On a Fundamental Task of Bidiophantine Geometric Figures
OOn a Fundamental Task of Bidiophantine Geometric Figures
Zurab Agdgomelashvili (Georgian Technical University, 77,M. Kostava str., Tbilisi, 0175, Georgia)
Abstract:
The goal of the work is to take on and study one of the fundamental tasks studying Bidiophantine polygons (let us call a polygon Diophantine, if the distance between each two vertex of those is expressed by a natural number and we say that a Diophantine polygon is Bidiophantine if the coordinates of its each vertex are integer numbers). Task **( n; k ): is there a
Bid iophantine n-gon ( n ≥ 3 ) with any side or diagonal equal to k for each fixed natural number k ? In case if it exists then let us find each such n . As a result of fundamental studies we have obtained a full answer to the above mentioned question for k=1, k=2, k=3 and k=4. The paper shows that for k=1 and k=2 such Bidiophantine n-gon does not exist, and for k=3 as well, as for k=4 definitely n is equal to 3 or 4. It is also shown that when k>2 there is always a Bidiophantine rectangle with the length of one of the sides equal to k. INTRODUCTION
Let us call a polygon Diophantine, if the distance between each two vertex of those is expressed by a natural number. Let us call as bidiophantine a Diophantine polygon those coordinates of each vertex are integers. In this paper, we will consider one of assigned by us task that undoubtedly represents one of the fundamental tasks of studying the properties of bidiophantine polygons.
Task** (n;k) exists if for not each fixed k natural number, bidiophantine n-gon ( n ≥3) whose length of arbitrary side or diagonal is equal to k, and if it exists, then let’s find all such n . Let’s firstly consider the task* (n; k). From the tasks of studying D iophantine geometric figures, one of the most important obviously is the following: Task* (n;k) exists or not for each fixed k natural number, Diophantine n-gon ( n ≥3) whose length of arbitrary side or diagonal is equal to k, and if it exists, then let’s find all such n . It is shown by us that there does not exist such D iophantine n-gon, both for convex and concave, whose length of arbitrary sides or diagonals are equal to 1. I.e. the above issue has been solved for . BASIC PART
To prove this, let’s consider a few tasks.
Lemma 1. If lengths of one of the side of Diophantine triangle is equal to 1, then the rest of its sides represent equal legs of the triangle.
Task: ; A B C ..1 ;, ACBCAB NACBC :u.d
Without limiting of generality let us assume that
ACBC . By 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 bidi ophantine geometric figure’s research apparatus.
Lemma 2. The length of each side and each diagonal of the convex Diophantine rectangle is greater than 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 ACBC and ADBD . This is impossible, since then C and D the points should be located on the median AB . I.e. our assumption is false. Therefore acceleration 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 . Using the inequality of triangle 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 . Fig. 1
Fig.2
I.e. finally we have that the length of each side and diagonal of each convex D iophantine 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 four points on the plane, none of that 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: ; A B C
2) Task:
A B A C B C m . ;;;1
NCD NDBAD NBCACAB This case is considered in the
Lemma . 2) From AOC and DOB we have: .,,,,, ;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.
Theorem 1. The length of each side and each diagonal of convex Diophantine n -gon ( n ) is greater than 1. .,,, , NlnmlCD nDBAD
Fig.4 Fig.3
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 polygon 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 ( ), none of the three point 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). Fig. 5
The first fours have no solution according to
Theorem 1 all fours will be as 1) and since 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.
From the tasks of studying bidiophantine geometric figures is obvious that one of the most important is the following. AC AD AD AD
1) 2) 3) 4) 5) 6) 7) 8) AC AC AD AD Task** (n;k)
Exists or no for each fixed k integer the bidiophantine n -gon ( ), whose length of arbitrary side or diagonal is equal to k , and if exist, then find all such n . At solution by us to the task* (n;1) , it was shown that there is not exist such Diophantine n-gon (n> 3) those length of arbitrary side or diagonal is equal to 1. Proceeding from this, there can be not exist also such bidiophantine n -gon ( n >3) whose length of arbitrary side or diagonal is equal to 1. As for the bidiophantine triangle, one of the sides of that is equal to 1, it should be isosceles accordingly of Lemma1, but then the vertex of this Diophantine triangle (the common point of the legs) cannot be Diophantine. I.e. there is not exist such bidiophantine triangle. This completely solved the task** ( n ;1) for k =1, or Theorem** (n;1) does not exist such a bidiophantine n-gon (n ≥ Let us now show that even for k = 2 the
Task** ( n ; k ) does not find a bidiophantine n-gon. Theorem** (n;2) does not find any bidiophantine n -gon ( n ≥
3) whose distance between arbitrary two vertexes is equal to 2.
