INTEGER WITH MEDIANS

Nicolae Nicolae, Secondary School Number 6, Vulcan, county of Hunedoara, Romania

ABSTRACT: This study introduces the triangles with sides of integer numbers and the results referring to their medians which have as a length integer numbers, too.

KEYWORDS: integer triangles, integer medians

MSC (2010): 51M25

INTRODUCTION AEF rational numbers where E, F are the feet of the bisectors from B and C. The study here tries to present A notion of in general can be interpreted some remarkable results of the geometry of the figures in many different ways. We think of geometry as of a of the integer values. set of objects and a relation which is normally defined by some group of transformations. For 1.TRIANGLES WITH INTEGER SIDES instance, in Euclidean geometry in the plane we study points, lines, segments, polygons, circles, etc, the Historically speaking, the is one of the congruence relation is defined by a group of all length mathematical objects which have attracted the attention preserving transformations (2, ) or the orthogonal of the man since antiquity. The collection of 84 group. problems on the papyrus Rhind (discovered in 1858 in The between the푂 geometricalℝ figures and the Valley of the Middle Nile and achieved by the arithmetic clearly leads us to taking in consideration Scottish Egyptologist HenrzRind ) dated from 1780- those which have the values expressed through integer 1700 B.C . which contains problems of geometry where numbers, or , generally through rational numbers. It is the scribe Ahmes presents the calculus formula of the well-known the formula that generates the integer-sided of an with the basis of 4 units and triangles which are not congruent and having the the sides of 10 units . In Book 4 from the Arithmetic of equal to a given number n. According to Diophantus it appears the problem of finding the right Alcuin from York (735-804) { ( )} the Alcuin’s triangle with integer sides and the length of the bisector sequence that expresses the number of non-congruent of one of the sharp , having as a triangles which have the perimeter n;푡 푛for the first values solution the triangle with the sides : 7, 24, 25 which cuts of n we have : the 24 long leg having the length of . We should also n 0 1 2 3 4 5 6 7 8 9 10 consider the works of Giovanni Ceva35 from the 17th ( ) 0 0 0 1 0 1 1 2 1 3 2 4 century who considers the triangles with integer sides It was proved that the function that generates the and analyses the Cevians problems ( closed intervals Alcuin’s푡 푛 sequence is : which join a vertex with a point from the opposite side

) concurrent and rational. The first mathematician who tries to find integer = ∞ ( ) (1 )(1 3 )(1 ) triangles with integer medians is Euler, the smallest 푥 푛 2 3 4 � 푡 푛 푥 triangles being 2( 68,85, 87) and 2 ( 127, 131, 158) − 푥 − 푥 − 푥 푛=0 An important role in the developing of the Geometry subject is taken by the computer, many of the results being the consequences of the calculus made with the help of some scientific programs or being part of some scientific projects : for example in project Euler problem 75 https://projecteuler.net/problem=75) which seeks to determinate the numbers of the right triangles with the perimeter equal to n or the problem 257 (http://projecteuler.net/problem=257) which seeks the numbers of the integer triangles ABC with the integer bisectors and the ratio between the area ABC and area Figure 1.1

