Ceva s triangle inequalities

Transcription

1 arxiv: v [math.mg] 7 Jun 202 Ceva s triangle inequalities Árpád Bényi Department of Mathematics, Western Washington University, 56 High Street, Bellingham, Washington 98225, USA Arpad.Benyi@wwu.edu Branko Ćurgus Department of Mathematics, Western Washington University, 56 High Street, Bellingham, Washington 98225, USA Branko.Curgus@wwu.edu Corresponding author 200 Mathematics Subject Classification: Primary 26D07, 5M04; Secondary 5M5, 5B48

2 Abstract. We are interested in the natural question about which three cevians form a triangle independent of the triangle where they are constructed. This apparently simple question is equivalent to solving a three-parameter family of inequalities which we call Ceva s triangle inequalities. The parametrization of the solution set is rather surprising and requires a combination of techniques from algebraic inequalities, synthetic geometry and linear algebra.. Introduction In this article we investigate a class of inequalities of the form fa,b,c,ρ fb,c,a,σ < fc,a,b,τ < fa,b,c,ρ+fb,c,a,σ where a,b,c belongs to T = { a,b,c R 3 : a b < c < a+b }. and f : T R R + is some function with positive values. We will refer to as Ceva s triangle inequalities. Clearly, expresses the fact that for some parameters ρ,σ,τ R the lengths fa,b,c,ρ, fb,c,a,σ, fc,a,b,τ form a triangle whenever the lengths a,b,c form a triangle. This problem, albeit much more general, is in the spirit of Bottema, Djordjević, Janić, Mitrinović and Vasić [, Chapter 3], in particular the statement 3.3 which is attributed to Brown [2]. For example, consider the function ha,b,c,ρ = ρ a 2a 2 b 2 +b 2 c 2 +c 2 a 2 a 4 +b 4 +c 4. Recall that ha,b,c,2 is the length of the altitude orthogonal to the side of length a in a triangle with sides a,b,c. It is easy to see that, with a = 3,b = 8,c = 9, the altitudes do not form a triangle. Thus, fails to hold for all a,b,c T with ρ = σ = τ for the function h. However, if we replace f with /h, then it turns out that does hold for all ρ = σ = τ R \ {0} and all a,b,c T. For a connection between the reciprocals of the altitudes and a Heron-type area formula, see Mitchell s note [7]. In this article we will study with 2 fa,b,c,ρ = ρρ a 2 +ρb 2 + ρc 2. Date: June, 202. Key words and phrases. triangle inequaltiy, triangle inequality for cevians, cone preserving matrices, common invariant cone.

3 2 CEVA S TRIANGLE INEQUALITIES By Stewart s theorem fa,b,c,ρ gives the length of the cevian AA ρ in a triangle ABC with sides BC = a,ca = a, and AB = c; the point A ρ lies on the line BC with BAρ = ρ BC. Similarly, fb,c,a,σ = BBσ and fc,a,b,τ = CC τ where B σ and C τ lie on the lines CA and AB respectively, with CBσ = σ CA and AC τ = τ AB. Indeed, it is this connection with cevians that inspired the name of the inequalities. Now, it is well known that the medians of a triangle form a triangle as well. This implies that is satisfied for all a,b,c T when ρ = σ = τ = /2. In fact, is true for all a,b,c T whenever ρ = σ = τ R. That is, for any triangle ABC and for all ξ R, the cevians AA ξ,bb ξ, and CC ξ always form a triangle. This statement is proved in the book by Mitrinović, Pečarić and Volenec [8, Chapter I.3.4] where it is attributed to Klamkin [6]; see also the articles of Hajja [3, 4] for a geometric proof. Interestingly, for an arbitrary triangle ABC, the triple ρ,σ,τ = 2, 2,0 produces the sides AA 2,BB 2, and CC 0 which also form a triangle. In other words, for the given f : T R R +, also holds for all a,b,c T and some non-diagonal triple ρ,σ,τ { r,s,t R 3 : r = s = t }. This is proved in Figure. The property does not hold, however, for all triples of real numbers. Consider, for example, ρ,σ,τ = /4,/2,5/6. In the right triangle ABC, we let AB =,BC = 8,CA = 3. Then, we easily find AA ρ = 3/2.225, BB σ = 3/2 =.5, and CC τ = 7/ Clearly, AA ρ + BB σ < CC τ, thus we cannot talk about a triangle formed by the three cevians in this case. However, for the same triple but the right triangle ABC in which AB =,BC =,CA = 2, we find AA ρ = 7/4.03,BB σ = 2/ , and CC τ = 37/6.04, which are now clearly seen to form a triangle see Figures 2 and 3. The cautionary examples above lead us to ask the natural question about which triples are universally good to yield triangles formed by the corresponding cevians. More precisely, we are interested in the following problem. Problem. Characterize the set A R 3 of all triples ρ,σ,τ R 3 such that, for the function f given by 2, Ceva s triangle inequalities hold for all a,b,c T. In other words, characterize the set of all triples ρ,σ,τ such that for all nondegenerate triangles ABC the cevians AA ρ,bb σ and CC τ form a non-degenerate triangle as well. The main result of this paper is the complete parametrization of the set A. Theorem. Let φ = + 5/2 denote the golden ratio. The set A is the union of the following three sets { } ξ,ξ,ξ : ξ R, { } ξ,2 ξ,ξ, ξ, ξ,2 ξ, 2 ξ,ξ, ξ : ξ R\{ φ,φ} and { ξ, ξ,ξ, ξ, ξ, ξ, ξ,ξ, ξ : ξ φ,φ 2 }.

