Angle trisection

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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. This requires taking a cube root; see below.

Perspective and relationship to other problems

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.

Nevertheless, three problems proved elusive, specifically:

Angles may not in general be trisected

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; again, see constructible number. 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. For example, $2pi/5$ radians (72°) may be constructed, and may be trisected, Also there are angles, while non-constructable, but (if somehow given) are trisectable, for example $3pi/7$

One general theorem

Again, denote the rational numbers

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 $Q(cos(theta))$ .

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

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.

There are other constructions (references).

Notes

External references

Other means of trisection

Trisecting via the limacon of Pascal; see also Trisectrix
Trisecting via an Archimedean Spiral
Trisecting via the Conchoid of Nicomedes
sciencenews.org site on using origami

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Last updated on Tuesday March 11, 2008 at 13:55:11 PDT (GMT -0700)
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