Originally posted by sarathiani dont think this is possible, in fact, it is impossible to get a perfect answer even with trig. i seem to remember that the vitruvian man had something to do with this (da Vinci). in any case, it can probably be done w/out trig if you allow calc.
Here is a geometrical puzzle. Can you square a given circle without using trigonometry or without any mathematical measurement of lengths or doing any calculations. You are to use only a ruler and a compass on which there is no calibration of centimetres or inches.
Originally posted by fearlessleaderYou seem to be correct. That is strictly and exactly speaking. But I think it is possible to construct a square equal in area to a given circle to within as high an accuracy as you desire. The method I have found uses the principles of trigonometry( and of course calculus) which is inherent in the particular construction that I have used. My method involves iterative construction, and within two iterations (iterative constructions) , an accuracy of 0.00001% error is reached.
i dont think this is possible, in fact, it is impossible to get a perfect answer even with trig. i seem to remember that the vitruvian man had something to do with this (da Vinci). in any case, it can probably be done w/out trig if you allow calc.
I dont know if there are other better methods.
Originally posted by cosmic voiceThat's impressive! How many steps does each construction take?
You seem to be correct. That is strictly and exactly speaking. But I think it is possible to construct a square equal in area to a given circle to within as high an accuracy as you desire. The method I have found uses the principles of trigonometry( and of course calculus) which is inherent in the particular construction that I have used. ...[text shortened]... curacy of 0.00001% error is reached.
I dont know if there are other better methods.
It IS possible, IF you allow a dubious step. You have your straightedge, and you can mark a distance on that with a pen. THe distance to be marked on it is one you create in the construction. Then you move your straightedge to set the same distance somewhere else.
If you allow distances to be moved like that, you CAN square a circle.
Originally posted by TheMaster37I thought you could reposition any length using the compass 😕
It IS possible, IF you allow a dubious step. You have your straightedge, and you can mark a distance on that with a pen. THe distance to be marked on it is one you create in the construction. Then you move your straightedge to set the s ...[text shortened]... ou allow distances to be moved like that, you CAN square a circle.
Originally posted by AcolyteWe are talking Euclidian geometry here. Euclides only had a pole with a rope attached to it, with which he could make circles, but not copy distances. Also, TheMaster says 'straightedge' and not 'ruler'. Maybe he doesn't know the word, but more likely it is because that was all Euclides had.
I thought you could reposition any length using the compass 😕
But, TheMaster has not paid enough attention in class, because Euclid's elements, Book I, Preposition 2, states exactly what he wants.
With Euclidian geometry however, it is impossible to square a circle.
Originally posted by pidermanAh. The pole with a rope is indeed more restrictive than a compass, if you can't adjust the radius. It means, for example, that even integer-length triangles are unconstructible unless they have 'special' angles like pi/2.
We are talking Euclidian geometry here. Euclides only had a pole with a rope attached to it, with which he could make circles, but not copy distances. Also, TheMaster says 'straightedge' and not 'ruler'. Maybe he doesn't know the word, but more likely it is because that was all Euclides had.
But, TheMaster has not paid enough attention in class, beca ...[text shortened]... exactly what he wants.
With Euclidian geometry however, it is impossible to square a circle.
Originally posted by AcolyteWell, you can adjust the radius, but you can't keep it. Like, if you pick up your compass it automatically folds.
Ah. The pole with a rope is indeed more restrictive than a compass, if you can't adjust the radius. It means, for example, that even integer-length triangles are unconstructible unless they have 'special' angles like pi/2.
BTW, I don't know what you mean by 'integer-length triangles', but Euclides never spoke about lengths, everything is relative. But you would be surprised at what you could construct using thos simple tools. Even a pentagram, though that is quite difficult.
Originally posted by TheMaster37dose it have anything to do w/ the vitruvian man? (please?):'(
It IS possible, IF you allow a dubious step. You have your straightedge, and you can mark a distance on that with a pen. THe distance to be marked on it is one you create in the construction. Then you move your straightedge to set the same distance somewhere else.
If you allow distances to be moved like that, you CAN square a circle.
Originally posted by pidermanThe convention, if we're looking at it from a modern perspective, would be to draw an arbitrary line segment and declare it to be of unit length - IOW pick a line and make everything relative to that. From there, lines of 'rational' length are constructible by means of similar triangles and making multiple end-to-end copies of a line segment. I'm sure Euclid had a concept of 'this is twice as long as that'.
Well, you can adjust the radius, but you can't keep it. Like, if you pick up your compass it automatically folds.
BTW, I don't know what you mean by 'integer-length triangles', but Euclides never spoke about lengths, everything is relative. But you would be surprised at what you could construct using thos simple tools. Even a pentagram, though that is quite difficult.
If you can adjust the radius as long as you don't pick up the compass, you could still do the trick of 'walking' the compass across the plane. I don't know what limitations that would have, though.
Originally posted by AcolyteThe iterative method which I have found , is as follows:
That's impressive! How many steps does each construction take?
Let ACB be the given circle with centre O, AB its (horizontal) diameter, and C is a point on its circumference such that OC is perpendicular on diameter AOB.
With ruler and compass, draw /_BOP1 = 22.5 deg. Take point P1 on OP1=BC in length. Draw perpendicular P1Q1 at P1 on OP1 which meets extended line OB at Q1. Now draw bisector OP2 of the /_BOP1 & take point P2 on this bisector such that OP2=OQ1.
Now draw perpendicular P2Q2 at P2 on the line OP2 which meets the extended line OB at point Q2. Next draw bisector OP3 of the /_BOP2. And go on doing this iterative construction and locating the successive points Q1, Q2. Q3....on the extended line OB.
In fact the points Q1, Q2, Q3,..converge so fast that Q3 is practically indistinguishable from Q2. The final point Q of convergence is practically reached after the second iteration .
Next extend OA backwards upto A1 so that AA1=OA, i.e. A1O=2OA. Now draw a semi-circle A1XQ with A1Q as the diameter.
Now let the extended line OC meet this cicle at point X.
Now draw a square on OX. This square is the required square equl to the given circle in area.All this can be done with a ruler & compass .
This iterative construction uses the trigonometrical identity
Z = sin Z*{sec Z/2 * sec Z/4 * sec Z/8......sec Z/(2^n).....to infinity}.
Originally posted by fearlessleaderHow can it be done, using calculus?
i dont think this is possible, in fact, it is impossible to get a perfect answer even with trig. i seem to remember that the vitruvian man had something to do with this (da Vinci). in any case, it can probably be done w/out trig if you allow calc.
Originally posted by cosmic voiceYes it will work towards successive approximation from iteration to iteration. But the accuracy is not as high as claimed by you in your earlier post. After 2 iterations the accuracy will be only of the order of 0.1% error.
The iterative method which I have found , is as follows:
Let ACB be the given circle with centre O, AB its (horizontal) diameter, and C is a point on its circumference such that OC is perpendicular on diameter AOB.
With ruler and compass, draw /_BOP1 = 22.5 deg. Take point P1 on OP1=BC in length. Draw perpendicular P1Q1 at P1 on ...[text shortened]... ometrical identity
Z = sin Z*{sec Z/2 * sec Z/4 * sec Z/8......sec Z/(2^n).....to infinity}.