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
Triangles in Regular Nonagons
Banyat Sroysang
Department of Mathematics and Statistics,
Faculty of Science and Technology,
Thammasat University, Pathumthani 12121 Thailand
banyat@mathstat.sci.tu.ac.th
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.
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2
Banyat Sroysang
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.
D
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
E
3663
A
F
B
C
Fig. 2. A triangle T in the regular nonagon R where the vertex A is in R
A
F
E
B
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
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Banyat Sroysang
Let x be the length of the segment FH. Then 0 ≤ x < 0.07 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
⎛ 4π ⎞
⎛ 4π ⎞ ⎞
2 ⎛
Define L( x) = 1 − x + x tan ⎜
⎟ + (1 − x) + ⎜ x tan ⎜
⎟ ⎟ . Then L( x) is
⎝ 9 ⎠
⎝ 9 ⎠⎠
⎝
the total length of the perimeter of the right triangle GHC. By the calculation on
L( x) , we obtain that L( x) ≥ 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 L( x) where 0 ≤ x < 0.07
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
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Received: May 11, 2013