# Dividing the Perimeter of a Triangle into Unequal Proportions

Authors: Nawinda Amadtohed; Thitipon Chaidee; Phonthakorn Racha-in; Thunwa Theerakarn
Journal: International Journal of Mathematics and Mathematical Sciences
 (2022)

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## Abstract

We fully describe the envelope of all line segments that divide the perimeter of a triangle into the ratioα:1−α as α varies from 0 to 1/2. If α 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 α, 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.


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## Body

## 1. Introduction

Is there a pointP where every line passing through P 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).Figure 1
Every line tangent to the perimeter-bisecting deltoid bisects the perimeter of the triangle.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 ratioα:1−α, for 0<α≤1/2. For example, there are two possibilities for an equilateral triangle. If α≤1/3, then the envelope of all line segments that divide the perimeter of a triangle into the ratio α:1−α is the union of three parabolic arcs. On the other hand, if 1/3<α<1/2, then the envelope is a 12-sided closed curve consisting of six line segments and six parabolic arcs (see Figure 2).Figure 2
The envelope of all line segments that divide the perimeter of an equilateral triangle into the ratioα:1−α, for 1/3<α<1/2.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 α:1−α is given in Section 6.


## 2. Preliminaries

We recall basic facts about the envelope of a family of curves. A one-parameter family of curvesΓ is a collection of all curves Fx,y,t=0 on the xy plane, where t is a parameter. In other words,(1)Γ=Fx,y,t=0:t∈I,where I is an index set. In this article, we assume that the index set I is an interval, and the function F 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.
Theenvelope of the family Γ is the set E1 of points given by(2)E1=x,y:Fx,y,t=Ftx,y,t=0for somet∈I.
Informally, we “eliminate”t between the equations Fx,y,t=0 and Ftx,y,t=0.Definition 2.
Theenvelope of the family Γ is a curve E2 that touches each curve Fx,y,t=0 at some point. More precisely, E2 is the set of points xt,yt that belong to Fx,y,t=0 and where E2 and Fx,y,t=0 have the same tangent line there. The range of values of t may be smaller than the full interval where Γ is defined.
We are only concerned with a family of straight lines in this article. Definition1 and Definition 2 are equivalent in this case.Proposition 1 (Proposition 5 in [2]).
Assume thatFx,y,t=atx+bty+ct, that is, all curves in Γ are straight lines. If a′tbt−atb′t≠0, then E1=E2.


## 3. Single Angle Envelopes

Our goal is to describe the envelope of all line segments that divide the perimeter of a triangle into the ratioα:1−α. These segments are generated from two endpoints, M and N, 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 M or N meets a vertex. As a result, our problem reduces to studying the envelope of all segments MN¯ when M and N travel along each leg of a fixed angle ∠O such that the sum OM+ON is constant. In this section, we show that such envelopes are parabolic arcs and provide some properties of the envelopes.Definition 3.
Let∠AOC be a fixed angle that measures less than π and σ be a fixed positive real number. We define the single angle envelope E∠O,σ to be the envelope of all segments MN¯ such that M is on the ray OA⟶, N is on the ray OC⟶, and OM+ON=σ. Since envelopes of congruent angles are congruent, we denote by Eθ,σ the envelope of any angle of measure θ and sum σ.
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∠AOC be an angle of measure 2ϕ<π. In a coordinate system centered at O such that the y-axis is a symmetry axis of ∠AOC when the point A is in the second quadrant and the point C is in the first quadrant (see Figure 3), the single angle envelope E2ϕ,σ is a parabolic arc given by the following equation:(3)y=cosϕ2σsin2ϕx2+σcosϕ2,whenx∈−σsinϕ,σsinϕ.Figure 3
The single angle envelopeE2ϕ,σ.Proof.
LetM be a point on the ray OA⟶ and N be a point on the ray OC⟶ such that OM+ON=σ. Let OM=t where t∈0,σ. Then, ON=σ−t. As vectors, OM⟶=t−sinϕı⟶+cosϕj⟶ and ON⟶=σ−tsinϕı⟶+cosϕj⟶. Hence, the segment MN¯ is part of the line given by the following equation:(4)y=σ−2tcosϕσsinϕx+tsinϕ+tcosϕ.
LetΓ be the collection of all such lines. We have that(5)Γ=Fx,y,t=0:t∈0,σ,where(6)Fx,y,t=−2cosϕσ2t2−2xcosϕσ2sinϕt+xcosϕσsinϕ+2xcosϕσ−yσ.
The system of equationsFx,y,t=Ftx,y,t=0 holds when(7)t=σ2−x2sinϕ.
By substituting the above value oft into (4), we have that the envelope of Γ is given by the following equation:(8)y=σcosϕ2+cosϕ2σsin2ϕx2.
Sincet∈0,σ, we have x∈−σsinϕ,σsinϕ. □
The following lemma provides a method for calculating the tangential points of a single angle envelope.Lemma 2.
Let∠AOC be an angle of measure 2ϕ<π and σ be a fixed positive real number. Let M be a point on OA¯ and N be a point on OC¯ with OM+ON=σ. If MN¯ is tangent to the single angle envelope E∠AOC,σ at P, then(9)PMOM=PNON.Proof.
LetOM=t and λ=PM/MN. Then, ON=σ−t and PN/MN  =  1−λ. We want to show that PMOM=PNON, which is equivalent to(10)λt=1−λσ−t.
From (7) and (8), we can write the coordinates of P in terms of t as(11)P=σ−2tsinϕ,σ−2t+2t2σcosϕ.
Alternatively, as a vector,(12)OP⟶=1−λOM⟶+λON⟶=−t+λσsinϕı⟶+t−2λt+λσcosϕj⟶.
(11) and (12) describe the same point. We must have that(13)−t+λσsinϕ=σ−2tsinϕandσ−2t+2t2σcosϕ=t−2λt+λσcosϕ.
Both equations imply (10) as wanted.
Proposition1 implies that the single angle envelope E∠AOC,σ is tangent to the legs of the angle ∠AOC. The locations of the tangential points can be calculated by (3) at the endpoints. We get the following lemma.Lemma 3.
Let∠AOC be an angle of measure 2ϕ<π and σ be a fixed positive real number. The single angle envelope E∠AOC,σ is tangent to the ray OA⟶ at a point U and tangent to the ray OC⟶ at a point V where OU=OV=σ.


