Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 42 (2001), No. 2, 439-442.

Polygons with Hidden Vertices

Günter Ewald

Fakultät und Institut für Mathematik, Ruhr-Universität Bochum Universitätsstraße 150, D-44801 Bochum, Germany

Abstract. Given a point M in Euclidean 3-space we show that there exists a polygon without self-intersection and not containing M such that viewed from M each vertex of the polygon is hidden behind an edge of the polygon. As an application, we construct a toric 4-variety which has peculiar compactification properties.

Let P_{1}P_{2} ⋯ P_{m} be a polygon P without self-intersection in Euclidean space E^{n} (so P = [P_{1}, P_{2}] ∪ [P_{2}, P_{3}] ∪ ⋯ ∪ [P_{m−1}, P_{m}] where [P_{i}, P_{i+1}] is the line segment with end points P_{i}, P_{i+1}, i = 1, …, m − 1). Given a point M not on P we say that viewing from M the vertex P_{j} is hidden behind [P_{i},_{P i+1} ] if [M, P_{j}] ∩ [P_{i}, P_{i+1}] is relative interior to [P_{i}, P_{i+1}].

Problem 1. Given M, does there exist a polygon P = P_{1}P_{2} ⋯ P_{m} such that viewing from M each vertex P_{j} is hidden behind some edge [P_{i}, P_{i+1}] of P?

For n = 1 one can “see” from M either P_{1} or P_{m} , hence Problem 1 has no solution. For n = 2 we shall prove (Theorem 1) that also no solution exists. In Theorem 2 we present a solution for n = 3 and m = 14. It also provides an example for n > 3.

The following question remains open:

Problem 2. What is the minimal number m for which, in case n = 3, Problem 1 has a solution?

M must lie in the interior of conv P (convex hull), as one can see as follows: If this is not so, consider the point U of conv P next to M, and let H be a supporting hyperplane of conv P in U. If M = U we apply Theorem 1’ below to H ∩ conv P. If M ≠ U we choose H perpendicular to the line joining M and U. Now we can “see” from M the vertices of conv P lying in H ∩ conv P.

0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix}2001 Heldermann Verlag

As a consequence, P must have at least four vertices which are not coplanar. Since each of these vertices is hidden behind an edge, it is readily seen that four further vertices exist. So if m_{min} is the least possible m we obtain

8 ≤ m_{min} ≤ 14.

We conjecture that m_{min} is closer to 14 than to 8.

Theorem 1. Problem 1 has no solution for n = 2.

Proof. Suppose P = P_{1}P_{2} ⋯ P_{m} does solve Problem 1 for n = 2. Let κ be the largest circle with center M whose interior is not intersected by P. Since P is a closed set, κ contains a point P of P. If P were a vertex of P we could “see” this vertex from M. So P lies in the relative interior of an edge [P_{i}, P_{i+1}]. The edge behind which P_{i} is hidden, has one end point, P_{r} say, inside the triangle MP P_{i} (Figure 1).

Figure 1.

Among all vertices of P in this triangle there is one, P_{s} say, such that the line through M and P_{s} intersects [P_{i}, P] in a point U for which inside the triangle MUP there is no further vertex of P. There may be more such vertices on [M, P_{s}]; the one closest to M can be “seen” from M, a contradiction. The proof applies without change to the following more general statement:

Theorem 1’. Theorem 1 remains true if the polygon is replaced by a finite set of noncrossing line segments, their end points, and further points.

Theorem 2. Problem 1 has a solution for n = 3 and m = 14.

Proof. Let M lie in the origin of a Cartesian coordinate system. We present the coordinates of each P_{i} and express it as a positive linear combination of the end points of some edge of P (Figure 2):

\begin{matrix}P_{1} = & (0, 10, 10) = & \frac{5}{16}P_{10} & + & \frac{15}{16}P_{11} & P_{8} = & (0, 15, 15) = & \frac{15}{32}P_{10} & + & \frac{45}{32}P_{11} \\ P_{2} = & (−10, 0, −10) = & \frac{5}{16}P_{5} & + & \frac{15}{16}P_{6} & P_{9} = & (−12, −3, −9) = & \frac{6}{5}P_{2} & + & \frac{3}{10}P_{3} \\ P_{3} = & (0, −10, 10) = & \frac{15}{16}P_{10} & + & \frac{5}{16}P_{11} & P_{10} = & (0, −16, 8) = & \frac{4}{15}P_{13} & + & \frac{4}{3}P_{14} \end{matrix}

