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1 This aticle was downloaded by: [Paul Yiu] On: 11 Novembe 2011, At: 09:46 Publishe: Taylo & Fancis Infoma Ltd Registeed in England and Wales Registeed Numbe: Registeed office: Motime House, Motime Steet, London W1T 3JH, UK Intenational Jounal of Mathematical Education in Science and Technology Publication details, including instuctions fo authos and subsciption infomation: Polygonal tiples and the double uling of a hypeboloid Paul Yiu a a Depatment of Mathematical Sciences, Floida Atlantic Univesity, 777 Glades Road, Boca Raton, FL , USA Available online: 11 Nov 2011 To cite this aticle: Paul Yiu (2011): Polygonal tiples and the double uling of a hypeboloid, Intenational Jounal of Mathematical Education in Science and Technology, DOI: / X To link to this aticle: PLEASE SCROLL DOWN FOR ARTICLE Full tems and conditions of use: This aticle may be used fo eseach, teaching, and pivate study puposes. Any substantial o systematic epoduction, edistibution, eselling, loan, sub-licensing, systematic supply, o distibution in any fom to anyone is expessly fobidden. The publishe does not give any waanty expess o implied o make any epesentation that the contents will be complete o accuate o up to date. The accuacy of any instuctions, fomulae, and dug doses should be independently veified with pimay souces. The publishe shall not be liable fo any loss, actions, claims, poceedings, demand, o costs o damages whatsoeve o howsoeve caused aising diectly o indiectly in connection with o aising out of the use of this mateial.

2 Intenational Jounal of Mathematical Education in Science and Technology, 2011, 1 9, ifist CLASSROOM NOTE Polygonal tiples and the double uling of a hypeboloid Paul Yiu* Depatment of Mathematical Sciences, Floida Atlantic Univesity, 777 Glades Road, Boca Raton, FL , USA (Received 11 Decembe 2010) Fo a given positive intege k 6¼ 4, let P k,n denote the n-th k-gonal numbe. We study k-gonal tiples (a, b, c) satisfying P k,a þ P k,b ¼ P k,c : A k-gonal tiple coesponds to a ational point on the ectangula hypeboloid x 2 þ y 2 ¼ z 2 þ 1. The simple obsevation that this is a doubly uled suface leads to an easy, complete detemination of k- gonal tiples. The main esult is expessed in tems of pimitive Pythagoean tiples. We also establish a coespondence between (2h þ 1)-gonal tiples and 4h-gonal tiples. Keywods: polygonal tiples; ectangula hypeboloid; pimitive Pythagoean tiples 1. Intoduction We conside polygonal numbes of a fixed shape. Fo a given positive intege k, the sequence of k-gonal numbes consists of the patial sums of the aithmetic pogession with beginning tem 1 and common diffeence k 1. These ae the integes P k,n :¼ 1 2 nððk 2Þn ðk 4ÞÞ: ð1þ See, fo example, [1, pp ] o [2, p. 1]. 4-gonal numbes ae the squaes: P 4,n ¼ n 2. The tiangula numbes ae P 3,n ¼ 1 2 nðn þ 1Þ, pentagonal numbes P 5,n ¼ 1 2 nð3n 1Þ, hexagonal numbes P 6,n ¼ n(2n 1), and heptagonal numbes P 7,n ¼ 1 2 nð5n 3Þ. Table 1 shows some beginning polygonal numbes. Genealizing the notion of Pythagoean tiple, we define a k-gonal tiple as a tiple of positive integes (a, b, c) satisfying P k,a þ P k,b ¼ P k,c : ð2þ * yiu@fau.edu ISSN X pint/issn online ß 2011 Taylo & Fancis

