International Journal of Mathematical Education in Science and Technology, Vol. 36, No. 8, 2005, 889–912

In search of more centres. A source of classroom projects in Euclidean geometry

S. ABU-SAYMEH and M. HAJJA* Department of Mathematics, Yarmouk University, Irbid – Jordan

(Received 25 June 2004)

A point E inside a triangle ABC can be coordinatized by the areas of the EBC, ECA, and EAB. These are called the barycentric coordinates of E. It can also be coordinatized using the six segments into which the cevians through E divide the sides of ABC, or the six angles into which the cevians through E divide the angles of ABC, or the six triangles into which the cevians through E divide ABC, etc. This article introduces several coordinate systems of these types, and investigates those centres of ABC whose coordinates, relative to a given coordinate system, are linear (or quasi-linear) with respect to appropriate elements of ABC, such as its side-lengths, its angles, etc. This results in grouping known centres into new families, and in discovering new centres. It also leads to unifying several results that are scattered in the literature, and creates several open questions that may be suitable for classroom discussions and team projects in which algebra and geometry packages are expected to be useful. These questions may also be used for Mathematical Olympiad training and may serve as supplementary material for students taking a course in Euclidean geometry.

1. Introduction and terminology

Throughout this paper, the letters A, B,andC stand for the measures of the angles of a triangle ABC, and the letters a, b, and c stand for its side-lengths BC, CA, and AB, respectively. The area of a triangle XYZ is denoted by ½XYZ, and its perimeter by p(XYZ). We draw the reader’s attention to the fact that some letters stand for different things in different sections. This holds in particular for the letters x, y, z, X, Y,andZ, which are used in almost all the sections. The sections in this paper are self-contained, and thus they can be read in any order. For terms that are used without being defined, and generally for any terms pertaining to triangle geometry, the reader is referred to the encyclopedic work [1], or to the standard textbook [2]. Let ABC be a general triangle, and let denote its area ½ABC. For a point E inside ABC, the barycentric or areal coordinates x, y,andz of E are defined by

½EBC ½ECA ½EAB x ¼ , y ¼ , z ¼

*Corresponding author. Email: [email protected]

International Journal of Mathematical Education in Science and Technology ISSN 0020–739X print/ISSN 1464–5211 online # 2005 Taylor & Francis http://www.tandf.co.uk/journals DOI: 10.1080/00207390500137928 890 S. Abu-Saymeh and M. Hajja and the , and of E are defined as the altitudes from E of the triangles EBC, ECA and EAB. Thus the barycentric and trilinear coordinates of E are related by

2x ¼ a,2y ¼ b,2z ¼ c ð1Þ

The barycentric and trilinear coordinates can be defined for all points in the plane of ABC by adopting the convention that ½EBC is taken positive or negative according as E and A lie on the same side or on different sides of the line BC, and so on. A similar convention is adopted for trilinear coordinates, and equations (1) are retained. Note that a point is interior if and only if its barycentric (equivalently, trilinear) coordinates are positive. Note also that a triple of real numbers ðx, y, zÞ qualifies as the barycentric coordinates of some point if and only if x þ y þ z ¼ 1. Both barycentric and trilinear coordinates are powerful tools in studying the geometry of the triangle. It is also transparent how to define them for simplices in any dimension. The of ABC is the point where the medians meet. It is also the centre of mass of three equal point masses placed at the vertices, and is also the centre of mass of the triangular lamina ABC with uniform density. The centre of mass of the triangular frame ABC having uniform density is a different point usually referred to as the Spieker centre. The incentre of ABC is the centre of its incircle. The Gergonne centre is the point of concurrence of the cevians AA0, BB0, and CC0, where A0, B0, and C0 are the points where the incircle touches the sides. Similarly, the Nagel centre is the point of concurrence of the cevians AA00, BB00,andCC00, where A00, B00, and C00 are the points where the excircles touch the sides. That such lines do indeed concur follows from Ceva’s Theorem according to which the cevians AA0, BB0, and CC0 concur if and only if

AC0 BA0 CB0 ¼ 1 C0B A0C B0A

Other centres will be mentioned in the sequel. For the time being, these centres were selected to illustrate certain properties of their barycentric coordinates. It is obvious that the barycentric coordinates of the centroid are ð1=3, 1=3, 1=3Þ, and it is not difficult to see that the barycentric coordinates of the incentre, the Spieker centre, and the Nagel centre are given by a b þ c a þ b þ c , , , , , , , , a þ b þ c 2ða þ b þ cÞ a þ b þ c where the second and third coordinates are filled in by applying the permutation ða ° b ° c ° aÞ, i.e. the cycle ðabcÞ. The reader is invited to find the barycentric coordinates of the Gergonne centre and to see that they cannot be expressed in the form f / g, where f and g are linear forms in a, b, and c. The family of centres whose barycentric coordinates can be so expressed is studied in [3, Theorem 2 and Note 3], where it is shown to form a straight line that turns out to be the first central line in the list of [1, p. 128]. In the next sections, we introduce other types of coordinates, and we investigate centres whose such coordinates can be nicely expressed. In search of more triangle centres 891

2. Side-intercept coordinates: the centroid, the Gergonne centre, the Nagel centre and more

Let ABC be a general triangle, and let A0, B0, and C0 be points on its sides. Let x, X, y, Y, z, and Z be defined (as shown in figure 1) by

x ¼ BA0, y ¼ CB0, z ¼ AC0 ð2Þ X ¼ a x, Y ¼ b y, Z ¼ c z ð3Þ

Ceva’s theorem states that the cevians AA0, BB0, CC0 are concurrent if and only if

xyz ¼ XYZ ð4Þ

The point E of concurrence is completely determined by the triple ðx, y, zÞ, which in turn is arbitrary except for having to satisfy the Ceva condition xyz ¼ða xÞðb yÞðc zÞ. Following [3], we call ðx, y, zÞ the side-intercept coordi- nates of E. Like the barycentric and trilinear coordinates, the side-intercept coordi- nates are allowed to be negative by thinking of BA0, CB0, and AC0 in (2) to be directed segments whose lengths change sign when their directions change. Thus for example, if A0 is such that B lies between C and A0, then BA0 is negative. With this understanding, everything above remains valid, and a point is interior if its side-intercept coordinates are positive. The most natural conditions that satisfy (4) are given by the following three systems of requirements:

