| 1) Given: Sides a, b, and c of a triangle, and its area, A, are all
integers.
Prove: A is even.
One formula for the area of a triangle is:
A = sqrt( s * (s-a) * (s-b) * (s-c) ),
where s = (a+b+c)/2. Call the product inside the square root function
P. Since P = A*A, P is an integer.
Lemma: s is an integer.
Suppose s be not an integer (this is proper English, see note
68 in SUMMIT""::SYS$NOTES:JOYOFLEX). Then s is in the form
i0/2, where i0 is an odd integer (i0 = a+b+c). Clearly, s-a,
s-b, and s-c also have forms i1/2, i2/2, and i3/2, for some
odd integers i1, i2, and i3. So P = i0*i1*i2*i3/16. But
i0*i1*i2*i3 must be odd, so P is not an integer, which is a
contradiction. Therefore s is an integer.
Lemma: P is even.
Suppose P be odd. Then none of s, s-a, s-b, and s-c could
be even integers. Since they are all integers, they must all
be odd. If s is odd and s-a is odd, then a must be even.
Similarly, b and c must be even.
If they are all even, then a = 2i, b = 2j, and c = 2k, for some
integers i, j, and k. Expanding P reveals
P = 2(i*i*j*j + j*j*k*k + k*k*i*i) - i**4 - j**4 - k**4.
Consider the remainder when P is divided by 4 when i, j, and k
take on even or odd values. Because i, j, and k are symmetric
in the above formula, it is only necessary to consider one case
where they are all even, one case where one is odd, one case
where two are odd, and so on.
i j k | formula, shown as congruences mod 4 | P mod 4
0 0 0 | 2(0+0+0)-0-0-0 | 0
0 0 1 | 2(0+0+0)-0-0-1 | 3
0 1 1 | 2(0+1+0)-0-1-1 | 0
1 1 1 | 2(1+1+1)-1-1-1 | 3
So P is congruent to either 0 or 3 modulo 4. If it is
congruent to 0, it is even, which contradicts the supposition
that P is odd. If it is congruent to 3, then A is not an
integer, because there is no integer whose square is congruent
to 3 modulo 4 (integers congruent to 0, 1, 2, and 3 have
squares congruent to 0, 1, 0, and 1, modulo 4). This is also
a contradiction.
Therefore the supposition that P is odd is false.
Since P is an even integer and A is an integer, then A is also even
(otherwise squaring A, an odd integer, would yield P, an even integer).
2) Is a Pythagorean triangle a right triangle with integral sides? If
so, one of the legs is necessarily even, so the area is obviously an
integer.
To see that one side is even, suppose they be both odd. Let the length
of the hypoteneuse be denoted by a and the lengths of the legs by b
and c. Then a*a = b*b + c*c. Since b and c are odd, b*b and c*c are
odd, and b*b + c*c is divisible by two but not by four. Clearly,
no integer has a square divisible by two but not by four. But a is
known to be an integer, so this is a contradiction. Therefore b and
c are not both odd.
-- edp (WHOAREYOU note 329)
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| Let the sides be a,b & c, angles A,B & C. The area of a triangle is
T = �absinC. Now, cosC= (a�+b�-c�)/2ab, so:
(*) T = _/((a+b+c)(a+b-c)(a-b+c)(-a+b+c)) /4
Any integral triangle must satisfy (*). Also, any solution to (*)
yields an integral triangle. For a,b,c must satisfy the triangle inequality,
or T will be imaginary. So the problem is entirely captured algebraically.
>1. The areas of integral triangles are always even: why?
If a+b+c is odd, so are the other three multiplicands under the
sqrt sign, and hence T� is not an integer (let alone T.) Contradiction. So
a+b+c is even, and similarly, the other three terms. So their product is a
multiple of 16, and we can deduce that T� is an integer.
Suppose then that one of {a,b,c} is odd. Wlog, c is odd. Then (a+b+c)
-(a+b-c) = 2c == 2 mod 4. So 4 divides one of (a+b+c) or (a+b-c). So 2 | T�.
So T is even.
Otherwise, all of {a,b,c} are even. Let a = 2a', etc. Then
T = _/((a'+b'+c')(a'+b'-c')(a'-b'+c')(-a'+b'+c'))
If a'+b'+c' is even, then so is T. Otherwise a'+b'+c' is odd, and computing
the rhs mod 4, have that T� == 3 mod 4. Contradiction.
>2. As a related problem, or as a warmup, try showing that all pythagorean
>triangles are integral triangles.
Pythagorean triangles are of the form:
k2mn
k(m�-n�)
k(m�+n�)
where m > n are coprime +ve integers, with m+n odd, and k is any
+ve integer. Pythagorean *triples* satisfy the above with the restriction
that k = 1. Plugging these values into (*) yields:
T = k�mn(m�-n�)
which is certainly an integer.
