Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 34 (1993), No. 2, 195-199.

On the Existence and Constructibility of Inscribed Polygons

Peter Schreiber

Fachrichtungen Mathematik/Infomatik, Ernst-Moritz-Arndt Universität, Jahnstr. 15a, D - 17489 Greifswald

In a little paper [1] for pupils in honour of the 450th birthday of François Viète in 1990 I revisited his construction of the inscribed quadrangle from its given four sides (first published in 1596 as an appendix on Vieta’s paper ”Pseudomesolabum” and reprinted 1646 in [2]). This has been the starting point for my further investigations on the existence and constructibility of inscribed polygons which I therefore want to dedicate to the memory of Vieta. My interest in this somewhat oldfashioned problem may be apologized by the variety of used proof techniques that include algebraical methods as well as geometrical variants of Bolzano’s intermediate value theorem and — last not least — by the fact that there are hitherto not too many really interesting construction problems from which we definitively know the unsolvability by compass and ruler.

Theorem 1. Let be n > 2 a natural number and a_{1},…, a_{n} positive reals. An inscribed polygon (i.e. a convex polygon with vertices on a circle) with sides a_{1},…, a_{n} in this cyclic order exists if and only if the n inequalities

a_{1} < a_{2} + a_{3} + ⋯ + a_{n},
a_{2} < a_{3} + ⋯ + a_{n} + a_{1},  (1)
⋮
a_{n} < a_{1} + a_{2} + ⋯ + a_{n−1}

hold true.

The necessity of these inequalities (even for arbitrary closed polygons) is obvious. We shall prove the sufficiency by induction. For this proof we need

Lemma. If n > 4 and a_{1},…, a_{n} satisfy (1) then there is at least one i such that a_{1},…, a_{i−1}, a_{i} + a_{i+1}, a_{i+2},…, a_{n} (where the indices are to understand modulo n) satisfy the respective inequalities (1) for n − 1 cyclic ordered reals.

0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix}1993 Heldermann Verlag, Berlin

Figure 1

Remark. For n = 4 and a_{1} = a_{3} and a_{2} = a_{4} this is obviously not true, i.e. it is impossible to deform a rectangle in a triangle by stretching two neighbouring sides in one side. Our lemma says: For all n > 4 such a deformation of an n-sided polygon into an (n − 1)-sided polygon is always possible.

Proof: Let us assume

a_{1} + a_{2} ≥ a_{3} + a_{4} + ⋯ + a_{n},
a_{2} + a_{3} ≥ a_{4} + ⋯ + a_{n} + a_{1},
⋮
a_{n} + a_{1} ≥ a_{2} + a_{3} ⋯ + a_{n−1}.

By adding these inequalities we obtain

2(a_{1} + ⋯ + a_{n}) ≥ (n − 2)(a_{1} + ⋯ + a_{n}),

i.e, n ≤ 4. Therefore there exists at least one i such that

a_{i} + a_{i+1} < a_{1} + ⋯ + a_{i−1} + a_{i+2} + ⋯ + a_{n}.

From (1) follows that also

a_{1} < a_{2} + ⋯ + (a_{i} + a_{i+1}) + ⋯ + a_{n}

and so on.

Proof of Theorem 1: For n = 3 we obtain from (1) the wellknown existence of a triangle with sides a_{1}, a_{2}, a_{3} which of course has a circumcircle. For n = 4 Vieta has shown the constructibility by compass and ruler but not strongly the existence under the only conditions (1). This existence (as a symmetrical trapezium) is obvious in the special case a_{1} = a_{3} (or a_{2} = a_{4} ). Let be a_{1} ≠ a_{3} and a_{2} ≠ a_{4}, say a_{1} < a_{3} and a_{2} < a_{4}, then a_{1} + a_{2} < a_{3} + a_{4}, i.e. we can stretch the quadrangle (assumed as a linkage) to a triangle with sides a_{1} + a_{2}, a_{3}, a_{4} (Fig. 1). Herein we have (with the denotations of Fig. 1) α + γ < 180^{o}, 180^{o} < β + δ < 360^{o}.

If we deform this special quadrangle continuously by diminishing β we arrive at a quadrangle with α = 180^{o} or γ = 180^{o} (or both if a_{1} + a_{4} = a_{2} + a_{3}). In each case we have β + δ < 180^{o} because of α + β + γ + δ = 360^{o}. The continuity of the motion ensures that there is a value of β for which β + δ = α + γ = 180^{o} and this is sufficient for being an inscribed quadrangle.

This step from the existence of (always inscribed) triangles to that of inscribed quadrangles gives the hint how to conclude from the existence of inscribed (n − 1)-angles that of inscribed n-angles using our lemma proved above. Let be n > 4 and a_{1},…, a_{n} positive reals which satisfy (1). According to the lemma let be a_{1} +a_{2} < a_{3} +⋯+a_{n}. By induction we may assume that there exist inscribed (n − 1)-angles with sides x, a_{3},…, a_{n} (Fig. 2) for each x such that x < a_{1} + a_{2} and

a_{3} < x + a_{4} + ⋯ + a_{n},
a_{4} < x + a_{3} + a_{5} + ⋯ + a_{n},  (2)
⋮
a_{n} < x + a_{3} + ⋯ + a_{n−1},

i.e. x, a_{3},…, a_{n} satisfy the conditions (1). We have to show that by diminishing of β from its initial value 180^{o} and shortening the chord x from its initial value a_{1} + a_{2} we may get an inscribed polygon with sides a_{1}, a_{2},…, a_{n}.