x ¼ X, y ¼ Y, z ¼ Z ð5Þ z ¼ Y, x ¼ Z, y ¼ X ð6Þ y ¼ Z, z ¼ X, x ¼ Y ð7Þ

A

z Y

B C

E

Z y

B xXA C Figure 1. 892 S. Abu-Saymeh and M. Hajja

(Possibilities such as ðx, y, zÞ¼ðX, Z, YÞ are excluded since they are satisfied only if ABC is isosceles.) It is clear that (5) and (6) define the centroid and the Gergonne centre of ABC, respectively, and it is not difficult to see that (7) defines the Nagel centre. We remind the reader that the Gergonne (respectively, the Nagel) centre is customarily defined by the equivalent geometric requirement that A0, B0, and C0 are the points where the sides touch the incircle (respectively, the excircles) [1]. Using (3), one finds that the solutions of (5), (6), and (7) are given by

a b c x ¼ , y ¼ , z ¼ 2 2 2 a b þ c b c þ a c a þ b x ¼ , y ¼ , z ¼ 2 2 2 a þ b c b þ c a c þ a b x ¼ , y ¼ , z ¼ 2 2 2 in the same order. The three centres mentioned above are symmetric in the sense that their side- intercept coordinates are invariant under any relabelling of the vertices A, B, and C. To state this algebraically, we note that every permutation on fA, B, Cg (together with the natural permutation it induces on fA0, B0, C0g) gives rise to a permutation on each of the sets fx, y, zg, fX, Y, Zg,andfa, b, cg in accordance with (2) and (3) above. Thus the cyclic permutations ðBCÞ and ðABCÞ stand for the permutations ðbcÞðxXÞðyZÞðzYÞ and ðabcÞðxyzÞðXYZÞ, respectively. With this understanding, the symmetry of a centre amounts to the requirement that the system of equations that define its coordinates is invariant under every permutation of fA, B, Cg. Here, a system () is invariant under a permutation if the system obtained by applying to () has the same solution as (). In our case, it is clear that applying any permutation of fA, B, Cg to each of the systems (5), (6), and (7) results in an equivalent (in fact, identical) system. The next theorem is Theorem 1 of [4]. It shows that the three centres defined above are the only symmetric centres whose side-intercept coordinates are linear in the side lengths. Theorem 1: Let the centre E of ABC be defined as the intersection of the cevians whose intercepts x, X, y, Y, z,andZ with the sides (as defined by (2) and (3)) are given by linear forms in a, b,andc. Then E must be either the centroid, the Gergonne centre or the Nagel centre. Proof: It is not difficult to see that x, y,andz must be of the form a b c ðx, y, zÞ¼ þ tðb þ cÞ, þ tðc þ aÞ, þ tða þ bÞ 2 2 2

Using the cevian condition xyz ¼ða xÞðb yÞðc zÞ, and simplifying, we obtain 1 1 ttþ t ða bÞðb cÞðc aÞ¼0 2 2

Clearly, the solutions t ¼ 0, 1=2, and 1=2 correspond to the centroid, the Gergonne centre, and the Nagel centre, respectively. œ In search of more triangle centres 893

Next we claim that the centroid can also be defined by the requirement

b þ c c þ a a þ b Y þ z ¼ , Z þ x ¼ , X þ y ¼ ð8Þ 2 2 2 or by the equivalent requirement

b þ c c þ a a þ b y þ Z ¼ , z þ X ¼ , x þ Y ¼ 2 2 2

Unlike (5), the system (8) is linearly dependent and does not have a unique solution. However, together with the Ceva condition (4), it yields the unique solution a b c ðx , y , z Þ¼ , , 0 0 0 2 2 2

To see the uniqueness, let ðx1, y1, z1Þ be another solution with x1 > x0, say. From x0 þ X0 ¼ x1 þ X1 it follows that X1 < X0. From y0 þ X0 ¼ y1 þ X1 it follows that y1 > y0.Fromy0 þ Y0 ¼ y1 þ Y1 it follows that Y1 < Y0.Fromz0 þ Y0 ¼ z1 þ Y1 it follows that z1 > z0.Fromz0 þ Z0 ¼ z1 þ Z1 it follows that Z1 < Z0. This leads to the contradiction x1y1z1 > x0y0z0 ¼ X0Y0Z0 > X1Y1Z1: Thus ðx1, y1, z1Þ¼ ðx0, y0, z0Þ. Similarly, one can show that the systems

y þ Z ¼ a, z þ X ¼ b, x þ Y ¼ c ð9Þ Y þ z ¼ a, Z þ x ¼ b, X þ y ¼ c ð10Þ re-define (with (4)) the Gergonne and Nagel centres (respectively). In contrast to (8), (9) and (10), the similar systems a þ b þ c Y þ z ¼ Z þ x ¼ X þ y ¼ ð11Þ 3 a þ b þ c y þ Z ¼ z þ X ¼ x þ Y ¼ ð12Þ 3 do not have transparent solutions. Thus the centres that are defined by (11) and (12) are expected to be new. Theorem 2: Each of the systems (11) and (12) defines (together with (4)) a unique centre. Proof: We first consider the system consisting of (11) and (4). To prove the existence of a solution, let p ¼ 3ða þ b þ cÞ and make the harmless assumption that c 5 b 5 a. Consider the triplets

ðx0, y0, z0Þ¼ð0, p a, c pÞ ( ða, p,2p bÞ if b 5 p ðx1, y1, z1Þ¼ ða þ b p, b, pÞ if b 4 p 894 S. Abu-Saymeh and M. Hajja

Then

ðX , Y , Z Þ¼ða, b þ a p, pÞ 0 0 0 ð0, b p, c þ b 2pÞ if b 5 p ðX1, Y1, Z1Þ¼ ðp b,0,c pÞ if b 4 p

These give rise to points A0, B0 and C0 that lie on the sides and they satisfy

x0y0z0 X0Y0Z0 < 0, x1y1z1 X1Y1Z1 > 0

Therefore there exists a solution for which xyz XYZ ¼ 0, as desired. The uniqueness can be established in the same way as was done for (8). Also, (12) can be treated similarly. œ Now consider any of the systems (8)–(12) and look at the three equations it consists of. It is clear that (i) the first equation is invariant under the permutation ðBCÞ, and that (ii) the remaining equations are the iterates of the first under the permutation ðABCÞ. Also, using (3), one can rewrite the system in such a way that (iii) the left-hand side of the first equation is a linear form in x, y, and z, and the right-hand side is a linear form in a, b, and c. It is not difficult to see that if a system satisfies (i), (ii) and (iii) above, then its first equation can be assumed to be of the form