>3. There are only 5 integral triangles whose sides add up to their area.
>Find these, and as a bonus, show they are the only ones.
Most puzzles concerning integral triangles are of this form. Posit an
extra restriction on top of (*), and find the solutions. Here (*) becomes:
16(a+b+c) = (a+b-c)(a-b+c)(-a+b+c)
(p) (-1 1 1)(a)
Let (q) = �( 1-1 1)(b). By the above, p,q,r are all +ve integers. Then:
(r) ( 1 1-1)(c)
4(p+q+r) = pqr
or:
1 1 1 1
--- + --- + --- = -
p.q p.r q.r 4
One of the 3 summands on the lhs must be >= 12. So wlog p.q =< 12. And we know
that p.q >= 5, so there are just a few cases. We have a solution if r is an
integer, ie: if 4(p+q)/(pq-4) is an integer.
p.q q p r a b c T
--------------------------------------
5 5 1 24 29 25 6 60
6 6 1 14 20 15 7 42
6 3 2 10 13 12 5 30
7 7 1 32/3 not integer
8 8 1 9 17 10 9 36
8 4 2 6 10 8 6 24
9 9 1 8 (see above)
9 3 3 24/5 not integer
10 10 1 44/6 not integer
10 5 2 28/6 not integer
11 11 1 48/7 not integer
12 => p.q = p.r = q.r = 12 => r = _/12, not integer.
So there are only 5 solutions, listed above.
>4. The larger integral triangles sometimes have areas that are integral
>multiples of the sums of their sides. For any integer M there are a finite
>number K>1 of integer triangles whose area = perimeter * M. I know of no
>way, short of enumeration, of evaluating f(M) = K. Maybe someone can find one.
>The first few terms:
>
> M K
> 1 5
> 2 15
> 3 33
>
>I haven't proved yet that I have f(2) and f(3) right: I only have upper and
>lower bounds. There may be triangles that satisfy the criteria where all
>the sides are > 100. (Although I doubt it)
The corresponding formula is:
1 1 1 1
--- + --- + --- = ---
p.q p.r q.r 4M�
As before, wlog, pq =< 12M�. There are a finite number of candidate
pairs (p,q) and if r = 4M�(p+q)/(pq-4M�) is an integer, then we have a solution.
I don't see any reason why there should be a neat formula for f(n), and I'm not
interested in checking the values of f(2) or f(3) given.
>5. Sometimes different integral triangles have the same area. What's the
>smallest area this is true for, and what are the two integral triangles that
>have this area?
Pluck out of the air: (5,5,6) & (5,5,8) have area 12.
Is there a smaller triangle? Yes, but only one. T� = (p+q+r)pqr &
T == 0 mod 6. (Easy to check possibilities for p,q & r mod 3 to see that 3|T.)
So any smaller example has T=6. The only integral triangle for T=6 is the
(3,4,5) right-angled one.
>6. As in the solution to #5, sometimes these equal area triangles have two
>sides out of three in common. Let's call these "linked equiareal triangles."
>Even excluding multiples of smaller l.e.t's, there are an infinite number of
>these. (I'm pretty sure, anyway: can anyone prove it?)
Equation * can be re-written:
c� = a�+b� �2_/(a�b� - 4T�).
So for any a,b,T for which there is one value of c�, there will be
another value of c�. The product of the two will be (a�-b�)� + 16T�, so since
one is positive, so will be the other.
>7. In l.e.t,'s, one of the two lengths that doesn't match up is always the
>largest length of its triangle. How come?
Examine:
c� = a�+b� �2_/(a�b� - 4T�).
When the sqrt is added a�+b�, the result will *always* be greater
than both a� & b�.
>8. The most acute integral triangle I have found so far has sides of 3, 160,
> [later changed to 162] and 163. Later I found an integral triangle with
> sides 3,865, and 866. Very acute. An area of 1224. I imagine that there are
> always integral triangles that are more acute, but I haven't investigated
> much. Are there? Is there a limit?
[paraphrase of question.]
I don't agree with (3,162,163), but (3,865,866) is OK.
Try * for the triangle (3,k,k+1)
T� = (2k+4).(k-1)
=> T� = 2k�+2k-4
=> 2T� = l�-9
where l = 2k+1. Take this equation mod 3. 2 is not a square mod 3, so 3 divides
both T & l. Write T=3y, l=3x.
x� - 2y� = 1.
which is Pell (who never lurks below the surface here.) The fundamental
solution is (0,1), and then we can derive all others by multiplying the (x,y)
vector by:
(3 4)
(2 3)
See 1756.3 for exactly this Pellian equation in action. x = 1,3,17,99,577...
yielding k = ?,4,25,148,865... So we pick up the case k=865, but there is
nothing remotely like the other example suggested by the poster. The cos of
the acutest angle here is 1 - 4/(k(k+1)) -> 1 as k -> %.