## 4.α-Splitters

From now on, we will assume that0<α≤1/2, and we will use the following notation for the envelope of all line segments that divide the perimeter of a triangle into the ratio α:1−α.Notation 1.
We denote the envelope of all line segments that divide the perimeter of a triangle into the ratioα:1−α by Dα.
We will also use the following notation for the side lengths of a triangle.Notation 2.
For a triangleΔABC, we write a=BC, b=CA, and c=AB. We denote the perimeter of ΔABC by p=a+b+c.
Recall that our investigation onDα reduces to the study of single angle envelopes, which are parabolic arcs. For example, for a triangle ΔABC, every line segment that is tangent to the single angle envelopes E∠A,αp divides the perimeter of ΔABC into the ratio α:1−α. However, the endpoints of such line segments are restricted to the sides of ΔABC. As a result, not every point of E∠A,αp is a part of Dα. It is cut by cevians that divide the perimeter of a triangle into the ratio α:1−α.Definition 4.
We call a cevian that divides the perimeter of a triangle into the ratioα:1−α an α-splitter. We call the endpoint of an α-splitter on the opposite side of a vertex an α-splitting point.
There can be up to twoα-splitters at each vertex. Thus, a triangle has at most six α-splitters. We will use the following notation for the six α-splitting points.Notation 3.
For a triangleΔABC, we let AB′ and AC′ denote the α-splitting points on the side BC¯ such that BA+BAB′=αp and CA+CAC′=αp, respectively. It follows that AAB′¯ and AAC′¯ are α-splitters. The notations for all α-splitting points are summarized. Figure 4 shows the locations of α-splitting points in a triangle.(14)AB′∈BC¯withBA+BAB′=αp;AC′∈BC¯withCA+CAC′=αp;BA′∈CA¯withAB+ABA′=αp;BC′∈CA¯withCB+CBC′=αp;CA′∈AB¯withAC+ACA′=αp;CB′∈AB¯withBC+BCB′=αp.
We note that whenα=1/2, AB′ coincides with AC′, BC′ coincides with BA′, and CA′ coincides with CB′. In plane geometry, a 1/2-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 ofα-splitting points.Figure 4
α-splitting points.Lemma 3.
LetΔABC be a triangle. The following statements hold.(1)
Theα-splitting points BC′ and CB′ exist if and only if α>a/p(2)
Theα-splitting points CA′ and AC′ exist if and only if α>b/p(3)
Theα-splitting points AB′ and BA′ exist if and only if α>c/pProof.
We give the proof of (1) here. The proofs of (2) and (3) are similar.
⇒ Assume that the α-splitting points BC′ and CB′ exist. It follows that BC+CBC′=αp and BC+BCB′=αp, where BCB′>0 and CBC′>0. We then have αp>BC=a. Hence, α>a/p.
⇐ Assume that α>a/p. Then, αp>a=BC. Thus, there exists x>0 such that x+BC=αp. From the triangle inequality, α≤1/2<c+a/p. This means x+BC=αp<AB+BC. Thus, x<AB. In particular, there exists a point CB′∈AB¯ such that BCB′=x; that is, BC+BCB′=αp. By a similar argument, there exists a point BC′∈AC¯ such that BC+CBC′=αp. The points CB′ and BC′ are α-splitting points. □
The following corollary determines the number ofα-splitters based on the value of α. It follows directly from the proof of Lemma 3.Corollary 1.
LetΔABC be a triangle such that a≤b≤c. Then, the following statements hold.(1)
Ifα≤a/p, then ΔABC has no α-splitters(2)
Ifa/p<α≤b/p, then ΔABC has exactly two α-splitters: BBC′¯ and CCB′¯(3)
Ifb/p<α≤c/p, then ΔABC has exactly four α-splitters: BBC′¯, CCB′¯, CCA′¯, and AAC′¯(4)
Ifc/p<α<1/2, then ΔABC has exactly six α-splitters: BBC′¯, CCB′¯, CCA′¯, AAC′¯, AAB′¯, and BBA′¯


