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Three Impossible Constructions

The three classical construction problems of antiquity are known as ``squaring the circle'', ``trisecting an angle'', and ``doubling a cube''. Here is a short description of each of these three problems:

It is beyond the scope of this paper to prove that these three problems are impossible to solve, but given the ideas that were presented in Section 5, we can at least indicate how the formal proofs work.

The easiest to show (to be impossible) is the problem of doubling the cube. Clearly if the volume of the required cube is double the volume of the original cube, if the length of the side of the original cube is $L$, the desired length is $\sqrt[3]{2}L$.

In other words, from an algebraic point of view, we can construct the cube root of 2, or equivalently, we can solve the following equation:

\begin{displaymath}L^3 = 2, \end{displaymath}

which is irreducible over the rationals, and is not going to be in any quadratic extension of the rationals.

The problem of trisecting an angle is similar, but what is usually done is to show that there is a particular angle that cannot be trisected, and that angle is typically chosen to be $60^\circ$. If we can't trisect this particular one, then we know that not all angles can be trisected. (Remember that a solution to the problem requires a general method that will work for any angle. Certain angles, such as $90^\circ$ angles, can obviously be trisected, since we can construct a $30^\circ$ angle from scratch. Also, we can construct a $60^\circ$ angle from scratch, so if this one cannot be trisected, the general problem of trisection is clearly unsolvable.)

If we can trisect a $60^\circ$ angle, that is equivalent to constructing a $20^\circ$ angle from scratch, which is equivalent to constructing the cosine of $20^\circ$.

Now the cosine of $60^\circ$ is $1/2$, so

\begin{displaymath}\cos 60^\circ = \cos (20^\circ + 20^\circ + 20^\circ) = 1/2. \end{displaymath}

Using the standard formulas for the sine and cosine of sums of angles, this equation above can be converted to:

\begin{displaymath}4\cos^3 20^\circ - 3\cos 20^\circ = 1/2, \end{displaymath}

or if $x = \cos 20^\circ$, to

\begin{displaymath}8x^3 - 6x - 1 = 0,\end{displaymath}

which is also an irreducible cubic equation that cannot possibly have a solution in any quadratic extension of the rationals.

Much more difficult is to show that a circle cannot be squared. If the original circle has radius 1, its area will be $\pi$, so squaring a circle is equivalent to the construction of a length equal to $\sqrt{\pi}$. It turns out (but is not easy to prove) that $\pi$ is transcendental--it is not the solution to any polynomial equation, so it is surely not the solution to any combination of quadratic polynomials.

next up previous
Next: Problems Up: construct Previous: Simple Algebraic Concepts
Zvezdelina Stankova-Frenkel 2000-11-13