4 CEVA S TRIANGLE INEQUALITIES 3 A = C τ B C A ρ A ρ B σ A Figure. Cevians always form a triangle B σ C τ B A ρ C A Figure 2. Cevians do not form a triangle B σ C τ B A ρ C Figure 3. Cevians do form a triangle

5 4 CEVA S TRIANGLE INEQUALITIES Notice that the following six points are excluded in the second set: φ,φ 2, φ, φ,φ,φ 2, φ 2, φ,φ, 3 φ,φ 2,φ, φ, φ,φ 2, φ 2,φ, φ. These points are the only common accumulation points of both the second and the third set in the parametrization of A. This claim follows from the identities φ,2 φ,φ = φ, φ,φ, and φ,2 φ, φ = φ 2,φ2, φ 2, which follow from φ 2 φ = 0. As we will see in Section, these six points are exceptional since for an arbitrary triangle they correspond to degenerate triangles formed by the three cevians. A picture of the solution set A is given in Figure 4. It consists of four lines and three arc segments with the six points excluded. One of the lines is the diagonal which is the black line in Figure 4. The other three lines are in different shades of gray. The arc segments are in the dark shade of gray, almost black. The six excluded points are represented by small circles. There is a beautiful symmetry in the set A. By relabeling the vertices A B,B C,C A or A C,B A,C B we see that ρ,σ,τ A if and only if σ,τ,ρ A if and only if τ,ρ,σ A. These symmetries correspond to the rotations about the diagonal by 2π/3 and 4π/3 radians. Furthermore, relabeling the vertices A A,B C,C B shows that ρ,σ,τ A if and only if ρ, τ, σ A. Similarly, ρ,σ,τ A if and only if τ, σ, ρ A if and only if σ, ρ, τ A. Thesesymmetries correspondto therotations by π radians about one of the lines that pass through the point 2, 2, 2 in directions,,0,,0, and 0,,, respectively. The three lines of symmetry lie in the plane ρ + σ + τ = 3 2. In Figure 4 they are shown as thin dashed black lines. By now we observed that the set A has five rotational symmetries. The four corresponding axes of rotation intersect at the point 2, 2, 2. In particular, we see that the six excludedpoints lieon aspherecentered at 2, 2, 2 andofradius 23/2. InFigure 5 we show the orthogonal projection of the set A onto the plane ρ+σ+τ = 3 2. Here, the diagonal projects onto a single point. The points of the remaining three lines of the set A are shaded according to their distance from the plane ρ +σ +τ = 3 2. The points in the plane are medium gray, the points above the plane are dark gray becoming almost black when they are far away and the points below the plane are light gray becoming almost white when they are far away. Interestingly, the problem of identifying the admissible set A lends itself nicely to other questions which have an operator theoretic flavor. Given ρ,σ,τ A and a triangle T = ABC, we can always talk about a Ceva s triangle of T formed by the cevians AA ρ,bb σ,cc τ. We denote this triangle by C ρστ T. Thus, for each triple ρ,σ,τ in A we can define an operator T C ρστ T acting on triangles.

6 CEVA S TRIANGLE INEQUALITIES τ σ ρ Figure 4. The set A Figure 5. The projection of A onto ρ+σ +τ = 3/2

7 6 CEVA S TRIANGLE INEQUALITIES An interesting problem in this context is to study the limiting behavior in shape of triangles resulting from iterations of such operators. Such questions were considered by many authors, see for example Ismailescu and Jacobs[5], Hajja[3] and Nakamura- Oguiso [9] where one can also find other references. The remainder of the paper is devoted to the proof of Theorem. Although our proof is rather involved, it has a simple structure. First, in Section 2, we introduce a new one-parameter family of inequalities whose solution set B is guaranteed to contain A. In Sections 3 to 8 we use a combination of arguments from analytic and synthetic geometry to explicitly find B. Finally, in Sections 9 through 2, we use linear algebra to prove that the bigger solution set B from which the six points listed in 3 are excluded is in fact contained in A; this provides the parametrization given in Theorem. It is worthwhile noting that the discussion about the set B reduces to the observation that an uncountable family M of 3 3 matrices has a common invariant cone; that is, for a particular cone Q we have MQ Q for all M M. The question about when does a finite family of matrices shares an invariant cone is of current interest and already non-trivial. For example, in the simplest case of a finite family that is simultaneously diagonalizable one has a characterization for the existence of such an invariant cone, but this criteria does not seem to be easily applicable; see Rodman, Seyalioglu, and Spitkovsky [, Theorem 2] and also Protasov [0]. 2. The limiting inequalities Recall that, for a given a,b,c T and a triple ρ,σ,τ R 3, we have the following expressions for the lengths of the cevians: Specifically, AA ρ = fa,b,c,ρ, BB σ = fb,c,a,σ, CC τ = fc,a,b,τ. AA ρ = ρρ a 2 +ρb 2 + ρc 2, BB σ = σa 2 +σσ b 2 +σc 2, CC τ = τa 2 + τb 2 +ττ c 2. Clearly, a triple ρ,σ,τ belongs to A if and only if the following three-parameter family of inequalities is satisfied: 4 AA ρ BB σ < CC τ < AA ρ +BB σ for all a,b,c T. We now introduce the one-parameter family of inequalities emerging from 4 by letting a,b,c belong to the faces of the closure of the infinite tetrahedron T. First, let a = b+c, b,c > 0. Then Similarly, AA ρ = ρρ a 2 +ρa c 2 + ρc 2 = ρa c 2 = ρa c. BB σ = σb a, CC τ = τc+b.

