us the answer:
- Take a pen and lay it along the top edge pointing right
- Turn it about the top right vertex through the orange angle so that it points down to the lower right
- Move the pen down that side of the pentagon to the next vertex and turn it through the next orange angle
- Repeat moving it along the sides and turning through the rest of the orange angles until it lies back on the top edge
- The pen is now back in its starting position, pointing to the right so it has turned through one compete turn.
- It has also turned through each of the 5 orange angles
- So the sum of the 5 orange angles is one turn or 360°
- Each orange angle is therefore 360/5=72°
The green angle is the same size as the orange angle so that the two "base" angles of the blue triangle are both 72°.
Since the angles in a triangle sum to 180° the yellow angle is 36° so that 72° + 72° + 36° = 180°. The basic geometrical facts we have used here are: The external angles in any polygon sum to 360°.
The angles on a straight line sum to 180°.
The angles in a triangle sum to 180°.
So the pentagram triangle has angles of 36 °, 72° and 72°. Now let's find out how long its sides are. The 36°-72°-72° triangle In this diagram, the triangle ABC is isosceles, since the two sides, AB and AC, are equal as are the two angles at B and C.
[Also, angles ABC and ACB are twice angle CAB.] If we bisect the base angle at B by a line from B to point D on AC then we have the angles as shown and also angle BDC is also 72°. BCD now has two angles equal and is therefore an isosceles triangle; and also we have BC=BD. Since ABD also has two equal angles of 36°, it too is isosceles and so BD=AD. So in the diagram the three sides BC, BD and DA are all the same length. We also note that the little triangle BCD and the whole triangle ABC are similar since they are both 36°-72°-72° triangles. Let's call the smallest segment here, CD, length 1 and find the lengths of the others in relation to it. We will therefore let the ratio of the smaller to longer sides in triangle BCD be r so that if CD is 1 then BC is r. In the larger triangle ABC, the base is now r and as it is the same shape as BCD, then its sides are in the same ratio so Ab is r times BC, e.d. AB is r2.
Also, we have shown BC=BD=AD so AD is r (and CD is 1).
From the diagram we can see that AC=AD+DB.
But AC=1+r and in isosceles triangle ABC, AB (which is r2) is the same length as AC (which is 1+r), so
r2 = 1 + r
and this is the equation which defined the golden ratio. r is Phi or -phi and since lengths are positive, we therefore have that r is Phi!
So the triangle with angles 36°, 72°, 72° has sides that are proportional to Phi, Phi and 1 (which is the same as 1,1,phi). Pentagrams and the 36°-72°-72° triangle If we look at the way a pentagram is constructed, we can see there are lots of lines divided in the golden ratio: Since the points can be joined to make a pentagon, the golden ratio appears in the pentagon also and the relationship between its sides and the diagonals (joining two non-adjacent points). The reason is that Phi has the value 2 cos (π/5) where the angle is described in radians, or, in degrees, Phi=2 cos (36°).
[See below for more angles whose sines and cosines involve Phi!]
Since the ratio of a pair of consecutive Fibonacci numbers is roughly equal to the golden section, we can get an approximate pentagon and pentagram using the Fibonacci numbers as lengths of lines:
There is another flatter triangle inside the pentagon here. Has this any golden sections in it? Yes! We see where further down this page, but first, a quick and easy way to make a pentagram without measuring angles or using compasses:
作者: ping88 时间: 2012-5-18 20:27
If we let T stand for t2 then we have a quadratic equation in T which we can easily solve: t4 + t2 – 1 = 0
T2 + T – 1 = 0
T = ( –1 ± √(1+4) ) / 2
SinceT is t2 it must be positive, so the value of T we want in this case is T = ( √5 – 1 ) / 2 = phi
Since T is t2 then t is √phi.
Since 1 + Phi = Phi2 then the hypotenuse √(1+T2) = Phi as shown in the triangle here.
From the 1, √Phi, Phi triangle, we see that tan(A)=cos(A)
tan(B)=1/sin(B) or tan(B)=cosec(B)
What is the angle whose tangent is the same as it cosine?
