Angle trisection is the construction of an angle equal to one third of a given arbitrary angle, using only two tools: an unmarked straightedge and a compass. It is a classical problem of straightedge and compass construction of ancient Greek mathematics.
In 1837, Pierre Wantzel proved that the problem, as stated, is impossible to solve for arbitrary angles. However, some special angles can be trisected: for example, it is trivial to trisect a right angle.
It is possible to trisect an arbitrary angle by using tools other than straightedge and compass. For example, neusis construction, also known to ancient Greeks, involves simultaneous sliding and rotation of a marked straightedge, which cannot be achieved with the original tools. Other techniques were developed by mathematicians over the centuries.
Because it is defined in simple terms, but complex to prove unsolvable, the problem of angle trisection is a frequent subject of pseudomathematical attempts at solution by naive enthusiasts. These "solutions" often involve mistaken interpretations of the rules, or are simply incorrect.
Three problems proved elusive, specifically, trisecting the angle, doubling the cube, and squaring the circle. The problem of angle trisection reads:
Construct an angle equal to one-third of a given arbitrary angle (or divide it into three equal angles), using only two tools:
The problem of constructing an angle of a given measure is equivalent to constructing two segments such that the ratio of their length is . From a solution to one of these two problems, one may pass to a solution of the other by a compass and straightedge construction. The triple-angle formula gives an expression relating the cosines of the original angle and its trisection: = .
It follows that, given a segment that is defined to have unit length, the problem of angle trisection is equivalent to constructing a segment whose length is the root of a cubic polynomial. This equivalence reduces the original geometric problem to a purely algebraic problem.
Every rational number is constructible. Every irrational number that is constructible in a single step from some given numbers is a root of a polynomial of degree 2 with coefficients in the field generated by these numbers. Therefore, any number that is constructible by a sequence of steps is a root of a minimal polynomial whose degree is a power of two. The angle (60 degrees, written 60°) is constructible. The argument below shows that it is impossible to construct a 20° angle. This implies that a 60° angle cannot be trisected, and thus that an arbitrary angle cannot be trisected.
Denote the set of rational numbers by . If 60° could be trisected, the degree of a minimal polynomial of over would be a power of two. Now let . Note that = = . Then by the triple-angle formula, and so . Thus . Define to be the polynomial .
Since is a root of , the minimal polynomial for is a factor of . Because has degree 3, if it is reducible over by then it has a rational root. By the rational root theorem, this root must be or , but none of these is a root. Therefore, is irreducible over by , and the minimal polynomial for is of degree .
So an angle of measure cannot be trisected.
For a positive integer , an angle of measure is trisectible if and only if does not divide .MacHale, Desmond. "Constructing integer angles", Mathematical Gazette 66, June 1982, 144–145. In contrast, is constructible if and only if is a power of or the product of a power of with the product of one or more distinct .
Theorem: An angle of measure may be trisected if and only if is reducible over the field extension .
The proof is a relatively straightforward generalization of the proof given above that a angle is not trisectible.
Many incorrect methods of trisecting the general angle have been proposed. Some of these methods provide reasonable approximations; others (some of which are mentioned below) involve tools not permitted in the classical problem. The mathematician Underwood Dudley has detailed some of these failed attempts in his book The Trisectors. (Originally published 1987 as A Budget of Trisections)
The construction begins with drawing a circle passing through the vertex of the angle to be trisected, centered at on an edge of this angle, and having as its second intersection with the edge. A circle centered at and of the same radius intersects the line supporting the edge in and .
Now the set square is placed on the drawing in the following manner: one cathetus of its right angle passes through ; the vertex of its right angle is placed at a point on the line in such a way that the second leg of the ruler is tangent at to the circle centered at . It follows that the original angle is trisected by the line , and the line perpendicular to and passing through . This line can be drawn either by using again the right triangular ruler, or by using a traditional straightedge and compass construction. With a similar construction, one can improve the location of , by using that it is the intersection of the line and its perpendicular passing through .
Proof: One has to prove the angle equalities and The three lines , , and are parallel. As the and are equal, these three parallel lines delimit two equal segments on every other secant line, and in particular on their common perpendicular . Thus , where is the intersection of the lines and . It follows that the and are congruent, and thus that the first desired equality. On the other hand, the triangle is isosceles, since all of a circle are equal; this implies that One has also since these two angles are alternate angles of a transversal to two parallel lines. This proves the second desired equality, and thus the correctness of the construction.
This requires three facts from geometry (at right):
Let be the horizontal line in the adjacent diagram. Angle (left of point ) is the subject of trisection. First, a point is drawn at an angle's ray, one unit apart from . A circle of radius is drawn. Then, the markedness of the ruler comes into play: one mark of the ruler is placed at and the other at . While keeping the ruler (but not the mark) touching , the ruler is slid and rotated until one mark is on the circle and the other is on the line . The mark on the circle is labeled and the mark on the line is labeled . This ensures that . A radius is drawn to make it obvious that line segments , , and all have equal length. Now, triangles and are isosceles, thus (by Fact 3 above) each has two equal angles.
Hypothesis: Given is a straight line, and , , and all have equal length,
Conclusion: angle .
Proof:
and the theorem is proved.
Again, this construction stepped outside the framework of allowed constructions by using a marked straightedge.
Hutcheson constructed a cylinder from the angle to be trisected by drawing an arc across the angle, completing it as a circle, and constructing from that circle a cylinder on which a, say, equilateral triangle was inscribed (a 360-degree angle divided in three). This was then "mapped" onto the angle to be trisected, with a simple proof of similar triangles.
While a tomahawk is constructible with compass and straightedge, it is not generally possible to construct a tomahawk in any desired position. Thus, the above construction does not contradict the nontrisectibility of angles with ruler and compass alone.
As a tomahawk can be used as a set square, it can be also used for trisection angles by the method described in .
The tomahawk produces the same geometric effect as the paper-folding method: the distance between circle center and the tip of the shorter segment is twice the distance of the radius, which is guaranteed to contact the angle. It is also equivalent to the use of an architects L-Ruler (Carpenter's Square).
A regular polygon with n sides can be constructed with ruler, compass, and angle trisector if and only if where r, s, k ≥ 0 and where the p i are distinct primes greater than 3 of the form (i.e. greater than 3).
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