543 It can be noticed that the sides of the big triangle are Definition 1.2. Let > 1 free of squars. We say that the double of the medians of the other triangle. In 1779 A belongs to the class n iff . If Euler investigates the triangles with the above- we say that푛 angle A belongs to class 1. mentioned property, obtaining for its sides the following 푠푖푛 퐴 ∈ 푸�√푛� expressions: 푠푖푛Observation퐴 ∈ 푸 1.1. The rational triangle ABC belongs to class n cu n , > 1 , n free of squares , if ( ) = 9 + 26 + , , ∗ . If , , 4 2(92 4 6 + ) we say that angle∈ 푵 A푛 belongs to class 1. 푎 = 푚(9 푚 + 26 푚 푛 +4푛 ) 2 2 4 푠푖푛 퐴 푠푖푛퐵 푠푖푛퐶 ∈ 푸�√푛� 푠푖푛 퐴 푠푖푛퐵 푠푖푛퐶 ∈ 4 −+ 2 푛(92 푚 −4 6푚 푛 + 푛 ) Example푸 : For = 7, = 5, = 8 the triangle belongs 푏 = 2푚 (9푚 10푚 푛 4푛3 ) 2 2 4 to class 3 because = 3. 4 푛2 2푚 − 4푚 푛 푛 푎 푏 1 푐 푐 In 푚1813푚 N.− Fuss푚 푛gives− 푛the first Example of an Through direct calculus we find the class of some 푠푖푛퐴 2 √ integer triangle with the three rational bisectors and, triangles: necessarily, the rational area as well ( = 14, = a b c Class 25, = 25) with the bisectors of = 24 and 푎 푏 1 2 2 15 푐 푖푎 = ( 1 = = 2 2 3 7 ( ) 4푎푐 2 560 2 푖푏 � 푎+푐 푝 푝 − 푏� 39 푖푐 2 3 4 15 Definition 1.1. The triangle ABC is called an integer triangle if its sides have the length integer numbers. The 3 4 5 1 triangle ABC is called rational triangle if its sides have the length rational numbers. 3 7 8 3

Theorem 1.1. For every rational triangle ABC there is 5 7 8 3 a similar integer triangle A’ B’ C’. 2 7 7 3 Proof: If ABC has the sides: = , = , = , 푎1 푏1 푐1 5 5 6 1 , , , , , , then the triangle with the sides 푎 푎2 푏 푏2 푐 푐2 , , is similar ( we multiplied the 1 2 1 2 1 2 sides푎 푎 of푏 the푏 ABC푐 푐 with∈ 풁 ) to ABC. Consequence 1.2. The rational right triangles belong 1 2 2 2 1 2 2 2 1 푎 푏 푐 푎 푏 푐 푎 푏 푐 to class 1. 2 2 2 Theorem 1.2 In a rational푎 푏 푐 triangle the values of the Proof: Obviously sin A=1 , = . cosines of the angles are rational numbers. 푏 푐 Proof: Applying the law of the cosines in ABC with Theorem 1.4. If a triangle퐵 has ∈the 푸 cosines for two the classical notation we have: angles rational numbers , then the cosine of the third one is rational if and only if the two angles belong to the = + 2 same class. 2 2 2 Proof: For C, the third angle, we have : = 180 푎 푏 푐 − 푏푐 푐표푠퐴 where = and the others as analogue = sin 0 2 2 2 where . In , 푎. −푏 −푐 conclusion if and only if 퐶 −is 2푏푐 푐표푠퐴 ∈ 푸 퐴equivalent − 퐵 to A and푐표푠퐶 B from the퐴 푠푖푛퐵 same− class.푐표푠퐴푐표푠퐵 Theorem푐표푠퐵 푐표푠퐶 1.3∈ 푸In a rational triangle the ratio between 푐표푠퐶 ∈ 푸 푠푖푛퐴푠푖푛퐵 ∈ 퐐 the of two angles is a rational number. 2.TRIANGLES WITH INTEGER MEDIANS Proof: From the sine rule: In order to find the expressions of the we will apply the law of the cosine for ABC and ADC with = = = 2 , sin sin sin AD as a median : 푎 푏 푐 = + 2 푅 We퐴 deduce퐶 that : 퐶 = . 2 2 2 퐴퐷 퐴퐶 퐷퐶 − 퐴퐶 ∙ 퐷퐶A ∙ 푐표푠퐶̂ sin 퐴 푎 sin 퐵 푏 Consequence 1.1. If ABC is ∈a 푸rational triangle which has sin = then 2c 2b k sin = , sin = F 1 E 퐴where푟 √ 푛, , and . 2 3 G 퐵 푟 √푛sin =퐶 sin푟 √푛= = Proof: We have1 2: 3 . l 푟 푟 푟 ∈ 푸 푏 푛 ∈ 푵푏 m Analogous for sin C. 1 2 푎 푎 B 퐵 퐴 푟 √푛 푟 √푛 C 2a D

