On a Fundamental Task of Diophantine Geometric Figures
OON A FUNDAMENTAL TASK OF DIOPHANTINE GEOMETRIC FIGURES
Zurab Aghdgomelashvili (Georgian Technical University, 77, M. Kostava str., Tbilisi, 01785, Georgia)
On a Fundamental Task of Diophantine Geometric Figures Zurab Agdgomelashvili Abstract
The goal of the work is to take on and study one of the fundamental tasks studying Diophantine n-gons (the author of the paper considers an integral n-gon is Diophantine as far as determination of combinatorial properties of each of them requires solution of a certain Diophantine equation (equation sets)). Task*( n; k ): is there a Diophantine n-gon ( n ≥ 3 ) with any side or diagonal equal to k for each fixed natural number k . In case it exists then let us find each such n . It is shown that there is no such Diophantine n-gon ( n > 3 ), neither convex nor concave, the length of any side or diagonal of which is equal to 1. It means that for k=1 the above mentioned task is solved. The studies made it possible to find certain Diophantine rectangles, one of the sides of which is equal to 2 (it is noteworthy that all of them appeared to be inscribed in a circumcircle). The studies showed that there is no such Diophantine rectangle inscribed ina circumcircle, the diagonal length of which is equal to 2. It is shown that for any natural ( k ≥ 3 ) there is a Diophantine rectangle with the side length equal to k . The fundamental studies showed that for k=2 and n = {3; 4; 5}for convex n-gons (though such Diophantine pentagon has not been found yet. In the author’s opinion such pentagon does not exist) and n= {3; 4; 5; 6} for concave n-gons (here as well for n=5 and n=6 no such n-gon has been found yet and for the case of its existence all probable types of such figures are presented). The paper considers task *( n; k ) for k=3 and shows that in this case 3≤ n ≤7. However, neither such Diophantine pentagon, nor Diophantine hexagon, nor Diophantine heptagon have been found yet. In the author’s opinion such Diophantine figures do not exist and in case they do then he presents the probable types of them. INTRODUCTION Since ancient times, mathematicians have been interested in the study of geometric figures. It is worthy to mention that great mathematicians, such as: P. Ferma, K.F. Gauss, L. Euler, V. Serpinski, H. Steinhouse et al., were interested in the study of such figures properties. We call such figures as Diophantine due that to determine the properties of each of them they need to be solved by a certain Diophantine equation (system of equations). By V. Serpinski and H. Steinhouse were stated and solved number of problematic issues related to this topic. Here we will stop on consideration of one of them. Task: For each n N on the plane will be find not locatedon a line such n points, for that between each two of them the distance will be expressed in natural numbers. For solving this task, they have shown that for n N always exist a Diophantine n -gon, the question arises here: exists or not for each fixed k N the Diophantine n -gon, the distance between any two vertexes of that is equal to k , and if so, then find all such n . From the tasks under study on Diophantine geometrical figures, one of the most important is obviously the following: Task* (n;k).Exist or not for any fixed k natural number the Diophantine n-gon (𝒏 ≥ 𝟑) those length of arbitrary side or diagonal is equal to k , and if so, then find all such n . It is shown by us that is not exist such Diophantine n -gon, both convex and concave, those lengths of arbitrary side or diagonal are equal to 1. I.e. for k the above-mentioned issue has been solved. BASIC PART To prove this, let us consider a few tasks. Lemma 1. If one of the sides of a Diophantine triangle is equal to 1, then its remaining sides represent equal legs of this triangle.
Task: ; ABC ..1 ;, ACBCAB NACBC :proven be to
Without limiting the generality let’s say
ACBC . Due the inequality of triangle ACBCAB or ACBC . I.e. ..1;, ACBCBCACBC NACBC
Q.E.D.. This remarkable property is one of the cornerstones of the Diophantine and Bidiophantine geometric figure’s research apparatus.
Lemma 2. The length of each side and each diagonal of the convex Diophantine rectangle is greater than 1.
Fig. 1
Let’s assume the opposite. I.e. let's say that in rectangle
ABCD the length of each side and diagonal is expressed by the natural number, and also the length of one of the sides, without limiting of generality let's say that AB . According to Lemma 1 ACBC and ADBD . This is impossible, because then C and D the points should be located on the median AB . I.e. our assumption is false. Therefore accordingly of this condition, the length of each side of ABCD is greater than 1. Now let's say that length of the
ABCD diagonal is equal to 1, without limitation of generality let's say AC . According to Lemma 1 BCAB and ADCD . At the same time using the inequality of triangles BOC and AOD , it is easy to show, that ADBCACBD or BDADAB . From
ABD we have BDADAB . I.e. .1;,, BDADABBD NABBDAB is impossible. Thus the assumption is false, or AC . I.e. finally we have that the length of each side and diagonal of each convex Diophantine rectangle is greater than 1. Q.E.D. Lemma 3. If none of the located on the plane four points are not located on one straight line and at the same time distance between each of them is expressed by a natural number, then each of these distances is greater than 1.