ð1 tÞðb þ cÞ Y þ z ¼ ta þ 2 or equivalently of the form

ð1 tÞðb þ cÞ y þ Z ¼ ta þ 2 for some t; see [3, section 3]. Denoting the system consisting of either of these equations and its iterates by St, and the centre it gives rise to (if any) by Ct, we see that C0, C1 and C1 are nothing but the centroid, the Gergonne centre and the Nagel centre, respectively. The two new centres described by (11) and (12) correspond to t ¼1=3 and t ¼ 1=3, and may be duly called the perimeter-trisecting centres. The centre C1=3 doesn’t seem to appear anywhere, while the centre C1=3 appears as Y9 in [5, p. 182] and as X369 in [1, p. 267], where it is called the trisected perimeter point; see also [6]. According to [1, p. 267] and [7], the trilinear coordinates ð, , Þ of C1=3 are given by

: ¼ cðc þ 2a vÞ : bðv c þ aÞ where v is the unique real zero of

2x3 3ða þ b þ cÞx2 þða2 þ b2 þ c2 þ 8ab þ 8bc þ 8caÞx ðb2c þ c2a þ a2b þ 5bc2 þ 5ca2 þ 5ab2 þ 9abcÞ

A more symmetric form is given in [3, Note 12] as

: ¼ cðc b þ 3a þ UÞ : bðb c þ 3a þ UÞ In search of more triangle centres 895 where U is the unique real zero of

x3 þð10ab þ 10bc þ 10ca a2 b2 c2Þx 8ða bÞðb cÞðc aÞ

We also mention that it is proved in [3] that the system St has a unique solution Ct for all t 2½0, 1, and the trilinear equation of the curve t ° Ct, t 2½0, 1, is given there as

ð1 2 cos AÞð2 2Þþð1 2 cos BÞð2 2Þþð1 2 cos CÞð2 2Þ¼0

We end this section with a brief word on non-symmetric centres. The most well-known of these are the ones defined by

x ¼ y ¼ z ð13Þ X ¼ Y ¼ Z ð14Þ

These are known in the literature as the (first and second) Yff centres after Peter Yff who discovered them in 1965 [8]. Due to their being analogues (in a sense described in the next section) to the older and much studied Brocard centres, the Yff centres have been studied fairly extensively [2]. It is interesting to note that the centre defined by (12) above turns out to be the analogue (in the same sense as above) to the well-known Fermat–Torricelli centre.

3. Angles between cevians and neighbouring sides: the incentre, the orthocentre, the circumcentre and more

Now we re-do the previous section with a new understanding of the variables involved. As before, we start with a general triangle ABC and we take points A0, B0 and C0 on its sides. This time, however, we let x, X, y, Y, z,andZ be defined as shown in figure 2 by

x ¼ffBAA0, y ¼ffCBB0, z ¼ffACC0 ð15Þ X ¼ A x, Y ¼ B y, Z ¼ C z ð16Þ

Also the Ceva’s condition (4) is easily seen to be equivalent to

sin x sin y sin z ¼ sin X sin Y sin Z ð17Þ see [9, Theorem 1.15.3, p. 56]. Following [3], we call ðx, y, zÞ the angle-intercept coordinates of E. Negative values of x, y and z are more delicate to handle, and one can confine oneself to interior points; see [4, Remark 1]. One sees that every triple of non- negative numbers ðx, y, zÞ qualifies as the angle-intercept coordinates of an interior point E if and only if they satisfy the trigonometric version (17) of Ceva’s Theorem and the conditions

0 < x < A,0< y < B,0< z < C 896 S. Abu-Saymeh and M. Hajja

A

x X

B C

E

Y z y Z B A C Figure 2.

Here again, the most natural conditions that satisfy (17) are given by (5), (6), and (7), and it is clear that these define the incentre, the orthocentre, and the circum- centre of ABC, respectively. The statement regarding (6) follows from the simple observation that if AA0, BB0 and CC0 are cevians in ABC such that the three quadrilaterals BCB0C0, CAC0A0 and ABA0B0 are cyclic, then the cevians are concurrent by (17), and they intersect at the orthocentre; see Theorem 4. Also, it is proved in [4, Theorem 2] that the incentre, the orthocentre, and the circumcentre are the only centres for which x, y, and z are linear forms in A, B,andC. The analogues of (8)–(12) are the systems given by B þ C C þ A A þ B Y þ z ¼ , Z þ x ¼ , X þ y ¼ ð18Þ 2 2 2 y þ Z ¼ A, z þ X ¼ B, x þ Y ¼ C ð19Þ Y þ z ¼ A, Z þ x ¼ B, X þ y ¼ C ð20Þ Y þ z ¼ Z þ x ¼ X þ y ð¼ 60Þð21Þ y þ Z ¼ z þ X ¼ x þ Y ð¼ 60Þð22Þ

To investigate the centres defined by these systems, we find it convenient to work with the angles , , and defined by

¼ffBEC, ¼ffCEA, ¼ffAEB, ð23Þ where E is the point of concurrence of the cevians AA0, BB0,andCC0, as shown in figure 3. It is fairly clear (and serves as an exercise suitable for classroom discussions) that a triple ð, , Þ with þ þ ¼ 360 defines, via (23), a unique point E inside a given triangle ABC if and only if

A <<180, B <<180, C <<180 ð24Þ In search of more triangle centres 897

A

ζ η ξ

BC Figure 3.

Using the obvious relations

¼ A þ Y þ z, ¼ B þ Z þ x, ¼ C þ X þ y one rewrites (18)–(22) in terms of , , and . Thus (18) takes the simple form

A B C ¼ 90 þ , ¼ 90 þ , ¼ 90 þ 2 2 2 which obviously is satisfied by, and hence defines, the incentre. The systems (19) and (20) take the forms

¼ 180 A, ¼ 180 B, ¼ 180 C ¼ 2A, ¼ 2B, ¼ 2C respectively, and they are satisfied by the orthocentre and the circumcentre, respec- tively. Hence these are the centres they define. Note that in view of (24), these two centres are interior if and only if ABC is acute-angled. The system (22) takes the form

¼ ¼ ¼ 120 ð25Þ and hence defines what is known as the Fermat–Torricelli point. This is defined as the point whose distances from the vertices has a minimal sum, and is characterised by the equiangular property (25) if no angle of ABC exceeds 120; see [10]. Note that 898 S. Abu-Saymeh and M. Hajja this restriction is also dictated by (24). It also follows from (22) that the Fermat–Torricelli point E can be characterized by the property

BC00 þ CB00 ¼ CA00 þ AC00 ¼ AB00 þ BA00 where A00, B00, and C00 are the points where the cevians AA0, BB0, and CC0 meet the circumcircle of ABC. Thus, the Fermat–Torricelli point is a circumcircle trisecting point. The point defined by the system (21) is also a circumcircle trisecting point since (21) is equivalent to the system

