Triangles in Regular Nonagons

Applied Mathematical Sciences, Vol. 7, 2013, no. 74, 3661 - 3665 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/10.12988/ams.2013.35264

Banyat Sroysang

Department of Mathematics and Statistics, Faculty of Science and Technology, Thammasat University, Pathumthani 12121 Thailand [email protected]

Copyright © 2013 Banyat Sroysang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

A cover for a family of arcs is a region containing a congruent copy of every arc in the family. In this paper, we seek the smallest regular 9-gon cover for the family of all triangles of perimeter two.

Mathematics Subject Classification: 52C15 Keywords: Cover, Nonagon, Enneagon, Triangle, Worm Problem

1 Introduction

The Moser’s worm problem [2] searches the region of smallest area which contains a congruent copy of every unit arc on the plane. A cover for a family of arcs is a convex region containing a congruent copy of every arc in the family. A popular shape of arcs is a triangle. The maximum and the minimum of the distance between the two parallel support lines of the set X are called the diameter of the set X and the thickness of the set X, respectively. Note that the diameter of any regular n-gon is the length of the diagonal for all n > 3. Moreover, the diameter and the thickness of any triangle are the length of the longest side and the length of the altitude to the longest side, respectively. Many smallest covers for the family of all triangles of perimeter two were posed (see [1,3-10]). In this paper, we seek the smallest regular 9-gon cover for the family of all triangles of perimeter two. 3662 Banyat Sroysang

2 Results

Theorem The smallest regular nonagon cover for the family of all triangles of perimeter two is the regular nonagon of diameter one (the area is approximately equal to 0.74562).

Proof. Let R be the regular nonagon of diameter one and let T be a triangle of perimeter two with the vertices A, B and C. WLOG, we can assume that the angle ∠A is greater than or equal to the angle ∠B, and the angle ∠B is greater than or equal to the angle ∠C. Let D be a vertex of the regular nonagon R and let P and Q be two points on the perimeter of the regular nonagon R such that the distance between the vertex D and the segment PQ is equal to 0.6 and the segment PQ is parallel to a diagonal of the regular nonagon R as shown in Fig. 1. We note that |PQ| < 0.8945. Now, we divide the triangle T into two cases. Case 1. The diameter of the triangle T is at most |PQ|. We note that the distance between the vertex D and the vertex P is greater than |PQ| as shown in Fig. 1. WLOG, we can put the triangle T into the regular nonagon R where the segment BC lies on the segment PQ, and C = P. This implies that the regular nonagon R contains a congruent copy of the triangle T.

Q P

Fig. 1. The segment PQ in the regular nonagon R

Case 2. The diameter of the triangle T is greater than |PQ|. We note on this case that the thickness of the triangle T is less than 0.325. WLOG, we can put the triangle T into the regular nonagon R where the segment BC lies on a diagonal of the regular nonagon R and the vertex A is above than the segment BC as shown in Fig. 2 or Fig. 3. Now, the vertex A may be in the regular nonagon R or not in the regular nonagon R.

Triangles in regular nonagons 3663

E A

F B C

Fig. 2. A triangle T in the regular nonagon R where the vertex A is in R

E A B F C

Fig. 3. A triangle T in the regular nonagon R where the vertex A is not in R

Suppose for a contradiction that the regular nonagon R does not contains the triangle T as shown in Fig. 3. Let G be the intersection of the segment AC and the segment EF and let H be the point on the segment BC such that the segment GH is perpendicular to the segment BC as shown in Fig. 4.

E A G F H B C

Fig. 4. A right triangle GHC in the triangle T and the regular nonagon R

3664 Banyat Sroysang

Let x be the length of the segment FH. Then 00.07≤ x < and the length of 4π the segment HC is 1− x . Note that the angle HFG is equal to . Then the 9 ⎛⎞4π length of the segment GH is x tan ⎜⎟. ⎝⎠9 2 ⎛⎞44ππ2 ⎛⎞ ⎛⎞ Define Lx( )=− 1 x + x tan⎜⎟ + (1 − x ) +⎜⎟ x tan ⎜⎟. Then Lx() is ⎝⎠99⎝⎠ ⎝⎠ the total length of the perimeter of the right triangle GHC. By the calculation on Lx(), we obtain that Lx ( )≥= L (0) 2 as shown in Fig. 5.

2.3

2.25

2.2

2.15

2.1

2.05

0.01 0.02 0.03 0.04 0.05 0.06 0.07

Fig. 5. The graph of Lx() where 00.07≤ x <

Since the total length of the perimeter of the triangle ABC is greater than the total length of the perimeter of the right triangle GHC, it follows that the total length of the perimeter of the triangle T is greater than two. This is a contradiction. Hence, the regular nonagon R is a cover for the family of all triangles of perimeter two. Next, we note that every cover for the family of all triangles of perimeter two must cover the line segment of length one. Hence, the diagonal of any regular nonagon cover for the family of all triangles of perimeter two must have length at least one. The diagonal of the regular nonagon R has length one. Therefore, the regular nonagon R is a smallest cover for the family of all triangles of perimeter two.

Acknowledgement. This research was supported by the Thammasat University Research Fund 2012, Thailand. Triangles in regular nonagons 3665

References

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Received: May 11, 2013