## 5. Cutting Single Angle Envelopes withα-Splitters

As discussed earlier, the problem of findingDα reduces to studying the single angle envelopes E∠A,αp and E∠A,1−αp. However, it is not necessary for all points of E∠A,αp and E∠A,1−αp to be Dα. They are cut by α-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 Dα in Section 6.


### 5.1. CuttingE∠A,αp with α-Splitters

Figure5 illustrates the results of Lemma 4 to Lemma 7.Figure 5
E∠A,αp∩Dα in (a) Lemma 5.1, (b) Lemma 5.2, (c) Lemma 5.3, and (d) Lemma 5.4.
(a)(b)(c)(d)Lemma 4.
LetΔABC be a triangle. If the α-splitters BBA′¯ and CCA′¯ do not exist, then E∠A,αp⊂Dα and E∠A,αp is tangent to the sides AB¯ and AC¯ at the points U and V where AU=AV=αp, respectively.Proof.
From Lemma1 and Lemma 3, E∠A,αp is a parabolic arc which is tangent to the rays AB⟶ and AC⟶ at the points U and V where AU=AV=αp, respectively. Because the α-splitters BBA′¯ and CCA′¯ do not exist, Lemma 4.5 implies that αp<AB and αp<AC. This means that the points U and V are on the sides AB¯ and AC¯, respectively. Therefore, E∠A,αp⊂Dα. □Lemma 5.
LetΔABC be a triangle. If the α-splitter BBA′¯ exists but the α-splitter CCA′¯ does not exist, then E∠A,αp∩Dα is a parabolic arc with endpoints BAα on BBA′¯ and V on the side AC¯ such that(15)BAαBBAαBA′=ABA′AB=αp−cc,AV=αp.Proof.
From Lemma1 and Lemma 3, E∠A,αp is a parabolic arc that is tangent to the rays AB⟶ and AC⟶ at the points U and V where AU=AV=αp, respectively.
Since theα-splitter BBA′¯ exists, Lemma 3 implies that AB<αp. Thus, the point U is not on the side AB¯. Therefore, the single angle envelope E∠A,αp is cut by the α-splitter BBA′¯ at a point BAα. From Lemma 2, BAαBAB=BAαBA′ABA′. Hence,(16)BAαBBAαBA′=ABA′AB=αp−cc.
Because theα-splitter CCA′¯ does not exist, Lemma 4.5 implies that αp<AC. Hence, the point V is on the side AC¯ and AV=αp. □
The proofs of Lemma6 and Lemma 7 are similar to the proof of Lemma 5.Lemma 6.
LetΔABC be a triangle. If the α-splitter CCA′¯ exists but the α-splitter BBA′¯ does not exist, then E∠A,αp∩Dα is a parabolic arc with endpoints CAα on CCA′¯ and U on the side AB¯ such that(17)AU=αp,CAαCCAαCA′=ACA′AC=αp−bb.Lemma 7.
LetΔABC be a triangle. If the α-splitters BBA′¯ and CCA′¯ exist, then E∠A,αp∩Dα is a parabolic arc with endpoints BAα on BBA′¯ and CAα on CCA′¯ such that(18)BAαBBAαBA′=ABA′AB=αp−cc,CAαCCAαCA′=ACA′AC=αp−bb.


