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# Class Note for MATH 410 with Professor Martin at KU

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This 2 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at Kansas taught by a professor in Fall. Since its upload, it has received 16 views.

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Date Created: 02/06/15

The three Classical geometric problems 1 Constructible numbers Suppose that you are given a line segment of length l and the Euclidean tools of compass and unmarked straightedge What other lengths can you construct It s easy to construct other integer lengths n by attaching 71 copies of your unit length back to back along a line A little more generally if you can construct lengths x and y then you can construct x y and x 7 y What about multiplication and division The trick is to use similar triangles Given lengths a and b you can construct ab as follows First draw a triangle AXYZ with XY 1 and XZ 1 Then draw a segment X Y of length I and construct a new triangle AX Y Z similar to the rst one Then X Z ab Z If instead you start by making X Z b then the new triangle will have X Y ba So the set K of all constructible numbers is closed under the four arithmetic operations of addition subtraction multiplication and division1 In particular every rational number is constructible But there are certainly nonrational numbers that are constructible For example can be constructed as the length of the hypotenuse of an isosceles right triangle So can W for any integer n in fact So the question arises What exactly is the set lK That is which real numbers can be constructed as the lengths of line segments 2 The three Classical problems There are three famous problems of classical geometry that the Greeks were unable to solve for good reason They were 1 Squaring the Circle Given a circle construct a square of the same area 2 Trisecting the angle Given an arbitrary angle of measure oz construct an angle of measure 013 3 Squaring the Circle Given a cube construct a new cube whose volume is double that of the rst cube All these problems can be rephrased in terms of constructible numbers A circle of radius 1 has area 7r so we can square the circle ill we can construct A cube Q of side length 1 has volume 1 to double it we d need to construct a line segment of length Finally it turns out that trisecting an arbitrary angle is equivalent to being able to construct a root of a certain cubic equation It turns out that all these things are impossible 7 the set of constructible numbers is known not to include transcendentals like or things like cube roots 1 Algebrajcally this says exactly that the set K is a eld 3 How to trisect the angle The following solution of the angle trisection problem is attributed to Archimedes While the construction works it is not Euclidean because it uses a new tool a straightedge with two points marked on it Let ZPOQ be the angle to be trisected Draw a circle Z centered at O and call the radius of the circle r Let A and A be the points where O iP meets Z with A between 0 and P and let B be the point where m meets Z Now here s the nonEuclidean step Mark two points on your straightedge at distance r Then move the straightedge so that the two marked points lie on O iP and Z 7 say at points C and D 7 and so that the line of the straightedge goes through Label the angles oz y 6 E 9 w as shown Remember that oz is the angle we want to trisect As we will see 40 013 Here s why By construction we know that OAOBOCCD72 In particular AOBC and ACOD are isosceles triangles with vertices O and C respectively Since the base angles of an isosceles triangle are equal we infer that 6 E and 1a w 1b Next we apply the theorem that the angles of a triangle add up to 180 In particular y 6 E 180 and 2a 6w180 i 2b Third because A O A are collinear and B C D are collinear we know that oz y 180 and 3a a 9 180 3b Now putting these six equations together and using some algebra it is possible to show that w 013 The details are left to the reader

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