Figure 2

Indeed, by diminishing β the vertex B(β) moves in direction of the exterior of the circumcircle k(β) (which continuously depends on β) until the polygon stops to be convex (or one of the conditions (2) becomes false). At this moment we have that a_{2} (or a_{1} or both, but let us assume a_{2} as in Fig. 2) is the straight line prolongation of its neighbouring side a_{3} and hence B(β) obviously is located outside of the circles k(β). Because of the continuity of this motion there exists a value of β for which B(β) lies exactly on k(β). It remains to show that it is impossible that one of the conditions (2), say the last one, becomes false before B(β) reaches k(β). If a_{n} is near to the sum x + a_{3} + ⋯ + a_{n−1} then we have a situation as shown in Fig. 3 and obviously the radius of k(β) goes to infinity, but the distance of the vertex B(β) from the chord x(β) is growing continuously.

Figure 3

Theorem 2. The inscribed quintangle is in general not constructible from its five sides by compass and ruler.

Proof: We show the nonconstructibility even in special cases if a_{1} = a_{3} = a, a_{2} = a_{4} = b, a_{5} = c (Fig. 4). Then three diagonals x_{1}, x_{2}, x_{3} have the same length x. According to tho wellknown Ptolemaean theorem on inscribed quadrangles we have (quadrangle ABCD) xy = ac + xb and hence

bxy = abc + b^{2}x,  (3)

(quadrangle ACDE) x^{2} = a^{2} + by and hence

bxy = x^{3} + a^{2}x.  (4)

From (3) and (4) we get

x^{3} − (a^{2} + b^{2})x − abc = 0.  (5)

If the inscribed quintangle ABCDE would be constructible from given a, b, c (which satisfy (1)) by compass and ruler then the diagonal x would be constructible and hence it would be a quadratic irrationality in a, b, c. Therefore it remains to show that in some cases (5) is not solvable by quadratic irrationals. We assume the reader to be familiar with the fact that we only must show that for special integer values of a, b, c (5) is not factorizable in integer coefficients. Indeed, for a = 3, b = 4, c = 5 we get from (5)

x^{3} − 25x − 60 = 0.

Setting x^{3} − 25x − 60 = (x^{2} + Ax + B)(x + D) leads to A + D = O, AD + B = −25, BD = −6O, i.e.

AB = 60 and A^{2} = B + 25.

This system is not solvable in integers: For A < 6 we have B > 1O, i.e A^{2} < 36 and B + 25 > 35. For A > 5 we have B < 11 and hence A^{2} > 35 and B + 25 < 36.

Figure 4  Figure 5

Theorem 3. For n > 5 the n-sided inscribed polygon is in general not constructible from its sides by compass and ruler.

Proof: Because of Theorem 2 it is sufficient to show that from a general solution for any n we could get a general solution for n − 1. Let us assume for the n-sided inscribed polygon a special co-ordinate system as given in Fig. 5.

If we assume the constructibility by compass and ruler in general then x_{1}, y_{1},…, x_{n−2}, y_{n−2} (and of course also x_{n}) are quadratic irrationalities R_{i} (i = 1,…, 2n − 4) depending on the variables a_{1},…, a_{n} and as such they are continuous functions of their n variables. On the other hand, the geometrical dependence of x_{i}, y_{i} (i = 1,…, n−2) on the variables a_{1},…, a_{n} is described by continuous functions f_{i} (i = 1,…, 2n − 4) of the variables a_{1},…, a_{n}. Because for a_{n} → 0 the n-sided inscribed polygon is converging to the (n − 1)- sided inscribed polygon with sides a_{1},…, a_{n−1} and for each i both functions R_{i} and f_{i} are identical for a_{n} ≠ 0 and both are continuous, the R_{i} must have the same limit values for a_{n} → 0 as the f_{i}, i.e. for a_{n} = 0 the quadratic irrationalities R_{i} describe the constructibility of the (n − 1)-sided inscribed polygon as depending on its sides a_{1},…, a_{n−1}.

Finally let us remark that a_{1} = a_{2} = ⋯ = a_{n} forces that the inscribed polygon with these sides is regular. Hence from the wellknown nonconstructibility of the regular heptagon by compass and ruler Theorem 3 would follow for n > 6. But this does not provide an answer to the problem for n = 5, 6.

References

[1] Schreiber, P.: Der gallische Apollonios. Zum 45O. Todestag von François Viète. Math. Schülerzeitschrift alpha 199O, Heft 6, 122-124.

[2] Francisci Vietae Opera omnia (ed. F. van Schooten). Lugduni Batavorum 1646.