### 5.2. CuttingE∠A,1−αp with α-Splitters

Lemma 8.
LetΔABC be a triangle. If the α-splitters BBC′¯ and CCB′¯ do not exist, then E∠A,1−αp and Dα are disjoint.Proof.
Since theα-splitters BBC′¯ and CCB′¯ do not exist, Lemma 3 implies that αp<BC, that is, 1−αp>b+c.
Recall that the single angle envelopeE∠A,1−αp is the envelope of all segment MN¯ such that M is on ray AB⟶ and N is on ray AC⟶ such that AM+AN=1−αp. Suppose for the sake of contradiction that M∈AB¯ and N∈AC¯. It follows that 1−αp=AM+AN≤b+c. This is a contradiction. Therefore, there is no such segment MN¯. In other words, E∠A,1−αp and Dα are disjoint. □
The proof of the following lemma is similar to the proof of Lemma5. Figure 6 illustrates the result of this lemma.Figure 6
E∠A,1−αp∩Dα in Lemma 5.6.Lemma 9.
LetΔABC be a triangle. If the α-splitters BBC′¯ and CCB′¯ exist, then E∠A,1−αp∩Dα is a parabolic arc with endpoints BC1−α on BBC′¯ and CB1−α on CCB′¯ such that(19)BC1−αBBC1−αBC′=ABC′AB=1−αp−cc,CB1−αCCB1−αCB′=ACB′AC=1−αp−bb.


## 5.1. CuttingE∠A,αp with α-Splitters

Figure5 illustrates the results of Lemma 4 to Lemma 7.Figure 5
E∠A,αp∩Dα in (a) Lemma 5.1, (b) Lemma 5.2, (c) Lemma 5.3, and (d) Lemma 5.4.
(a)(b)(c)(d)Lemma 4.
LetΔABC be a triangle. If the α-splitters BBA′¯ and CCA′¯ do not exist, then E∠A,αp⊂Dα and E∠A,αp is tangent to the sides AB¯ and AC¯ at the points U and V where AU=AV=αp, respectively.Proof.
From Lemma1 and Lemma 3, E∠A,αp is a parabolic arc which is tangent to the rays AB⟶ and AC⟶ at the points U and V where AU=AV=αp, respectively. Because the α-splitters BBA′¯ and CCA′¯ do not exist, Lemma 4.5 implies that αp<AB and αp<AC. This means that the points U and V are on the sides AB¯ and AC¯, respectively. Therefore, E∠A,αp⊂Dα. □Lemma 5.
LetΔABC be a triangle. If the α-splitter BBA′¯ exists but the α-splitter CCA′¯ does not exist, then E∠A,αp∩Dα is a parabolic arc with endpoints BAα on BBA′¯ and V on the side AC¯ such that(15)BAαBBAαBA′=ABA′AB=αp−cc,AV=αp.Proof.
From Lemma1 and Lemma 3, E∠A,αp is a parabolic arc that is tangent to the rays AB⟶ and AC⟶ at the points U and V where AU=AV=αp, respectively.
Since theα-splitter BBA′¯ exists, Lemma 3 implies that AB<αp. Thus, the point U is not on the side AB¯. Therefore, the single angle envelope E∠A,αp is cut by the α-splitter BBA′¯ at a point BAα. From Lemma 2, BAαBAB=BAαBA′ABA′. Hence,(16)BAαBBAαBA′=ABA′AB=αp−cc.
Because theα-splitter CCA′¯ does not exist, Lemma 4.5 implies that αp<AC. Hence, the point V is on the side AC¯ and AV=αp. □
The proofs of Lemma6 and Lemma 7 are similar to the proof of Lemma 5.Lemma 6.
LetΔABC be a triangle. If the α-splitter CCA′¯ exists but the α-splitter BBA′¯ does not exist, then E∠A,αp∩Dα is a parabolic arc with endpoints CAα on CCA′¯ and U on the side AB¯ such that(17)AU=αp,CAαCCAαCA′=ACA′AC=αp−bb.Lemma 7.
LetΔABC be a triangle. If the α-splitters BBA′¯ and CCA′¯ exist, then E∠A,αp∩Dα is a parabolic arc with endpoints BAα on BBA′¯ and CAα on CCA′¯ such that(18)BAαBBAαBA′=ABA′AB=αp−cc,CAαCCAαCA′=ACA′AC=αp−bb.