B00C00 ¼ C00A00 ¼ A00B00 where A00, B00, and C00 are as defined above. The next theorem shows that the point defined by (21) is nothing but the first isodynamic point, defined as one of the two points where the three Apollonian circles of ABC intersect. Here, the A-Apollonian circle of ABC is the circumcircle of ABC, where B* and C* are the points where the internal and external angle-bisectors of A meet BC; see [1, p. 68]. Theorem 3: Let E be a point inside ABC, and let x, y, z, X, Y, Z, , , and be as defined in (15), (16), and (23). Let a0 ¼ AE, b0 ¼ BE, c0 ¼ CE: Then the following conditions are equivalent: (i) Y þ z ¼ 60, Z þ x ¼ 60,andX þ y ¼ 60. (ii) ¼ A þ 60, ¼ B þ 60,and ¼ C þ 60. (iii) aa0 ¼ bb0 ¼ cc0: (iv) E is the first isodynamic point of ABC. Proof: The equivalence of (i) and (ii) follows from the Exterior Angle Theorem by extending AE. The implication (ii) ¼) (iii) is established in [3], and the implication (iii) ¼) (ii) is established in [11]. In fact, observing that

ffEBA00 ¼ C, ffECA00 ¼ B, ffEA00B ¼ C, ffEA00C ¼ B and applying the Law of Sines to the triangles EBA00 and ECA00, we see that bb0 : cc0 ¼ sinð BÞ : sinð CÞ, and therefore

aa0 : bb0 : cc0 ¼ sinð AÞ: sinð BÞ: sinð CÞ

From this, it follows that (ii) and (iii) are equivalent. Thus it remains to show that (i), (ii), and (iii) imply (iv). Referring to figure 4, let the internal and external angle- bisectors of A meet BC at P and Q, respectively. Note that EP bisects ffBEC, since AB=AC ¼ EB=EC, by (iii). Therefore

ffAQP þffAEP ¼ ðÞþff90 ffAPQ ðÞAEC þffCEP A ¼ 90 B þ þ þ 2 2 A A ¼ 90 B þ ðÞþ60 þ B 30 þ ¼ 180 2 2 In search of more triangle centres 899

A

E

B P C Q Figure 4.

Hence AEPQ is cyclic. Therefore the A-Apollonian circle, being nothing but the circumcircle of APQ, passes though E. It follows from symmetry that E lies on the intersection of the three Apollonian circles of ABC, and hence is the first isodynamic point. œ We finally remark that the systems (13) and (14), where x, y, z, X, Y,andZ are as defined in (15) and (16), define the well-known Brocard points that were discovered and studied by A. L. Crelle in 1816, and that have attracted the attention of several mathematicians and generated a vast amount of research since their rediscovery by H. Brocard in 1875; see [12] and [13]. Note: It is proved in [3, section 4] that the system consisting of the equation

ð1 tÞðB þ CÞ Y þ z ¼ tA þ 2 and its iterates (together with (3) and (17)) defines a unique centre Ct for all t in ½1, 1. We have seen that C1, C1=3, C0, C1=3 and C1 are the orthocentre, the Fermat–Torricelli centre, the incentre, the isodynamic centre, and the circumcentre, respectively. Professor Clark Kimberling kindly drew our attention to the fact that these five points lie on what is known as the Neuberg cubic, listed as ZðX30Þ in [1, p. 240]. That our curve is not identical with the Neuberg cubic is shown in [3, Theorem 15]. It also follows from [14] since the Neuberg curve is algebraic.

4. Angles between cevians and opposite sides: more defining properties of the orthocentre

Let E be a point inside triangle ABC, and let AA0, BB0,andCC0 be the cevians through E. Let x, X, y, Y, z and Z be defined, as in figure 5, by

x ¼ffAA0B, y ¼ffBB0C, z ¼ffCC0A X ¼ 180 x, Y ¼ 180 y, Z ¼ 180 z

Replacing a, b,andc in (5)–(14) by 180, it is easy to prove, using appropriate instances of Theorem 4, that each of the resulting systems leads to the orthocentre. 900 S. Abu-Saymeh and M. Hajja

A

Y B' C' z y Z E

xX BCA' Figure 5.

Theorem 4: Let AA0, BB0,andCC0 be cevians in ABC that intersect at E. If any two of the six quadrilaterals

BCB0C0, CAC0A0, ABA0B0, AC0EB0, BA0EC0, CB0EA0 are cyclic, then all of them are cyclic and E is the orthocentre. Proof: (i) If A0EB0C and A0EC0B are cyclic, then AE  AA0 ¼ AC0  AB and AE  AA0 ¼ AB0  AC, and therefore AC0  AB ¼ AB0  AC,andBCB0C0 is cyclic. It now follows from x ¼ y, X ¼ Z, and y ¼ Z that x ¼ X ¼ 90. (ii) The same argument applies if A0EB0C and BCB0C0 are cyclic. (iii) If A0EB0C and ABA0B0 are cyclic, then X þ y ¼ 180 and x ¼ Y, and therefore X ¼ y ¼ 90. (iv) If BCB0C0 and ABA0B0 are cyclic, then y ¼ Z and x ¼ Y, and therefore Z þ x ¼ y þ Y ¼ 180. Hence BA0EC0 is cyclic, and this case reduces to (ii). Note that the case (iv) is treated in [15, Problem 10, pp. 207–208 and 221–222]; see also [40]. œ

5. Areas of subtriangles formed by cevians, I: more defining properties of the centroid

Here, we take a point E inside ABC, we let AA0, BB0, and CC0 be the cevians through E. We let u, v and w stand for the areas of the triangles EBC, ECA and EAB respectively, and we note that these are proportional to the barycentric coordinates In search of more triangle centres 901

A

z Y B' C'

E Z y

x X

BCA' Figure 6.

of E. We also let x, X, y, Y, z and Z stand for the areas of the six triangles into which the cevians through E divide ABC, as shown in figure 6. Thus

x ¼½EBA0, y ¼½ECB0, z ¼½EAC0, X ¼½ECA0, Y ¼½EAB0, Z ¼½EBC0

In contrast with the situation in the previous sections, x þ X, y þ Y, and z þ Z depend on the point E, being nothing but u, v,andw, respectively. Thus we do not expect the systems obtained from (5)–(14) by setting ða, b, cÞ¼ðu, v, wÞ to give rise to unique legitimate centres. This is illustrated in the next theorem, which should provide a source of classroom exercises where purely geometric solutions are invited. Theorem 5: Let E be a point inside the triangle ABC and let AA0, BB0 and CC0 be the cevians through E. Let x, X, y, Y, z, Z, u, v, and w be as defined above. Let the systems obtained by setting ða, b, cÞ¼ðu, v, wÞ in (5)–(14) be denoted by (50)–(140). Then (120) holds if and only if either E is the centroid, or E is the mid-point of one of the medians; while each of the remaining systems defines the centroid. Proof: We use the relations

uw uv vu vw wv wu x ¼ , X ¼ , y ¼ , Y ¼ , z ¼ , Z ¼ v þ w v þ w w þ u w þ u u þ v u þ v which follow from simple area considerations such as w x x ¼ ¼ v X u x 902 S. Abu-Saymeh and M. Hajja