Let’s assume the opposite. I.e. we assume that the length of each side and each diagonal of ... n A A A A n-gon is expressed by a natural number, and at the same time the length of one of them is equal to 2. Then for all of triangles those base is this equal to 2 segment, and the third vertex of this n-polygon, the modulus of difference of leg’s length is: 0 or 1 (this is easily illustrated by using the triangle inequality).
As there is not exist non-parallel to coordinate axis equal of 2 length bidiophantine segment (this leads due to the fact that there is no Pythagorean triangle with hypotenuse equal to 2). Therefore for each of the above mentioned triangles we have only one of the cases listed below (see Fig. 6 and Fig. 7).
Fig. 6 Fig. 7 I. In this case, due that kp AA and kqp AAA - are bidiophantines, therefore qp MAA - also will be bidiophantine, but this is not possible, because there is not exist such Diophantine rectangular triangle those length of any cathetus is equal to 1, so if there is exist such bidiophantine triangle then it must not be isosceles. II I
II.
In this case from the bidiophantine qp MAA and qk MAA due the Pythagorean theorem we have: .,, ;4412 ;.,, ;4412 ;.,, ;21 ; Nhba ba hbaNhba hbbaa hbaNhba hba hba
But this is impossible.
So our assumption is false, or is not exist such bidiophantine n -gon ( ) both convex and concave, the length of any side or diagonal of that is equal to 2. Q.E.D.
Now let’s consider the task** (n;k) for k . Task** ( n ;3) Is there exist such bidiophantine n- polygon ( n ) whose length of side or diagonal is equal to 3 . Let’s assume that is exist such bidiophantine n-gon ( n ) whose length of any side or diagonal is equal to 3. Then for all the triangles whose base represents this equal to 3 segment and the third vertex - any vertex of this n-gon, the modulus of difference of leg’s length is equal to 0; 1 or 2 (this is easily illustrated by using the triangle inequality). As there is not exist non-parallel to coordinate axis equal of 3 length bidiophantine segment, therefore for each of such triangles we have only one of the cases listed below (see Fig. 8, Fig. 9, Fig. 10). I. In this case because kp AA and kqp AAA - are bidiophantine, therefore qp MAA will be bidiophantine. But this is impossible, because MA p is not bidiophantine. II.
In this case from the bidiophantine qp MAA and qk MAA due the Pythagorean theorem we have: .,, ;31 ;
222 222
Nhba hba hba ., ;43 ;212.,, ;9612 ;
Nhb ba bbhNhba ba hba (1)
I II III
Fig. 8
Fig. 9
Fig. 10
III.
In this case from the bidiophantine qp MAA and qk MAA due the Pythagorean theorem we have: .,, ;14322 ;.,, ;9644 ;.,, ;32 ; Nhba ba hbaNhba hbbaa hbaNhba hba hba (2)
But that is impossible . Therefore for all such triangles kp AA is parallel to the coordinate axis and bbhMAbMAbAAbAAAA pkqpkp kq , where Nb and bb represents the square of the natural number. For pb tppbb . (3) and for pb the tppbb (4) (3) and (4) have infinitely many solutions in natural numbers. Indeed, because pp therefore (3) will be found such Nnm , , for that mnnp mp . (5) a b c d a c b d a d b ca b c d a c b d a d b c Due the (6) and (7) by combining solutions of equations dc and ba we can obtain all new solutions of (5). Below are presented some results of computations: n m mn b bb
65 918 160
93 222 358
It is much easier to obtain the solutions of (4). Indeed, as pp , therefore, we can more easily obtain the solution of (4). It follows that there will be existing such Nnm , for that
21 2 21 121 ; 2 12 1 . p m n mp n . (8)
Thus we can obtain the solution of (4) by solving Pell's equation (8). In addition mb . II I . type set of biophysical triangles, as it is shown above, is not an empty set . b bMAAAAA KqqKp .
53 3cos bbAAA kpq . Therefore by increasing b decreases qkp AAA cos and. kpq AAA cos . I.e. are increasing qkp AAA and kpq AAA . Therefore, if )(, bdNdb , then, for type II I triangles we would have case IV or V. Let’s consider case IV.