## 5.2. CuttingE∠A,1−αp with α-Splitters

Lemma 8.
LetΔABC be a triangle. If the α-splitters BBC′¯ and CCB′¯ do not exist, then E∠A,1−αp and Dα are disjoint.Proof.
Since theα-splitters BBC′¯ and CCB′¯ do not exist, Lemma 3 implies that αp<BC, that is, 1−αp>b+c.
Recall that the single angle envelopeE∠A,1−αp is the envelope of all segment MN¯ such that M is on ray AB⟶ and N is on ray AC⟶ such that AM+AN=1−αp. Suppose for the sake of contradiction that M∈AB¯ and N∈AC¯. It follows that 1−αp=AM+AN≤b+c. This is a contradiction. Therefore, there is no such segment MN¯. In other words, E∠A,1−αp and Dα are disjoint. □
The proof of the following lemma is similar to the proof of Lemma5. Figure 6 illustrates the result of this lemma.Figure 6
E∠A,1−αp∩Dα in Lemma 5.6.Lemma 9.
LetΔABC be a triangle. If the α-splitters BBC′¯ and CCB′¯ exist, then E∠A,1−αp∩Dα is a parabolic arc with endpoints BC1−α on BBC′¯ and CB1−α on CCB′¯ such that(19)BC1−αBBC1−αBC′=ABC′AB=1−αp−cc,CB1−αCCB1−αCB′=ACB′AC=1−αp−bb.


## 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 ratioα:1−α. Figure 7 illustrates Dα as α varies from 0 to 1/2. For convenience, we will denote a parabolic arc with endpoints P and Q by PQ^.Figure 7
Dα for a triangle ΔABC with a≤b≤c when (a) α≤a/p, (b) a/p<α≤b/p, (c) b/p<α≤c/p, and (d) c/p<α<1/2.
(a)(b)(c)(d)Theorem 1.
LetΔABC be a triangle such that a≤b≤c. Then, the following statements hold.(1)
Ifα≤a/p, then Dα is the union of three parabolic arcs. In particular,(20)Dα=E∠A,αp∪E∠B,αp∪E∠C,αp.The single angle envelopeE∠X,αp is tangent to the sides XY¯ and XZ¯ at the points UX and VX, respectively, such that XUX=XVX=αp, when X,Y,Z is a permutation of A,B,C.(2)
Ifa/p<α≤b/p, then Dα is the union of E∠A,αp and a 5-sided nonclosed curve whose sides alternate between two line segments and three parabolic arcs. In particular,(21)Dα=VBCBα^∪CBαCB1−α¯∪CB1−αBC1−α^∪BC1−αBCα¯∪BCαUC^∪E∠A,αp,where(a)
the single angle envelopeE∠A,αp is tangent to the sides AB¯ and AC¯ at the points UA and VA, respectively, such that AUA=AVA=αp(b)
VBCBα^⊂E∠B,αp and BCαUC^⊂E∠C,αp have endpoints VB and UC on the sides AB¯ and AC¯, respectively, with BVB=CUC=αp, and(c)
CB1−αBC1−α^⊂E∠A,1−αp(3)
Ifb/p<α≤c/p, then Dα is a 9-sided nonclosed curve whose sides alternate between four line segments and five parabolic arcs. In particular,(22)Dα=UACAα^∪CAαCA1−α¯∪CA1−αAC1−α^∪AC1−αACα¯∪ACαBCα^∪BCαBC1−α¯∪BC1−αCB1−α^∪CB1−αCBα¯∪CBαVB^,where(a)
UACAα^⊂E∠A,αp and CBαVB^⊂E∠B,αp have endpoints UA and VB on the side AB¯ with AUA=BVB=αp(b)
ACαBCα^⊂E∠C,αp, and(c)
BC1−αCB1−α^⊂E∠A,1−αp and CA1−αAC1−α^⊂E∠B,1−αp(4)
Ifc/p<α<1/2, then Dα is a 12-sided closed curve whose sides alternate between six line segments and six parabolic arcs. In particular,(23)Dα=BAαCAα^∪CAαCA1−α¯∪CA1−αAC1−α^∪AC1−αACα¯∪ACαBCα^∪BCαBC1−α¯∪BC1−αCB1−α^∪CB1−αCBα¯∪CBαABα^∪ABαAB1−α¯∪AB1−αBA1−α^∪BA1−αBAα¯,where(a)
BAαCAα^⊂E∠A,αp,ABαCBα^⊂E∠B,αp, andBCαACα^⊂E∠C,αp, and(b)
BC1−αCB1−α^⊂E∠A,1−αp,CA1−αAC1−α^⊂E∠B,1−αp, andAB1−αBA1−α^⊂E∠C,1−αp
The locations of endpoints of sides ofDα can be identified exactly. In all cases, the endpoints YZα and YZ1−α are on the α-splitter YYZ′¯ so that(24)YZαYYZαYZ′=αp−YZYZ,YZ1−αYYZ1−αYZ′=1−αp−XYXY,when X,Y,Z is a permutation of A,B,C.We note that whenα=1/2, E∠X,αp coincides with E∠X,1−αp. As a result, six pairs of sides of Dα coincide. In this case, we may consider Dα as a 6-sided closed curve. Berele and Catoiu [1] call this curve the perimeter-bisecting deltoid.