8 CEVA S TRIANGLE INEQUALITIES 7 Thus, for all b,c > 0, we have ρb+c c σb b c τc+b ρb+c c + σb b c. Equivalently, by letting b/c = t, we have 5 ρt+ σt t τ +t ρt+ + σt t, for all t > 0. Second, let b = a+c, a,c > 0. Then AA ρ = ρa+c, BB σ = σb a, CC τ = τc b. So, letting c/a = t, we have 6 ρ+t σt+ τt t ρ+t + σt+, for all t > 0. Finally, let c = a+b, a,b > 0. Then AA ρ = ρa c, BB σ = σb+a, CC τ = τc b, and setting a/b = t, gives 7 ρt t σ +t τt+ ρt t + σ +t, for all t > 0. As pointed out in the introduction, we denote by B the set of all ρ,σ,τ R 3 which satisfy all three limiting inequalities 5, 6, and 7 for all t > 0. We have just shown that A B. In the next six sections we parametrize B. 3. Four important points We will now prove that there are four relevant families of points ρ,σ,τ in R 3 that belong to B. For τ R, they are and P τ := τ,τ,τ, P 2 τ := 2 τ, 2+τ,τ, P 2 τ := τ,2 τ,τ, P 22 τ := 2+τ, τ,τ. We will only provide the arguments that show the points P ij τ,i,j {,2}, satisfy the limiting inequality 5. The limiting inequalities 6 and 7 are treated in a similar way. Let t > 0. To show that P τ satisfies 5 is equivalent to τt+ τt t τ +t τt+ + τt t. Letting u = τt+, v = τt t, the previous inequality is equivalent to u v u v u + v, which is clearly true by virtue of the triangle inequality. Now, P 2 τ satisfies 5 because τt+ 2 τt t τ +t τt+ + 2 τt t

9 8 CEVA S TRIANGLE INEQUALITIES is equivalent again, via u = τt +, v = 2 τt t, to the triangle inequality u v v u u + v. For P 2 τ, we have that 2 τt+ 2+τt t τ +t 2 τt+ + 2+τt t is equivalent, via u = 2 τt+,v = 2+τt t, to u v u v u + v. Finally, for P 22 τ, the inequality 2+τt+ τt t τ +t 2+τt+ + τt t is equivalent, via u = 2+τt+,v = τt t, to u v u+v u + v. It is worthwhile noting that the triangle inequalities considered above hold in a stronger form. Namely, if u,v 0,0, then u v < u±v = u + v or u v = u±v < u + v. As we shall see shortly, this means that the four special points considered here always lie on the boundary of certain crosses determined by them. We will employ this fact later to show that, for a fixed τ R, besides these four points there is at most one other point that belongs to B. For now, in Section 4, we aim at a less ambitious goal of proving that if a point belongs to B, then it must be collinear either with P τ and P 2 τ, or P 2 τ and P 22 τ for some τ R. 4. Two important lines Let τ R be arbitrary. Now we prove that if a triple ρ,σ,τ B, then 8 ρ τ + = στ +τ. We break our discussion into three cases, depending on the range of the parameter τ. If τ 0, and we let t τ + in 5, then the rightmost inequality gives 8. If τ τ 0,], then we let t in the second part of 7 to get τ ρ τ τ = σ + τ, τ τ τ which implies 8. Finally, if τ >, let t = in the second inequality of 6 to τ obtain ρ+ = σ τ τ +, which implies 8.