From qqp AAA )(3)53()53( bdbdc || qqqp qqpqkpq EANAAAMAEAAMNAAAMNA )2)(1(22 )2)(1(22)2)(1(22)53( dd bbdddMA q d bdddbddbdbd .2)(31 )(2)(3 bdd bdbd .1)(3., ;2)(3)(3 bdcNdb bdcbd qq NAA -დან qqqq NANAAA )2)(1()2)(1(81)(3 bbddbdba )2)(1()2)(1(81)(6)(8 bbddbdbd , II I ნახ. 11 IV ნახ. 12 that is impossible, because the left side of the last equation is odd, and the right side if it is natural, will be even. Now let’s consider the case V. It is obvious that (3 4; 3 5 ) q A O b b and )53;43( ddOA l , therefore (3 3 8; 3 3 1 0 ) q l A A b d b d and if q l
A A N , then .)3(3933 dbdbAA lq ( From the rectangle
NMEA l ) . ;3 NAME dbNMEA ll , NEANNAMA qlq , , as . NAMAEA lqq From the rectangle triangle ql EAA l q l q A A A E A E , from that we will obtain q q b d b d A E b d A E . This is impossible, because .,, NEAdb q I.e. are not existing IV and V type bidiophantine rectangles, so all bidiophantine n -gons those length of side is equal to 3 should only be of type VI (on one side of a straight line containing equal to 3 the side length) and whose diagonal length is equal to 3 would be only of type VII only. Fig. 14
Fig. 15
Fig. 16 VI VII VI I Fig. 17
Fig. 18 VI III VI II VII I Fig 19 V Fig. 13 For the case VI from ql EAA due the Pythagorean Theorem: we have: . l q l q A A A E A E from that we obtain ddbbdbc . (9)
Let’s consider (9) for d (this is the case VI ). For this case we have : bMAbAAbAAAAAAAA KqKqPKlKPlP ,43,53,5,3,4 , bbEAbbMA qq . Therefore teh (9) will be as: bbbc . (10) From the blunt triangle qlP
AAA due the inequality of triangles we have: bbcbc bc . As Nc , thus bbbc . (11) By transformation of (10) we will have: cbbbb . (12)
Due the taking into account (11) and (12) we will have: VI III VI IV Fig. 20
Fig. 21
Fig. 22
Fig. 23 VI V VI VI . .25646216 ;321246216 ;371846216. .434130946216 ;334130946216 ;234130946216 Zb bbb bbb bbbZb bbbbb bbbbb bbbbb (13)
The first and third equations of the system (13) did not have the solution in positive integers, because the left side of these equations, if it is natural, is even, as well as the right side is odd. Thus from (13) we will obtain : . ;834624. ;321246216
020 2
Zb bbbZb bbb .0..0 ;2348. ;04823. ;6448946216 bZbbbZb bbZb bbbb
In this case we obtain a VI II type rectangle. Now let’s consider (9) for bd . In this case we have VI III . Such bidiophantine rectangles exist for those Nb , for that bb - is a natural number. As it is mentioned above, such numbers are numerous. As for (9), its solution in case when
Ndb , and db by us in Basic software was tested from b and d up to 1,000,000, but were not find such natural c , b and d . In our opinion such c , b and d do not exist. It is noteworthy that this is possible to be considered as a task on a bidiophantine ellipse, those focuses are located in l A and q A points and length of major diameter is equal to db . Therefore, the existence of such bidiophantine rectangle is equivalent to the existence on a bidiophantine ellipse of a bidiophantine cord that is not parallel to the coordinate axis. Similarly to (9) for VII from the lq AEA rectangle we obtain : ddbbbdc . (14) For db we obtain case VIII and (14) will be as ., ;2122., ;2124 Nbc bbcNbc bbc (15)
As it is shown above the (15) has lots of solutions, and as for (14), similarly to (9) has been tested on
Basic software was tested from b and d up to 1,000,000, but were not find such natural c , 2 b and d and up to 1,000,000, but no such c, and not found. (In our opinion there is no such rectangle). VI III has not solution because the value of [EA q ] will be natural. VI IV has not solution because l p A N A and k p A N A at the same time will be Diophantine right triangles that is impossible because the Diophantine equation x +3 = y has the unique solution in natural numbers x =4 and y =5. VI V is impossible because klplplkl AAAAAAAA , this contradicts the well-known task accordingly of that sum of the lengths of diagonals of convex rectangle will be greater than the sum of the lengths of the opposite sides. VI VI is impossible because we would then have that BD =3 c+ BD =3 c+ . The first case ( BD =3c+3) is impossible because we would have that BD + AC = BC + AD that contradicts to the mentioned in VI V provision . In the second case we obtain that ∆ BAD is a right triangle and (this was considered in VI II ). Let's get back to V . .;0)31arccos;0( );31arccos;0(., ;433 43143cos ;)53(3 43153 3cos qplkpl kpqkpl kpq AAAAAA AAANbd dd dAAA bbbAAA
Similarly .;0 qkl AAA We obtain that for k =
3, the bidiophantine n-gon will only be either a triangle, or a rectangle, because it cannot have the following shapes (see Fig. 24, Fig. 25, Fig. 26).
