Angle trisection

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s]] s. The displayed ones are marked — an ideal straightedge is un-marked]] ]]

The problem of trisecting the angle is a classic problem of compass and straightedge constructions of ancient Greek mathematics. Two tools are allowed

An un-marked straightedge, and
a compass.

Problem: construct an angle one-third a given arbitrary angle.

With such tools, it is generally impossible, as shown by Pierre Wantzel (1837). This requires taking a cube root, impossible with the given tools. The fact that there is no way to trisect an angle in general with just a compass and a straightedge does not mean that it is impossible to trisect all angles so. Also, it is possible to trisect any angle analytically.

Perspective and relationship to other problems

of arbitrary angles has long been solved.]] Using only an unmarked straightedge and a compass, Greek mathematicians found means to divide a line into an arbitrary set of equal segments, to draw parallel lines, to bisect angles, to construct many polygons, and to construct squares of equal or twice the area of a given polygon.

Three problems proved elusive, specifically:

Trisecting the angle,
Doubling the cube, and
Squaring the circle

Angles may not in general be trisected

The geometric problem of angle trisection can be related to algebra – specifically, the roots of a cubic polynomial – since by the triple-angle formula,

cos(3theta)=4cos^3(theta)-3cos(theta).

Denote the rational numbers $mathbb{Q}$ .

Note that a number constructible in one step from a field $K$ is a solution of a second-order polynomial. Note also that $pi/3$ radians (60 degrees, written 60°) is constructible.

However, the angle of $pi/3$ radians (60 degrees) cannot be trisected. Note $cos(pi/3) = cos(60^circ) = 1/2$ .

If 60° could be trisected, the minimal polynomial of $cos(20$ ^{circ) over $mathbb{Q}$ would be of second order. Note the trigonometric identity $cos(3alpha) = 4cos$}{3}(alpha) - 3cos(alpha). Now let $y = cos(20^circ)$ .

By the above identity, $cos(60$ ^{circ) = 1/2 = 4y}{3} - 3y. So $4y$ ^{{3} - 3y - 1/2 = 0. Multiplying by two yields $8y$}{3} - 6y - 1 = 0, or $(2y)$ ^{{3} - 3(2y) - 1 = 0. Now substitute $x = 2y$ , so that $x$}{3} - 3x - 1 = 0 . Let $p(x) = x^{3} - 3x - 1$ .

The minimal polynomial for x (hence $cos(20$ ^{circ)) is a factor of $p(x)$ . If $p(x)$ has a rational root, by the rational root theorem, it must be 1 or −1, both clearly not roots. Therefore $p(x)$ is irreducible over $mathbb{Q}$ , and the minimal polynomial for $cos(20$}circ) is of degree 3.

So an angle of $60^circ = pi/3$ radians cannot be trisected.

Some angles may be trisected

However, some angles may be trisected. Given angle $theta$ , angle $3theta$ trivially trisects to $theta$ . More notably, $2pi/5$ radians (72°) may be constructed, and may be trisected.»http://www.physicsforums.com/showthread.php?t=160571 Also there are angles, while non-constructable, but (if somehow given) are trisectable, for example $3pi/7$ .Five copies of $3pi/7$ combine to make $15pi/7$ which is a full circle plus $pi/7$ .

One general theorem

Again, denote the rational numbers

mathbb{Q}

Theorem: The angle $theta$ may be trisected if and only if $q(t) = 4t^{3}-3t-cos(theta)$ is reducible over the field extension $mathbb{Q}(cos(theta))$ .

Proof. The proof would take us afield, but it may be derived from the above trigonometric identity.

Means to trisect angles by going outside the Greek framework

Origami

Trisection, like many constructions impossible by ruler and compass, can easily be accomplished by the more powerful (but physically easy) operations of paper folding, or origami. Huzita's axioms (types of folding operations) can construct cubic extensions (cube roots) of given lengths, whereas ruler-and-compass can construct only quadratic extensions (square roots). See mathematics of paper folding.

Auxiliary curve

There are certain curves called trisectrices which, if drawn on the plane using other methods, can be used to trisect arbitrary angles.[http://www.jimloy.com/geometry/trisect.htm#curves Trisection of an Angle]

With a marked ruler

Another means to trisect an arbitrary angle by a "small" step outside the Greek framework is via a ruler with two marks a set distance apart. The next construction is originally due to Archimedes, called a Neusis construction, i.e., that uses tools other than an un-marked straightedge.

This requires three facts from geometry (at right):

Any full set of angles on a straight line add to 180°,
The sum of angles of any triangle is 180°, and,
Any two equal sides of an isosceles triangle meet the third in the same angle.

Look to the diagram at right; note angle a left of point B. We trisect angle a.

First, a ruler has two marks distance AB apart. Extend the lines of the angle and draw a circle of radius AB.

"Anchor" the ruler at point A, and move it until one mark is at point C, one at point D, i.e., CD = AB. A radius BC is drawn as obvious. Triangle BCD has two equal sides, thus is isosceles.

That is to say, line segments AB, BC, and CD all have equal length. Segment AC is irrelevant.

Now: Triangles ABC and BCD are isosceles, thus by Fact 3 each has two equal angles. Now re-draw the diagram, and label all angles:

Hypothesis: Given AD is a straight line, and AB, BC, and CD are all equal length,

Conclusion: angle $b = (1/3) a$ .

Proof:

Steps:

From Fact 1) above, $e + c = 180$ °.
Looking at triangle BCD, from Fact 2) $e + 2b = 180$ °.
From the last two equations, $c = 2b$ .
From Fact 2), $d + 2c = 180$ °, thus $d = 180$ ° $- 2c$ , so from last, $d = 180$ ° $- 4b$ .
From Fact 1) above, $a + d + b = 180$ °, thus $a + (180$ ° $- 4b) + b = 180$ °.

Clearing, $a - 3b = 0$ , or $a = 3b$ , and the theorem is proved.

Again: this construction stepped outside the framework of allowed constructions by using a marked straightedge. There is an unavoidable element of inaccuracy in placing the straightedge.

With a string

Hutcheson published an article in Mathematics Teacher, vol. 94, No. 5, May, 2001 that used a string instead of a compass and straight edge. A string can be used as either a straight edge (by stretching it) or a compass (by fixing one point and identifying another), but can also wrap around a cylinder, the key to Hutcheson's solution.

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.

For the detailed proof and its generalization, see the article cited: Mathematics Teacher, vol. 94, No. 5, May, 2001, pp. 400–405.

There are other constructions (references).

»Trisecting via (»Archived 2009-10-25) the limacon of Pascal; see also Trisectrix
»Trisecting via an Archimedean Spiral
»Trisecting via the Conchoid of Nicomedes
»sciencenews.org site on using origami
»Hyperbolic trisection and the spectrum of regular polygons

home | This article is licensed under the GNU Free Documentation License. See full license termsIt uses material from the Wikipedia article "Angle_trisection ". | compliance | March 19th 2010