In the triangle here, it is angle A:
A = 38.1727076° = 30° 10' 21.74745" = 0.1060352989.. of a whole turn = 0.666239432.. radians.
The other angle, B has its tangent equal to its cosecant (the reciprocal of the sine):
B = 90° – A = 51.8272923..° = 51° 49' 38.2525417.." = 0.143964701.. of a whole turn = 0.90455689.. radians. Notice how, when we apply Pythagoras' Theorem to the triangle shown here with sides 1, Phi and root Phi, we have Phi2 = 1 + (√Phi)2 = 1 + Phi
which is one of our classic definitions of (the positive number) Phi.....
and we have already met this triangle earlier on this page! The next section looks at the other trigonometrical relationships in a triangle and shows that, where they are equal, each involves the numbers Phi and phi. Notation for inverse functions A common mathematical notation for the-angle-whose-sin-is 0.5 is arcsin(0.5) although you will sometimes see this written as asin(0.5).
We can prefix arc (or a) to any trigonometrical function (cos, cot, tan, etc.) to make it into its "inverse" function which, given the trig's value, returns the angle itself.
Each of these inverse functions is applied to a number and returns an angle.
e.g. if sin(90°)=0 then 0 = arcsin(90°).
What is arccos(0.5)?
The angle whose cosine is 0.5 is 60°.
But cos(120°)=0.5 as does cos(240°) and cos(300°) and we can add 360° to any of these angles to find some more values!
The answer is arccos(0.5) = 60° or 120° or 240° or 300° or ...
With all the inverse trig. functions you must carefully select the answer or answers that are appropriate to the problem you are solving. "The angle whose tangent is the same as its cosine" can be written mathematically in several ways:
tan(A)=cos(A) is the same as arctan( cos(A) ) = A
Can you see that it can also be written as arccos( tan(A) )= A? More trig ratios and Phi (sec, csc, cot) In a right-angled triangle if we focus on one angle (A), we can call the two sides round the right-angle the Opposite and the Adjacent sides and the longest side is the Hypotenuse, or Adj, Opp and Hyp for short.
| | | You might wonder why we give a name to the ratio Adj/Opp (the tangent) of angle A but not to Opp/Adj.
The same applies to the other two pairs of sides:
we call Opp/Hyp the sine of A but what about Hyp/Opp?
Similarly Adj/Hyp is cosine of A but what about Hyp/Adj?
|
In fact they do have names: - the inverse ratio to the tangent is the cotangent or cot i.e. Adj/Opp; cot(x)=1/tan(x)
- the inverse ratio to the cosine is the secant or sec i.e. Hyp/Adj; sec(x)=1/cos(x)
- the inverse ratio to the sine is the cosecant or cosec or sometimes csc i.e. Hyp/Opp; csc(x) = 1/sin(x)
You'll notice that these six names divide into two groups: - secant, sine, tangent
- cosecant, cosine, cotangent
and show another way of choosing one representative for each of the 3 pairs of ratios (x/y and y/x where x and y are one of the three sides). Here is a graph of the six functions where the angle is measured in radians:
作者: ping88 时间: 2012-5-18 20:28
http://www.maths.surrey.ac.uk/ho ... /phi2DGeomTrig.html
作者: ping88 时间: 2012-5-18 20:29
Heesch's Problem
Heesch's Problem asks the following question: How many times can a tile be surrounded by congruent copies of itself? That is, how many layers made of copies of the tile can you place around the tile. The layers are called coronas, and the maximum number of coronas that can surround a tile is called the Heesch number of the tile.
Here are some examples of tiles with Heesch number 1. The first one is by Heesch himself.
