andrews@yale.ARPA (Thomas O. Andrews) (11/12/85)
Enough silly polar bear problems.
Here's a problem that requires a little more sophistication:
The 3-4-5 triangle has integral area (area=6).
The 13-14-15 triangle has integral area (area=84).
Find all triangles with sides n-(n+1)-(n+2) that have
integral area.
Note: This isn't an elementary problem.
I'll send the answer to anybody who wants it, and maybe even post
it . . .
--
Thomas Andrews
andrews-thomas@yalebs@faron.UUCP (Robert D. Silverman) (11/15/85)
> > Enough silly polar bear problems. > Here's a problem that requires a little more sophistication: > > The 3-4-5 triangle has integral area (area=6). > The 13-14-15 triangle has integral area (area=84). > Find all triangles with sides n-(n+1)-(n+2) that have > integral area. > > Note: This isn't an elementary problem. > I'll send the answer to anybody who wants it, and maybe even post > it . . . > -- > Thomas Andrews > andrews-thomas@yale Not to contradict you but the solution IS elementary. One need only know a little elementary number theory including simple diophantine equations. The area of a triangle is: SQRT( s (s-a) (s-b) (s-c)) where a,b,c are the lengths of the sides and s is the semi-perimeter. (Heron's formula) Substitute n, n+1, n+2 and we obtain: 1/4 (n+1) sqrt( 3 (n+3) (n-1) ) Thus we require that 3(n+3)(n-1) be a square, say A^2. Now it is easy to see that either n+3 or n-1 must be divisible by 3, so n = 0 or 1 mod 3. Further, A must be an multiple of 3, say 3h. ------------------------------------------------------------------------------ Case 1: n = 1 mod 3 Let n = 3 k + 1 for some k we obtain: k(3k+4) = h^2 Solving for k we obtain via the quadratic formula: (-4 +/- sqrt(16 + 12h^2))/6 And we require that 4 + 3h^2 be a square ------------------------------------------------------------------------ Case 2: n = 0 mod 3 Let n = 3 k we obtain: (k+1)(3k-1) = h^2 again we obtain 4 + 3h^2 must be a square by solving for k. ----------------------------------------------------------------------- Combining these we find: k = (sqrt(4 + 3h^2) -1) / 3 or k = (sqrt(4 + 3h^2) - 2) /3 We now must have 4 + 3h^2 = s^2 or s^2 - 3h^2 = 4 This is a (modified) Pells equation. Its solution is given by: P (1/2 (s + sqrt(3) h ) P = 0, 1, 2, ... 0 0 where s and h are the least positive solution. (4,2) 0 0 One can also obtain new solutions via the linear recurrence relation: s = 4s - s i i-1 i-2 This yields the sequence of solutions: s = (2, 4, 14, 52, 214, ...) h = (0, 2, 8, 30, 112, ...) Note: One can derive the solution to the above by elementary means using a little theory about recurrence relations OR if you know some algebraic number theory by finding units in the quadratic extension field Q(sqrt(-3)). If we substitute the values given for h, we can solve for k and hence get n. This solution is essentially complete. Bob Silverman (they call me Mr. 9)
leimkuhl@uiucdcsp.CS.UIUC.EDU (11/18/85)
bs is correct in his analysis, but he did not need to break the problem down to cases. We seek solutions in integers to n^2 + 2n - 3 = (r^2)/3. Obviously this means r is divisible by three--r=3q. So r^2=9q^2, and we can replace our first equation by n^2 +2n - 3 = 3q^2. Now we can just complete the square: q^2 + 2q +1 = (q+1)^2 = 3r^2 + 4 Which is the same equation bs derives via his case by case analysis. -Ben Leimkuhler
leimkuhl@uiucdcsp.CS.UIUC.EDU (11/18/85)
Sorry, my notation got confused there in transit to the net. The final equation should have been (n+1)^2 = 3*q^2 +4. My apologies.
pumphrey@ttidcb.UUCP (Larry Pumphrey) (11/19/85)
> Enough silly polar bear problems. > Here's a problem that requires a little more sophistication: > > The 3-4-5 triangle has integral area (area=6). > The 13-14-15 triangle has integral area (area=84). > Find all triangles with sides n-(n+1)-(n+2) that have > integral area. > > Note: This isn't an elementary problem. > I'll send the answer to anybody who wants it, and maybe even post > it . . . Yes, this is quite a problem! The answer is given below in rot13 format but I also won't post the proof as it is rather lengthy. Guvf ceboyrz unf na vasvagr ahzore bs fbyhgvbaf juvpu pna or qvivqrq vagb gjb pngrtbevrf jurer gur v-gu zrzore pna or erphefviryl qrsvarq va grezf bs gur 2 cerivbhf ragevrf. pngrtbel 1 pngrtbel 2 ---------- ---------- e = 1 e = 1 1 1 e = 3 e = 4 2 2 Abj gur v-gu (sbe v>2) zrzore bs rnpu pngrtbel pna or sbhaq ol gur sbyybjvat erphefvir sbezhyn: e = 4*e - e v+1 v v-1 Gura n gevnatyr bs fvqrf a, (a+1), (a+2) unf vagrteny nern vs naq bayl vs 2 a = 12*e + 1 jurer e vf n zrzore bs pngrtbel 1 v v 2 a = 6*e - 3 jurer e vf n zrzore bs pngrtbel 2 v v Nf fgngrq ol gur bevtvany cbfgre, gur cebbs vf dhvgr vaibyirq naq vaibyirf gur fbyhgvba bs fbzr Qvbcunagvar rdhngvbaf. Vg abj nccrnef ubjrire, gung univat gur nobir erphefviryl qrsvarq fbyhgvba(f) vg fubhyq or rnfl gb hfr rvgure vaqhpgvba be vasvavgr qrfprag gb rfgnoyvfu gur erfhyg. V unira'g gevrq guvf nccebnpu.
ins_adsf@jhunix.UUCP (David S Fry) (11/20/85)
> This yields the sequence of solutions: s = (2, 4, 14, 52, 214, ...) > h = (0, 2, 8, 30, 112, ...) > This is incorrect. The solutions for s are s = (2, 4, 14, 52, 194, 724, 2702, 10084,...) Since one finds easily that n = s - 1, then n is n = (1, 3, 13, 51, 193, 723, 2701, 10083,...). Also, Mr. Silverman has only shown that these are necessary conditions, not sufficient. In other words, these numbers make sqrt(3 (n+3) (n-1)) an integer, but we do not necessarily know that they make 1/4 (n+1) sqrt(3 (n+3) (n-1)) an integer. To see this, note that 2 always divides s (since s = 2 and s = 4s - s ) 0 i i i-2 so 4 divides s^2. Also, s^2 - 3h^2 = 4, or 3h^2 = s^2 - 4 so 4 divides 3h^2, and hence 2 divides sqrt(3 (n+3) (n-1)) = 3h. And, n is always odd (n = s -1) so n+1 is even and divisible by 2. Hence the area is an integral. NOW the solution is complete. David Fry
ins_adsf@jhunix.UUCP (David S Fry) (11/20/85)
> Hence the area is an integral.
Excuse me...the area is an integer. I have calculus on the brain apparently.
-David Fry