Ideas in Geometry/Constructible Polygons and Impossibilities

Lesson Twenty-Nine: Constructible Polygons and Impossibilities


While a compass and a straightedge allow for almost any constructible polygons, there are several impossible constructions. There are three main problems that are known as the “impossibilities” which cannot be solved with a compass and a straightedge.


The first impossibility is doubling the cube.


The goal of doubling the constructed cube is to double its volume. To do this, we must construct the roots of the equation.


To demonstrate this, we can begin with a cube that has a volume of 1. In order to “double the cube” we must take v=1*2 (v=2). This means we must construct a cube with a volume of 2. For a cube with this volume, each side length must equal We know that the cube root of any number is not constructible so therefore we cannot double the cube. This is true for all volumes of a cube because each will have in its equation.


The second impossibility is squaring the circle.


The goal of squaring a constructed circle is to construct a circle with a given radius and then construct a square with the same area as the original circle.


To demonstrate this, we can begin with a circle that has a radius of 1 unit. Area of a circle is A=πr^2 so we know that the area of this circle is A=π(1^2) or A=π. In order to square the circle, we must construct a square with an area of π. We know that π is not a constructible number so this square cannot be constructed. This is true for all constructed circles because each will have π in its radius.


The third impossibility is trisecting an angle.


The goal of this construction is to trisect an angle. Though bisecting an angle is simple enough, to trisect one is impossible. Trisecting an angle is defined as an impossibility because there is no possible way to do it with just a compass and a straightedge. Though the concept of trisecting an angle cannot be constructed, there are certain angles in which it is possible to trisect.


One of these angles is a 45° angle. Given an angle ABC that equals 45°, we can construct an equilateral triangle at point B. This gives us a new angle DBE that is equal to 60°. Now all we need to do is bisect the 60° angle, giving us one 15° angle (formed between the 45° angle and the 60° angle) and a 30° angle. The last step is to bisect the 30° angle giving us three 15° angles, thus successfully trisecting a 45° angle.