\begin{matrix}P_{4} = & (10, 0, −10) = & \frac{15}{16}P_{5} & + & \frac{5}{16}P_{6} & P_{11} = & (0, 16, 8) = & \frac{4}{3}P_{12} & + & \frac{4}{15}P_{13} \\ P_{5} = & (16, 0, −8) = & \frac{4}{3}P_{7} & + & \frac{4}{15}P_{8} & P_{12} = & (−3, 12, 9) = & \frac{6}{5}P_{1} & + & \frac{3}{10}P_{2} \\ P_{6} = & (−16, 0, −8) = & \frac{4}{15}P_{8} & + & \frac{4}{3}P_{9} & P_{13} = & (15, 0, −15) = & \frac{45}{32}P_{5} & + & \frac{15}{32}P_{6} \\ P_{7} = & (12, −3, −9) = & \frac{3}{10}P_{3} & + & \frac{6}{5}P_{4} & P_{14} = & (−3, −12, 9) = & \frac{3}{10}P_{2} & + & \frac{6}{5}P_{3} \end{matrix}

Figure 2.

Let Σ be a fan in R^{n} , that is, a system of finitely many rational convex polyhydral cones with apex 0 such that if a cone is in Σ also its faces lie in Σ, and such that the intersection of two cones is always a common face of the cones (conditions of a cell complex). To Σ an n-dimensional algebraic variety X_{Σ} is assigned, called a toric variety (see, for example, [1], [2], [3]). We remind in the following facts:

(a) X_{Σ} is complete (compact) if Σ covers all of R^{n}.

(b) If ρ is a k-dimensional cone of Σ, then to the star of ρ in Σ there corresponds an (n−k)- dimensional subvariety X_{Σ/ρ} of X_{Σ} which is again toric. If Σ/ρ is isomorphic to the fan {{0}} we obtain an algebraic torus (K\{0})^{n−k} if K is the field under consideration.

We say X_{Σ′} is a special partial completion of X_{Σ} if Σ is a subfan of Σ′ and if Σ, Σ′ have the same 1-cones. By the positive hull pos X of a set X we mean the set of all linear combinations of elements of X with nonnegative coefficients. |Σ| denotes the set of all points contained in some cone of Σ.

Theorem 3. There exists a 4-dimensional toric variety X_{Σ} with pos|Σ| = R^{4} and a 3-dimensional algebraic torus X_{Σ/ρ} (ρ a 1-cone) in it such that for each special partial completion X_{Σ′} of X_{Σ} we have X_{Σ′/ρ} = X_{Σ/ρ}.

Remark 1. As a consequence of Theorem 3 we find: X_{Σ} cannot be completed (compactified) without increasing the number of 1-cones in Σ. This, however, can also be shown by using simpler fans, for example one easily constructed from a polyhedron presented in [4].

Remark 2. Theorem 3 and Remark 1 have no analogues for three-dimensional toric varieties.

Proof of Theorem 3. We construct a fan Σ as follows. Let the polygon P lie in the hyperplane {x_{4} = 1} of R^{4}. We set P′_{i} := (P_{i}, 1), i = 1, …, 14, and M′ := (M, 1). Let pos{P′_{i}}, i = 1, …, 14, pos{M′}, and pos{−M′} be the 1-dimensional cones of Σ, and let pos[P′_{i}, P′_{i+1}], i = 1, …, 14, be the 2-dimensional cones of Σ. No 3- and 4-dimensional cones are assumed to belong to Σ.

If we look for a fan Σ′ which contains Σ and has the same 1-cones as Σ has, we see that a new cone cannot contain both, pos{M′} and pos{−M′} (it would not have 0 as apex), also not pos{M′} and pos{P′_{i}}, by construction of P. So choosing ρ = pos{M′} we obtain the theorem.

References

[1] Ewald, G.: Combinatorial Convexity and Algebraic Geometry. Graduate Texts in Math., Springer, New York, Berlin, Heidelberg 1996.

[2] Fulton, W.: Introduction to Toric Varieties. Ann. Math. Studies 131, Princeton University Press, Princeton 1993.

[3] Oda, T.: Convex Bodies and Algebraic Geometry. An Introduction to the Theory of Toric Varieties. Springer, Berlin, Heidelberg, New York 1988.

[4] Schönhardt, E.: Über die Zerlegung von Dreieckspolyedern in Tetraeder. Math. Ann. 98 (1928), 310–312.

Received March 5, 2000; revised version July 24, 2000