10 CEVA S TRIANGLE INEQUALITIES 9 Notice that, for a fixed τ R\{0,}, 8 represents two lines in the ρσ-plane: l τ : σ = τ ρ+, τ l 2 τ : σ = τ ρ+ τ 2. τ τ The line l τ contains the pair of points P τ,p 2 τ, while l 2 τ contains both points P 2 τ,p 22 τ. Note also that the lines l τ and l 2 τ have distinct opposite slopes mτ and mτ. The mapping τ mτ = τ τ is a bijection between R\{0,} and R\{,0}. For a fixed τ R, denote by B τ the horizontal cross section of B at the level τ. Then B = { B τ : τ R }. In the next four sections we will prove that only one of the following two situations can occur: B τ = { P ij τ,i,j {,2} } or B τ = { P ij τ,i,j {,2} } l τ l 2 τ. In the following section we consider the simple case when τ {0,}. To deal with other values of τ, for fixed τ and t, we develop a geometric understanding of the solution set of each of the inequalities 5, 6 and The cases τ = 0 and τ = Let τ {0,}. We show that B τ = { P ij τ,i,j {,2} }. Recall that the points ρ,σ,τ must satisfy 8. Substituting τ = 0 in this equation gives ρ =, thus ρ = 0 or ρ = 2. If ρ = 0, then 6 is v t+ t+ σt+, t > 0. If we let t 0 +, we get σ σ, thus σ =, that is σ = 0 or σ = 2. So our points are 0,0,0 and 0,2,0. If ρ = 2, then 7 is t σ +t t + σ +t, t > 0. Substituting t = gives σ + σ +, that is σ = 0 or σ = 2. This gives the points 2,0,0 and 2, 2,0. All in all, the four points found are exactly P ij 0,i,j {,2}. Substituting τ = in 8 gives = σ. Thus σ = or σ =. If σ =, then 6 is ρ+t t ρ+t +t, t > 0, and letting t 0 + gives ρ =. We have obtained the points,, and,,. Finally, if σ =, then 7 is ρt t t ρt t + t t > 0. We let now t = to obtain ρ 2 =, that is ρ = or ρ = 3, which produce the points,, and3,,. Thefourpointsfoundareexactly P ij,i,j {,2}.

11 0 CEVA S TRIANGLE INEQUALITIES 6. Crosses Let τ R\{0,} and t > 0 be fixed. Notice that inequalities 5, 6 and 7 can be written in the form atρ bt ctσ dt gt atρ bt + ctσ dt, where a,b,c,d,g are all linear functions of t; one of the coefficients of g depends on the fixed value τ. Since gt = 0 yields 8, we will only consider the case gt 0. Renaming ρ = x, σ = y, we are interested in fully understanding the geometric representation in the xy-plane of inequalities in the generic form 9 ax b cy d g ax b + cy d, where a,b,c,d,g R\{0}. Note that 9 is equivalent to 0 a x b c y d g a g c a x b + c y d. g a g c The canonical form of such inequalities is x y x + y. Elementary considerations show that the solution set of is the set represented in Figure 6. We will refer to it as the canonical cross. It is centered at the origin and its marking points are, in counterclockwise direction,, 0,0,,, 0, and 0,. Note also that there are two pairs of parallel lines that define the boundary of the canonical cross, and the slopes of the lines that are not parallel have values and. Figure 6. The canonical cross Figure 7. Shifted and skewed The change of coordinates x a x b g a, y c g y d c,

12 CEVA S TRIANGLE INEQUALITIES transforms into 0 and the canonical cross is transformed into a shifted and skewed version of it, illustrated in Figure 7. Hence, a shifted and skewed cross is the solution set of 9. Its center is at b a, d c, and its marking points are b 2 a + g, d b, a c a, d c + g b, c a g, d b, and a c a, d c g. c The interior of the parallelogram with vertices at the marking points will be referred to as the nucleus of the cross. As mentioned above, there are two pairs of parallel lines defining the boundary of the cross, and the slopes of the non-parallel lines are of the form m and m, with m > 0. Specifically, m = a. 7. A family of crosses Let τ R\{0,} be fixed throughout this section. For a given t > 0, the limiting inequality 5 has the form 9. Hence its solution set is a cross in the ρσ-plane at the fixed level τ. We denote this cross by Cross 5 t,τ. Similarly Cross 6 t,τ and Cross 7 t,τ refer to the solution sets of 6 and 7. Denote by Cτ the family of all these crosses: Cτ = { Cross 5 t,τ : t > 0 } { Cross 6 t,τ, t > 0 } { Cross 7 t,τ, t > 0 }. As shown in Section 3, the points P ij τ,i,j {,2} lie on the boundary lines of all of the crosses in C. Next, we will show that, for arbitrary positive slope m, there exists a cross in Cτ whose boundary lines through the points P j τ,j {,2} have slope m consequently the boundary lines through P 2j τ,j {,2} have slope m, and thereexistsacrossincτwhoseboundarylinesthroughthepointsp 2j τ,j {,2} have slope m consequently the boundary lines through P j τ,j {,2} have slope m. First, we show that, for a given positive slope m, there exist two distinct crosses in Cτ with this slope. For m >, the two crosses are Cross 7 m,τ and Cross 5 /m,τ, while for 0 < m < the crosses are Cross7 m,τ and Cross 6 +/m,τ. To see that these crosses are truly distinct we just calculate their marking points. The marking points of Cross 7 m,τ are m+ m + τ m+ m+ m τ m+ m m m m, m, m+ m, m+ τm+,, m, m+ m, m τm+. These points are in the set { x,y : x > or y < 0 }. Further notice that the second marking point, following the order in 2, of both crosses Cross 5 /m,τ and Cross 6 +/m,τ is m m,m+ τm +. Since the first coordinate of this point is strictly less than and the second one is positive, this point is not in the set { x,y : x > or y < 0 }. Hence it differs from all four marking points of Cross 7 m,τ. c