3 2 Classoom Note Table 1. Polygonal numbes. n P 3,n P 5,n P 6,n P 7,n Fom Table 1, we have the following: (i) tiangula tiples: (2, 2, 3), (3, 5, 6), (5, 6, 8), (4, 9, 10),..., and (ii) pentagonal tiples: (4, 7, 8), (5, 5, 7),..., (iii) hexagonal tiples (5, 11, 12), (13, 14, 19),..., and (iv) heptagonal tiples (6, 16, 17), (24, 27, 36),... We shall assume in this aticle that k 6¼ 4. Since the definition of a polygonal numbe in (1) can be ewitten as we ewite (2) as P k,n ¼ 1 8ðk 2Þ ð2ðk 2Þn ðk 4ÞÞ2 ðk 4Þ 2, ð2ðk 2Þa ðk 4ÞÞ 2 þð2ðk 2Þb ðk 4ÞÞ 2 ¼ ð2ðk 2Þc ðk 4ÞÞ 2 þðk 4Þ 2 : ð3þ Dividing thoughout by (k 4) 2, we obtain a ational point on the suface Pðk; a, b, cþ :¼ ðga 1, gb 1, gc 1Þ, S : x 2 þ y 2 ¼ z 2 þ 1: g ¼ 2ðk 2Þ k 4, ð4þ This is the suface of evolution of a ectangula hypebola about its conjugate axis. Fo k ¼ 3, 5, 6, 8, the point P(k; a, b, c) is always an intege point, with coesponding g ¼ 2, 6, 4, 3. Fo k ¼ 3 (tiangula numbes), we shall change signs, and conside instead the point P 0 ð3; a, b, cþ :¼ ð2a þ 1, 2b þ 1, 2c þ 1Þ, ð5þ whose coodinates ae all odd integes exceeding Double uling of the hypeboloid S It is well-known that the suface S, being a hypeboloid of one sheet, has a double uling [3, p. 11]. This means that though each point on the suface, thee ae two

4 Intenational Jounal of Mathematical Education in Science and Technology 3 staight lines lying entiely on the suface. Let P(x 0, y 0, z 0 ) be a point on the suface S. A line though P with diection numbes p : q : has paametization : x ¼ x 0 þ pt, y ¼ y 0 þ qt, z ¼ z 0 þ t: Substitution of these expessions into the equation of S shows that the line is entiely contained in the suface S if and only if Since x 2 0 þ y2 0 z2 0 ¼ 1, we have This means px 0 þ qy 0 ¼ z 0, p 2 þ q 2 ¼ 2 : 2 ¼ 2 ðx 2 0 þ y2 0 z2 0 Þ ¼ 2 ðx 2 0 þ y2 0 Þ ðpx 0 þ qy 0 Þ 2 ¼ðp 2 þ q 2 Þðx 2 0 þ y2 0 Þ ðpx 0 þ qy 0 Þ 2 ¼ðqx 0 py 0 Þ 2 : ð6þ ð7þ Solving Equations (6) and (8), we get qx 0 py 0 ¼ ", " ¼1: ð8þ p ¼ ðx 0z 0 "y 0 Þ x 2 0 þ y2 0 and q ¼ ð y 0z 0 þ "x 0 Þ x 2 0 þ : y2 0 Fom these, we have the diection numbes of the line as p : q : ¼ ðx 0z 0 "y 0 Þ x 2 0 þ : ð y 0z 0 þ "x 0 Þ y2 0 x 2 0 þ : y2 0 ¼ x 0 z 0 "y 0 : y 0 z 0 þ "x 0 : x 2 0 þ y2 0 : We summaize this in the following poposition.