Let’s assume the opposite. I.e. let's say that there are such four points on the plane, none of them are located on one straight line, the distance between each of them is expressed by a natural number, and at the same time some of them are equal to 1. Then, according to Lemma 1, the remaining both points must be located on a median of segment with equal to 1 length. Here we would have two cases
1) Task: ; ABC
2) Task:
1, ;
AB AC BC m . ;;;1
NCD NDBAD NBCACAB
1) This case is considered in
Lemma 2 . 2) From
AOC and DOB we have: Fig.2 .,,, ,
NlnmlCD nDBAD
Fig.4 Fig.3 .,,,,, ;5,0, ;;
222 222
NlnmlCDnDBADmCBAC OBAOCDOCOD ODOBBD OCAOAC .,, ;14 ;.,, ;214 ;2.,, ;21 ;21
22 22222 222222 222
Nlnm qm Qql lmnOCNlnm OCm l lmnOCNlnm OCln OCm .12 ;12 ;,.122 ;,.14 ;, qm qm Nqmqmqm Nqmqm Nqm This is impossible. Thus the distance between arbitrary two points from these given points is greater than 1. Q.E.D.
Theorem 1. The length of each side and each diagonal of convex Diophantine n -gon ( n ) is greater than 1. Let’s assume the opposite. I.e. assume a convex Diophantine n -polygon ( n ), the distance between any two vertexes of that is equal to 1. By virtue of Lemma 1 , the remaining vertexes should be located on the median of the connecting these two vertexes segment. Because this n gon should be convex, so obviously is n = 4, but according to Lemma 2 such rectangle does not exist. I.e. our assumption is false, or the length of each side and each diagonal of arbitrary convex Diophantine n -gon ( n ) is greater than 1. Q.E.D. Theorem 2. If from located on the plane n-point ( 𝒏 > 3 ), none of the three poins is located on one straight line and the distance between each of them is expressed by a natural number, then each of these distances is greater than 1.
As in the previous theorem, if the distance between arbitrary two vertexes of a Diophantine n -polygon is equal to 1, then the its remaining vertexes should be located on the median of connecting these two vertexes segment. If we take into account that none of these vertexes are located on a one straight line, then we will have only the following cases (see Fig. 5). AC AD AD AD
1) 2) 3) 4) AC AC AD AD Fig. 5
The first fours have no solution according to
Theorem 1.
All fours will be as 1) and because there is not exist such Diophantine rectangle, therefore are not exist 2), 3) and 4) Diophantine rectangles, and the last four have no solution by virtue of
Theorem 2 (Here, too 6), 7) and 8) by fill will be reduced to 5). I.e. our assumption is false or each of the distances given by the task conditions is greater than 1. Thereby fully is proved. I.e. we have proved the following theorem.
Task* (n;1)does not exist such a Diophantine n-gon (n≥3), both convex and concave, those lengths of arbitrary sides or diagonals are equal to 1
Task1. In Diophantine ∆ABC - ში AC and NaaAB ; let’s find BC . Is given: ∆ ABC; ;,2; aABACNBC .1, aNa ___________________________________________ Must be found. BC We have two cases: ; 2; ; 2;1) 2) . . a N a a N aBC a BC a
1) From∆ABC accordingly of inequality of triangle a BC .1; ;2;.1;.2;2;; aBC aBC aNaBCa BCa NaBCaBC aNaBC
2) From ∆
ABC accordingly of inequality of triangle aBC .1; ;2;.2 ;2;; aBC aBC aNaaBCa aNaBC By us below are stated all such Diophantine tetragon. length of those side or diagonal would be equal to 2
5) 6) 7) 8)
Is given : .;2 ;;;,, dBDAC bCDAD aBCAB Ndba We have already shown in Lemma 3 that such a Diophantine tetragon does not exist. Is given: .2; ;; ;;,,
BDdAC bCDAD aBCAB baNdba
From
ABC we have
12 ;,.2; ;, babab Nbaba ba Nba
It is known that the sum of the diagonals length a convex tetragon is greater than the sum of the lengths of the opposite sides.Thus bad or bd . From ABC ;222 ;, bad Nbd i.e. bdb Nbd this is impossible. Thus such Diophantine tetragon is not exist. Is given: .;2 1; ;;,, dBDAC bCDbAD aBCABNdba From
ABD d a b . From the tetragon ABCD
CDABBDAC , or bad . We have: Ndba badbaNdba bad bad ,, 1,, 12 this is impossible. Thus such Diophantine tetragon is not exist. Is given: .;2 ;,1 ;1, ;,, dBDAC bCDbAD aBCaAB Ndba .,,, ; ;
NBDADBCAB ADABBD ADBCACBD i.e. badba Ndbabad bad Ndba
This is impossible. Thus such Diophantine tetragon is not exist. Is given:
2; ;;;,,
ACdBD bCDAD aBCAB Ndba
From the rectangle
AED and rectangle AEB we have:
22 222222 22222 222 xbxb Nxbbx Qd dbax Ndbaadxdx bx Ndbaaxd bx Ndba xb xb Nxb
This is impossible. Thus such Diophantine tetragon is not exist.
Is given:
Ndba ADdBC bCDACaBDAB ,, .2; ;;
Proceeding from the task conditions the B and C points are located on the middle perpendicular of AD , thus ADEDAE . From the rectangle
AEB and rectangle CED we have: NBEa BEaNdba Qd adbBE BEaNdba BEdb BEa , 1.,, 21.,, ;11 ., ;1;1., ;1
NBEa BEa BEaNBEa BEaBEa
This is impossible. Thus such Diophantine tetragon is not exist. Is given: .2;1 ;;;;1 ;,,,
ADbBD bCDcBCaACaAB Ncdba
Such Diophantine tetragon is not exist, because the sum of the lengths of diagonal of the convex tetragon will be greater than the sum of the lengths of the opposite sides of this tetragon, i.e. it should be
CDABACBD , but proceeding from this we would have baba , that is impossible. Is given: , , , ;1; 1; ; ;; 2. a b d c NAB a AC b BC c CD bBD a AD In VIII, similarly to VII would be
CDABACBD , but proceeding from this we would have baba . Thus such Diophantine tetragon is not exist. Is given: .1; ;;;2;1 ;,,, aADdBD aCDbBCACbAB Ncdba Let’s circumscribe from the point A by radius AD and point B by radius BD the arcs and say they intersect BC at points K and M, respectively. From AKC we have aCKaCK . .11.;1 ; abdbdadBMaCK bBMCK BDC - დან abdbad . i.e. Ndba abdabNdba abd abd ,, ;1,, 1;
This is impossible. Thus such Diophantine tetragon is not exist. Is given: ..3;2 ;1;;;1 ;3;3;,,
ACAC bCDbADaBCaAB baNBDba
222 222 bDOb aBDaCAD BCAbbAC aaACACACba Nba babaBDbaODBOba . From
BCD we have BD a b i.e.