We start with (5), (6), and (7). It is easy to see that the each of the statements x ¼ X, z ¼ Y, and Z ¼ y is equivalent to the statement v ¼ w. Thus each of (5), (6), and (7) defines the centroid. Regarding (80), it follows that the equation Y þ z ¼ðv þ wÞ=2 is equivalent to the equation

2vwðu þ v þ wÞ¼ðu þ vÞðv þ wÞðw þ uÞ2uvw which in turn shows that u is an expression symmetric in fu, v, wg. Therefore (80) yields u ¼ v ¼ w. The equation y þ Z ¼ u in (90) is equivalent to u2 ¼ v2 þ w2 vw. This in turn is equivalent to the statement that fu, v, wg are the sidelengths of a triangle whose angle opposite to u is 60. Hence the system (90) yields u ¼ v ¼ w. The equation Y þ z ¼ u reduces to vw ¼ u2, and therefore (100) implies that uvw ¼ u3 ¼ v3 ¼ w3, and hence u ¼ v ¼ w. The equation Y þ z ¼ Z þ x simplifies into u þ v þ w 1 ðu vÞ þ ¼ 0 ðu þ wÞðv þ wÞ u þ v and is hence equivalent to u ¼ v. Thus (110) implies u ¼ v ¼ w. The equation y þ Z ¼ z þ X in (120) is equivalent to w u v ðu vÞ ¼ 0 u þ v v þ w u þ w

Thus (120) is equivalent to

ðu vÞðW UVÞ¼0, ðv wÞðU VWÞ¼0, ðw uÞðV UWÞ¼0 where u v w U ¼ , V ¼ , W ¼ v þ w w þ u u þ v If w 6¼ u and w 6¼ v, then U ¼ VW and V ¼ UW and therefore W ¼ 1andU ¼ V. Thus w ¼ u þ v and u ¼ v. This happens if and only if

AC0 ¼ C0B, BA0 ¼ 2A0C, AB0 ¼ 2B0C

We conclude that ða, b, cÞ is proportional to either one of the following :

ð1, 1, 1Þ, ð2, 1, 1Þ, ð1, 2, 1Þ, ð1, 1, 2Þ

Thus (120) is satisfied by the centroid and by the three points given by the areal coordinates (1,1,2), (1,2,1), (2,1,1). These correspond to the midpoints of the medians. It remains to treat (13) and (14). The system x ¼ y ¼ z is equivalent to u2 þ uv ¼ v2 þ vw ¼ w2 þ wu. But u2 þ uv ¼ v2 þ vw implies ðu vÞðu þ vÞ¼ vðw uÞ, which in turn implies that u v and w u have the same sign. Similarly for v w. Since ðu vÞþðv wÞþðw uÞ¼0, it follows that u ¼ v ¼ w. (14) is treated similarly. We remark that the system x ¼ y ¼ z is treated in [16] and [17]. œ In search of more triangle centres 903

6. Areas of subtriangles made by cevians, II: still more defining properties of the centroid

Here again, we take a point E inside ABC, we let AA0, BB0, and CC0 be the cevians through E, but we let x, X, y, Y, z and Z be defined by

x ¼½ABA0, y ¼½BCB0, z ¼½CAC0, X ¼½ACA0, Y ¼½BAB0, Z ¼½CBC0 as shown in figure 7. It is easy to see that Ceva’s concurrence condition (4) retains the form xyz ¼ XYZ. Since x þ X ¼ y þ Y ¼ z þ Z ¼ , where is the area of ABC, the analogues of (5)–(14) are obtained by replacing a, b, and c by . The next theorem describes the centres defined by each of the resulting systems. Theorem 6: Let E be a point inside the triangle ABC and let AA0, BB0 and CC0 be the cevians through E. Let x, X, y, Y, z, Z, and be as defined above, and let the systems obtained by replacing a, b, and c in (5)–(14) by be denoted by (50)–(140). Then each of the systems (50)–(140), together with the concurrence condition xyz ¼ XYZ, leads to the centroid. Proof: We start with (5). It is clear that the statement x ¼ X is equivalent to the statement BA0 ¼ CA0. Therefore the system (5) defines the centroid. We next consider the systems (6) and (7). It is easy to see that the statements z ¼ Y and y ¼ Z are equivalent to the statements B0C0 k BC. Thus each of the systems (6) and (7) is equivalent to the conditions

A0B0 k AB, B0C0 k BC, C0A0 k CA ð26Þ

A

x X

BCA' Figure 7. 904 S. Abu-Saymeh and M. Hajja

Simple similarity properties show that these hold if and only if AA0, BB0, and CC0 are the medians of ABC. Thus each of the systems (6) and (7) defines the centroid. The equation Y þ z ¼ U is equivalent to y ¼ z. Hence each of the systems (80)–(140) degenerates into x ¼ y ¼ z, X ¼ Y ¼ Z. Together with the concurrence condition xyz ¼ XYZ, this leads to the centroid. œ Note: If A0, B0, and C0 are points on the sides BC, CA,andAB of triangle ABC, respectively, and if A, B, and C are the angles of ABC, and A0, B0,andC0 are the angles of A0B0C0, then it is immediate that (26) implies that

A ¼ A0, B ¼ B0, C ¼ C0 ð27Þ

However, the converse is not true without extra assumptions. The next theorem provides such appropriate assumptions, and may prove to be an instructive classroom challenge [18]. Theorem 7: Suppose that the cevians AA0, BB0, and CC0 of triangle ABC are concurrent, and let E be their intersection point. Then (27) holds if and only if (26) holds, i.e. if E is the centroid. Proof: Define t by setting ffA0C0B ¼ A þ t and find the other angles in terms of t, as shown in figure 8. Thus

ffA0C0B ¼ A þ t, ffB0C0A ¼ B t, ffC0B0A ¼ C þ t ffCB0A0 ¼ A t, ffB0A0C ¼ B þ t, ffC0A0B ¼ C t

Since

AC0 sinðC þ tÞ ¼ , etc: AB0 sinðB tÞ

A

C ′ B − t

A + t C + t B′

A − t

C + t B + t BCA′ Figure 8. In search of more triangle centres 905 it follows that the cevian condition

