Geometric Construction (bisecting an angle with a compass and straightedge)

In summary, Snow's co-worker believes that all angles can be bisected or trisected with a protractor. However, this is not the case. Geometric constructions must be present in a geometry course in order to be able to bisect or trisect an angle.
  • #36
well i meant kids who like to be actively engaged, good point. kids who want to do rather than listen. but where i noticed it particularly was with that group that was selected for, yes all three: ability and motivation and interest.

here is their current website with their description of who is a likely candidate.
https://epsiloncamp.org/is-epsilon-a-good-fitactually maybe most students would prefer doing to listening, but as a college professor i wasn't trained to teach that way. actually i wasn't trained at all, but i did have examples to imitate, and they were not like that either.
 
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  • #37
mathwonk said:
Probably you know Hartshorne discusses the very point you raise about copying line segments on page 20 of his book
I had a vague recollection of reading about that and I figured that if I mentioned it you'd be able to confirm for me if it was in Hartshorn or elsewhere. Thanks!
 
  • #38
actually that was my class of 10 year olds that soved this problem in under 5 minutes!
 
  • #39
@brainpushups: I would be grateful if you, as an actual teacher, were to take a look at my notes:
https://www.math.uga.edu/sites/default/files/inline-files/10.pdf

They are just mostly comments on Euclid, books 1-4, but contain three contributions I invented while teaching the course.

1) Without enough time to cover chapters V and VI, hence to cover the concept of similarity of triangles, i.e. Prop VI.2, I observed that Prop III.35 implied that result if I defined proportionality of segments in terms of areas. I.e. define two ratios of segments to be equal, A:B = C:D if and only if the rectangle with sides A and D, is equal to the rectangle with sides B and C. Then observe that in a circle, two intersecting secants define a pair of similar triangles, and every such pair of triangles can be so embedded. Then the fact that two secants meet in segments that define equal (area) rectangles, does the trick.

2) The volume of a "bicylinder" is computed in calculus classes as a very difficult problem. I noticed that Archimedes had announced that he had the answer in his famous work: On Method, but the solution had been removed by medieval monks coveting the parchment used in his book. So I resuscitated the solution by analogy with his solution of the volume of a 3-ball, only to learn later that it had been found by a famous topologist over 100 years ago. (But no faculty colleague I knew was aware of this solution.)

3) (This is perhaps a bit over the edge, but I never want to underestimate the potential of bright kids and wanted to give them everything I had. This was not included in the actual course, but added at the end of the class notes, so I do not know how the kids would have responded to the idea of 4 dimensional volume. Since these kids included the "profoundly gifted" I suspect some would have liked it.) Since Archimedes showed that the volume of a half - ball is the difference of the volumes of a cylinder and a cone, it follows that revolving these around a (planar) axis in 4-space, that the 4 dimensional volume generated by revolving half a 3-ball, i.e. the volume of a 4-ball, equals the difference of the volumes obtained by revolving a cylinder and a cone. These can be computed by knowing their centers of gravity, which Archimedes knew, hence Archimedes' methods allow one to easily compute the volume of a 4 dimensional ball, much more easily than one can do by usual calculus methods!

anyway, thanks for your on - the - job insights, and everyday contributions. My goal is to support you, learn from you, and I aspire occasionally to suggest some things that you may want to implement.
 
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  • #40
  • #41
mathwonk said:
They are just mostly comments on Euclid, books 1-4, but contain three contributions I invented while teaching the course.
This is great. Thank you. I've downloaded it and added it to my folder that contains all of my notes from studying these early works (which includes the works of Archimedes). I'm looking forward to reviewing it.
 
  • #42
mathwonk said:
Given your answer, which seems to confirm some basic geometric knowledge, (the topics on area mentioned, plus some facts about triangles, consist basically of book 1 of euclid), what do you mean by saying "almost all 9th and 10th graders do not have an intuition for geometry"? I.e. what are they missing?
Having TA/tutored HS students for 7 years now. The geometry education is extremely lacking. Often times, there is no proofs in the actual geometry course. It is mostly akin to a analytic geometry course, where students just plug and chug numbers into an equation, or remember properties of parallel lines to solve problems. It is a very sad state of affairs.

You know those 2 column proofs (I hate them btw), are not present 90% of the time. You will be hard pressed to find a high school where the majority of students know how to prove two triangles are congruent...

I recently got my BS in math, and could not afford the cost of graduate school (family obligations). So I accepted a position to teach at a fancy catholic school on westside (k-8th ). Needless to say, the kids do not even learn division in the 4th/5th grade. I brought this issue up in a recent meeting. Why do my students (7th grade) do not know long division? I received angry stairs from the elementary teachers...
 