NBDNBD babaBD ;1; . Thus such Diophantine tetragon is not exist.
BFBNABMBABFMBN
12 12 9 12 12 9 2 1 2 311 2 1 2 312 12 92 11 .12 1 2
MB b b a a a ab b bb ba b ab b
We have abb ababMB . abBDNBD abMBBD From
ADB we have: abBDbBDa . i.e. .1,, ;1 abBDNBDba abBDab From the rectangle
DNB we have BDBNDN . i.e. .14 91212912122 abaabbab From that we will obtain: (*)01340281337622424 abbaababba
Proceeding from this
Nkp kb paNkp kb pababa , 2 12, 122)2(mod01)2(mod0131337 If b=2k then from (*) we have: ),4(mod01337)1(3741337 ppa that is impossible. if a =2p, then from (*) we have ),4(mod0)2)1(4(131313 ppb that also is impossible. I.e. such Diophantine rectangle is not exist. Is given: .;,,, ;2,, BDCDNcaCDAC BCcADaBDAB
From the conditions of task we have: c<2a ;
1; (*)1; (**)
AC aCD a AC a
Let’s consider each of them separately: (*) From
ABC and ACD due the cosine theory we have: cos)1(21)1( cos)1(212
222 222 acaca aaaa )1(2 )4)(4(sin;)1(2 4cos ;)1(2 91212sin;2 322cos ac cacac ac aa aaaa
From
ABD ( 4)(4 )(12 12 9)( 4 )(2 2 3)cos( )2 4 ( 1) 4 ( 1) c a c a ac c a a aa ac a ac a )1288()52()91212)(4)(4( aaacaaacac acacaacaaca . Proceeding from this
Nppc ,4 . Due the introduction we will obtain: apapapap From the last equation we have:
Ntpa ptaNpa pa ,,, )(mod0
Due the introduction of obtained in the last equation and it’s simplification we will obtain: ., .14310 ;14310, 0)964(20 tNtp ppt pptNtp pptt
Remark : .., ;12 ;12, ;1)2)(2(, ;14 nNnp np npNnp npnpNnp np i. e. such Diophantine tetragon is not exist. (**) Is given: ., ;;2;1 ; Nca cADBCaCD aACBDAB
From the rectangle
BED and rectangle CED we have: aCECEaCEa ; aaaaDE From the rectangle
AKD due the Pythagorean theorem: aaacaacaaaac where caNca . From the obtained equation leads that: a=2p; Np . By introduction we will obtain: (16 4 3) (4 1) 0 (1), , , 1 c p p c p pp c N p c From (1) for p =2 we will have: .721, 019253 cccNc cc For c =7the ACD is degenerated, because ADCDAC . For c =2 we will obtain the yet known to us Diophantine trapezoid. The arbitrary other solution of (1) currently is unknown. In our other natural solution is not exist. Is given: ., ;;2;1 ; Nca cADBCaAC aCDBDAB
From the rectangle
AEC and rectangle AEB we have: aEBEBaEBa ; aaaaAE . From the rectangle AKD due the Pythagorean Theorem: aaaac , from that we will obtain: aacaac . Similarly to XII
III will be obtained: pa cpNcp ppcppc (2) Currently is not found even one solution of (2). In our opinion it will not have the solution on natural numbers. Is given: .,, ;;2; ;1;1; ;180
Ncba cACBDbAB bADaCDaBC F From ; BCD ABD and
ABC due the cosine’s theorem we have : .2)cos(cos 4 91212sin;4 23cos 4 91212sin;4 23cos).cos())(360cos(cos ;cos2 ;cos42)1( ;cos42)1(
222 22222 222 222 ab cba b bbb b a aaa aabbac bbb aaa )(8)23)(23()91212)(91212( 16 )91212)(91212()23)(23()cos(2 cbababbaa ab bbaabaab cba c c b a ab a b a b a ab b a b If we in (1)introduce a=b - ს , we will obtain: aaaaaaaaaaaacc aac , Which has no solution in natural numbers, because otherwise its left side would be flat and the right side would be odd. As for (1), at present we do not know at least one in the natural solutions, because, in contrary, it’s left side will be even and the rigjt side – the odd. As for (1), Currently for us is unknown even one from triplet natural solutions. Is given: .,, ;2;;1 ;;
Ncba ACcBMbAM bMCaBCAB ;4 124 321 bbMK bbMEKDbcBK ; bbbccbbbcBD cBD bbbbcc .112)12(3)1(3013013 cbcbbcbbbb cbb c (1) cbdadbcadcba . (2) (1) represents the Pelli equation. By it’s solution we obtain the natural solutions of (1). Amongst them first three pairs are pairs: .390;675;.28;48;.2;3.390 ;135112;.28 ;9712;.2 ;712 cbcbcbcbcbcb Is given: ..,,;1 ;1;;;
NcbaaBD bACbCDcBCaAB
ACD - და ნ bNbb bCDA . Accordingly of one of the Kumer theorem (theorem: if the lengths of the sides and diagonals of a convex tetragon are expressed by rational numbers, then the diagonals at the intersection point are divided by rational length sections ) QODAO , . From AOD due the law of sines we will have: bba aabbbaaDAC ADBDAC ADBDAO ADOODAO . Proceeding from this Qtbb aa
344 344 or )344(344 bbtaa . i.e.this solution of takes is reduced on the solution of following task. Let’s solve in natural a, b, m და n numbers the equation )344()344( bbmaan . aaAM From the tetragon
MAAA lqp
MAAAAAAM pllp bcbccaba .