5 4 Classoom Note Poposition 1: The two lines lying entiely on the hypeboloid S : x 2 þ y 2 ¼ z 2 þ 1 and passing though P(x 0, y 0, z 0 ) have diection numbes x 0 z 0 "y 0 : y 0 z 0 þ "x 0 : x 2 0 þ y2 0, fo " ¼1. In paticula, if P is a ational point, these diection numbes ae ational. 3. Pimitive Pythagoean tiple associated with a k-gonal tiple Let P be the ational point detemined by a k-gonal tiple (a, b, c), as given by (4) fo k 5, and by (5) fo k ¼ 3 (tiangula numbes). We fist note that the coodinates of P all exceed 1. This is clea fo k ¼ 3. Fo k 5, it follows fom the fact that g ¼ 2ðk 2Þ k 4 ¼ 2 þ 4 k The diection numbes of the uling lines on S though the point P, as given in Poposition 1, ae all positive. In view of (7), we may theefoe choose a pimitive Pythagoean tiple (p, q, ) fo these diection numbes. Poposition 2: Let k 6¼ 4. Evey k-gonal tiple (a, b, c) detemines two infinite sequences of k-gonal tiples coesponding to the two lines on S though P(k; a, b, c) (o P 0 (3; a, b, c)) each with diection numbes given by a pimitive Pythagoean tiple. Example 1: The tiangula tiple (3, 5, 6) coesponds, accoding to (5), to the point (2 3 þ 1, 2 5 þ 1, 2 6 þ 1) ¼ (7, 11, 13) on S. Though this point the two lines on S have diection numbes (i) fo " ¼ 1, 7 13 þ 11 : :7 2 þ 11 2 ¼ 102 : 136 : 170 ¼ 3 : 4 : 5, and (ii) fo " ¼ 1, : þ 7:7 2 þ 11 2 ¼ 80 : 150 : 170 ¼ 8 : 15 : 17. Along the line with diection numbes 3 : 4 : 5, we have the tiangula tiples ð3 þ 3t, 5þ 4t, 6þ 5tÞ, t ¼ 0, 1, 2,..., and along the line with diection numbes 8:15:17 we have the tiangula tiples ð3 þ 8t, 5þ 15t, 6þ 17tÞ, t ¼ 0, 1, 2,..., Example 2: The pentagonal tiple (4, 7, 8) coesponds, accoding to (4), to the point (23, 41, 47) on S. The two lines on S though this point have diection numbes 8 : 15 : 17 and 33 : 56 : 65, espectively. These lead to two infinite sequences of pentagonal tiples: (i) (4 þ 8t, 7þ 15t, 8þ 17t), and (ii) (4 þ 33t, 7þ 56t, 8þ 65t) fo t ¼ 0, 1, 2, Tiangula tiples fom Pythagoean tiples We study the convese question of detemining k-gonal tiples fom (pimitive) Pythagoean tiples. As is well-known, evey pimitive Pythagoean tiple is geneated by two elatively pime integes m 4 n of diffeent paity: p ¼ m 2 n 2, q ¼ 2mn, ¼ m 2 þ n 2 : ð9þ

6 Intenational Jounal of Mathematical Education in Science and Technology 5 Given one such pimitive Pythagoean tiple (p, q, ), we want to detemine a tiangula tiple (a, b, c) coesponding to it. Given an odd intege z 0 4 1, we obtain, fom (6) and (8), x 0 ¼ pz 0 þ "q, y 0 ¼ qz 0 "p : ð10þ We claim that it is possible to choose z so that x 0 and y 0 ae also odd integes 4 1. By the Euclidean algoithm, thee ae odd integes u and v such that qu þ v ¼ 1. (Note that v must be odd, since q is even. If u is even, we eplace (u,v)by(u, v þ q), in which both enties ae odd.) Clealy, fo " ¼1, the intege z 0 ¼ "pu is such that qz 0 "p ¼ "p(qu 1) ¼ "pv. This makes y 0 an intege. The coesponding x 0 is also an intege. Replacing z 0 by z 0 þ t fo a positive intege t if necessay, the integes z 0, x 0 and y 0 can be chosen geate than 1. Fom (10), the integes x 0 and y 0 ae both odd, since p and q ae of diffeent paity and z 0 is odd. We summaize this in the following theoem. Theoem 3: Let ( p, q, ) be a pimitive Pythagoean tiple. Thee ae two infinite families of tiangula tiples (a " (t), b " (t), c " (t)), " ¼1, such that one of the two lines on S containing the point P 0 (3; a " (t), b " (t), c " (t)) has diection numbes p : q :. Example 3: Conside the pimitive Pythagoean tiple 3 : 4 : 5. With x 0 ¼ 3z 0þ4" 5 and y 0 ¼ 4z 0 3" 5, we choose fo " ¼ 1, z 0 ¼ 13 and obtain the point (x 0, y 0, z 0 ) ¼ (7, 11, 13) on S leading to the tiangula tiples (3 þ 3t, 5þ 4t, 6þ 5t). (The othe line though (7,11,13) has diection numbes 8 : 15 : 17). Fo " ¼ 1, z 0 ¼ 7 and obtain the point (5, 5, 7) on S leading to the tiangle tiples (2 þ 3t, 2þ 4t, 3þ 5t). (The othe line though (5,5,7) has diection numbes 4 : 3 : 5). Table 2 gives some futhe examples. 5. k-gonal tiples fom Pythagoean tiples Now, we conside k 5. We shall adopt the notation h 0 h if h is odd, :¼ h 2 if h is even, fo an intege h. Table 2. Tiangula tiples fom pimitive Pythagoean tiples. (m, n) (p, q, ) (a þ (0), b þ (0), c þ (0)) (a (0), b (0), c (0)) (2, 1) (3, 4, 5) (2, 2, 3) (3, 5, 6) (4, 1) (15, 8, 17) (9, 4, 10) (5, 3, 6) (3, 2) (5, 12, 13) (4, 9, 10) (5, 14, 15) (6, 1) (35, 12, 37) (20, 6, 21) (14, 5, 15) (5, 2) (21, 20, 29) (6, 5, 8) (14, 14, 20) (4, 3) (7, 24, 25) (6, 20, 21) (7, 27, 28) (8, 1) (63, 16, 65) (35, 8, 36) (27, 7, 28) (7, 2) (45, 28, 53) (35, 21, 41) (9, 6, 11) (5, 4) (9, 40, 41) (8, 35, 36) (9, 44, 45)