  • #43
MidgetDwarf said:
Having TA/tutored HS students for 7 years now. The geometry education is extremely lacking. Often times, there is no proofs in the actual geometry course. It is mostly akin to a analytic geometry course, where students just plug and chug numbers into an equation, or remember properties of parallel lines to solve problems. It is a very sad state of affairs.

You know those 2 column proofs (I hate them btw), are not present 90% of the time. You will be hard pressed to find a high school where the majority of students know how to prove two triangles are congruent...

I recently got my BS in math, and could not afford the cost of graduate school (family obligations). So I accepted a position to teach at a fancy catholic school on westside (k-8th ). Needless to say, the kids do not even learn division in the 4th/5th grade. I brought this issue up in a recent meeting. Why do my students (7th grade) do not know long division? I received angry stairs from the elementary teachers...
Extremely Sad! Public Education has lost its way.
 
  • #44
MidgetDwarf said:
Why do my students (7th grade) do not know long division? I received angry stairs from the elementary teachers
I should think long division is largely unpracticed because computers and even calculators are around everywhere. I hope angry stairs is a misprint for angry stares. Unless they literally threw you down the stairs.

The two column proof brings back memories. Your idea that the geometry curriculum is plug-and chug analytic geometry would be supported with fashoning geometry as something to be done with a programmable calculator. I note that back to school items often include a particular scientific/graphing calculator.
 
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  • #45
How would the history of geometry or geometry itself be different if the Greeks has not insisted on construction with only a straightedge and compass? Are we better off or worse off and how does that compare to not training modern students in classical construction vs. giving them different concepts?
 
  • #46
bob012345 said:
How would the history of geometry or geometry itself be different if the Greeks has not insisted on construction with only a straightedge and compass? Are we better off or worse off and how does that compare to not training modern students in classical construction vs. giving them different concepts?
Construction with straight-edge and compass probably was the logical, best way to develop Geometry and what actually was found. What alternatives would anyone suggest? Would carpentry, or related activity have had a strong influence on those early geometers?
 
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  • #47
Some courses in Abstract Algebra have as one of their goals to show it is impossible to trisect a (e.g. a 60 degree) angle, using only straight edge and compass. This does not answer the above question proposed, but, I can see why we teach geometry and include a little historical perspective with construction. In the same way we teach (very) elementary astronomy in jr high schools, and this is likely because Greek philosophers felt this should be part of education.
Another reason to present geometric construction is that some students might even think it's fun. This is the same reason, that I think that schools should at least have a class on the slide rule. Maybe students are a little tired of the computer, and might want to form some appreciation for the way calculation was done in the old days.
 
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  • #48
mpresic3 said:
Another reason to present geometric construction is that some students might even think it's fun. This is the same reason, that I think that schools should at least have a class on the slide rule. Maybe students are a little tired of the computer, and might want to form some appreciation for the way calculation was done in the old days.
Beyond just your reasoning is the possibility of increased understanding. Also, the slide rule only needs ordinary shaded illumination during daylight hours, but we have electrically powered lighting whenever we want it for most practical purposes.
 
  • #49
It seems like Newton basically did everything using geometry. Open Principia and that's all you see!
 
  • #50
@mpresic3: in case of possible interest, here is my presentation on the topic of impossible constructions, to my class of bright 8,9,10 year olds.

Remarks on impossible constructions:
I want to say something about why the only regular p-gons that can be constructed with
prime p, are for the Fermat primes p = 2^(2^n) + 1, or equivalently p – 1 = 2^(2^n).
First of all, note that a number of form 2^(nm)+1 where m is odd is never prime, because it
equals (2^n)^m + 1, and any number of form x^m+1 where m is odd can be factored.
E.g., you probably know the basic example of x^3+1 = (x+1)(x^2+x+1). Then x^5 +1 =
(x+1)(x^4+x^3+x^2+x+1), and so on…. Wait a minute, why am I going to so much
trouble? Probably you know the factor theorem from algebra implies that a polynomial
f(x) is divisible by x+1 if and only if f(-1) = 0. Since plugging x = -1 into x^m + 1 does
give zero when m is odd, x+1 always divides x^m+1 for m odd. So 2^n +1 always
divides (2^n)^m + 1 when m is odd, by taking x = 2^n.
Thus if a prime has form 2^n + 1, then it has form 2^(2^n) + 1. Hence it suffices to show
that every constructible prime must have form 2^n + 1.

Theorem: If p is a constructible prime, then it has form p = 2^n+1.
Description of proof: Notice that every construction involves finding points by
intersecting lines and circles. This is the key to understanding what constructions are
possible.