### 6.2. Proof of theorem1

Theorem1 can be proven by combining the results of appropriate lemmas in Section 5 based on the existence of α-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 thata/p<α≤b/p. From Corollary 1, the α-splitters BBC′¯ and CCB′¯ exist, but AAB′¯, AAC′¯, BBA′¯, and CCA′¯ do not exist.First, we consider the angle∠A. Because the α-splitters BBC′¯ and CCB′¯ exist, Lemma 4 implies that E∠A,αp⊂Dα, and it is tangent to the sides AB¯ and AC¯ at the points UA and VA where AUA=AVA=αp, respectively. Lemma 9 implies that E∠A,1−αp∩Dα is a parabolic arc that is cut from E∠A,1−αp by BBC′¯ and CCB′¯. The endpoints are BC1−α∈BBC′¯ and CB1−α∈CCB′¯ such that(25)BC1−αBBC1−αBC′=ABC′AB=1−αp−ABAB,CB1−αCCB1−αCB′=ACB′AC=1−αp−ACAC.Next, we consider the angle∠B. Since the α-splitter CCB′¯ exists, Lemma 6 implies that the intersection of E∠B,αp and Dα is a parabolic arc which is cut from E∠B,αp by the α-splitter CCB′¯ and the side AB¯. The endpoints are CBα∈CCB′¯ and VB∈AB¯ such that(26)CBαCCBαCB′=BCB′BC=αp−BCBC,BVB=αp.Because the α-splitters AAC′¯ and CCA′¯ do not exist, Lemma 5.5 implies that E∠B,1−αp and Dα are disjoint.Lastly, we consider the angle∠C. Since the α-splitter BBC′¯ exists, Lemma 5.3 implies that the intersection of E∠C,αp and Dα is a parabolic arc that is cut from E∠C,αp by the α-splitter BBC′¯ and the side AC¯. The endpoints are BCα∈BBC′¯ and UC∈AC¯ such that(27)BCαBBCαBC′=CBC′CB=αp−CBCB,CUC=αp.Because the α-splitters AAB′¯ and BBA′¯ do not exist, Lemma 5.5 implies that E∠C,1−αp and Dα are disjoint.We have shown thatDα contains parabolic arcs VBCBα^, CB1−αBC1−α^, and BCαUC^, which are cut from appropriate single angle envelopes by α-splitters. Since CBα and CB1−α are two points on the same α-splitter, CCB′¯, which divides the perimeter of ΔABC into the ratio α:1−α, the segment CBαCB1−α¯ is also contained in Dα. Similarly, the segment BC1−αBCα¯ is also a part of Dα. As a result, we have that(28)VBCBα^∪CBαCB1−α¯∪CB1−αBC1−α^∪BC1−αBCα¯∪BCαUC^⊂Dα.Therefore,(29)Dα=VBCBα^∪CBαCB1−α¯∪CB1−αBC1−α^∪BC1−αBCα¯∪BCαUC^∪E∠A,αp.In other words, Dα is the union of E∠A,αp 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 ofDα.Corollary 2.
Ifα>a/p, then adjacent sides of Dα either connect smoothly or make a cusp.Proof.
This follows directly from the properties of the envelopes of families of curves (see Proposition1).Corollary 3.
Parabolic sides ofDα have the angle bisectors as their axes of symmetry.Proof.
The corollary follows directly from Lemma1. □
The corollaries below are special cases of Theorem1.Corollary 4 (equilateral case).
LetΔABC be an equilateral triangle. Then, the following statements hold.(1)
Ifα≤1/3, then Dα is the union of three parabolic arcs(2)
If1/3<α<1/2, then Dα is a 12-sided close curve whose sides alternate between six line segments and six parabolic arcsCorollary 5 (isosceles case).
LetΔABC be an isosceles triangle.(1)
Ifa<b=c, then the following statements hold(a)
Ifα≤a/p, then Dα is the union of three parabolic arcs(b)
Ifa/p<α≤b/p, then Dα is the union of E∠A,αp and a 5-sided nonclosed curve whose sides alternate between two line segments and three parabolic arcs(c)
Ifb/p<α<1/2, then Dα is a 12-sided closed curve whose sides alternate between six line segments and six parabolic arcs(2)
Ifa=b<c, then the following statements hold(a)
Ifα≤b/p, then Dα is the union of three parabolic arcs(b)
Ifb/p<α≤c/p, then Dα is a 9-sided nonclosed curve whose sides alternate between four line segments and five parabolic arcs(c)
Ifc/p<α<1/2, then Dα is a 12-sided closed curve whose sides alternate between six line segments and six parabolic arcs


