Geometrical constructions (really fun[not a lie]) Watch

Hathlan
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I have a lot of geometrical construction problems I can't do so I thought I'd make a thread of them.

Give geometrical constructions for
 \sqrt{2}, \sqrt{2+\sqrt{2}}, \sqrt{2+\sqrt{2+\sqrt{2}}}
Well, the first one's just your good old isosocles rectangular traingular, but I included it because it's probably a hint on how to do the others. I've found a few ways of doing the second but they all need Euclid's construction to multiply out a factor of root 2, and I can't see a way of generalizing it to the last bit, although I've been told it's "trivial". So I was wondering if there's a nicer way.

My way:
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Start with the black. An isosocles right angled triangle with sides 1, 1, \sqrt{2}. Make a circle radius \sqrt{2}. Then add one. Hypotenuse will have length \sqrt{2}\sqrt{2+\sqrt{2}}. So we need to divide out the root two, doing Euclid's construction (green) on the hypotenuse of a triangle whose other sides have length a half.




I don't have time to get a picture and explain what I've done so far with these 'cause I have a lecture.
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The next questions are all about finding the roots of the quadratic equation with rational (or integer if you please, just multiply by the lcm of the denominator) coefficents, ax^2+2bx+c=0

Take a line AB of unit length. Draw BC=-2b/a perpendicular to AB, and CD=c/a perpedicular to BC and in the same direction as BA. On AD as diameter describe a circle cutting BC in X and Y. Then BX and BY are the roots.
I don't really have any idea what to do here.


Draw a circle of unit radius, a diameter PQ, and the tangents at the ends of the diameter. Take PP'=-2a/b and QQ'=-c/2b, having regard to sign. Join P'Q', cutting the circle in M and N. Draw PM and PN, cutting QQ' in X and Y. The QX and QY are the roots of the equation with their proper signs.


So what I need to find is QY and QX.

Ooh, I just realized they were parallel lines. So PMP' and XMQ' are similar, and Q'NY and P'NP are similar. Mmmm

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SimonM
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What we need is a method to add and a method to take square roots. Can you do these things?
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Hathlan
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hmm, I just can't seem to think of how to find the square root of a length, unless it's an integer.
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SimonM
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Square roots are the diagonals of squares (Pythagoras' theorem)...
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watari_pg
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I think is the construction Hardy was looking for. Draw the first 45-45-90 triangle with a hypotenuse of sqrt(2) as shown in yellow. Then, using a compass, measure the sqrt(2) amount to the baseline, and add 2 more to that to get a triangle of base 2+sqrt(2). Then you can subtract 1 from 2+sqrt(2) to get the base measurement of another right triangle, which by the Pythagorean theorem has a hypotenuse of sqrt(4+2*sqrt(2)). Divide this new hypotenuse in half and build another 45-45-90 triangle with two sides of length (½)*sqrt(4+2*sqrt(2)), as shown in green. Then, when you calculate the hypotenuse from this triangle, you'll find that it’s sqrt(2+sqrt(2)), which is what we were looking for. You can repeat the process from this triangle, to build another 45-45-90 triangle, as shown in blue, with a hypotenuse of sqrt(2+sqrt(2+sqrt(2))), which of course, is the next iteration.
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watari_pg
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I'm sorry. Apparently I posted the wrong link for the image.

Just in case, it's: http://i411.photobucket.com/albums/p...i_pag/asdf.gif
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watari_pg
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I don't know if people still look at this thread occasionally, but perhaps so, and indeed the questions posed by the youngster who started it (from G.H. Hardy's "A Course of Pure Mathematics") piqued my interest, so I thought I'd continue on nonetheless.

The second topic brought up in Hardy's book concerns geometrical constructions that provide solutions to quadratic equations. The one thing I found odd about Hardy's description of the problem was the way he set forth the basic quadratic equation as being of the form ax^2 + 2bx + c = 0. Why did he put the 2 there? Moreover, he tells us that the roots of this equation are determined by x = \frac{-b \pm \sqrt{b^2-ac}}{a}, whereas most of us remember this as x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}. Honestly, it's easier if you simply take the basic quadratic equation as being of the form ax^2 + bx + c, and then, in the geometrical construction, take PP' = -4a/b and QQ' = -c/b

To grasp the full range of these constructions, I suggest taking a look at F. Klein's "Famous Problems of Elementary Geometry," pp. 32-41, which is free in Google Books:

Regarding these geometrical constructions and why they work, Hardy notes, "The proof is simple and we leave it as an exercise to the reader." I confess though, for me, only after seeing the proof in Klein's book did I realize that yes it is fairly straightforward.

At any rate, to give one simple example of such a construction for x^2 - x - 6

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watari_pg
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Oh no, my initial construction of nested square roots of 2 was not correct. Although the green triangle is correct, you cannot go on duplicating the process with similar and larger triangles, because the hypotenuse will just keep getting bigger, i.e., it already exceeds 2 in the blue triangle, whereas the nested square roots of 2 are supposed to converge to 2.
In other words, it's a geometric progression, with larger and larger hypotenuses, which does not correspond to the convergence to 2 of
\sqrt{2}, \sqrt{2+\sqrt{2}}, \sqrt{2+\sqrt{2+\sqrt{2}}},  ...
as it should.

A correct construction (I'm sure there are others), which should be obvious can be created with a straightedge and compass, is shown below.

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