544 = + 2 relation (2.1.) we have k which is even and from the 2 2 2 other relations from (2.1.) that l, m are odd. 퐴퐵We eliminate퐴퐶 퐵퐶 − 퐴퐶from∙ 퐵퐶 the∙ two푐표푠 퐶equalitieŝ and we obtain the well-known relation of the median: Theorem 2.4. Only one of the numbers a, b, c, k, l, m is 푐표푠퐶̂ devisable by 4. 4 = 2 + 2 Proof: According to the previous theorem we can 2 2 2 2 suppose that = 2 , = 2 + 1, = 2 + 1 and 퐴퐷If the previously퐴퐵 퐴퐶 mentioned− 퐵퐶 equality stays true for = 2 , = 2 + 1, = 2 + 1. Replacing the first the natural numbers it is necessary that BC=c to be relation with (2.1.)푎 we ∝get:푏 훽 푐 훾 even and their analogous as well so that the other 푘 훿 푙 휆 푚 휇 medians to be integer. Therefore we say that the triangle + = 2 + 2 ABC has even sides with the lengths: 2a, 2b, 2c.We 2 2 2 2 mark their medians from A,B, C with k, l, m. The 푘or 푎 푏 푐 relations that give these lengths become : 4 + 4 = 8 + 8 + 2 + 8 + 8 + 2 = 2 + 2 2 2 2 2 2 = 2 2+ 2 2 2 (2.1.) from훿 where훼 : 훽 훽 훾 훾 푘2 = 2푏2 + 2푐2 − 푎2 � 푙 2 푐 2 푎 2− 푏 2 + = 2 + 2 + 2 + 2 + 1 푚 푎 푏 − 푐 2 2 2 2 By adding them we obtain : + + = 3( + 훿meaning훼 훽that 훽 훾 + 훾 1( 2) which + ) and the first equation can2 be 2written2 like this2 : demonstrates that or or 2 are 2even and or a or k are 2 = 22 + 2 = 2( +푘 +푙 ) 푚 3 =푎 divisible by 4. 훿 훼 ≡ 푚표푑 푏(2 푐+ 2 + 2) 32 . It 2follows2 that2 the relations2 ∝ 훿 2푘 푏 푐 − 푎 푎 푏 푐 − 푎 Theorem 2.5. (Euler) If the triangle ABC with the sides that 2give 2the expressions2 2 of the semi sides according to 3 (2a, 2b, 2c) is in MED and has the medians (k, l, m) the푘 medians푙 are푚 : − 푎 then the triangle A’B’C’ with the sides (2k, 2l, 2m) is in

9 = 2 + 2 MED. Proof: By marking the medians of the triangle A’B’C’ 9 2 = 2 2 + 2 2 2 (2.2.) with , , we have: 9푎2 = 2푙 2 + 푚 2− 푘2 � 푏 푚 푘 − 푙 ′= 2′ (2 + 2 ) + 2(2 + 2 ) 2 2 2 2 푘 푙 푐 푘 푙 − 푚 ′2 2(2 +2 2 2 ) = 92 2 2 푘 푐 푎 − 푏 푎 푏 − 푐 − Definition 2.1. We name a med- triangle an integer ⎧ = 2(2 + 22 2 ) +2 (2 +2 2 ) 푏 푐 − 푎 푎 triangle with integer medians. We mark the set of the ⎪ ′2 2(2 +2 2 2 ) =2 9 2 2 med- triangles with MED. 푙 푎 푏 − 푐 푏 푐 − 푎 − = 2(2 + 2 2 ) +2 2(2 +2 2 )