7 6 Classoom Note Theoem 4: Let k 5. A line on S with diection numbes defined accoding to (9) by m 4 n geneates k-gonal tiples if and only if (k 2) 0 divides one of n and m n. Poof: As in (10) above, the ational points though which the suface S contains a line of diection numbes p : q : ae of the fom pz þ "q, qz "p, z : ð11þ Suppose this coesponds, accoding to (4), to a k-gonal tiple (a, b, c). Fo " ¼ 1, we obtain fom (10), This equation simplifies to ga 1 ¼ pz þ q, gb 1 ¼ qz p, z ¼ gc 1: ð12þ a ¼ gpc p þ q þ, b ¼ g Substituting Equation (9) into (13), we get a ¼ gqc p q þ : ð13þ g m þ n ðk 2Þ 0 ðm 2 þ n 2 Þ ðk 2Þ0 ðm nþc þðk 4Þ 0 n, ð14þ n b ¼ ðk 2Þ 0 ðm 2 þ n 2 Þ ðk 2Þ0 2mc ðk 4Þ 0 ðm nþ : ð15þ Note that (k 2) 0 and (k 4) 0 ae always elatively pime, since gcd(k 2, k 4) ¼ 1 o 2 accoding as k is odd o even. Fom these expessions, a 2 þ b 2 c 2 2ðk 4Þ 0 n ¼ ðk 2Þ 0 2ðm 2 þ n 2 Þ ðk 2Þ0 ðm nþc þðk 4Þ 0 n : ð16þ We claim that n must be divisible by (k 2) 0 fo a, b, c to be integes. Let d :¼ gcd(n, (k 2) 0 ), so that n ¼ d n, ðk 2Þ 0 ¼ d ðk 2Þ fo elatively pime integes n* and (k 2)*. It follows fom (16) that a 2 þ b 2 c 2 2ðk 4Þ 0 n ¼ ðk 2Þ 2 ðm 2 þ n 2 Þ ðk 2Þ ðm nþc þðk 4Þ 0 n : Since (k 2)* is pime to each of (k 4) 0 and n*, the only possible pime diviso of (k 2)* is 2. This means that (k 2)* is a powe of 2, (possibly 1). If (k 2)* is even, then afte cancelling a common diviso 2, the numeato of a 2 þ b 2 c 2 is odd, and the denominato is even. This cannot be an intege. It follows that (k 2)* ¼ 1, justifying the claim that n must be divisible by (k 2) 0. Since g ¼ 2ðk 2Þ0 ðk 4Þ, the condition that n be divisible by (k 0 2)0 is equivalent to 2n g being an intege. Unde this condition, thee is a unique positive intege c 0 5 m 2 þ n 2 fo which a 0 defined by (14) is an intege. Note that a 2 0 þ b2 0 c2 0 is also an intege.