Finally we have
Theorem** (n;3)
If for bidiophantine n -gon k , then n . Fig. 24 Fig. 25 Fig. 26
Task** (n;k) for k . Task** (n;4)
Is there existing or not such bidiophantine n-gon ( n ≥
3) those side or diagonal length is equal to 4, and if exists let’s find all such n . Let’s suppose that there is existing such bidiophantine n -gon ( n ) whose length of any side or diagonal is equal to 4. Then for all those triangles the base of that is represented by this equal to 4 length of segment, and the vertex is any of these n -polygon’s vertex the modulus of difference of a legs lengths is equal to 0; 1; 2 or 3 (This is easily to show by using the triangle inequality). As there is not existing non-parallel to the coordinate axis bidiophantine segment those length is equal to 4, (this implies that there is no Pythagorean triangle with hypotenuse length equal to 4), we have one of the following four cases for all such triangles (see Fig. 27, Fig. 28, Fig. 29, Fig. 30). In the case VIII I , as kp AA and kqp AAA - are bidiophantine , thus qp MAA also will be bidiophantine, but this is impossible because is not exist bidiophantine right triangle, the length of those cathetus is 2. In the case VIII II we have from the bidiophantine triangles qp MAA and qk MAA accordingly of Pythagorean theorem: .,, ;16812 ;.,, ;16812 ;.,, ;41 ; Nhba ba hbaNhba bhbaa hbaNhba hba hba
This is impossible.
In the case VIII IV , similarly to VIII II , from the right bidiophantine triangles qp MAA and qk MAA we have: .,, ;16896 ; Nhba ba hba this is impossible . Fig. 27 Fig. 28 Fig. 29
VIII I VIII II VIII
III
VIII IV Fig. 30
VIII
III bidiophantine triangles qp MAA and qk MAA accordingly of Pythagorean theorem we have : ., ;32 ;313.,, ;32 ;.,, ;42 ; Nhb ba bbhNhba ba hbaNhba hba hba
So for all such triangles Kp AA are parallel to the coordinate axis. In addition: ;4 Kp AA bbhMAbAAbAA qKqp q , where Nb and bb represents the full square of the natural number. We tested up to 25,000 and such b are found 54 in total. b bb b bb IX X XI Fig. 32 Fig. 33 Fig. 31
XII
XII I XII II Fig. 36 Fig. 35 Fig. 34 The set of type IX Bidiophantine triangles, as it is shown above, is not an empty set. bbMAAAAA bbAAA
KqqKp kpq . Therefore, by increasing in b decreases bbMAAAAA KqqKp . Here, similarly to the previous paragraph, by the increasing of b decreases qKp
AAA cos and XII
III
XIII I XIII
Fig. 37 Fig. 39 Fig. 38 XV XIV
Fig. 40 Fig. 41
XVI
XVII
Fig. 42 Fig. 43 .cos kpq AAA I.e. increases qkp
AAA and kpq AAA . Therefore for )(, bdNdb type IX triangles we can only have cases X and XI. Let’s consider the case X From qqp AAA .)(2)52()52( bdbdc qqqqpkp MAEANAdMAEAqAMNAqAAMN dbdbddd bbddd .1)(2.,, ;2)(2)(2 ;2)(25.1)(25.1 ))(52(5.1 5.11)52( bdcNcdb bdcbd bdd bdbdd bdddbd
From the right triangle qq NAA qqqq NANAAA i.e. )3)(1()3)(1(3)(1)(2 bbddbdbd )3)(1)(3)(1(61)8843(2 ddbbbddb . (16) This is impossible, because its left side is odd, and its right side if it will be natural - even. From the case XI is obvious that .52;32 ;52;32 ddOA bbOA lq l q A A b d b d . In addition, if we take into account that from
Kql
AAA l q l K q K A A A A A A d b b d , and NAA lq , then we obtain dbAA lq . From the right triangle ql EAA accordingly of Pythagorean theorem we have EAEAAA qlql from that we derive ddbbdbdb ddbbdbbd . (16 ). (16 ) has no solution in natural numbers, because the left side is odd and the right is even In the case
XII from the right triangle ql EAA accordingly of Pythagorean theorem we have EAEAAA qlql from that we derive ddbbdbc . (17) In (17) for d =0 (XIII case) we will have: ;32;52;5;4;3 bAAbAAAAAAAA qkqpklkplp .3)3)(1(3;)3)(1(3; bbEAbbMAbMA qqk and in this case (17) will be as : bbbc . (18) From the blunt triangle qlp
AAA , due the inequality of triangles we have: )52;22(.52 ;523 bbcbc bc , b ut as , Nc thus bbc . (19) By transformation of (18) we obtain cbbbb . (20)