⎨ 2 푐 푎 − 푏 푏 Theorem21. If a, b, c, k, l, m, are natural numbers with ′ 2(2 +2 2 2 ) = 92 2 2 ⎪푚 푏 푐 − 푎 푐 푎 − 푏 − ( , , , , , ) = 1 then only one of the halves of the ⎪ 2 2 2 2 ⎩ sides is an even number. meaning that the medians푎 푏 of −the 푐 new triangle푐 are exact 푎 푏 푐 푘 푙 푚 (3a,3b,3c). The process can be clearly continued , Proof: Suppose that two halves of the sides are even meaning that if (2 , 2 , 2 ) with the medians and one of them is odd meaning that : 0( 2), (k, l, m) then (2 , 2 , 2 ) with the medians 푎 푏 푐 ∈ 푀퐸퐷 0( 2) 1( 2) . Then from the (3a,3b,3c) and therefore (6 , 6 , 6 ) is a 푘 푙 푚 ∈ 푀퐸퐷 equation (2.1.) we get: 2 푎+ ≡ 2 푚표푑 triangle similar to the initial one. 푎 푏 푐 ∈ 푀퐸퐷 푏2( ≡ 푚표푑 4) which푎푛푑 is 푐false ≡ 푚표푑because2 a perfect2 square2 2 is congruent with 0 or 1 modulo 4.푘 ≡ 푏 푐 − 푎 ≡ Theorem 3.6. There are no isosceles triangles in MED. Suppose푚표푑 that all the halves of the sides are odd meaning Proof: If a=b then k=l and the equations (2.1.) that 1( 2), 1( 2) become: 1( 2) . Then the same relation becomes: = + 2 2 + 2 푎 ≡3( 푚표푑 4), again푏 ≡ impossible.푚표푑 푎푛푑 푐2 ≡ (2.3.) 2 = 42 2 It 푚표푑stays2 true2 the2 fact that only one half of the sides푘 ≡is 푘 푎 푐 even푏 , the푐 other− 푎 two≡ being푚표푑 odd. � 2 2 2 The푚 second푎 equation− 푐 tells us that m and c have the same

parity and if they are odd 1( 4) and Theorem 2.2. The semi perimeter of the triangle MED 4 3( 4) the equality2 is impossible . is even . Hence2 both2 m and c are even or 푚=≡ 2 , 푚표푑= 2 the Proof: Pointed in the theorem 2.1. relations푎 − 푐 (3.3.)≡ 푚표푑 being:

푐 퐶 푚 푀 Theorem 2.3. Only one median is even the other two = + 8 being odd. 2 = 2 2 Proof: From theorem 2.1 we find out that only one half �푘 2 푎 2 퐶2 of the side is even , being marked with a. From the first 푀 푎 − 퐶

545 Or replacing the first relation with their difference we Proof: Supposing we have the relation : + = 2 get: which, replaced in (2.1), gives us: 2 2 2 푎 푐 푏 = + 9 = 3 (2.4.) 2 = 2 + 2 2 = 3 2