Assume we are in an Archimedean geometry, i.e. Euclid’s geometry plus the extra
axiom of Archimedes, that says any segment can be laid off repeatedly until it reaches
any other point. Then we can introduce real numbers as coordinates, by introducing a
pair of perpendicular axes in the plane. Then every point of the plane can be described
by a pair of real coordinates (x,y). Conversely we may assume that every pair of real
coordinates (x,y) represents a point of the plane, but they need not all be constructible.
We want to examine which ones among the many points of the plane are constructible
by ruler and compass.

We start from only two points (0,0) and (1,0), and ask what points can be constructed
from these. E.g. we can lay off as many copies of these as we wish along the x axis, so
we get all points of form (n,m) where n,m are integers. Then we can also subdivide the
interval between (0,0) and (1,0) into n equal parts for every natural number n, and then
lay off copies of those, so we also get all rational points on the x-axis of form (n/m, 0)
where n,m, are integers, and m ≠ 0.

Since we can construct the perpendicular to the x-axis we also get the y axis, and then
we can lay off rational points on the y axis. Now we can construct perpendiculars to
both x and y axes and intersect them, so we also get all “rational points” of the plane,
i.e. all points of form (n/m, a/b), with a,b,n,m, integers and mb ≠ 0. What else can we
get?

Well we also get points that arise from intersecting lines determined by two rational
points, with other such lines, or with circles with rational centers and a rational point on
the circumference, or two such circles.

Now intersecting sets defined by equations means solving the equations
simultaneously, so we want to know what kind of numbers occur as simultaneous
solutions of equations of lines and circles. In fact all we need to know is the degree of
the equations. Solving two linear equations with rational coefficients just gives rational
solutions, so intersecting such rational lines does not give any new points. Moreover
intersecting two rational circles gives two points which are also obtained by intersecting
one circle with a rational line (subtract the equation for the two circles to get the
equation of the line). So we look at points obtained by intersecting rational lines and
rational circles.

An equation for a line through two rational points looks like ax+by = c, where a,b,c, are
rational numbers. An equation for a circle determined by a rational center and rational
point on its circumference looks like (x-u)^2 + (y-v)^2 = w^2, where u,v,w, are rational
numbers. To solve ax+by = c, and (x-u)^2 + (y-v)^2 = w^2, simultaneously, we solve
the linear equation for y, getting: by = c – ax, then y = c/b – ax/b, and substitute this for y
in the quadratic equation. The result is some complicated quadratic equation in x,
Maybe (x-u)^2 + ([c/b – ax/b] – v)^2 = w^2. I don’t care what it is exactly, as I am only
interested in its degree, or the fact that it is quadratic, i.e. of degree two. We call
solutions of quadratic equations with rational coefficients “quadratic” numbers.
Conversely we know how to use Pythagoras to solve quadratic equations whose
coefficients are segments, or numbers, that we have already constructed. So we can
construct points in the plane whose coefficients are solutions of quadratic equations with
rational coefficients, and any algebraic combination of those numbers. E.g. we can
construct the point (sqrt(2)-sqrt(5), 1+sqrt(7/3)). Similarly we can construct all quadratic
points, i.e. points whose coefficients are quadratic numbers. These quadratic points are
the points that only require one “quadratic step” to construct, i.e. one use of the
compass.

What next? Well once we have those quadratic guys we can intersect more lines and
circles. So now we are solving quadratic equations whose coefficients are quadratic
numbers. We call these biquadratic numbers. Thus we can construct all biquadratic
points I claim the solution of such an equation is also the solution of an equation of
degree 4, but with rational coefficients. I.e. we claim all biquadratic numbers are also
quartic numbers, or degree 4 numbers.

E.g. if we have an equation like X^2 – sqrt(3).X + 1 = 0, with quadratic coefficients, we
can rewrite it as X^2 + 1 = sqrt(3)X, and square both sides, to get (X^2+1)^2 = 3X^2, or
X^4 + 2XC^2 + 1 = 3X^2, which becomes X^4 –X^2 + 1 = 0. Thus our number X
becomes a solution of a 4th degree equation with rational coefficients. In general every
biquadratic number is a quartic number. The idea is that biquadratic numbers are those
that only require two quadratic steps to construct, or two uses of the compass. As in
this example, they are all quartic numbers, i.e. they satisfy degree 4 equations with
rational coefficients. Now what about points that require three quadratic steps to
construct?

If we have X^2 – 2^(1/4)X -3 = 0, where the coefficient 2^(1/4) is biquadratic hence
quartic, we get X^2 -3 = 2^(1/4)X, and raising both sides to the 4th power gives (X^2-
3)^4 = 2X^4, which is an equation of degree 8 with rational coefficients. So the
triquadratic number X, which requires three uses of the compass, is of degree 8, or an
“octic” number.