## 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 ratioα:1−α. Figure 7 illustrates Dα as α varies from 0 to 1/2. For convenience, we will denote a parabolic arc with endpoints P and Q by PQ^.Figure 7
Dα for a triangle ΔABC with a≤b≤c when (a) α≤a/p, (b) a/p<α≤b/p, (c) b/p<α≤c/p, and (d) c/p<α<1/2.
(a)(b)(c)(d)Theorem 1.
LetΔABC be a triangle such that a≤b≤c. Then, the following statements hold.(1)
Ifα≤a/p, then Dα is the union of three parabolic arcs. In particular,(20)Dα=E∠A,αp∪E∠B,αp∪E∠C,αp.The single angle envelopeE∠X,αp is tangent to the sides XY¯ and XZ¯ at the points UX and VX, respectively, such that XUX=XVX=αp, when X,Y,Z is a permutation of A,B,C.(2)
Ifa/p<α≤b/p, then Dα is the union of E∠A,αp and a 5-sided nonclosed curve whose sides alternate between two line segments and three parabolic arcs. In particular,(21)Dα=VBCBα^∪CBαCB1−α¯∪CB1−αBC1−α^∪BC1−αBCα¯∪BCαUC^∪E∠A,αp,where(a)
the single angle envelopeE∠A,αp is tangent to the sides AB¯ and AC¯ at the points UA and VA, respectively, such that AUA=AVA=αp(b)
VBCBα^⊂E∠B,αp and BCαUC^⊂E∠C,αp have endpoints VB and UC on the sides AB¯ and AC¯, respectively, with BVB=CUC=αp, and(c)
CB1−αBC1−α^⊂E∠A,1−αp(3)
Ifb/p<α≤c/p, then Dα is a 9-sided nonclosed curve whose sides alternate between four line segments and five parabolic arcs. In particular,(22)Dα=UACAα^∪CAαCA1−α¯∪CA1−αAC1−α^∪AC1−αACα¯∪ACαBCα^∪BCαBC1−α¯∪BC1−αCB1−α^∪CB1−αCBα¯∪CBαVB^,where(a)
UACAα^⊂E∠A,αp and CBαVB^⊂E∠B,αp have endpoints UA and VB on the side AB¯ with AUA=BVB=αp(b)
ACαBCα^⊂E∠C,αp, and(c)
BC1−αCB1−α^⊂E∠A,1−αp and CA1−αAC1−α^⊂E∠B,1−αp(4)
Ifc/p<α<1/2, then Dα is a 12-sided closed curve whose sides alternate between six line segments and six parabolic arcs. In particular,(23)Dα=BAαCAα^∪CAαCA1−α¯∪CA1−αAC1−α^∪AC1−αACα¯∪ACαBCα^∪BCαBC1−α¯∪BC1−αCB1−α^∪CB1−αCBα¯∪CBαABα^∪ABαAB1−α¯∪AB1−αBA1−α^∪BA1−αBAα¯,where(a)
BAαCAα^⊂E∠A,αp,ABαCBα^⊂E∠B,αp, andBCαACα^⊂E∠C,αp, and(b)
BC1−αCB1−α^⊂E∠A,1−αp,CA1−αAC1−α^⊂E∠B,1−αp, andAB1−αBA1−α^⊂E∠C,1−αp
The locations of endpoints of sides ofDα can be identified exactly. In all cases, the endpoints YZα and YZ1−α are on the α-splitter YYZ′¯ so that(24)YZαYYZαYZ′=αp−YZYZ,YZ1−αYYZ1−αYZ′=1−αp−XYXY,when X,Y,Z is a permutation of A,B,C.We note that whenα=1/2, E∠X,αp coincides with E∠X,1−αp. As a result, six pairs of sides of Dα coincide. In this case, we may consider Dα as a 6-sided closed curve. Berele and Catoiu [1] call this curve the perimeter-bisecting deltoid.