32 3212 12 babaab
From the right triangle l AFA due the Pythagorean Theorem:
16 )32)(12(32 132 312 12116 912122 1 91212912121612 1 bbbabababbba aabbbac
32 312 121 baba (*). From (*) we have: )(12 32432 1)12( )(416 )32)(12(32 1 abbbbab abbbbac ; )(32 12432 1)32( )(416 )32)(12(32 1 abbbbab abbbbac . I.e. if exists such Diophantine tetragon, then c will satisfy the conditions and . Is given: ., ; ;2; 1; Nca cAA AAaAA aAAAAAA AAAA lq kplk lpkqqp klqp
From the rectangles lp EAA and lk EAA we have: EAaEAa kk EAaEAEAaa kkk aEA k . aaEFaaaFA q . aaaaEA l . From lq MAA due the Pythagorean Theorem we have: aaaaaaaaaaac . Due the simple transformations we will obtain the equation: aaaaaac . (1) )2(9)91212)(2()2(9912123 aaaaaaaaaaa acaaaaac
50 36 0 48 48 36 49 98 a a a a a a a a a a a a a a a a ( 2 )(12 12 9) 3.5( 2 ) a a a a a a . I.e.
22 2 22 a a a a a a a ac c . Finally we have that if exists such Diophantine tetragon, then .2 1;2 1 aac Let’s now consider the issue of inscribed in a circle and circumscribed on a circle for such Diophantine tetragons, the length of one of those sides is equal to 2. As we have seen, in some cases on the XIV type Diophantine tetragon will be circumscribed the circle. Now let's study whether or no inscribe in this kind of Diophantine tetragonthe circle?
Task2 . Is given: In ABCDwill be inscribed the circle .,, 2 ;1 ;;
NBCbaAD bAC bCDaBDAB (In ABCDwill be inscribed the circle) baADCDABBC . aBEEDBEBD .4 912124 12)2( bbbbaCFKCBCKEBKBE I.e. bbbbaa bbabba )91212)(1(8912121616636864321216 0912121616 bbabbabaabba bba
16 12 12 74 8 10 11 ( 1)(12 12 9). a b bab a b a b b (1) (*)125)2(2131899212212)(12(1110884.2;2 bbaaabba
1( 2) 2.5 2.5.2 b a b i.e. a . Due the introduction of a =2 in (1) we will obtain: . 34431110168 Nb bbbb bNb bbNb bbb Due the introduction of b =2 in (1) we will obtain: .2; )1(73112088 aNa aaa aaNa a .2; )1(79 . I.e. such Diophantine tetragon is not exist. Let’s now consider the issue of inscribed in a circle and circumscribed on a circle for XIV type Diophantine tetragons. Task3 . Is given:
ABCD is inscribed in the circle; .,, ;;2;1 ;;1;
Ncba cBCADbAC bCDaBDaAB (□ ABCD is inscribed in the circle)
ADBCCDABACBD
1( 1)( 1) 2 2 a ba b ab c c ; From the
AEB and BED right triangles EDBDEAAB .4 32)2()1( aEAEAaEAa
Similarly bDF . babaDFADEAEF . i.e. . BCBKEF We will obtain that
ABCD is rectangle. Thus bbaaCFBE .)1()1( babbaa I.e.
Naaac . This is impossible. Thus such Diophantine rectangle is not exist.
Task 4 . Is given: □
ABCD is circumscribed in the circle; ..;,, ;;2;1 ;;1; abNcba cBCADbAC bCDaBDaAB (In □
ABCD is inscribed the circle). bacbac bababaEF aabbBECFbac aabbbac )91212)(91212(2 9121291212)1()2(44
22 22222 aabb aabbbaba baabaabb baabaabb )1414413()344)(3124( baabaabb baabbabaabbaba aababbababba aababbabab aababbabab (*)10301313222288 abababbaab If ., ;3 Nba ab then .103223222222 ;3032)(88 ;131326328 ba ababababa babbab .10301313222288 abababaab
I.e. .3;2;1; ;, a ab Nba
Due the introduction of a = 1 , a = 2 and a = 3 in (*)we will obtain: .. ;.061413 ;062 ;015. ;.10901311766227224 ;1060135244223216 ;10301313222288
22 222 22 22 bNb bb bbbNb bbbbb bbbbb bbbbb
I.e. such Diophantine tetragon is not exist.
Task 5 . Is given:
ABCD is circumscribed on the circle; .,, .2 ;2 )1(; ;;;;;
Ncba dBD dcba dACcBCaABdADbCD ,2 BD then due the inequality of triangles and consideration of (1)we have: I. .1;1 ad cb ; II. .; ad cb ; III. b cd a Let’s consider each of them I. This is impossible. II . This case is considered in I of CHAPTER III. III.
