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Chapter 6: Regular Polygons. Constructability This section will contain a number of theorems that we will not prove, because their proofs would involve a lot of extraneous theory, especially from abstract algebra.

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Chapter 6 regular polygons
Chapter 6: Regular Polygons

Constructability

  • This section will contain a number of theorems that we will not prove, because their proofs would involve a lot of extraneous theory, especially from abstract algebra.

  • An n-gon (n-sided polygon) is said to be regular if all n sides are congruent to each other and all angles are congruent to each other.


  • Two well-known examples are the square (n = 4) and the equilateral triangle (n = 3). The first question we will treat is how regular n-gons are constructed.

  • It is traditional in Euclidean Geometry to try to do constructions with an unmarked straightedge and a compass. Here we will mainly discuss the regular polygons, which are constructible with this equipment.

  • For this moment, however, let us assume that we have additional equipment available to us – such as a protractor – and that we can draw angles of any size. Using such equipment we can now construct regular n-gons for any . .


  • Let take n = 10. First we calculate , which in this case is . We choose any point O which will be the center of our regular 10-gon and draw a ray with initial point O.


  • Next, we draw nine more rays each with initial point O and making angles of with the ray . Now we draw a circle with center O and any radius, which intersects the ten rays in the ten points . Then will be the required 10-gon.

  • To prove this, you would show that the ten triangles , , etc. are all congruent to each other by SAS. It would then follow that and that . .


  • It follows from the concepts introduced in this discussion that regular n-gons exist for each and that we can construct them if we can construct angles of size . The converse is also true. So the problem of constructing a regular n-gon with straightedge and compass is equivalent to the problem of constructing a angle with straightedge and compass. Stated formally, we have sketched a proof of the following theorem.

  • Theorem. For a given integer , it is possible to construct a regular n-gon if and only if it is possible to construct an angle of size .


  • Question. that regular For which values of n is it possible to construct a regular n-gon?

  • We will call a regular n-gon constructible if we can construct it with straightedge and compass.

  • Corollary 1. Let n = 2m, where . If the regular m-gon is constructible, then the regular n-gon is also constructible.

  • Proof. Let and . If we use our theorem, we see that the statement “a regular m-gon is constructible” is equivalent to the statement “an angle of size is constructible”; and the statement “a regular n-gon is constructible” is equivalent to the statement “an angle of size is constructible.”


  • Hence, in order to prove our corollary we may prove, instead, the equivalent statement: If the angle is constructible, then the angle is constructible. But, . .The corollary is now immediate, because if we can construct we can certainly construct , since we have a construction for bisecting angles.

  • Corollary 1 has a number of consequences. For example, since we know that a rectangular 4-gon is constructible, we can conclude that a regular 8-gon is also constructible.

  • Also, by repeating applications of the corollary, a regular 16-gon, a regular 32-gon, etc. would also be constructible.


  • Likewise, since a regular instead, the equivalent statement: If the angle is constructible, then the angle is constructible. But, . .The corollary is now immediate, because if we can construct we can certainly construct , since we have a construction for bisecting angles.3-gon is constructible, so is a regular 6-gon, 12-gon, 24-gon, etc.

  • We also see from this that there are infinitely many values of n for which regular n-gons are constructible.

  • We will see in a moment that the converse of Corollary 1 is also true. This reduces the question of constructability of regular n-gons to the case of odd values of n. Why?

  • Suppose you are given an even value of n, such as n = 120 and you wanted to know if it was possible to construct a regular 120-gon.You could factor 120 into a product of a power of two times an odd number, .


  • By applying Corollary 1 and its converse three times, we see that a regular 120-gon is constructible if and only if a regular 15-gon is constructible. In a similar manner, for any even n such there is an odd m such that the constructability of a regular n-gon is equivalent to the constructability of a regular m-gon.

  • Hence, we will turn to the case of odd n. It will turn out that there are only finitely many odd numbers n such that regular n-gons are constructible.

  • Corollary 2. Let n = mk where and k are integers. If a regular n-gon is constructible, then a regular m-gon and a regular k-gon ( if ) are constructible.


  • Proof. that a regular Let and . Then Corollary 2 is equivalent to the statement that if an angle of size is constructible then an angle of size is constructible. Now . It is immediate that if we can construct an angle of size , then we can construct any integral multiple of .

  • Note that the converse of Corollary 2 is false. It is also not true that if a regular m-gon is constructible and if a regular k-gon is constructible, then a regular mk-gon is constructible.

  • For example, a regular 3-gon is certainly constructible, but it is impossible to construct a regular 9-gon with straightedge and compass.


  • However, this statement will be true if that a regular k and m are relatively prime, but the proof involves some techniques of number theory.

  • Let first see what happens when n is a prime number. In this case the constructible n-gons are the 3-gon, the 5-gon, the 17-gon, the 257-gon and the 65537-gon.

  • The primes 3, 5, 17, 257 and 65537 are called Fermat primes, after Pierre Fermat. Note that , , , and ; moreover, the exponents are all powers of 2.