AC0 Â BA0 Â CB0 ¼ C0B Â A0C Â B0A of concurrence is equivalent to the condition

sinðA þ tÞ sinðB þ tÞ sinðC þ tÞ¼sinðA tÞ sinðB tÞ sinðC tÞð28Þ

Using the identity

sin A sin B cos C þ sin B sin C cos A þ sin C sin A cos B ¼ 1 þ cos A cos B cos C obtained by expanding 1 ¼ cosðA þ B þ CÞ, (28) in turn simplifies into

sin t ðcos A cos B cos C þ cos2 tÞ¼0 ð29Þ

Take any two acute angles of the triangle, say A and B. Since t < A and t < B,it follows that cos2 t > cos A cos B 5 cos A cos Bj cos Cj, and that cos A cos B cos C þ cos2 t cannot be 0. It follows from (29) that sin t ¼ 0 and hence t ¼ 0. Thus C0AB0A0 and C0B0CA0 are parallelograms and A0 is the midpoint of BC. Similarly for B0 and C0, and hence E is the centroid. œ Notes: (i) A more geometric proof of Theorem 7 can be obtained as follows. Let A*, B*, and C* be the midpoints of BC, CA, and AB, and assume without loss of generality that A0 and B0 lie on the closed line segments BA* and CB*, respectively; see figure 9. From A0, we draw lines parallel to AB and AC that meet AC and AB

A

B″

C* B*

C ″ B′

BCA′ A* Figure 9. 906 S. Abu-Saymeh and M. Hajja at B00 and C00, respectively. Then C0 lies on the closed line segment C00A because ffB0A0C0 ¼ffB00A0C00. By Ceva’s Theorem,

BA0 CB0 AC0 BA0 CB0 AC00 BA0 CB0 A0C CB0 CB 1 ¼ 4 ¼ ¼ 4 ¼ 1 A0C B0A C0B A0C B0A C00B A0C B0A BA0 B0A BA

Therefore we have equality everywhere, i.e. C0 ¼ C00 and B0 ¼ B. Hence ffB0A0C0 ¼ffBAC ¼ffBAC ¼ffBA0C00. Since ffB0A0C0 ¼ffB00A0C00, it follows that B00 ¼ B, and therefore A0 ¼ A. Since B0 ¼ B, the result follows. (ii) Theorem 7 follows also immediately from a theorem of Seebach [19] stating that if ABC and UVW are any triangles, then there exist unique points U0, V0,andW0 on the segments BC, CA, and AB such that U0V0W0 and UVW are similar and such that the cevians AU0, BV0, and CW0 are concurrent. To prove Theorem 7, one takes the special case when ABC ¼ UVW and invokes the uniqueness. Another interesting special case of Seebach’s Theorem is when UVW is equilateral. This case is treated in [20] and is quoted in [1], where the point of concurrence of AU, BV, and CW is denoted by X370 and is referred to as the equilateral cevian triangle point of ABC. A proof of Seebach’s Theorem that is much simpler than the proof in [19] is given in [21].

7. Perimeters of subtriangles made by cevians: the isoperimetric and equal detour points

Here again, we take a point E inside ABC, and we first consider the six triangles formed by these cevians in the manner desribed in section 5, and then we consider the six triangles formed by them in the manner desribed in section 6, paying attention this time to their perimeters instead of their areas. A reference that is relevant is [22], where the author calls a point E an isoperimetric point if the three triangles EBC, ECA and EAB have equal perimeters, i.e., if

EB þ BC þ CE ¼ EC þ CA þ AE ¼ EA þ AB þ BE and an equal detour point if

EB BC þ CE ¼ EC CA þ AE ¼ EA AB þ BE

The following theorem records a corrected version of a result of Veldkamp [22] regarding the existence and uniqueness of such points. Theorem 8: For a triangle ABC, let

A B C Q :¼ tan þ tan þ tan 2 2 2 2

Then ABC has exactly one isoperimetric point, no isoperimetric points, exactly one equal detour point, or exactly two equal detour points, according as Q is negative, non-negative, non-positive, or positive, respectively. In search of more triangle centres 907

Proof: See [23]. The proof makes use of what is usually referred to as Descartes Circle Theorem, and it turns out that the isoperimetric and equal detour points are the centres of the circles commonly known as the Soddy circles. œ Other relevant results are summarized in the next theorem. Theorem 9: Suppose that the cevians AA0, BB0, and CC0 of triangle ABC are concurrent, and let E be their intersection point. Then (i) E is an equal detour point of ABC if and only if pðBEA0Þ¼pðCEA0Þ. (ii) E is an isoperimetric point of ABC if and only if any of the three statements

pðAC0EÞ¼pðAB0EÞ, pðABEÞ¼pðACEÞ,andpðABB0Þ¼pðACC0Þ holds. Proof: (i) It follows from [24, Problem 3, section 2.7, p. 68] that pðBEA0Þ¼pðCEA0Þ if and only if AB þ CE ¼ AC þ BE. The last statement is clearly equivalent to AE þ EB AB ¼ AE þ EC CA, which in turn is the definining property of point(s) of equal detour. (ii) The three statements are equivalent by Theorem 1 of [25], and the statement pðABEÞ¼pðACEÞ is the defining property of the isoperimetric point. It is worth mentioning that the implication pðAC0EÞ¼pðAB0EÞ¼)pðABEÞ¼pðACEÞ is the theorem of Urquhart that Dan Pedoe refers to in [26] as the most elementary theorem of Euclidean geometry. œ Going back to the question of what centres are defined by perimeter considera- tions, we set

x ¼ pðEBA0Þ, y ¼ pðECB0Þ, z ¼ pðEAC0Þ X ¼ pðECA0Þ, Y ¼ pðEAB0Þ, Z ¼ pðEBC0Þ x0 ¼ pðABA0Þ, y0 ¼ pðBCB0Þ, z0 ¼ pðCAC0Þ X0 ¼ pðACA0Þ, Y0 ¼ pðBAB0Þ, Z0 ¼ pðCBC0Þ

Since x þ X, x0 þ X0, etc., do not stand anymore for anything sensible, we confine ourselves to the systems (5), (6), and (7), and to the systems (50), (60), and (70) obtained by replacing x, y, z, X, Y, and Z by x0, y0, z0, X0, Y0, and Z0, respectively. We shall explore what points E these six systems give rise to. It follows immediately from Theorem 9 that the system (70) gives rise to the point of equal detour, while each of the systems (6) and (60) gives rise to the isoperimetric point. The system (50) trivially defines the , since the equation x0 ¼ X0 is equivalent to x ¼ða þ b cÞ=2. This is treated in [27], where other interesting questions are raised. We leave it to the interested reader to explore what centres correspond to the systems (5), (7), (13), (14), (130), and (140). We expect such a venture to involve considerable challenge. We also mention that it has been proved in [28] that if the triangles EBA0, ECB0, and EAC0 have equal perimeters and equal areas, then ABC is equilateral. It is also announced, without a proof, that if any three of the six triangles EBA0, EA0C, ECB0, EB0A, EAC0, and EC0B have equal perimeters, and if any three of them have equal areas, then ABC is equilateral. A related result is obtained in [29], where it is proved 908 S. Abu-Saymeh and M. Hajja that if E is the centroid, then the circumcentres of these six triangles lie on a circle, and where the editors announce, without proof, that the converse is also true, and invited the reader to supply a proof that is coordinate-free and not too heavily computational.

