Four problems of
antiquity
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Three
geometric questions raised by the early Greek
mathematicians attained the status of classical problems in Mathematics.
These are:
The
problems are legendary not because they did not have solutions, or the
solutions they had were unusually hard. No - numerous simple
solutions have been found yet by Greek mathematicians. The problem was in that
all known solutions violated by an important condition for this kind of
problems, one condition imposed by the Greek mathematicians themselves: Valid
solutions to the construction problems are assumed to consist of a finite
number of steps of only two kinds: drawing a straight line with a ruler (or
rather a straightedge as no marks are allowed on the ruler) and drawing a
circle. You are referred to solutions of
problems 2
and 3 as examples of existent solutions. That no solution exists
subject to the self-imposed constraints have been proven only in the 19th
century. Doubling the Cube
Doubling
the Cube, the most famous of the collection, is often referred to as the Delian
problem due to a legend that the Delians had consulted Plato on the
subject. In another form, the story asserts that the Athenians in 430 B.C.
consulted the oracle at Delos in the hope to stop the plague ravaging their
country. They were advised by Apollo to double his altar that had the form of
a cube. As a result of several failed attempts to satisfy the god, the
pestilence only worsened and at the end they turned to Plato for advice.
(According to Rouse Ball and Coxeter, p 340, an Arab variant insists that the
plague had broken between the children of Israel but the name of Apollo had
been tactfully omitted.) According to a letter from the mathematician
Eratosthenes to King Ptolemy of Egypt, Euripides mentioned the Delian problem
in one of his (now lost) tragedies. Angle trisection
That there may be a difficulty trisecting an
angle is rather surprising. In the classical framework, one draws two kinds
of lines: straight lines and circles. The same easy approach applies to
bisecting both straight line segments and circular arcs:
The problem of trisecting a line segment is
no more difficult than finding its n-th part for an
arbitrary n. However, the general problem of trisecting and angle (i.e.,
trisecting an arbitrary angle) is not solvable in a finite number of steps.
Two points must be made. Firstly,
some angles are trisectable with a ruler and a compass. An obvious example is
supplied by the three quarters of the straight angle - 67.5o. (The
angle itself is constructible as it is obtained by two consecutive angle
bisections. Its third is obtained along the way.) Angles of 30o
(draw a right triangle with a side 1 and hypotenuse 2) and 45o
(bisect the right angle) are both constructible. Therefore, the latter also
admits a classical trisection. However, and this is where the difficulty
lies, an arbitrary angle can't be trisected with a ruler and a compass, as
stated above. Secondly,
consider the geometric series 1/4 + 1/16 + 1/64 + ... that adds up to 1/3.
Once we know how to bisect an angle, we may also find its 2-n-th
part, for any n. In particular, one may construct 1/4, 1/16, ... of any angle
and, in principle, find its third after an infinite number of steps. This
solution is universal but requires a forbidden (infinite) number of steps. The problem had been settled in 1837 by Pierre Laurant Wantzel
(1814-1848) who had proven that there was no way to trisect a 60o
angle in the classical framework. Constructing a regular heptagon
3-, 4-, 5-, and 6-gons (i.e., regular
triangles, squares, pentagons, and hexagons) are all and easily
constructible. Octagons are constructible on the heels of squares with a
single angle bisection. All polygons obtained from the above four by doubling
the number of sides are also constructible. Not so a heptagon, a 7-sided
polygon. In 1796, at the age of 19, Gauss have shown that a regular
heptadecagon (a 17-sided polygon) is constructible. As Gauss showed, the
heptadecagon was only a particular case of a family of constructible regular
polygons. Any N-gon, where N is in the form 2np1...pm,
where pi's are distinct Fermat primes (i.e. primes in the
form 22k + 1), is constructible. Wantzel later proved
that no other regular polygon has this property. Squaring a circle
Surprisingly,
the impossibility proofs for the three cited problems: doubling the cube,
trisecting an angle, and constructing a regular heptagon, all fall into the
same framework. Each depends on showing that roots of certain third degree
polynomial equations are not constructible. The problem of squaring of a
circle found itself in a completely different category. It has been solved
later in the 19th century when the number has been shown to be transcendental.
Not being algebraic, it is automatically not constructible. However, the area
of a circle with radius 1 is exactly
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