The left side of (18) must necessarily be a natural number and then it will be even, thus in order for the right side also to be even, from (19) would be used only bc (21) i.e. by substitution of (21) in (20) we will obtain : .0. ;01443. ;)42(3420491236
020 222 bZb bbZb bbbbb
In this case we obtain the
XII II rectangle . Now let’s consider (17) for bd . In this case we have
XII
III . Such bidiophantine rectangles exist only such Nb for that bb - is a natural number. As we have shown, such numbers are numerous. As for the solutions of (17), in the case when Ndb , and db , by us in Basic software was tested from b and d up to 1,000,000, but were not find such natural c , b and d . In our opinion such b and d do not exist. It is noteworthy that this case is possible to be considered as a task on bidiophantine ellipse, those focuses are located in A l and A q points and those major diameter is equal to 2 b+ d+
8. Therefore, the existence of such kqlp
AAAA bidiophantine rectangle is equal to the existence on an ellipse of such bidiophantine cord that is not parallel to the coordinate axis.
Similarly of (17) in case XIII from right triangle lq AEA we have ddbbbdc . (21) For db we obtain XIII I and (21) will be as 8 ., ;3132., ;3134 Nbc bbcNbc bbc (22)
As it was shown above, (22) has several solutions, and as for (21) for db by us in Basic software was tested from b and d up to 1,000,000, but were not find such natural c , b and d . XIV does not have solution in natural numbers, because then A k MA q must be bidiophantine, but there is not exist Pythagorean triangle with one of the cathetus equal to 2. XV does not have solution in natural numbers because otherwise for the □ p l k M A A A would be violated the inequality k p l l p k
A A A M A A A M . The XVI case is impossible for the same reason as for XV.
In the XVII variant we have that BD = c+
1 or BD = c +5. In the first case, the true inequality BD + AC > BC + AD is violated that is impossible, and in the second case we obtain that BAD is a right triangle and c =0 (this was considered in XIIII). Let's go back to XI: );0(.32;0 ;3;0.32 5.12132cos ;52 5.12152 4cos qplkpl kpqkpl kpq
AAAAAA AAAdddAAA bbbAAA
Similarly .);0( qkl AAA We obtain that for k =4 the bidiophantine n -polygon would be either a only triangle or a rectangle because it cannot have the following shapes: Finally we have:
Theorem** (n;4) If for bidiophantine n -gon k , then n . We investigate the
Task** (n;k) as well as for k , k , k . We have some cases to checked. The reader will find them in the following work. Fig. 45 Fig. 44 Fig. 46 )3( kNk bidiophantine triangle, the length of any side of that is equal to k , the answer is positive. See Fig. 47. To the question whether there exist a bidiophantine rectangle for each of )3( kNk the length of any sides of those is k , the answer is also positive. See Fig. 48 Fig.48
CONCLUSIONS
The goal of the work is to take on and study one of the fundamental tasks studying Bidiophantine n-gons (the author of the paper considers an integral n-gon is Bidiophantine if the coordinates of its each vertex are integer numbers). Task **( n; k ): is there a
Bid iophantine 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 . As a result of fundamental studies we have obtained a full answer to the above mentioned question for k=1, k=2, k=3 and k=4. The paper shows that for k=1 and k=2 such Bidiophantine n-gon does not exist, and for k=3 and k=4 definitely n is equal to 3 or 4. It is also shown that when k>2 there is always a Bidiophantine rectangle with the length of one of the sides equal to k REFERENCES Adamar J. Elementary geometry. Moscow, State pedagogical-training publishing house of the Ministry of Education of RSSR, 1937 (In Russian).
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