8 Intenational Jounal of Mathematical Education in Science and Technology 7 Since b 0 is ational, it too must be an intege. Evey k-gonal tiple associated with the pimitive Pythagoean tiple (p, q, ) is of the fom fo a positive intege t. Fo " ¼ 1, we obtain fom (10), a t ¼ a 0 þ pt, b t ¼ b 0 þ qt, c t ¼ c 0 þ t, ga 1 ¼ This equation simplifies to a ¼ pz q, gb 1 ¼ qz þ p, z ¼ gc 1: ð17þ gpc p q þ, b ¼ g gqc þ p q þ : ð18þ g Substituting equation (9) into (18), we get a ¼ b ¼ m n ðk 2Þ 0 ðm 2 þ n 2 Þ ðk 2Þ0 ðm þ nþc ðk 4Þ 0 n, ð19þ m ðk 2Þ 0 ðm 2 þ n 2 Þ ðk 2Þ0 2nc þðk 4Þ 0 ðm nþ : ð20þ In this case, a simila easoning as above shows that m n must be divisible by (k 2) 0 fo a, b, c to be integes. œ Since m and n ae elatively pime, the intege (k 2) cannot divide both n and m n. This means that a pimitive Pythagoean tiple (p, q, ) coesponds to at most one line on S associated with k-gonal tiples (fo k 5). Example 4: The Pythagoean tiple fom (m, n) geneates pentagonal tiples if and only if one of m n and n is divisible by 3. Since the Pythagoean tiple ( p, q, ) ¼ (3, 4, 5) esults fom (m, n) ¼ (2, 1), fo which none of m n and n is divisible by 3, thee is no pentagonal tiples on uling lines of S with diection numbes 3 : 4 : 5. Table 3 gives some examples. The pentagonal tiples (a þ (0), b þ (0), c þ (0)) ae computed using (14) and (15) with (k 2) 0 ¼ 3 and (k 4) 0 ¼ 1 fo a suitable choice of c þ (0) 5 ¼ m 2 þ n 2. Likewise, the tiples (a (0), b (0), c (0)) ae computed using (19) and (20), also with (k 2) 0 ¼ 3 and (k 4) 0 ¼ 1. Example 5: The Pythagoean tiple fom (m, n) geneates heptagonal tiples if and only if one of m n and n is divisible by 5. The examples in Table 4 ae constucted using (14) and (15) fo n divisible by 4 with (k 2) 0 ¼ 5 and (k 4) 0 ¼ 3, and using (19) and (20) fo m n divisible by 5, also with (k 2) 0 ¼ 5, and (k 4) 0 ¼ 3. Example 6: Fo k ¼ 4h þ 2, k-gonal tiples can be obtained only fom pimitive Pythagoean tiples with even n only since (k 2) 0 ¼ 2h does not divide the odd numbe m n. Table 5 gives some examples of hexagonal tiples geneated by pimitive Pythagoean tiples with even n, using (14) and (15), with (k 2) 0 ¼ 2 and (k 4) 0 ¼ 1.

9 8 Classoom Note Table 3. Pentagonal tiples fom pimitive Pythagoean tiples. (m, n) (p, q, ) (a þ (0), b þ (0), c þ (0)) (a (0), b (0), c (0)) (4, 1) (15, 8, 17) (7, 4, 8) (4, 3) (7, 24, 25) (7, 23, 24) (5, 2) (21, 20, 29) (5, 5, 7) (7, 4) (33, 56, 65) (4, 7, 8) (7, 6) (13, 84, 85) (13, 82, 83) (8, 3) (55, 48, 73) (22, 19, 29) (8, 5) (39, 80, 89) (35, 72, 80) Table 4. Heptagonal tiples fom pimitive Pythagoean tiples. (m, n) (p, q, ) (a þ (0), b þ (0), c þ (0)) (a (0), b (0), c (0)) (6, 1) (35, 12, 37) (16, 6, 17) (6, 5) (11, 60, 61) (11, 57, 58) (7, 2) (45, 28, 53) (33, 21, 39) (8, 3) (55, 48, 73) (27, 24, 36) (8, 5) (39, 80, 89) (39, 79, 88) (9, 4) (65, 72, 97) (24, 27, 36) Table 5. Hexagonal tiples fom pimitive Pythagoean tiples. (m, n) (p, q, ) (a, b, c) (3, 2) (5, 12, 13) (5, 11, 12) (5, 2) (21, 20, 29) (14, 13, 19) (5, 4) (9, 40, 41) (9, 38, 39) (7, 2) (45, 28, 53) (18, 11, 21) (7, 4) (33, 56, 65) (11, 18, 21) (7, 6) (13, 84, 85) (13, 81, 82) (9, 2) (77, 36, 85) (11, 5, 12) (9, 4) (65, 72, 97) (13, 14, 19) (9, 8) (17, 144, 145) (17, 140, 141) 6. Coespondence between (2h Y 1)-gonal and 4h-gonal tiples Let k 1 5 k 2 be two positive integes 5. Theoem 4 suggests that thee is a oneto-one coespondence between k 1 -gonal tiples and k 2 -gonal tiples, povided (k 1 2) 0 ¼ (k 2 2) 0. This is the case if and only if k 1 ¼ 2h þ 1, k 2 ¼ 4h, fo some h 2: ð21þ In this case, (k 1 2) 0 ¼ (k 2 2) 0 ¼ 2h 1, while (k 1 4) 0 ¼ 2h 3, and (k 2 4) 0 ¼ 2h 2. Recall that the existence of (2h þ 1)-gonal and 4h-gonal tiples coesponding to the Pythagoean tiple geneated by m 4 n equies 2h 1 dividing n o m n. If n is divisible by 2h 1, then the smallest c 4 0 fo a (2h þ 1)-gonal tiple (a, b, c)