Heesch number 2 examples:
Heesch number 3 examples (The first one was discovered by Robert Ammann, and it was the record holder for a long time):
The proof that Amman's tile has Heesch number 3 is quite nice, and this proof stimulated some research on my part. The basic idea of the proof is to notice that each tile has 3 "innies" and 2 "outies" so each tile has an excess "charge". In any configuration of these tiles, the excess charge must appear on the boundary of the configuration. The area of the configuration is propotional to the amount of excess charge, and so grows on the order of a quadratic polynomial (area is quadratic). But, the number of edges on the boundary of a configuration, which is where the excess charges are stored, grows linearly (think circumference). Thus eventually there is not enough room on the boundary to put the excess charges. By simply counting in this example you can see that a fourth corona is impossible because the boundary of the fourth corona doesn't have enough edges to hold all of the innies that would have to be there.
I have extended this basic argument to other more complicated tiles made of hexagons. This work, which I won't go into here, actually predicts the following examples. This work should be submitted soon, though.
作者: ping88 时间: 2012-5-18 20:30
Heesch number 4 example (First discovered by W.R. Marshall, and then independently by me)
Heesch number 5 example (discovered by me):
This last example has the largest known finite Heesch number. Of course, a prototile that can be used to tile the entire plane has infinite Heesch number, but I am interested in those prototiles that cannot be used to tile the plane.
Also, it has been customary in the past to require that the coronal configurations be simply connected (that is, there are no holes), but if we relax this condition, we see some can have even higher Heesch numbers. For example, several people have noticed that Ammann's example under the relaxed rules has Heesch number 4:
The Heesch number 5 example from above cannot, however, attain 6 coronas, even if we relax the simple connectivity requirement.
So the big question is this: Is there a maximum Heesch number? That is, is there some number N so that when any tile surrounds itself N times, then it must tile the plane? If so, what is N?
At present, there is not an answer to this question. It has been conjectured by well-known and talented people in the field of tiling theory that there is a maximum (although they cannot say what that maximum is). I have also heard it conjectured that there is no bound on the Heesch number! I am reluctant to make a conjecture myself. But I can say that after Ammann's example was given, many people thought that there would not be another example with higher Heesch number.
Heesch's problem has connections to a few well-known unsolved problems -- the domino problem and the "Einstein" problem. The domino problem asks if there exists an algorithm that, when given a prototile as input, decides if the prototile can be used tile the entire plane. If the Heesch number is in fact bounded, this gives a simple algorithm for deciding if a prototile can be used to tile the plane: Suppose the maximum Heesch number is N. One just places tiles in a systematic fashion until either he cannot proceed any further, in which case the prototile cannot be used to tile the plane, or until he has placed more than N coronas of tiles, in which case the tile must tile the plane. The domino problem in turn has a deep connection with the "Einstein" problem. The Einstein problem asks if there exists a single aperiodic prototile (Ein = one, stein = tile). The nonexistence of a single aperiodic prototile implies the the existence of a decision method for the domino problem.
作者: ping88 时间: 2012-5-18 20:30
http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/phi2DGeomTrig.html
作者: 活建鬼 时间: 2012-5-18 21:33
感谢楼主的分享~~~看完帖我回去补课先
作者: zhouningyi1 时间: 2012-5-18 21:49
好贴 顶一下
彭罗斯镶嵌啥的也可以带下吧 另外据说09年出了一种单组元的非周期镶嵌，叫taylor镶嵌..
http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/phi2DGeomTrig.html
ping88 发表于 2012-5-18 20:28
作者: yinny 时间: 2012-5-18 21:54
看着头大掉。。。
作者: 0378zy 时间: 2012-5-20 09:47
这个有没有中文版呀?
作者: 白牙王子 时间: 2012-5-20 17:31
受教！有没有公式可用？
作者: jason 时间: 2012-5-20 20:54
学习了。。
作者: wqdesky 时间: 2012-5-22 19:53
顶顶顶顶顶顶顶顶顶顶顶顶顶顶顶顶顶顶顶顶顶顶，英文不好啊
作者: guihuashizyl 时间: 2012-5-23 21:31
好好学英语啊童鞋们
作者: guihuashizyl 时间: 2012-5-23 21:31
好好学英语啊童鞋们
作者: bbbiiii 时间: 2012-5-24 15:51
大好，一元拼装的形式啊
作者: s.k. 时间: 2012-5-25 12:13
犀利，高数没学好的表示很崇拜。。
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