Quadrature the polygon

Convex polygon with five unequal sides

The quadrature of the polygon or the quadrature of the rectilinear figure is a task from ancient geometry . It consists in using the Euclidean tools, compasses and ruler, to draw a square with the same area from a given convex or concave polygon .

In the following, this article only deals with the irregular polygon. The squaring of the rectangle is described in detail in a separate article.

initial situation

Johann Friedrich Lorenz describes in his book Euclid's Elements , from 1781, the solution of this quadrature in a given straight-line figure, A, to make a square equal .

The following mathematical theorems of Euclid are used to work on the task .

1. Conversion of a triangle into a parallelogram with the same area.
2. Conversion of a straight figure into a parallelogram with the same area.
3. The set of heights in the right triangle.

Squaring the straight figure using the theorem of heights, according to Euclid.
For the transformation of the triangle MNO see also the animation.

Triangle MNO (for a clear illustration with a different shape) converted into a rectangle with the same area for a given rectangle side T ₁ W (animation with pause 31 s) .
In each step, the parallelograms concerned agree on one side and the associated height and are therefore each of the same area:
F (U ₁ ZC ₁ C) = F (U ₁ ZDA ₁ ) = F (ZDWB ₁ ) = F (C ₁ DWM ₁ )

method

The irregular pentagon KLMNO was chosen to illustrate the problem with the squaring of convex polygons with more than four sides .

First, the area of ​​the polygon is broken down into triangles, that is, starting from a freely selectable polygon corner, such as. B. O , the diagonals are drawn. This results in the smallest possible number of triangles; in the example there are the three triangles KLO (yellow), LMO (red) and MNO (green).

It continues with drawing in the triangular heights PK , QM and RN and dividing them in half; the intersection points S , T and U are obtained .

Next, the diagonal LO (alternatively diagonal MO ) is plotted as a segment BE on a straight line.

From the formula for determining the triangular area for the yellow triangle , a corresponds to the diagonal in LO and h the distance PK , the yellow rectangular area can be derived and as F = BE · BS ₁ can be constructed. Accordingly, this also applies to the red triangle LMO and the red rectangle T ₁ WVS ₁ . ${\displaystyle F={\frac {a\cdot h}{2}}}$

The conversion of the green triangle MNO requires a little more effort , because its base line MO is shorter than the side length T ₁ W now to be taken into account . After the first conversion of the green triangle MNO into the red-blue rectangle U ₁ ZM ₁ T ₁ , a second conversion into a rectangle with the side length T ₁ W follows .

First the point U _{1 is} connected to the point W and the distance EW is lengthened somewhat. A subsequently constructed parallel to the route U ₁ W from the point Z gives the intersection point D .

The following parallel to the segment T ₁ W from point D generates the green rectangle CDWT ₁ with the same area as the green triangle MNO ; see the explanation in the adjacent animation triangle MNO converted into a rectangle with the same large area at a given side of the rectangle T ₁ W . The green triangle MNO has a different shape in the animation to illustrate the step-by-step conversion .

The squaring of the thus assembled rectangle CDEB begins with the extension of its side BE and a quarter circle with the radius ED around the point E ; so there is the intersection F .

After halving the line BF at point G , draw a Thales circle around G and extend the side of the rectangle DE to the Thales circle; the point of intersection H is thus obtained . The segment EH is the first side of the square sought, the area of ​​which is the same as that of the given irregular pentagon.

Web links

• Uwe Förster: Prof. Johann Friedrich Lorenz Biography uni-magdeburg.de
• Decomposition of polygons into triangles ( Memento from October 18, 2016 in the Internet Archive ) Visual dictionary of mathematics

Individual evidence

1. ^ Johann Friedrich Lorenz: Euclid's elements . tape 2 . Verlag der Buchhandlung des Waysenhauses, Halle 1781, section: The 14th sentence. To make a given straight figure, A, a square. , S. 33 ( e-rara.ch [accessed on October 16, 2016]).
2. ^ Johann Friedrich Lorenz: Euclid's elements . First book. Verlag der Buchhandlung des Waysenhauses, Halle 1781, section: The 42nd sentence. There is a triangle, ABC, given; one should make a parallelogram ... like it. , S. 21 ( e-rara.ch [accessed on October 16, 2016]).
3. ↑ John M. Lee: Construction Problem 16:19 (Parallelogram with the Same Area as a Triangle) . In: American Mathematical Society (Ed.): Axiomatic Geometry . Rhode Island 2013, p. 302 ff . ( books.google.de [accessed October 16, 2016]).
4. ^ Johann Friedrich Lorenz: Euclid's elements . First book. Verlag der Buchhandlung des Waysenhauses, Halle 1781, section: The 45th sentence. There is a straight line figure, ABCD; one should make a parallelogram ... the same. , S. 22nd ff . ( e-rara.ch [accessed on October 16, 2016]).
5. ^ John M. Lee: Construction Problem 16.20 (Rectangle with a Given Area and Edge) . In: American Mathematical Society (Ed.): Axiomatic Geometry . Rhode Island 2013, p. 303 ff . ( books.google.de [accessed October 16, 2016]).