This is impossible. I.e. does not existing the circumscribed on the circle Diophantine tetragon those length of diagonal is equal to 2.
Remark : A case when the length the diagonal of inscribed in the circle Diophantine tetragon is equal to 2 is considered in Task 7. As already was mentioned, mathematicians were interested by the study of integer geometric figures since BC, despite this only some of the tasks occurs in this scope as oasis in the Gobi desert. The reason for this result is the absence of a unified method that would give the possibility us to study the properties of each Diophantine geometrical equation from the characteristic equations of this figure in a refined laconic way, to obtain in as simple as possible, the Diophantine equation (system of equations) andsimultaneously description in the optimal way for solving the equation. In this paper we will consideralso this issue. We have proved in § 3 that there is no Diophantine n-gon (n> 3), both convex and concave, those lengths of arbitrary side or diagonal are equal to 1 Therefore, the task * (n; 1) is completely solved. (From ABCD ) (From ADC ) (From ABCD ) (From ABC ) .,, ;1.,, ; ;)1(2 Ndca cadcaNdca cad cad .,, 23.,, );1()1( ;12
Ndca acdacNdca acd acd As we have seen above, solving the task *(n; 2) for k = 2 is the quiet complex problem even for n = 4, or for case of rectangles. In particular, as we will see below, even in the simplified case when the two sides of the Diophantine tetragon are parallel (the other two are parallel or not), the task *(n; k) for k = 2 is rather complex.
Task 6. Find all Diophantine parallelograms and Diophantine trapezoids those length of one side or diagonal is equal to 2.
It is known that for each Diophantine tetragon the length of each side and diagonal is more than 1, and if in Diophantine
MNP MN and kNPMP , then k . Let’s say that conditions of task is satisfied by ABCD : ;;;|| bCDaABADBC ; BC c ; ; . AD d AC d BD d .1\,,,,, Ndddcba Let’s drawn up: .;;||;|| AEDADECDBEBDCE Without the limiting of generality let’s say ba and cd . It is easy to show that cdbadd (1) and dd (2) Let’s consider the following cases: I. .2 c From
ABC and BDC we have: .1;;1;1;;1 bbbdaaad In this case for solution of issue will be considered 9 cases, and as we will see further for a we will consider 27cases. This is so labor consumptive that even at the solution of most complex problem, will be lost any desire to continue the work. There we shall propose the method that drastically reduces the number of considered at solution of this type tasks variants. Let’s say kad and tbd (3). It is obvious that k and t . Due the introduction of (3) in (1) dbatbka , from that by simplification we will obtain tbbtakd . (4) It is obvious that tk . Therefore from (4)due the taking into account that tk , we have: tk . i.e. tk . Due the introduction the (4) we will obtain the four equations: bad (5); bad (6); bad (7); bad (8). From ABE : bacd ; abcd (equalities take place when cd და ba ). I.e. bad and bad , or from (5)- ს , (6) and (7), by given linmitations, has not the solution in natural numbers set. As for (8), in that case tk . I.e. ad and bd . It iseasy to show that ddddcdbabacd . (9) Fig. 6 (This formula is valid as for ABCD trapezoid, as well as for
ABCD parallelogram). If we consider that: bdadc and dba , then from (9) we will obtain: dbaddbad a b d d d d a b d a b , but in that case they left part of (8) will be even, and right part the odd that is impossible. I.e. c . II. From dca , ABEABC and ABD we will obtain that: cccd ; cdcdcdb ; dddd . (10) ABC and BAD are not acute, thus cd or dd . If we consider that from BCD dcb and dddcb , we will obtain that dd . I.e. we have cd . I.e. (10) will be as following: dca ; cd ; cdcdcdb ; ddd ;1 . (11) Similarly to I let’s say that mcdb and ndd (12). It is obviously that: m and n (13). Due the introduction of (12) in (1) we will obtain: c d n dcmcd mndnccmdm . (14) i.e. mn (15). From (13) and (15) we have: n ; m (16) (if mn , then cd , that is impossible. I.e. mn , from that will be obtained (16)). I.e. .02 ;1;1,0.022 ;01 ccmdm mndnccmdmmn .22 ;2.02 ;02 cddccd ccd d is impossible, because acd . For cd we will obtain: acdcdcb . Due introduction of these data in (9)we will obtain: .3.1\.3 ;21.1\ ;13321323 cNcccNc cccccc Thus we will obtain expressed on XII trapezoid. (See Fig.7). (Let’s notice that t6hhis was found (in 1989) the first Diophantine tetragon, those length of side is equal to 2). III. dd . Then from BDCABC , and ACE we have: ccca ; dddb ; cdcdd ;1 . In addition cdd because cdcbd . There also as in II let’s say: pdblca ; and Fig. 7 kcdd (17). Obviously there also pl ; k ; (18). Due the introduction of (17) in (1) we will obtain: cdpdlckcd kpldplcdkck . (19) From (19)we have: ..1;1;0.1;1;0.1 ;1;0,0)2(mod0 plkklpkplkplkpl ananan (20) For these values in the first three cases dplcdkck (21 ), and in the fourth case dplcdkck (21 ). Due the try we will easy confirm that (21 ) and (21 ), accordingly of conditions (20), does not exist. Finally we have that such Diophantine parallelogram does not exist, and from the Diophantine trapezoid only one satisfies to the conditions of task (seeFig.7). Let’s consider once more task. Task7 . Let’s find all such inscribed in circle Diophantine tetragon, those length of side or diagonal is equal to 2. Let’s say that
ABCD satisfies the condition of task and the length of it’s any side without limitation of generality will be: lbnkamc ;;2 . (22) Obviously lk . Accordingly of Ptolemy theorem (23) we have: cdabmn (24). Due the introduction of (22) in (24) and simplification we will obtain dlkbkal (25). Due the taking into account of (23) from (25) we have: dab (26); dab (27); db (28); da (29); d (30); da (31); dab (32); db (33); dab (34). Obviously by stated in (26), (28) (29) and (30) limitations, have not the solutions in natural numbers. Also (27) has not the solution in natural numbers because in this case from ABD ADBDaAB dbdb . Similarly we will obtain that (32) has not the solution in natural numbers. Let’s now show that by stated limitations (34) also has not the solution in natural numbers. Let’s assure the contrary. Let’s say
ABCD satisfies the condition of task. In this case we have:
1; ;2 a bCD b AD d ; AB a BC
1; 1.