## 6.2. Proof of theorem1

Theorem1 can be proven by combining the results of appropriate lemmas in Section 5 based on the existence of α-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 thata/p<α≤b/p. From Corollary 1, the α-splitters BBC′¯ and CCB′¯ exist, but AAB′¯, AAC′¯, BBA′¯, and CCA′¯ do not exist.First, we consider the angle∠A. Because the α-splitters BBC′¯ and CCB′¯ exist, Lemma 4 implies that E∠A,αp⊂Dα, and it is tangent to the sides AB¯ and AC¯ at the points UA and VA where AUA=AVA=αp, respectively. Lemma 9 implies that E∠A,1−αp∩Dα is a parabolic arc that is cut from E∠A,1−αp by BBC′¯ and CCB′¯. The endpoints are BC1−α∈BBC′¯ and CB1−α∈CCB′¯ such that(25)BC1−αBBC1−αBC′=ABC′AB=1−αp−ABAB,CB1−αCCB1−αCB′=ACB′AC=1−αp−ACAC.Next, we consider the angle∠B. Since the α-splitter CCB′¯ exists, Lemma 6 implies that the intersection of E∠B,αp and Dα is a parabolic arc which is cut from E∠B,αp by the α-splitter CCB′¯ and the side AB¯. The endpoints are CBα∈CCB′¯ and VB∈AB¯ such that(26)CBαCCBαCB′=BCB′BC=αp−BCBC,BVB=αp.Because the α-splitters AAC′¯ and CCA′¯ do not exist, Lemma 5.5 implies that E∠B,1−αp and Dα are disjoint.Lastly, we consider the angle∠C. Since the α-splitter BBC′¯ exists, Lemma 5.3 implies that the intersection of E∠C,αp and Dα is a parabolic arc that is cut from E∠C,αp by the α-splitter BBC′¯ and the side AC¯. The endpoints are BCα∈BBC′¯ and UC∈AC¯ such that(27)BCαBBCαBC′=CBC′CB=αp−CBCB,CUC=αp.Because the α-splitters AAB′¯ and BBA′¯ do not exist, Lemma 5.5 implies that E∠C,1−αp and Dα are disjoint.We have shown thatDα contains parabolic arcs VBCBα^, CB1−αBC1−α^, and BCαUC^, which are cut from appropriate single angle envelopes by α-splitters. Since CBα and CB1−α are two points on the same α-splitter, CCB′¯, which divides the perimeter of ΔABC into the ratio α:1−α, the segment CBαCB1−α¯ is also contained in Dα. Similarly, the segment BC1−αBCα¯ is also a part of Dα. As a result, we have that(28)VBCBα^∪CBαCB1−α¯∪CB1−αBC1−α^∪BC1−αBCα¯∪BCαUC^⊂Dα.Therefore,(29)Dα=VBCBα^∪CBαCB1−α¯∪CB1−αBC1−α^∪BC1−αBCα¯∪BCαUC^∪E∠A,αp.In other words, Dα is the union of E∠A,αp 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 ofDα.Corollary 2.
Ifα>a/p, then adjacent sides of Dα either connect smoothly or make a cusp.Proof.
This follows directly from the properties of the envelopes of families of curves (see Proposition1).Corollary 3.
Parabolic sides ofDα have the angle bisectors as their axes of symmetry.Proof.
The corollary follows directly from Lemma1. □
The corollaries below are special cases of Theorem1.Corollary 4 (equilateral case).
LetΔABC be an equilateral triangle. Then, the following statements hold.(1)
Ifα≤1/3, then Dα is the union of three parabolic arcs(2)
If1/3<α<1/2, then Dα is a 12-sided close curve whose sides alternate between six line segments and six parabolic arcsCorollary 5 (isosceles case).
LetΔABC be an isosceles triangle.(1)
Ifa<b=c, then the following statements hold(a)
Ifα≤a/p, then Dα is the union of three parabolic arcs(b)
Ifa/p<α≤b/p, then Dα is the union of E∠A,αp and a 5-sided nonclosed curve whose sides alternate between two line segments and three parabolic arcs(c)
Ifb/p<α<1/2, then Dα is a 12-sided closed curve whose sides alternate between six line segments and six parabolic arcs(2)
Ifa=b<c, then the following statements hold(a)
Ifα≤b/p, then Dα is the union of three parabolic arcs(b)
Ifb/p<α≤c/p, then Dα is a 9-sided nonclosed curve whose sides alternate between four line segments and five parabolic arcs(c)
Ifc/p<α<1/2, then Dα 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 ratioα:1−α, for 0<α≤1/2. This is the very first step in an exciting avenue that will lead to interesting questions about the properties and generalizations of these objects.



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*Source: 2751666-2022-06-09.xml*