  • Fermat conjectured that was always a prime number, for any value of n.


  • It is known that Fermat was wrong! For example, is not prime; it is divisible by 641. It is not currently known whether there are any prime numbers of the form for n > 4, and this is an important open question.

  • Theorem. If n is a prime number, then a regular n-gon is constructible if and only if n is a Fermat prime.

  • Theorem. (a) If n is odd, then a regular n-gon is constructible if and only if n is a product of distinct Fermat primes. (b) In general, a regular n-gon is constructible if and only if n is a power of 2 times a product of distinct Fermat primes.


  • Now, for various values of is not prime; it is divisible by n we tell whether or not is it possible to construct regular n-gons.

  • n = 75. The regular 75-gon is not constructible; , and although 3, 5 and 5 are Fermat primes, they are not distinct.

  • n = 120. The regular 120-gon is constructible; . which is a power of 2 times two distinct Fermat primes.

  • n = 23. The regular 23-gon is not constructible; 23 is a prime which is not a Fermat prime.


In the footsteps of archimedes
In the Footsteps of Archimedes is not prime; it is divisible by

  • Here we will use regular polygon to approximate the areas and circumferences of circles. The method we use was known to the ancient Greeks. Archimedes' approximation of using regular polygons is considered to be one of the masterpieces of ancient mathematics.

  • Assume we are given a circle of radius r and that we wish to estimate the area A and circumference C. To get lower bounds we will use inscribed polygons.


  • A polygon is said to be inscribed in a circle if every vertex of the polygon lies on the circle or, equivalently, every side of the polygon is a chord of the circle.

  • Given a circle and an inscribed polygon, the polygon must have a smaller area and a smaller perimeter. We will use these facts to get lower bounds for A and C.

  • Given any circle with center O, it is fairly easy to construct a regular inscribed hexagon (6-gon) as follows:


  • Choose any point on the circle and construct points and on the circle such that

  • Without giving a detailed proof, the reason this construction works is that the triangles , etc. are equilateral triangles.

  • From our observations, the perimeter of the inscribed hexagon gives a lower bound for the circumference of the circle and the area of the hexagon gives a lower bound for the area of the circle.


  • Since each side of the hexagon has length and on the circle such that r, the perimeter is 6r. Hence,

  • To calculate of the area of the hexagon we will use the fact that the six triangles are all congruent, so that the area of the hexagon is six times the area of . Let be the altitude, OD = h.

  • Since is equilateral, D is the midpoint of and we may now use the Pythagorean theorem to calculate h.


  • so . Hence, has area and the hexagon has area

  • Therefore,

  • We remark that

  • We may now improve these crude estimates to obtain whatever accuracy we like by using larger values of n.

  • We will illustrate with the case of n = 12. The regular 12-gon (dodecagon) can be constructed using the regular 6-gon. We construct the angle bisector of each of the six central angles. These lines meet the circle at the points , , and , which bisect the arcs


  • Then the polygon will be a regular 12-gon. The perimeter of this 12-gon will be 12 times the length of the side . Call this length s.



  • Hence, and the Pythagorean theorem. And we already calculated that . So

  • We may calculate this as

  • The area will be 12 times the area of . If we take as the base, the height will be . So .

    Multiplying by 12 yields

  • Rather than continuing in this vein, we turn to the question of computing upper bounds for C and A. To do this we will use regular polygons circumscribed about the circle.


  • A polygon is said to be circumscribed about the circle if every side of the polygon is tangent to the circle.

  • It is clear in this case that the area of the polygon will be greater than the area of the circle. It is also true that the perimeter of the polygon will be greater than the circumference of the circle.

  • We observe as a technical point that this fact is neither obvious nor easy to prove. We will assume it in order to do our calculations.




  • We now calculate the perimeter and area of this figure. First, is the hypotenuse of a right triangle, each of whose sides is r. Hence, . Since OA = r weget that . But , so the side . If we multiply by 8 we get

    or

  • Finally, the area of the octagon equals the area of the square minus four times the area of . This triangle has base and height . Hence the area of the octagon is

    or .


  • Combining First, is the hypotenuse of a right triangle, each of whose sides is our results, we see that

    or , and or

  • You likely learned in a previous course that and . It is not possible to prove these results without saying quite a bit about limits. We do not intend to do so. However, you may observe that the bounds for C were all of the form some constant times r and that by taking n larger we could increase the lower bound and decrease the upper bound. Likewise, we can sandwich A between expressions with constants times .


Pascal s theorem
Pascal's Theorem First, is the hypotenuse of a right triangle, each of whose sides is

  • The dual of Brianchon's theorem (Casey 1888, p. 146), discovered by B. Pascal in 1640 when he was just 16 years old (Leibniz 1640; Wells 1986, p. 69).

  • It states that, given a (not necessarily regular, or even convex) hexagon inscribed in a conic section, the three pairs of the continuations of opposite sides meet on a straight line, called the Pascal line.


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