10 Intenational Jounal of Mathematical Education in Science and Technology 9 Table 6. Coesponding (2h þ 1)-gonal and 4h-gonal tiples. (h,2h þ 1, 4h) (m, n) (p, q, ) (2h þ 1)-gonal (a, b, c) 4h-gonal (a 0, b 0, c 0 ) (2, 5, 8) (4, 1) (15, 8, 17) (7, 4, 8) (14, 8, 16) (4, 3) (7, 24, 25) (7, 23, 24) (7, 22, 23) (5, 2) (21, 20, 29) (5, 5, 7) (10, 10, 14) (7, 4) (33, 56, 65) (4, 7, 8) (8, 14, 16) (7, 6) (13, 84, 85) (13, 82, 83) (13, 80, 81) (8, 3) (55, 48, 73) (22, 19, 29) (44, 38, 58) (8, 5) (39, 80, 89) (35, 72, 80) (31, 64, 71) (3, 7, 12) (6, 1) (35, 12, 37) (16, 6, 17) (33, 12, 35) (6, 5) (11, 60, 61) (11, 57, 58) (11, 56, 57) (7, 2) (45, 28, 53) (33, 21, 39) (44, 28, 52) (8, 3) (55, 48, 73) (27, 24, 36) (36, 32, 48) (8, 5) (39, 80, 89) (39, 79, 88) (26, 52, 58) (9, 4) (65, 72, 97) (24, 27, 36) (32, 36, 48) and the smallest c fo a 4h-gonal tiple (a 0, b 0, c 0 ) ae elated by ðm nþðc c 0 Þ n 2h 1 ðmod m2 þ n 2 Þ: On the othe hand, if 2h 1 divides m n, then these satisfy 2nðc c 0 Þ m n 2h 1 ðmod m2 þ n 2 Þ: Fo example, given the pentagonal tiple (13, 82, 83) (see Example 4), we find the coesponding octagonal tiple. In Table 6, this pentagonal tiple coesponds to (m, n) ¼ (7, 6) and the Pythagoean tiple (13, 84, 85). Hee, h ¼ 2 and 2h 1 ¼ 3 divides n ¼ 6. Theefoe, we find c 0 satisfying ð7 6Þðc c 0 Þ 6 3 ¼ 2ðmod 85Þ. This yields c 0 ¼ 81. Now, fom (14) and (15), we have (a 0, b 0 ) ¼ (13, 18); hence the octagonal tiple (13, 80, 81), as shown in Table Concluding emak We have demonstated how tiples of k-gonal numbes (P k,a, P k,b, P k,c ) satisfying P k,a þ P k,b ¼ P k,c can be oganized accoding to the double uling of the hypeboloid suface S : x 2 þ y 2 ¼ z 2 þ 1. The diection numbes of uling lines though ational points can be egaded as Pythagoean tiples. Fo k ¼ 3, given a lattice point on S coesponding to a tiangula tiple, the two uling lines though the point contain two infinite sequences of tiangula tiples. Fo a fixed k 5, we have shown how to decide if a line of given diection numbes contains ational points coesponding to k-gonal tiples, and how to explicitly identify such tiples. Refeences [1] J.H. Conway and R.K. Guy, The Book of Numbes, Spinge, New Yok, [2] L.E. Dickson, Histoy of Numbes, Vol. II, Chelsea, New Yok, 1920, Dove epint, [3] D. Hilbet and S. Cohn-Vossen, Geomety and the Imagination, 1932, AMS epint, 1999.

Heronian Triangles of Class K: Congruent Incircles Cevian Perspective

Heronian Triangles of Class K: Congruent Incircles Cevian Perspective Foum Geometicoum Volume 5 (05) 5. FORUM GEOM ISSN 534-78 Heonian Tiangles of lass K: onguent Incicles evian Pespective Fank M. Jackson and Stalislav Takhaev bstact. We elate the popeties of a cevian that