AC a DB b
Let’s plot: , ; ;
E F CB DF CB AE K EA DK . Fig. 8 From the right triangles
AEB and AEC CEACAEBEAB . I.e. a BE aa BE BE . Similarly bFC . EF FC CB BE a b a b AD . I.e.
ADDK და AK or AEDF . FCCDBEAB bbaa a a b b a b . i.e.
Nabad . This is impossible. I.e. our assumption is false or by stated limitations (34) has not the solution in natural numbers.Remains to be considered the identical equations (31) and (33). adDKaAE . aaKFAE . KFDKDFFBDB aaad hence we have: daaa d (*). This case reduces to considered by us XIV. Now let’s say that the circle is inscribed in Diophantine ABCD those lengths of diagonal without limitations let’s say is: tbADaCDkaBCbABDB ;;;;2 . Is obvious that kt (35). Due the Ptolemy theorem we have: tbkaabm (36) ktbkatabm . From ADC tbam I.e. ktkbtaabktbkatabtba . baabb bbaba I.e. ktkbtaabbaab kbtakt .(37) The (37) due the (35) condition has not the solution in natural numbers. I.e. ba (37 ). From AOD and ACB due the application of inequality of triangles is easy to show that CDABBDAC i.e. bambam ;. m am b i.e. Fig. 9
Fig. 10 bbaam . (38) With taking into account (38) from (36) we have: tbkaab ; (39) tbkaab ; (40) tbkaba ; (41) tbkaba . (42) It is easy to show that these equations have not the solution in natural numbersif: .1;0;1, ;1\;2 tk Nba ა ნ .1;0;1, ;1\;2 tk Nab I.e.finally we have that length of each diagonal of inscribed in circle Diophantine tetragon is more that 2 and all having equal to 2 side Diophantine tetragon by the solution of Diophantine da (43) equation aACdDBdADdCDBCaAB . (44) Let’s mention that from the found Diophantine tetragons those length of any side is equal to 2, all will be inscribed in circle. Currently is not known this type of Diophantine tetragon on that would not circumscribed on circle. Even worse is situation related to the Diophantine tetragons those length of any diagonal is equal of 2. We have shown that if they exist, on them will not be circumscribed the circle, but what happens is other case is unknown to us, although as we have seen from the discussion of Task-2 (I-XVII), we have made attempts.
The method shown by us in Task-2 and Task-3 does indeed significantly reduce the number of variants under consideration. However, the method of solving the obtained Diophantine equations is clearly expressed. All of this, for task*(n;k) , generally would be summarized as follows: Consider n-gon vertices for all Diophantine triangles thhhose distance between any two vertices is equal to k , and taking into account that the modulus of difference of the other two sides represents an element of k set, let us express the length of one of the sides by the length of other added t , where kkkt ; By introducing the characteristic equation(s) of a given n-gon of this data, we obtain all possible Diophantine equations; The excluding from the obtained Diophantine equations of equations which that due the stated limitations do not have solution in natural numbers, is mainly achieved by the additional limit6ations to the Diophantine equations obtained by using the triangle inequalities and also by the result of the task*( n ; k ) we have already obtained nkkk . By ((((k; n)) is denoted condition that satisfies by the Diophantine n-gon sides and diagonals when the length of any side is equal to k ). Let’s return to the task*( n ; k ) for k = 2. Finally, we have that for the task * ( n ;2) convex n n-gon (though none of the Diophantine n-gons are found for n = 5 yet. In our opinion, such a pentagon does not exist) and for concave n-gons (here also for n = 5 and n = 6 we have the same picture as for convex for n = 5). Fig. 11 Fig. 12 Fig. 13
Let’s now consider the task * ( n ;3). Similarly to k =2, If the obe side of Diophantine triangle is equal to 3 then modulus of difference in other sides would be equal to 0, 1 or 2. With taking it into account all possible Diophantine tetragons for k -3 are presented below: Let’s consider from them the case when is obligatory for solution of issues of task* ( n ;3: From the right triangles AOB and AOD we have: Is given: ABCD ..,, ;;3 ;;
