#### Abstract

We fully describe the envelope of all line segments that divide the perimeter of a triangle into the ratio as varies from 0 to . If is larger than the ratio of the longest side length to the perimeter, then the envelope is a 12-sided closed curve consisting of six line segments and six parabolic arcs. For other values of , the envelope is the union of one to three parabolic arcs and possibly a 5- or 9-sided nonclosed curve consisting of line segments and parabolic arcs.

#### 1. Introduction

Is there a point where every line passing through bisects the perimeter of a triangle? The answer is “*no*.” These lines are not concurrent. Instead, they “wrap” around a curve called the *envelope* of perimeter-bisecting lines. Berele and Catoiu [1] show that the envelope of perimeter-bisecting lines, called the *perimeter-bisecting deltoid*, is a 6-sided closed curve consisting of three line segments and three parabolic arcs (see Figure 1).

In this article, we extend Berele and Catoiu’s result to imbalanced proportions. Particularly, we classify and describe the envelope of all line segments that divide the perimeter of a triangle into the ratio , for . For example, there are two possibilities for an equilateral triangle. If , then the envelope of all line segments that divide the perimeter of a triangle into the ratio is the union of three parabolic arcs. On the other hand, if , then the envelope is a 12-sided closed curve consisting of six line segments and six parabolic arcs (see Figure 2).

To tackle this problem, first, we reduce the problem of finding the envelope for a triangle to a smaller problem of finding the envelope for each angle. Then, we combine the results from the three angles together. The construction of “*single angle envelopes*” is given in Section 3, and the groundwork for combining single angle envelopes is given in Section 4 and Section 5. A complete description of the envelope of all line segments that divide the perimeter of a triangle into the ratio is given in Section 6.

#### 2. Preliminaries

We recall basic facts about the envelope of a family of curves. A one-parameter family of curves is a collection of all curves on the plane, where is a parameter. In other words,where is an index set. In this article, we assume that the index set is an interval, and the function is sufficiently smooth. There are multiple definitions of the envelope of a family of curves. Here, we introduce two definitions that are equivalent in our interesting situations. The reader should refer to [2, 3] for facts and examples of the envelope of a one-parameter family of the curve.

*Definition 1. *The of the family is the set of points given byInformally, we “eliminate” between the equations and .

*Definition 2. *The of the family is a curve that touches each curve at some point. More precisely, is the set of points that belong to and where and have the same tangent line there. The range of values of may be smaller than the full interval where is defined.

We are only concerned with a family of straight lines in this article. Definition 1 and Definition 2 are equivalent in this case.

Proposition 1 (Proposition 5 in [2]). * Assume that , that is, all curves in are straight lines. If , then .*

#### 3. Single Angle Envelopes

Our goal is to describe the envelope of all line segments that divide the perimeter of a triangle into the ratio . These segments are generated from two endpoints, and , which move along the sides of a triangle while dividing the perimeter into a fixed ratio. The trip is split into three parts where separations occur when either or meets a vertex. As a result, our problem reduces to studying the envelope of all segments when and travel along each leg of a fixed angle such that the sum is constant. In this section, we show that such envelopes are parabolic arcs and provide some properties of the envelopes.

*Definition 3. *Let be a fixed angle that measures less than and be a fixed positive real number. We define the single angle envelope to be the envelope of all segments such that is on the ray , is on the ray , and . Since envelopes of congruent angles are congruent, we denote by the envelope of any angle of measure and sum .

The term and concept of single angle envelopes were developed by Berele and Catoiu [1] when they studied the envelope of perimeter-bisecting lines of a triangle. Definition 3 and the statements of Lemma 1 and Lemma 2 are adapted from [1]. For completeness, we include alternate proofs of both lemmas here.

The next lemma shows that a single-angle envelope is a parabolic arc.

Lemma 1. *Let be an angle of measure . In a coordinate system centered at such that the -axis is a symmetry axis of when the point is in the second quadrant and the point is in the first quadrant (see Figure 3), the single angle envelope is a parabolic arc given by the following equation:*

*Proof. *Let be a point on the ray and be a point on the ray such that . Let where . Then, . As vectors, and . Hence, the segment is part of the line given by the following equation:Let be the collection of all such lines. We have thatwhereThe system of equations holds whenBy substituting the above value of into (4), we have that the envelope of is given by the following equation:Since , we have . □

The following lemma provides a method for calculating the tangential points of a single angle envelope.

Lemma 2. *Let be an angle of measure and be a fixed positive real number. Let be a point on and be a point on with . If is tangent to the single angle envelope at , then*

*Proof. *Let and . Then, and = . We want to show that , which is equivalent toFrom (7) and (8), we can write the coordinates of in terms of asAlternatively, as a vector,(11) and (12) describe the same point. We must have thatBoth equations imply (10) as wanted.

Proposition 1 implies that the single angle envelope is tangent to the legs of the angle . The locations of the tangential points can be calculated by (3) at the endpoints. We get the following lemma.