�푘 2 푀 2 퐶2 푘2 = 3푐2 This푎 way푀 we퐶 got to the well-known Euler’s problem � 푙 2 푏 2 referring to concordant numbers which is stated like F푚rom here푎 it results that the medians are proportional this : with its sides. The numbers a and b are told to be concordant if there Vice versa if the medians are proportional with the exist integer numbers x, y, z, t with 0 so that: sides we have : + = . 2 + 2 = 2 푥푦 ≠ 푥 푎푦 푧 = = = �On2 the contrary2 2 ( if the numbers x, y, z, t under the 푘 푙 푚 above푥 푏푦conditions푡 don’t exist ) the numbers are called 푥 DISCONCORDANT. Then푐 푏 the equations푎 (2.1.) become: The problem we got to shows again the fact that the numbers 1 and 9 are or aren’t concordant. It is = 2 + 2 demonstrated [2] that the numbers 1 and 9 are 2 2 = 2 2 + 2 2 2 DISCORCONDANT therefore there aren’t integer 푥2푐 2 = 2푏2 + 2푐2 − 푎2 numbers k, a, M, C to respect the equations (3.4.).The �푥2푏2 푐 2 푎 2 − 푏2 consequence is that there are no isosceles triangles in We푥 푎eliminate푎 X from푏 − 푐the first two equations and we the set MED. get: (2 + 2 ) = (2 + 2 ) Theorem 2.7. If the triangle ABC is right with + 2 2 2 2 2 2 2 2 푐 푐 푎 − 푏 푏 (푏2 + 푐 )(− 푎 + = then in the triangle with the sides ( 2a,2b, 22c) Through calculus we get : ) 2 2 2 2 we2 only2 have a median with the length a natural푏 2 = 0 and we also get the condition + 푐number.푎 2 2 = 0 . If we had started with푐 =푏 = 푐2= 푎 2we− 푏 푐 푎 − Proof : If + = the equations (2.1) become: would2 have got + 2 =푘 0 and푙 if푚 we had 2 2 2 푏 푎 푐 푏 푥 started with = =2 =2 then2 we would have found = 푏+ 푐 푎 푐 푏 − 푎 out that : +푘 푙 2푚 = 0. 2= 4 2 + 2 푏 푎 푐 2 2 2 푥 푘2 = 4푏 2 +푐 2 푏 푎 − 푐 � 푙 2 푐 2 푏 2 Theorem 2.9. (Euler 1779)The triangle with the sides: where푚 the푏 median푐 from A is obviously k=a. Considering (2a, 2b, 2c) gives us the expressions: the pythagoric numbers a, b, c given by: = ( + ) ( ) = + = ( ) + ( + ) =2 2 2 푎 푚= 2푛 푝 − 2푚 − 푛 푞 푎 = 푢 푣 �푏 푚 − 푛 푝 푚 푛 푞 � 푏 2 푢푣 2 where 푐 푚푝 − 푛푞 We푐 get푢 for− 푣l and m the following expressions : = ( + )(9 ), = 2 (9 + ) with , 2 2 has integer2 2 medians. 2 2 = 4 8 + 4 푝 푚 푛 푚 −푛 푞 푚푛 푚 푛 2 = 4 + 142 2 + 2 Observation:푚 푛 ∈ 풁 Not all the triangles from MED can be

�푙 2 푢4 − 푢 푣2 2 푣2 obtained through Euler parameter. For example for Mordell ( [7] pages 20-21) show that the only 푚 푢 푢 푣 푣 a=226, b=486, c=580 we have k=523, l=367 m=244 a solutions are for the pairs triangle which does not come from Euler’s parameter. ( , ) {(1,0), (1,1), (0,1)} , cases in which the triangle2 2 becomes confluent.

푢 푣 ∈ 3. REFERENCES Definition 2.2. We say that the triangle ABC is an automedian triangle if its medians are proportional with its sides. [1]. Bradley C.J., Challengers in geometry for Mathematical Olympians past and present, Oxford Theorem 2.8. The triangle with the sides 2a, 2b, 2c is University Press automedian if and only if one of the relations : [2]. Buchholz R.H., On triangles with rational , angle bisectors or medians , PhD Theses, University of + = 2 + = 2 + = 2 Newcastle [3]. Beauregard R., Suryanarayan E.R., The 2 2 2 2 2 2 2 2 2 is푎 true.푏 푐 푏 푐 푎 푎 푐 푏 Brahmagupta Triangles, The College Mathematics Journal, Vol. 29, No. 1, January 1998

546 [4]. Krier N., Manvel B., Counting integer triangles, [6]. Peterson B.E., Jordan J.H., Integer geometry: Mathematics Magazine Vol. 71, No. 4 (Oct., 1994) Some examples and constructions The Mathematical [5]. Dove K.L., Sumner J.S., Tetrahedra with integer Gazette, Vol. 81, No. 490 (Mar. 1997) edges and integer volume, Mathematics Magazine, Vol. [7]. Mordell L.J., DiophantineEquations, Academic 65, No. 2 (Apr. 1992) Press,1969.

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