Ndba dBDAC bCDADaBCAB
Nqd abd bd abd d abdODNdba bOD aODd
222 2222 222222 222 )2(92,, 5.1 )(5.1 bNqbqbNqb qbqb , 4 5.2, 9)2)(2(
I.e. such Diophantine tetragon does not exist. Is given: ABCD ..,, ;;3;2 ;;2;
Ndba dBDACbCD aBCaABbAD
From
ABC and ADC due the cosine law: bDAC aCABDACbbb CABaba
222 222 .;032arccos;0 32arccos;0
DABDAC CAB dNdba badbaNdba bad bad ,, 21,, 223 2 . I.e. such Diophantine tetragon does not exist. Is given:
ABCD ..,, ;;3; ;1;;1
Ndba dBDACbAD bCDaBCaAB
There also as in previous task we have: ;031arccos;0 31arccos;03431cos )1(3 431cos DACDAC CABbDAC aCAB badNdba badbaNdba bad bad ,, 11,, 113 1 . )3()1( )3()1( CMaCMb KAbKAb (From ABD ) (From ABCD ) (From ABD ) (From ABCD ) 8 aaEKBM bbbbDK aCMbKA
From the DEB right triangle
DEBEBD , i.e , 3 168816883 1 Nab bbaabaab
Nba bbaabaab , )1688)(1688(21)15558(2 . This is impossible because the left side is even, and if right side is natural,then will be odd. I.e. such Diophantine tetragon does not exist. Is given: ABCD ..,, ;3;; ;2;
Ndba ACdBDbAD bCDaBCAB
This is impossible. i.e.such Diophantine tetragon does not exist. Is given: DBC ..,, ;1;;3 ;;
Ndba bCDbADAC aBCABBDA
From
BAC and DAC accordingly of cosine law we have: ., ;3 4cos ;23cos., );cos(63)1( ;cos63
222 222
Nba bbCAB aCABNba CABbbb CABaaa (From ABD ) (From ABCD ) Ndba badbaNdba dba badBDABAD DCABBDAC ,, ;1,, ; ;23. ; .., ;928., ;3 423 Nba aabNba bba
Because a , thus a (mod(2a-9)) .992 aaa We can find that only two pairs satisfies the conditions of task: ab and .16;5 ba Is given: ABCD ; ..,, ;3;; ;2; Ndba ACdBDbAD bCDaBCAB
From
ABC and DAC we have: .32;2 ;2;43arccos.6532cos ;43coscos63)2( ;23cos bbbb a .67;43arccos2 i.e. A and C points related to BD straight line would be located in one half-plane as well as in different half-planes. The case when bad i.e. DACCAB coscos or
54 1026 5423 bbabba . If b ,then
154 10 bb and thus Na . If b , then
54 10 bb is natural number when ba და ba From the above mentioned for the
Task* ( n ;3) n . In addition for k =3 is not found neither Diophantine pentagon nor Diophantine hexagon and Diophantine heptagon. In our opinion such Diophantine figures does not exis and in they exist their probable kind will be as followings: REFERENCES Matiyasevich Y, Hilbert’s Tenth Problem. MLT Press Cambridge, Massachusetts, 1993. 2.
Yaglom A.E. Non-elementary Tasks of Elementary Statement. Moscow: Gostekhizdat, 1954. (In Russian) 3.
Boltianski V.G., Gokhberg I.G. The Theorems and Tasks of Combinatory Geometry. Mpscpe> Publishing Nauka, 1965, (In Russian) 4.
Boltianski V.G. Hilbert’s Third Problem. Moscow: Publishing Nauka, 1977. (In Russian) 5.
Adamar J. Elementary geometry. Moscow, State pedagogical-training publishing house of the Ministry of Education of RSSR, 1937 (In Russian). 6.
Hugo Steinhaus.
Problems and Reflections , Moscow, 1974 (In Russian). 7.
Sierpinski V., Pythagorean triangle. Moscow, State pedagogical-training publishing house of the Ministry of Education of RSSR, 1939 (In Russian). 8.
W. Sierpinski, Theory of numbers, State Publishing House, Warsaw, 1959 (In Polish). 9.
Vasilyev N.B., et al., Extramural mathematical Olympiads. Moscow, publishing house Nauka, 1986 (In Russian). 10.
Dickson L.E., Introduction to theory of numbers. Tbilisi, publishing house of Academy of Sciences of GSSR, 1941 (In Russian). 11.
Ozhigova E. P., What is theory of numbers. Moscow, publishing house Znaniye, 1970 (In Russian). 12.
Mikhelevich Sh.Kh., Theory of numbers. Moscow, publishing house Visshaya Shkola. 1967 (In Russian). 13.
Shklyarski D.O, et al., Selected problems and theorems of Elementary Mathematics. Moscow, state publishing house of technical-theoretical literature, 1954 (In Russian). 14.
Z. Aghdgomelashvili. Diophantine geometric figures. Problems and solutions from the mathematical tournament of gifted children “Pythagorean Cup 2001-2004”, Tbilisi, publishing house TsisNami, 2004 (In Georgian). Z. Aghdgomelashvili. Collection of mathematical problems with solutions. Tbilisi, publishing house Ganatleba, 1991 (In Georgian). 16.