Lemma 3. *Let be an angle of measure and be a fixed positive real number. The single angle envelope is tangent to the ray at a point and tangent to the ray at a point where .*

#### 4. -Splitters

From now on, we will assume that , and we will use the following notation for the envelope of all line segments that divide the perimeter of a triangle into the ratio .

*Notation 1. *We denote the envelope of all line segments that divide the perimeter of a triangle into the ratio by .

We will also use the following notation for the side lengths of a triangle.

*Notation 2. *For a triangle , we write , , and . We denote the perimeter of by .

Recall that our investigation on reduces to the study of single angle envelopes, which are parabolic arcs. For example, for a triangle , every line segment that is tangent to the single angle envelopes divides the perimeter of into the ratio . However, the endpoints of such line segments are restricted to the sides of . As a result, not every point of is a part of . It is cut by cevians that divide the perimeter of a triangle into the ratio .

*Definition 4. *We call a cevian that divides the perimeter of a triangle into the ratio an *-splitter*. We call the endpoint of an -splitter on the opposite side of a vertex an *-splitting point*.

There can be up to two -splitters at each vertex. Thus, a triangle has at most six -splitters. We will use the following notation for the six -splitting points.

*Notation 3. *For a triangle , we let and denote the -splitting points on the side such that and , respectively. It follows that and are -splitters. The notations for all -splitting points are summarized. Figure 4 shows the locations of -splitting points in a triangle.We note that when , coincides with , coincides with , and coincides with . In plane geometry, a -splitter is simply called a *splitter*. The three splitters concur at a point known as the Nagel point [4].

The following lemma gives necessary and sufficient conditions for the existence of -splitting points.

Lemma 3. *Let be a triangle. The following statements hold.*(1)*The -splitting points and exist if and only if *(2)*The -splitting points and exist if and only if *(3)*The -splitting points and exist if and only if *

*Proof. *We give the proof of (1) here. The proofs of (2) *and* (3) are similar.

Assume that the -splitting points and exist. It follows that and , where and . We then have . Hence, .

Assume that . Then, . Thus, there exists such that . From the triangle inequality, . This means . Thus, . In particular, there exists a point such that ; that is, . By a similar argument, there exists a point such that . The points and are -splitting points. □

The following corollary determines the number of -splitters based on the value of . It follows directly from the proof of Lemma 3.

Corollary 1. *Let be a triangle such that . Then, the following statements hold.*(1)*If , then has no -splitters*(2)*If , then has exactly two -splitters: and *(3)*If , then has exactly four -splitters: , , , and *(4)*If , then has exactly six -splitters: , , , , , and *

#### 5. Cutting Single Angle Envelopes with -Splitters

As discussed earlier, the problem of finding reduces to studying the single angle envelopes and . However, it is not necessary for all points of and to be . They are cut by *-splitters*. In this section, we find the locations of these cut points. The results in this chapter will serve as key lemmas for the classification of in Section 6.

##### 5.1. Cutting with -Splitters

Figure 5 illustrates the results of Lemma 4 to Lemma 7.

**(a)**

**(b)**

**(c)**

**(d)**

Lemma 4. *Let be a triangle. If the -splitters and do not exist, then and is tangent to the sides and at the points and where , respectively.*

*Proof. *From Lemma 1 and Lemma 3, is a parabolic arc which is tangent to the rays and at the points and where , respectively. Because the -splitters and do not exist, Lemma 4.5 implies that and . This means that the points and are on the sides and , respectively. Therefore, . □

Lemma 5. *Let be a triangle. If the -splitter exists but the -splitter does not exist, then is a parabolic arc with endpoints on and on the side such that*

*Proof. *From Lemma 1 and Lemma 3, is a parabolic arc that is tangent to the rays and at the points and where , respectively.

Since the -splitter exists, Lemma 3 implies that . Thus, the point is not on the side . Therefore, the single angle envelope is cut by the -splitter at a point . From Lemma 2, . Hence,Because the -splitter does not exist, Lemma 4.5 implies that . Hence, the point is on the side and . □

The proofs of Lemma 6 and Lemma 7 are similar to the proof of Lemma 5.

Lemma 6. *Let be a triangle. If the -splitter exists but the -splitter does not exist, then is a parabolic arc with endpoints on and on the side such that*

Lemma 7. *Let be a triangle. If the -splitters and exist, then is a parabolic arc with endpoints on and on such that*

##### 5.2. Cutting with -Splitters

Lemma 8. *Let be a triangle. If the -splitters and do not exist, then and are disjoint.*

*Proof. *Since the -splitters and do not exist, Lemma 3 implies that , that is, .

Recall that the single angle envelope is the envelope of all segment such that is on ray and is on ray such that . Suppose for the sake of contradiction that and . It follows that . This is a contradiction. Therefore, there is no such segment . In other words, and are disjoint. □

The proof of the following lemma is similar to the proof of Lemma 5. Figure 6 illustrates the result of this lemma.

Lemma 9. *Let be a triangle. If the -splitters and exist, then is a parabolic arc with endpoints on and on such that*

#### 6. Classifying the Envelopes

##### 6.1. Statement of the Main Theorem

The following theorem completely describes the envelope of all line segments that divide the perimeter of a triangle into the ratio . Figure 7 illustrates as varies from 0 to . For convenience, we will denote a parabolic arc with endpoints and by .