13 2 CEVA S TRIANGLE INEQUALITIES In conclusion, all crosses in Cτ contain the points P ij τ,i,j {,2} on their boundary lines, and for each slope in R\{,0,} there is a unique cross in Cτ with a boundary line going through P τ and having that particular slope. 8. Intersection of crosses determined by four points Let τ R be fixed. In this section we answer the following question: Are there any other points besides the P ij τ,i,j {,2}, that belong to all the crosses in the family Cτ? In Section 5 we showed that the answer is negative if τ {0,}. Therefore, for the remainder of our discussion, we assume τ {0,}. Surprisingly, the answer is that there is possibly only one extra point having the property of belonging to all the crosses. That point is P 0 τ := τ, τ τ which is the intersection of the lines l τ and l 2 τ defined in Section 4. Now, for which τ-s could one expect P 0 τ to belong to all the crosses in Cτ? Letting t 0 + in 5 and 7, we get ρ τ ρ +, and,τ, σ τ + σ. If we substitute ρ = / τ and σ = τ /τ, we obtain that τ must satisfy τ τ τ 2 τ τ + τ. Straightforward calculations show that the solution of the above inequalities is the following union of disjoint intervals: [ ] [ 5, 5 3 ] [ 5, ] 5, 3+ 5, or in terms of the golden ratio φ: [ 3 φ, φ ] [ φ 2,φ ] [ φ,φ 2]. In Section we will indeed show that, for τ in the interior of this set, the point P 0 τ belongs to all the crosses in the family Cτ. Therestofthissection is devoted toprovingthat nootherpointbesidesp 0 τ and P ij τ, i,j {,2}, belongs to all the crosses in the family Cτ. Since τ R\{0,} is fixed, henceforth in this proof we will not emphasize the dependence on τ. We denote by S j the open line segment determined by P j and P j2,j {,2}. We will break our discussion into three separate cases that take into account the relative positions of the five points P 0,P,P 2,P 2,P 22. Case. {P 0 } = S S 2. Case 2. P 0 S S 2. Case 3. P 0 S \S 2 S2 \S.

14 CEVA S TRIANGLE INEQUALITIES 3 Let us assume that the line l containing P and P 2 has slope m > 0 and that line l 2 containing P 2 and P 22 has slope m < 0. Case. Any point X that lies in S S 2 can be eliminated by a cross that has the boundary lines through P and P 2 of slope m, and boundary lines through points P 2 and P 22 of slope m; see Figure 8. To eliminate a point X outside of the segments S and S 2, consider the cross having boundary lines that pass through P and P 2 of slope m ǫ, and boundary lines passing through P 2, P 22 of slope m+ǫ, with ǫ 0,m sufficiently small; see Figure 9. P P P 2 P 2 P 0 X P 2 P 2 P 0 P 22 X P 22 Figure 8. Figure 9. Case 2. As in Case, we will show that there are no other points besides the P ij,i,j {,2}, that belong to all the crosses in C. Note that the lines l and l 2 divide the plane into four quadrants intersecting at P 0. Since P 0 is not in S S 2, the points P ij,i,j {,2}, belong to the boundary of the same quadrant. Let us assume that they are in the configuration pictured in Figure 0. Any point X that lies in l \S l 2 \S 2 is eliminated by a cross that has boundary lines through P and P 2 of slope m, and boundary lines through points P 2 and P 22 of slope m. Thus, it remains to eliminate the points inside either of the segments S or S 2. Consider X S. Select a point Q l and between X and P so that the line through Q and P 22 has a finite non-zero slope. Denote this slope by m 0. The cross with boundary lines through P and P 2 of slope m 0, and boundary lines through points P 2 and P 22 of slope m 0 contains X in its nucleus, thus eliminating it; see Figure. A similar argument works for X S 2. Case3. In this case, we cannot eliminate P 0. We will show that no other point on l and l 2, besides P 0,P ij,i,j {,2}, belongs to all crosses in C. Let us assume that P 0 S \S 2.

15 4 CEVA S TRIANGLE INEQUALITIES Clearly, the cross that has the boundary lines through P and P 2 parallel to l 2, and the boundary lines through points P 2 and P 22 parallel to l eliminates any point X in l \S, as well as the points in the interior of S 2 ; see Figure 2. Let now X S. Moreover, assume that X is between P and P 0. Select a point Q on l and between X and P 0 such that the slope of the line through Q and P 22 has non-zero finite slope. Denote it by m 0. The cross with boundary lines through P and P 2 of slope m 0, and boundary lines through points P 2 and P 22 of slope m 0 contains X in its nucleus, thus eliminating it. This way, we eliminate all the points inside the segment S with the exception of P 0 ; see Figure 3. P 0 P 2 P 2 P X P 0 P 2 X Q P P 22 P 2 P 22 Figure 0. Figure. P 2 P 2 P 0 P 22 P 0 P 22 X Q P P X P 2 P 2 Figure 2. Figure 3.