Z. Aghdgomelashvili. Mathematics (individual and group work). Tbilisi, publishing house Tsis-Nami, 2001 (In Georgian). ერთი ფუნდამენტური ამოცანის შესახებ დიოფანტურ გეომეტრიულფიგურებზე ზურაბაღდგომელაშვილი რეზიუმე ნაშრომში დასმულია და შესწავლილია დიოფანტური n -კუთხედები. ამ სტატიის ავტორი დიოფანტურს უწოდებს მთელრიცხვა n -კუთხედს იმ მოტივით, რომ თითოეული მათგანის კომბინატორული თვისებების დასადგენად საჭიროა გარკვეული დიოფანტური განტოლების (განტოლებათა სისტემის) ამოხსნა. შემსწავლელი ამოცანებიდან ერთ-ერთი ფუნდამენტური ამოცანა. ამოცანა (n;k): არსებობს თუ არა ყოველი ფიქსირებული k ნატურალური რიცხვისათვის ისეთი დიოფანტური n -კუთხედი ( n ), რომლის რომელიმე გვერდის ან დიაგონალის სიგრძე ტოლია k -სი, და თუ არსებობს, მაშინ იპოვეთ ყველა ასეთი n . ნაჩვენებია, რომ არ არსებობს ისეთი დიოფანტური n -კუთხედი ( 3) n , როგორც ამოზნექილი, ასევე ჩაზნექილია, რომლის რომელიმე გვერდის ან დიაგონალის სიგრძე ტოლია ++-ის. i.e. k -სათვის ზემოთხსენებული საკითხი გადაჭრილია. კვლევის მეშვეობით ნაპოვნია რიგი დიოფანტური ოთხკუთხედები, რომელთა ერთ-ერთი გვერდი ტოლია 2-ის (აქ უნდა შევნიშნოთ, რომ ყოველი მათგანი წრეწირში ჩახაზული აღმოჩნდა). ნაჩვენებია, რომ ყოველი მათგანი წრეწირში ჩახაზული ისეთი დიოფანტური ოთხკუთხედი, რომლის დიაგონალის სიგრძე ტოლია k-ის. ნაჩვენებია, რომ ნებისმიერი ნატურალური ( 3 k )-ისათვის მოიძებნება დიოფანტური ოთხკუთხედი, რომლის გვერდის სიგრძე ტოლია k -სი. {3; 4;5} n ფუნდამენტური კვლევის შედეგად ნაჩვენებია, რომ k -სათვის ამოზნექილი n -კუთხედებისათვის (თუმცა არც ერთი ასეთი დიოფანტური ხუთკუთხედი ჯერ არ არის ნაპოვნი. ავტორის აზრით ასეთი ხუთკუთხედი არ არსებობს) და {3; 4;5; 6} n ჩაზნექილი n -კუთხედებისათვის (აქაც n და n -სათვის არცერთი ასეთი n -კუთხედი არ არის ჯერ ნაპოვნი. და თუ არსებობს, შემთხვევისათვის მოყვანილია ასეთი ფიგურების ყველა შესაძლო სახე). ნაშრომში განხილული ამოცანა (n;k), k=3-სათვის და ნაჩვენებია, რომ ამ შემთხვევაში n , მაგრამ არ არის ნაპოვნი არც ერთი ასეთი დიოფანტური ხუთკუთხედი, არც დიოფანტური ექვსკუთხედის და არც დიოფანტური შვიდკუთხედი, ავტორის აზრით ასეთი დიოფანტური ფიგურები არ არსებობენ, და თუ არსებობენ. მოყავს მათი სავარაუდო სახე. ОБ ОДНОЙ ФУНДАМЕНТАЛЬНОЙ ЗАДАЧЕ ПРОДИОФАНТОВЫМ ГЕОМЕТРИЧЕСКИМ ФИГУРАМ Зураб Агдгомелашвили Резюме В труде поставлена и изучена одно из основным, ниже приведенных задач про диофантовым n -угольником (Автор статьи диофантовым называет целочисленным n -угольник, потому что, для установления комбинаторных свойства каждого из них требуется решить опредеденное диофантовон уравнение (систему уравнения). Задача (n;k): Существует или нет для каждого фиксированного натурального числа такой диофантовый n -угольник ( n , для которого длина некоторого стороны или диагонала равна k , и если существует, то найти все такие n . Показана, что не существует такой диофантовый n -угольник ( 3) n , т.е. для k вышепоставленная задача решена. Тщательным исследованием найдена ряд диофантовые четырехугольников сторона которых равна 2-м (Все они оказались вписанными в окружность). Показана, что не существует такой вписанный четырехугольник длина диагонадя которой равна 2-м. Показана что для любой k N ( k ) существует четырехугольник сторона которой равна k .Фундаментальным исследованием показана, что при k , {3; 4;5} n длявыпуклых n -угольников (но пока не найдено ни одно такой диофантовый пятиугольник). По мнению автора такой пятиугольник не существует) и {3; 4;5; 6} n , для вогнутых n -угольников (аналогично и здесь для n и n ). On a Fundamental Task of Diophantine Geometric Figures Zurab Agdgomelashvili Abstract
The goal of the work is to take on and study one of the fundamental tasks studying Diophantine n-gons (the author of the paper considers an integral n-gon is Diophantine as far as determination of combinatorial properties of each of them requires solution of a certain Diophantine equation (equation sets)). Task*( n; k ): is there a Diophantine n-gon ( n ≥ 3 ) with any side or diagonal equal to k for each fixed natural number k . In case it exists then let us find each such n . It is shown that there is no such Diophantine n-gon ( n > 3 ), neither convex nor concave, the length of any side or diagonal of which is equal to 1. It means that for k=1 the above mentioned task is solved. The studies made it possible to find certain Diophantine rectangles, one of the sides of which is equal to 2 (it is noteworthy that all of them appeared to be inscribed in a circumcircle). The studies showed that there is no such Diophantine rectangle inscribed ina circumcircle, the diagonal length of which is equal to 2. It is shown that for any natural ( k ≥ 3 ) there is a Diophantine rectangle with the side length equal to k . The fundamental studies showed that for k=2 and n = {3; 4; 5}for convex n-gons (though such Diophantine pentagon has not been found yet. In the author’s opinion such pentagon does not exist) and n= {3; 4; 5; 6} for concave n-gons (here as well for n=5 and n=6 no such n-gon has been found yet and for the case of its existence all probable types of such figures are presented). The paper considers task *( n; k ) for k=3k=3