**(a)**

**(b)**

**(c)**

**(d)**

Theorem 1. *Let be a triangle such that . Then, the following statements hold.*(1)*If , then is the union of three parabolic arcs. In particular,* *The single angle envelope is tangent to the sides and at the points and , respectively, such that , when is a permutation of .*(2)*If , then is the union of and a 5-sided nonclosed curve whose sides alternate between two line segments and three parabolic arcs. In particular,* *where(a) the single angle envelope is tangent to the sides and at the points and , respectively, such that (b) and have endpoints and on the sides and , respectively, with , and(c)*(3)

*If , then is a 9-sided nonclosed curve whose sides alternate between four line segments and five parabolic arcs. In particular,*

*where(a)*(4)

*and have endpoints and on the side with*(b)*, and*(c)*and**If , then is a 12-sided closed curve whose sides alternate between six line segments and six parabolic arcs. In particular,*

*where(a)*

*(b)**, and**, and**,**, and**The locations of endpoints of sides of can be identified exactly. In all cases, the endpoints and are on the -splitter so thatwhen is a permutation of .*

We note that when , coincides with . As a result, six pairs of sides of coincide. In this case, we may consider as a 6-sided closed curve. Berele and Catoiu [1] call this curve the perimeter-bisecting deltoid.

##### 6.2. Proof of theorem 1

Theorem 1 can be proven by combining the results of appropriate lemmas in Section 5 based on the existence of -splitters, which is given in Corollary 1. For conciseness, we only give a detailed proof of Theorem 1 (2) here. Theorem 6.1 (1), (3), and (4) can be proven with similar methods.

Proof of (2).

Assume that . From Corollary 1, the -splitters and exist, but , , , and do not exist.

First, we consider the angle . Because the -splitters and exist, Lemma 4 implies that , and it is tangent to the sides and at the points and where , respectively. Lemma 9 implies that is a parabolic arc that is cut from by and . The endpoints are and such that

Next, we consider the angle . Since the -splitter exists, Lemma 6 implies that the intersection of and is a parabolic arc which is cut from by the -splitter and the side . The endpoints are and such thatBecause the -splitters and do not exist, Lemma 5.5 implies that and are disjoint.

Lastly, we consider the angle . Since the -splitter exists, Lemma 5.3 implies that the intersection of and is a parabolic arc that is cut from by the -splitter and the side . The endpoints are and such thatBecause the -splitters and do not exist, Lemma 5.5 implies that and are disjoint.

We have shown that contains parabolic arcs , , and , which are cut from appropriate single angle envelopes by -splitters. Since and are two points on the same -splitter, , which divides the perimeter of into the ratio , the segment is also contained in . Similarly, the segment is also a part of . As a result, we have thatTherefore,In other words, is the union of and a 5-sided nonclosed curve whose sides alternate between two line segments and three parabolic arcs. This completes the proof of Theorem 1 (2).

##### 6.3. Properties and Special Cases

In this section, we list some properties and special cases of .

Corollary 2. *If , then adjacent sides of either connect smoothly or make a cusp.*

*Proof. *This follows directly from the properties of the envelopes of families of curves (see Proposition 1).

Corollary 3. *Parabolic sides of have the angle bisectors as their axes of symmetry.*

*Proof. *The corollary follows directly from Lemma 1. □

The corollaries below are special cases of Theorem 1.

Corollary 4 (equilateral case). *Let be an equilateral triangle. Then, the following statements hold.*(1)*If , then is the union of three parabolic arcs*(2)*If , then is a 12-sided close curve whose sides alternate between six line segments and six parabolic arcs*

Corollary 5 (isosceles case). *Let be an isosceles triangle.*(1)*If , then the following statements hold(a) If , then is the union of three parabolic arcs(b)If , then is the union of and a 5-sided nonclosed curve whose sides alternate between two line segments and three parabolic arcs(c)If , then is a 12-sided closed curve whose sides alternate between six line segments and six parabolic arcs*(2)

*If , then the following statements hold(a)*

*If , then is the union of three parabolic arcs*(b)*If , then is a 9-sided nonclosed curve whose sides alternate between four line segments and five parabolic arcs*(c)*If , then is a 12-sided closed curve whose sides alternate between six line segments and six parabolic arcs*#### 7. Conclusion

The envelopes of area bisectors and perimeter bisectors of triangles, polygons, and convex regions have been topics of interest in plane geometry (e.g., see [1, 5–7]). These objects have potential applications in computational hydrodynamics and naval engineering [8]. The envelope of area bisectors of a triangle was generalized to imbalanced proportions by Middleton [9].

This work generalizes the envelope of perimeter bisectors. We fully describe the envelope of all line segments that divide the perimeter of a triangle into the ratio , for . This is the very first step in an exciting avenue that will lead to interesting questions about the properties and generalizations of these objects.

#### Data Availability

No data were used to support this study.

#### Conflicts of Interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.

#### Acknowledgments

Theerakarn is partially supported by Srinakharinwirot University.