>9. The most obtuse integral triangle is probably the same as the most acute
>one, right? :-)
There is no "most acute triangle", we have established.
There is no "most obtuse triangle", either. Go for triangles of the
form (x,x,2x-2). If 2x-1 is a square (not the hardest thing to arrange.) then
T = (x-1)_/(2x-1). The cos of the obtuse angle = -1 + (8x-4)/(2x�) -> -1 as
x -> %. The acute angles have cos 1 - 1/x -> % as x -> %, but not as fast as
the case in 8.
Cheers,
Andy.
|
| >.3 TOOLS::STAN
>For quite a while, I have been researching the question of whether
>there is a formula that generates all Integral Triangles (also known as
>Heronian Triangles in the literature). [...] I believe the following
>is a valid formula:
>
> [a=] kmn(p^2+q^2)
> [b=] kpq(m^2+n^2)
> [c=] k(mq+np)(mp-nq)
>
>where gcd(m,n)=1 and gcd(p,q)=1, mp>nq,
>although the published proof seemed wrong.
>I hope you find this formula useful. Let me know if it misses any solutions.
>.5 RAINBO::GRANT
>The formula in .3 seems to miss some triangles. (5,5,6) is Heronian, with an
>area of 12, and cannot be generated by the formula if k is required to be an
>integer. It does come from the formula when m=p=2, n=q=1, and k = 1/2.
>
>If this is a valid use of the formula, then what is the domain of k? k=1/2
>does not always produce integral solutions.
>.6 TOOLS::STAN
>Oops, another candidate down the drain.
The formula given is sufficient but not necessary. The problem was with
the constant multiplier k.
Rather than deal with the sides a,b & c, it's cleaner to work with the
terms from Hero's formula for the area of a triangle:
w = �(a+b+c)
x = w-a
y = w-b
z = w-c
Then *any* four positive integers w,x,y,z which satisfy:
(i) w = x+y+z
(ii) wxyz is a square
yield an integral triangle (a,b,c)
******************************************************************************
*
* FORMULA TO GENERATE THE HERONIAN TRIANGLES
*
******************************************************************************
Now let m,n,p,q be +ve integers such that (m,n) = (p,q) = 1, and
mp > nq. Then set:
w = Kmpr
x = Kmqs
y = Knps
z = Knqr
where:
r = (mq+np)/(mq+np,mp-nq)
s = (mp-nq)/(mq+np,mp-nq)
and:
K = N/(mn,pq), N being any +ve integer we like.
Then I claim that
(1) given p,q,r,s,N then w,x,y,z are of the required form.
(2) given w,x,y,z, then we can derive p,q,r,s,N.
(3) given w,x,y,z, then the derived p,q,r,s,N are unique.
(1) w,x,y,z could only fail to be integers if (mn,pq) > 1. Wlog, (m,p) = C > 1.
C | mq+np, but C \ mp-nq. ("\" denotes "doesn't divide".) So C \ (mq+np,mp-nq).
So C | r. Thus C | mpr,mqs,nps,nqr, so w,x,y & z are all integers.
w=x+y+z is easy to check.
wxyz a square is easy to check.
(2) Given w,x,y,z of the required form, let N = (w,x,y,z), and w = NW, x = NX,
y = NY, z = NZ. Since W=X+Y+Z, no three of W,X,Y&Z share a factor. WXYZ is a
square.
Define m & n by _/(WX/YZ) = m/n (s.t. (m,n) = 1)
Similarly: _/(WY/XZ) = p/q
And: _/(WZ/XY) = r/s
K = N/(mn,pq)
(m,n) = (p,q) = 1 by construction
mp/nq = W/Z = X+Y+Z/Z > 1. So mp>nq.
Let:
w' = Kmpr
x' = Kmqs
y' = Knps
z' = Knqr
Our task is now to show that w' = w, etc.
w'/x' = Kmpr/Kmqs = pr/qs = _/(WY/XZ)*_/(WZ/XY) = W/X = w/x, &
similarly for other w'/y' etc. So w',x',y',z' = �w,�x,�y,�z.
w=x+y+z
=> w'=x'+y'+z'
=> Kmpr = Kmqs + Knps + Knpr
=> s(mq+np) = r(mp-nq)
=> r = (mq+np)/(mq+np,mp-nq)
s = (mp-nq)/(mq+np,mp-nq)
as required.
It remains to show that no three of w'/N, x'/N, y'/N, z'/N share a
factor. For then we know that �=1.
If three share a factor, then so do all four.
(w'/N, z'/N) = (mpr/(mp,nq),nqr/(mp,nq))) = r
Similarly:
(x'/N, y'/N) = s. But we know that (r,s) = 1. So �=1.
(3) The values of m&n are the only coprime numbers which allow WX/YZ to be in
the right proportion. Similarly p&q, r&s. And given these 6 values, only one
value of K will allow wxyz to be correct. So the solution is unique.
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