I don’t know how to make this entirely clear, but as we go on what happens is that the
points we get are solutions of equations of degree 2,4,8,16,32,….., i.e. degree 2^n, with
rational coefficients. And that is all we can get. More precisely, a point that can be
constructed in n steps satisfies an equation with rational coefficients, and has degree
dividing 2^n, (since steps not using the compass have degree one). It follows that a
point (x,y) cannot be constructed unless its coefficients are solutions of an equation of
some degree 2^k with rational coefficients.

Now this applies also to complex numbers x+iy corresponding to our points. I.e, if (x,y)
is a constructible point, then the complex number z = x+iy must satisfy an equation of
degree 2^n for some n. But the first vertex on the unit circle of a regular p - gon, after
the point (1,0), is exactly “1/p^ th” of the way around the circle. Since multiplying
complex numbers adds their angles and multiplies their lengths, it is a complex solution
of the equation z^p – 1 = 0.

Now this equation factors as z^p – 1 = (z-1)(z^(p-1) +….+z + 1) = 0, and since z=1 is
the only solution of the first factor, the complex number we want is a solution of the
second factor, which has degree p-1. For that point, which gives the first vertex of the
regular p-gon, to be constructible, we must have p-1 = 2^n, i.e. p = 2^n+1. Then of
course since p is prime, we have seen it must have form 2^(2^k) + 1, i.e. it must be a
Fermat prime.

These ideas are usually taught in a college abstract algebra course as an application of
linear algebra. One possible source is the book Abstract algebra, a geometric
approach, by Theodore Shifrin, or my math 4000 notes #4f, the last couple lectures, on
my web page at UGA. http://www.math.uga.edu/~roy/
 
  • #51
mathwonk said:
@mpresic3: in case of possible interest, here is my presentation on the topic of impossible constructions, to my class of bright 8,9,10 year olds.

...

These ideas are usually taught in a college abstract algebra course as an application of
linear algebra. One possible source is the book Abstract algebra, a geometric
approach, by Theodore Shifrin, or my math 4000 notes #4f, the last couple lectures, on
my web page at UGA. http://www.math.uga.edu/~roy/
Nice presentation but I'm just a bit skeptical an eight year old really understands you vs. saying they do. These ideas are taught at the college level for a reason.
 
  • #52
bob012345 said:
Nice presentation but I'm just a bit skeptical an eight year old really understands you vs. saying they do. These ideas are taught at the college level for a reason.
Well, @mathwonk did specify that they were bright 8, 9, and 10 year-old students. I think that what he meant by that implicitly included the notion that these young people are significantly more advanced than 'average' in their comprehension of mathematical concepts. Furthermore, if he says something that implies that he is confident that his students are following his trains of thought on these moderately abstruse mathematical matters , I see no reason to doubt him on that.
 
  • #53
sysprog said:
Well, @mathwonk did specify that they were bright 8, 9, and 10 year-old students. I think that what he meant by that implicitly included the notion that these young people are significantly more advanced than 'average' in their comprehension of mathematical concepts. Furthermore, if he says something that implies that he is confident that his students are following his trains of thought on these moderately abstruse mathematical matters , I see no reason to doubt him on that.
What's the probability of having a whole class full of Sheldon's unless one teaches at a special school for very advanced students?
 
  • #54
bob012345 said:
What's the probability of having a whole class full of Sheldon's unless one teaches at a special school for very advanced students?
I don't know the pre-teen normings for the requisite ability level.
 
  • #55
You could imagine, if you find the rare exceptionally motivated and smart young people, gather them, put them in a room in which someone like Mathwonk leads them, ... there you have it!
 
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  • #56
That's what it was. I did not design a presentation intended to suit all 8 year olds. I was given a class of about 30 bright, some brilliant, 8,9,10 year olds and asked to give a presentation appropriate to them. I would guess however that once that presentation exists it might suit some in a wider audience. At least it is worth trying.

One day in class after I discussed the geometry of an icosahedron, the most complex of the regular solids, one of the 8 year olds came up and showed me a "planar net" he had drawn of it, i.e. a diagram of contiguous triangles which if folded up along the edges of the triangles, would form an icosahedron! I was quite incapable of that myself.

Another of the ex scholars from this program, Espen Slettnes, won a spirit of Ramanujan award in 2020, and now has a minor planet named after him. Although apparently still in high school, the website says he currently teaches math at the Berkeley Math circle.
https://spiritoframanujan.com/home/
 
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