8. Distances of a centre from the vertices

Let E be a point inside ABC, and set

a0 ¼ EA, b0 ¼ EB, c0 ¼ EC

It is immediate that the isoperimetric point is defined by

a a0 ¼ b b0 ¼ c c0 ð30Þ and the point(s) of equal detour by

a þ a0 ¼ b þ b0 ¼ c þ c0 ð31Þ

It also follows from Theorem 3 that the first isodynamic point is defined by

aa0 ¼ bb0 ¼ cc0 ð32Þ

In view of (30), (31), and (32), it is natural to ask about what points, if any, are defined by the conditions

a b c ¼ ¼ ð33Þ a0 b0 c0 This question is indeed asked in [30], where a point satisfying (33) is called a good point and where it is proved that an obtuse-angled triangle has no good points, a right triangle has exactly one good point (namely the point that completes the triangle to a rectangle), and that an acute-angled triangle has two good points. The problem was investigated earlier in [31], where it is proved that the two good points lie on the . Less transparent is the fact that the orthocentre is characterized by the requirements

a2 þ a02 ¼ b2 þ b02 ¼ c2 þ c02 ð34Þ and that a point E satisfies

a2 a02 ¼ b2 b02 ¼ c2 c02 ð35Þ if and only if AA ¼ BB ¼ CC, where A*, B*, and C* are the of EBC, ECA, and EAB, respectively. The proofs of these statements can best be established in the context of tetrahedra, where E is allowed to be a point outside the plane of ABC. One then shows that the altitudes of a tetrahedron ABCE are concurrent if and only if (34) holds, and that the three medians AA*, BB*, and CC* are equal if and only if (35) holds; see [32] and [33]. In search of more triangle centres 909

We invite the interested readers to find characterizations for their favourite centres in terms of a, b, c, a0, b0, and c0.

9. Side-intercepts made by perpendiculars from a centre to the sides

Let A0, B0 and C0 be points on the sides BC, CA, and AB of a triangle ABC, respectively, and let x, X, y, Y, z,andZ be defined (as shown in figure 10) by (2) and (3). It is proved in [34] and in [41] that the perpendiculars from A0, B0,andC0 to the sides BC, CA, and AB, respectively, are concurrent if and only if x2 þ y2 þ z2 ¼ X2 þ Y2 þ Z2: ð36Þ Here again, the most natural conditions that satisfy (36) are given by (5), (6), and (7). It is clear that (5) and (6) define the circumcentre and the incentre of ABC, respectively. As for the centre defined by (7), we are indebted to Professor Clark Kimberling for drawing our attention to the fact that it is the centre catalogued in [1] as X40 and referred to sometimes as the Bevan point. It was argued in [3] that if AA0, BB0, and CC0 are the cevians through a centre E, and if x is a linear form in the side-lengths a, b, and c, then a a x ¼ þ tðb cÞ, X ¼ tðb cÞ ð37Þ 2 2 for t ¼ 0, 1=2or1=2, and where y, z, Y,andZ are defined by iteration. These limitations on the values of t are dictated by the cevian condition xyz ¼ XYZ. In contrast, it is interesting to note that (36) is satisfied for all t, since a 2 b 2 c 2 þ tðb cÞ þ þ tðc aÞ þ þ tða bÞ 2 2 2 a 2 b 2 c 2 ¼ tðb cÞ þ tðc aÞ þ tða bÞ 2 2 2

A

z Y C '

B ' Z y

BCxXA' Figure 10. 910 S. Abu-Saymeh and M. Hajja for all t. Thus if x, y, z, X, Y, and Z are given by (37) (and its iterates), then the perpendiculars from A0, B0,andC0 to the sides BC, CA, and AB are concurrent for all t. Denoting the centre defined this way by Pt, it would be interesting to study the shape of the curve P defined by t ° Pt, and to find its trilinear equation. It remains to consider the non-symmetric Brocard-like centres defined by (13) and (14), i.e. by

x ¼ y ¼ z and X ¼ Y ¼ Z

Plugging x ¼ y ¼ z in (36), one immediately obtains

a2 þ b2 þ c2 x ¼ 2ða þ b þ cÞ

The equation X ¼ Y ¼ Z is treated similarly. The two Brocard-like points thus obtained have already been studied in [35].

10. A note on central properties

Let E be a point inside the triangle ABC and let AA0, BB0 and CC0 be the cevians through E (figure 1). In [24], E was said to have the centroidal property at A if

BA0 ¼ A0C ð38Þ

(38) was so called because it satisfies the following three conditions : (i) (38) is symmetric in B and C. (ii) The centroid is completely defined by the iterates BA0¼A0C, CB0¼B0A and AC0¼C0B of (38) under the permutation ðABCÞ. (iii) Any two of the three iterates in (ii) imply the third. Gergonne, Nagel, incentral, circumcentral, orthocentral, and Fermat–Torricelli properties were defined by AB0 ¼ AC0, BC0 ¼ CB0, ffBAA0 ¼ffCAA0, ffB0BC ¼ ffC0CB, ffABB0 ¼ffACC0, and ffAEB ¼ffAEC, respectively. The previous sections tell us that these central properties could have been defined differently. For example, the property AB0 þ AC0 ¼ðAB þ ACÞ=2 is a centroidal property (at A) since it satisfies (i), (ii), and (iii), the centroid being defined also by (8). The table below lists different options for defining the various central properties. It refers to a triangle ABC with cevians AA0, BB0,andCC0 that meet at E, and where

a0 ¼ EA, b0 ¼ EB, c0 ¼ EC, ¼ffBEC, ¼ffAEC, ¼ffAEB

This list forms a source of questions regarding the family of triangles in which a certain centre has a certain central property. Some of these questions may turn out to be challenging and may give rise to more interesting families of what was referred to in [36] as balanced triangles. For example, the challenging Theorem 1 of [36] (or of [37]) states that the circumcentre of ABC has the Gergonne property AB0 ¼ AC0 if and only if AB ¼ AC. The readers are invited to discover by themselves the elegant equivalence of this theorem and the celebrated Steiner–Lehmus theorem. Another example is provided by [38], where the main result is equivalent to the statement that In search of more triangle centres 911 the centroid of ABC has the isodynamic property ffBAA0 þffBCC0 ¼ffCAA0 þffCBB0 if and only if the triangle is automedian in the sense that its medians are proportional in some order to the sidelengths. One may also ask about those triangles in which there is a point with two central properties for two different centres. For example, [39] states in effect that there exists P in ABC having the incentral property ffBAP ¼ffCAP and the Gergonne property AB0 ¼ AC0 if and only if AB ¼ AC. Questions along these lines may turn out to be suitable for classroom discussions, homeworks and examinations, and for team projects in which computer algebra and geometry packages may prove useful.