16 CEVA S TRIANGLE INEQUALITIES 5 To eliminate the points X l 2 \S 2, we consider the following three cases. If P 0 is between X and P 22. Let Q be a point on l 2 between X and P 0 such that the slope of the line through P 2 and Q is non-zero and finite. Denote this slope by m 0. The cross with boundary lines through P and P 2 of slope m 0, and boundary lines through points P 2 and P 22 of slope m 0 does not contain X; see Figure 4. If X is between P 0 and P 22 a similar argument eliminates X. Finally, if X and P 0 are on the opposite sides of S 2 then pick Q on l 2 between X and P 2 such that the slope of the line through P and Q is non-zero and finite. Denote this slope by m 0. The cross with boundary lines through P and P 2 of slope m 0, and boundary lines through points P 2 and P 22 of slope m 0 does not contain X; see Figure 5. P 2 P 2 P 0 P 22 X Q P P 0 P 22 P P 2 Q P 2 X Figure 4. Figure 5. In conclusion, the solution set B is a subset of the union of the following three sets: D = { P τ : τ R }, E = { P 2 τ, P 2 τ, P 22 τ : τ R }, and F = { P 0 τ : τ [ φ, φ ] [ φ 2,φ ] [ φ,φ 2]}. In addition, let G = {P 0 φ, P0 φ, P 0 φ 2, P 0 φ, P 0 φ, P0 φ 2 } be the set of the six excluded points in Theorem, see 3. In the remainder of the paper we prove that D E F \G = A. The next section sets the stage for the application of linear algebra which is used in Section 0 to prove that D A, in

17 6 CEVA S TRIANGLE INEQUALITIES Section to prove that F \ G A, G A =, and in Section 2 to prove that E\G A. 9. From the tetrahedron to the cone In this section, we show that a triple a,b,c T if and only if a 2,b 2,c 2 Q, where Q is the interior in the first octant of the cone That is x 2 +y 2 +z 2 2xy +yz +zx = 0. Q = x y : x,y,z > 0, x 2 +y 2 +z 2 < 2xy +yz +zx. z By definition, a,b,c T if and only if a b < c < a+b. The last two inequalities are equivalent to a 2 +b 2 2ab < c 2 < a 2 +b 2 +2ab, and these two inequalities are, in turn, equivalent to a 2 +b 2 c 2 < 2ab. Squaring both sides, followed by simple algebraic transformations, yields the following equivalent inequalities: a 2 +b 2 c 2 2 < 4a 2 b 2, a 4 +b 4 +c 4 +2a 2 b 2 2a 2 c 2 2b 2 c 2 < 4a 2 b 2, 2 a 4 +b 4 +c 4 < a 4 +b 4 +c 4 +2a 2 b 2 +2a 2 c 2 +2b 2 c 2, 2 a 4 +b 4 +c 4 < a 2 +b 2 +c 2 2. Now, taking the square root of both sides, we adjust the last inequality to look like an inequality for a dot product of two unit vectors: a 2 +b 2 +c 2 2 a 4 +b 4 +c 4 > 3 3. This means that the cosine of the angle between the vectors a 2,b 2,c 2 and,, is bigger than 2/3. In other words, a,b,c are lengths of sides of a triangle if and only if the point a 2,b 2,c 2 is inside the cone centered around the diagonal x = y = z and with the angle arccos 2/3. The algebraic equation of this cone is x 2 +y 2 +z 2 2xy +yz +zx = 0. The relevance of the above observation is that now we can characterize A in terms of a cone preserving property of a class of 3 3 matrices. More precisely,

18 CEVA S TRIANGLE INEQUALITIES 7 this observation, in combination with Section 2, yields the following sequence of equivalences: ρ,σ,τ A AA ρ,bb σ,cc τ T for all a,b,c T AA ρ 2,BB σ 2,CC τ 2 Q for all a 2,b 2,c 2 Q Mρ,σ,τQ Q, where Mρ,σ,τ = ρρ ρ ρ σ σσ σ τ τ ττ. In the next three sections we use the cone preserving property to show that D E F \ G A and G A =. Specifically, in Section 0 we will prove that M P τ Q Q for all τ R. Then, in Section, we demonstrate that M P 0 τ Q Q for τ in the interior of the set 3, while in Section 2 we show M P 2 τ Q Q for all τ R \{2 +φ,2 φ}, which is equivalent to showing that M P 2 2 ξ Q Q for all ξ R\{ φ,φ}. Consider the matrix Mτ,τ,τ = 0. Rotation τ τ τ τ τ τ τ τ τ τ τ τ. For simplicity, we write M instead of Mτ,τ,τ. A long but straightforward calculation shows that M T M, that is, τ τ τ τ τ τ τ τ τ τ τ τ τ τ τ τ τ τ τ τ τ τ τ τ evaluates to ττ + 2 I and detm = ττ + 3. Therefore the matrix ττ + M is orthogonal and its determinant is. Thus, ττ + M is a rotation. Clearly the eigenvector corresponding to the real eigenvalue is the vector,,. Therefore the transformation of R 3 induced my M leaves the cone Q invariant.

19 8 CEVA S TRIANGLE INEQUALITIES. Degenerate case Let τ R\{0,} and consider the matrix τ τ M = M τ, τ τ,τ τ 2 τ τ = τ τ τ τ 2 τ. τ τ τ τ A simple verification yields that the linearly independent vectors τ,0, and τ,τ,0 are eigenvectors corresponding to the eigenvalue 0. Since M has rank the third eigenvector is τ τ, τ,ττ τ and it corresponds to the eigenvalue τ 6 3τ 5 +3τ 4 τ 3 +3τ 2 3τ + τ 2 τ 2. The matrix M maps Q into Q if and only if the last eigenvector or its opposite is in Q and the corresponding eigenvalue is positive. To explore for which τ this is the case, we introduce a change of variable τ +x. 2 Then the eigenvector becomes x+ x, x x+, x2 4 and the eigenvalue x 6 3x 4 +5x x 2 2 The polynomial in the numerator is even and, since its second derivative 30x 4 36x is strictly positive, it is concave up. Thus, its minimum is 5. Therefore the third eigenvalue is always positive. Next, we use the dot product to calculate the absolute value of the cosine of the angle between x+ x, x x+, x2 and,,. 4 After simplifying, the absolute value of the cosine is x x 6 7x 4 +59x 2 +. This simplification is based on the following two identities 4x+ 2 +4x 2 +x 2 2 = 3+x x x 4 +x 2 4 = 3+x 2 x 6 7x 4 +59x 2 +.

20 CEVA S TRIANGLE INEQUALITIES 9 Next, we need to find those values of x for which 4 is greater than 2/3. This is equivalent to x x 6 7x 4 +59x 2 + > 2, and, after further simplification, to 5 x 6 +23x 4 9x 2 +5 > 0 Since 5 is a root of the polynomial y 3 +23y 2 9y+5, factoring the last displayed polynomial yields x 6 +23x 4 9x 2 +5 = 5 x 2 8x 2 +x 4 = 5 x 2 x x Since = 2+ 5 and = 2+ 5, the roots of the polynomial in 5 in the increasing order are 2 5, 5, 2 5, 2+ 5, 5, Thus, the solutions of 5 are the open intervals 2 5, 5, 2 5, 2+ 5, 5, The corresponding intervals for τ are φ, φ, φ 2, φ, φ, φ 2, with the approximate values being.68, 0.68, 0.382, 0.68,.68, Moreover, for τ { φ, φ,φ 2,φ,φ,φ 2}, the eigenvector of M lies on the boundary of the cone Q. This boundary corresponds to the squares of the sides of degenerate triangles. Thus, for an arbitrary triangle ABC, for these six triples, the corresponding cevians form a degenerate triangle. Consider the matrix M = Mξ, ξ,2 ξ = Eigenvectors of this matrix are, φ 2, φ 2 2. Real eigenvalues ξ ξ ξ ξ ξ + ξ +ξ ξ 2 ξ ξ ξ2 ξ φ 2 φ 2, where φ =

21 20 CEVA S TRIANGLE INEQUALITIES The corresponding eigenvalues are ξ +ξ 2, φ 2 ξ 2, φ 2 ξ 2. This can be verified by direct calculations. In the verification of these claims and in the calculations below, the following identities involving φ 2 are used: φ 2 2 = φ 2, φ 2 +φ 2 = 3, φ 2 φ 2 2 = 5. The matrix M is singular if and only if ξ +ξ 2 = 0. That is, for ξ = φ = φ 2 or ξ = φ = φ 2. But for ξ = φ the matrix M of this section coincides with the matrix M in Section with τ = φ 2 there. For ξ = φ the matrix M of this section coincides with the matrix M in Section with τ = φ 2 there. Therefore in the rest of this section we can assume that M is invertible, that is, we assume 6 ξ φ and ξ φ. The matrix diagonalizes M since B MB = D = B = φ 2 φ 2 φ 2 φ 2 ξ +ξ φ 2 ξ φ 2 ξ 2. Notice that only the top left diagonal entry in D might be non-positive. Also, the square of the top left diagonal entry in D is the product of the remaining two diagonal entries: 7 φ 2 ξ 2 φ 2 ξ 2 = ξ +ξ 2 2. Next we define another cone Q 0 = u v : v,w > 0, u 2 < 4vw w and prove that DQ 0 = Q 0 and BQ 0 = Q.