Centroidal properties (i) BA0 ¼ A0C (ii) AB0 þ AC0 ¼ðc þ bÞ=2 (iii) ½AC0EB0¼ð½ABB0þ½ACC0Þ=2 (iv) ½AC0EB0¼½EBC (v) ½AB0E¼½AC0E Gergonne properties (i) AB0 ¼ AC0 (ii) BC0 þ CB0 ¼ a (iii) BA0 ¼ða b þ cÞ=2 Nagel properties (i) BC0 ¼ CB0 (ii) AB0 þ AC0 ¼ a (iii) BA0 ¼ða þ b cÞ=2 Incentral properties (i) ffBAA0 ¼ffCAA0 (ii) ¼ 90 þ A=2 Circumcentral properties (i) b0 ¼ c0 (ii) ¼ 2A (iii) ffBAA0 ¼ðA þ B CÞ=2 ¼ 90 C Orthocentral properties (i) BCB0C0 is cyclic (ii) AC0EB0 is cyclic (iii) AA0 ? BC (iv) ffAB0E ¼ffAC0E (v) ¼ 180 A Fermat–Torricelli (i) ¼ 120 properties (ii) ¼ Isodynamic properties (i) ¼ A þ 60 (ii) ffBAA0 þffBCC0 ¼ffCAA0 þffCBB0

Acknowledgement

The authors would like to express their thanks to Yarmouk University for the financial support, and to Ahmad Hajja for drawing the figures.

References

[1] Kimberling, C., 1998, Triangle Centers and Central Triangles (Winnipeg, Canada: Congressum Numerantium, Vol. 129, Utilitas Mathematica Publishing Incorporated). [2] Coxeter, H.S.M., 1969, Introduction to Geometry (New York: John Wiley, Inc.). [3] Abu-Saymeh, S. and Hajja, M., Linearly defined triangle centers, preprint. [4] Abu-Saymeh, S. and Hajja, M., 2005, Forum Geom., 5, 33–36. [5] Kimberling, C., 1994, Math. Mag., 67, 163–187. [6] Bennett, G., Glenn, J. and Kimberling, C., 1986, Problem 3155, Amer. Math. Monthly, 93, 482 (Solution, ibid, 95 (1988), 874). [7] Yff, P., August 2004, Private Communications. [8] Yff, P., 1963, Am. Math. Monthly, 70, 495–501. [9] Baragar, A., 2001, A Survey of Classical and Modern Geometries (New Jersey: Prentice Hall). [10] Hajja, M., 1994, Math. Mag., 67, 30–36. [11] Poonen, B., 1994, Problem 10378, Am. Math. Monthly, 101, 363 (Solution, ibid, 104 (1997), 769–771). 912 S. Abu-Saymeh and M. Hajja

[12] Stroeker, R.J., 1984, Nieuw. Arch. Wisk., 2, 281–310. [13] Stroeker, R.J., 1988, Math. Mag., 61, 172–187. [14] Dirnbo¨ ck, H. and Schoißengeier, J., 2003, J. Geom. Graphics, 7, 23–39. [15] Honsberger, R., 2004, Mathematical Delights (Washington, DC: MAA). [16] Tsintsifas, G., 1983, Problem 1181, Math. Mag., 56, 326 (Solution, ibid, 57 (1984), 304–305). [17] Lee, H., 2000, Problem 1611, Math. Mag., 73, 403 (Solution, ibid, 74 (2001), 407). [18] Hajja, M., 2005, Problem 1711, Math. Mag., 78, 68. [19] Seebach, K., 1987, Elem. Math., 42, 132–139. [20] Huanxin, J., 1994, Problem 10358, Am. Math. Monthly, 101, 76 (Solution, ibid, 104 (1997), 567–570). [21] Hajja, M., The arbitrariness of the cevian triangle, Am. Math. Monthly, to appear. [22] Veldkamp, G.R., 1985, Am. Math. Monthly, 92, 546–558. [23] Hajja, M. and Yff, P., The isoperimetric point and the point(s) of equal detour in a triangle, preprint. [24] Andreescu, T. and Enescu, B., 2004, Mathematical Olympiad Treasures (Boston: Birkha¨ user). [25] Hajja, M., 2004, J. Geom. Graphics, 8, 17–22. [26] Pedoe, D., 1976, Math. Mag. 49, 40–42. [27] Goggins, J.R., 1986, Math. Gaz., 70, 133–134. [28] Paˆ rvaˆ nescu, F.S., 2000, Problem 10838, Am. Math. Monthly, 107, 950 (Solution, ibid, 109 (2002), 397–398). [29] van Lamoen, F., 2000, Problem 10830, Am. Math. Monthly, 107, 863 (Solution, ibid, 109 (2002), 396–397). [30] Furdui, O., 2002, Problem 10950, Am. Math. Monthly, 109, 569 (Solution, ibid, 110 (2003), 548–549). [31] Yiu, P., 2001, Math. Informatics Q., 11, 15–19. [32] Brown, B.H., 1924, Am. Math. Monthly, 31, 371–375. [33] The´ bault, V., 1946, Problem 4201, Amer. Math. Monthly, 53, 225 (Solution, ibid, 56 (1949), 114–117). [34] Clement, P.E., 1958, Am. Math. Monthly, 65, 601–605. [35] Goormaghtigh, M.R., 1918, Nouvelles Annales de Mathe´matique, 18, 417–424. [36] Hajja, M., 2001, Int. J. Math. Educ. Sci. Technol., 32, 21–36. [37] Hajja, M., 2001, Am. Math. Monthly, 108, 760–767. [38] Parry, C.F., 1990, Math. Gaz., 74, 151–154. [39] Izard, R., 1981, Problem 1132, Math. Mag., 54, 270 (Solution, ibid, 56 (1983), 47). [40] Chao, W.W., 2003, Problem 10996, Am. Math. Monthly, 110, 156 (Solution, ibid., 112 (2005), 88). [41] Bradley, C.J., 2005, Challenges in Geometry (New York: Oxford University Press).