22 CEVA S TRIANGLE INEQUALITIES 2 First notice that the definition of Q 0, 7 and 6 yield the following equivalences: u v Q 0 v,w > 0 and u 2 < 4vw w φ 2 ξ 2 v > 0, φ 2 ξ 2 w > 0 and ξ +ξ 2 2 u 2 <4 φ 2 ξ 2 v φ 2 ξ 2 w ξ +ξ 2 u φ 2 ξ 2 v φ 2 ξ Q 0 2 w D u v Q 0, w which, in turn, prove DQ 0 = Q 0. Second, we verify that the matrix B maps Q 0 onto Q. Set x y = B u u+v +w v = u+φ 2 v +φ 2 w, z w u+φ 2 v +φ 2 w and calculate x 2 +y 2 +z 2 2xy +yz +zx = x 2 2xy 2zx + y 2 +z 2 2yz = x x 2y +z + y z 2 = u+v +w 5u+ 2φ 2 +φ 2 v +w + φ 2 φ 2 2 v w 2 = 5 u+v +w u+v +w +5v w 2 = 5 u 2 v +w 2 +v w 2 = 5 u 2 4vw. Sincex 2 +y 2 +z 2 2xy+yz+zx < 0 represents the interior or the disconnected part of the circular cone with the axis x = y = z and with a directrix x = y,z = 0, all three coordinates x, y, z satisfying the last inequality must have the same sign. Clearly, if u 2 < 4vw, then v,w are nonzero and have the same sign. Assume x 2 +y 2 +z 2 2xy +yz +zx = 5 u 2 4vw < 0.

23 22 CEVA S TRIANGLE INEQUALITIES Then, if both v,w < 0, we have x = u+v +w < 2 v w v w = v w 2 0. The contrapositive of this implication is that, if x > 0, then at least one, and hence both, v,w > 0. Conversely, if v,w > 0, then x = u+v +w > 2 v w+v +w = v w 2 0. That is: v,w > 0 implies x > 0. Hence, the following equivalences hold: x y Q x > 0 and x 2 +y 2 +z 2 2xy +yz +zx < 0 z v,w > 0 and u 2 < 4vw u v Q 0. w This proves that BQ 0 = Q. Since B MB = D and DQ 0 = Q 0, we have MQ = MBQ 0 = BDQ 0 = BQ 0 = Q. This finally completes our proof of Theorem. 3. Closing comments This article has revealed the universal role played by some special triples in R 3 to yield triangles with cevians sides; we denoted the set of all such triples by A. We wish to end with a remark concerning an arbitrary fixed triangle ABC with the sides a,b,c. For such a triangle, denote by Aa,b,c the set of all triples ρ,σ,τ R 3 for which the cevians AA ρ, BB σ, and CC τ form a triangle. Clearly, A Aa,b,c. In fact, A = { Aa,b,c : a,b,c T }. The set Aa,b,c is a subset of R 3 bounded by the surface which is the union of the solutions to AA ρ +BB σ = CC τ or BB σ +CC τ = AA ρ or CC τ +AA ρ = BB σ. The general equation of this surface is quite cumbersome. To get an idea how this surface looks like, we plot it in Figure 6 for an equilateral triangle. By squaring twice any of the above three equalities and simplifying we obtain the implicit equation of the surface bounding A,, as ρ +σ +τ 6ρ +σ +τ 2 +64ρ σ +σ τ +τ ρ = 0, where ρ = ρ 2 2,σ = σ 2 2,τ = τ 2 2.

24 CEVA S TRIANGLE INEQUALITIES τ σ Figure 6. The set A and surface 8 ρ 2 Figure 6 also illustrates nicely the last sentence in Section : For an arbitrary triangle ABC, the six special points listed in 3 belong to the surface bounding Aa,b,c. References [] O. Bottema, R.Ž. Djordjević, R.R. Janić, D.S. Mitrinović, P.M. Vasić, Geometric inequalities. Wolters-Noordhoff Publishing, Groningen 969. [2] J.L. Brown Jr., Solution to E 366, Amer. Math. Monthly [3] M. Hajja, On nested sequences of triangles, Results in Mathematics [4] M. Hajja, The sequence of generalized median triangles and a new shape function, Journal of Geometry [5] D. Ismailescu, J. Jacobs, On sequences of nested triangles, Periodica Mathematica Hungarica [6] M.S. Klamkin, Second solution to Aufgabe 677, Elem. Math , [7] D. W. Mitchell, A Heron-type formula for the reciprocal area of a triangle, Mathematical Gazette 89 November 2005, 494. [8] D.S. Mitrinović, J.E. Pečarić, V. Volenec, Recent advances in geometric inequalities. Mathematics and its Applications East European Series, 28. Kluwer Academic Publishers Group, Dordrecht, 989. [9] H. Nakamura, K. Oguiso, Elementary moduli space of triangles and iterative processes, The University of Tokyo. Journal of Mathematical Sciences [0] V. Protasov, When do several linear operators share an invariant cone? Linear Algebra Appl , no. 4, [] L. Rodman, H. Seyalioglu, I. Spitkovsky, On common invariant cones for families of matrices. Linear Algebra Appl , no. 4, 9926.