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C
. .
. .
.
.
E
.
.
.
.
.
A---------D---------------B
with lengths
AD = 80
DB = 128
AE = 49
EC = 63
CB = 112
CD = 48
DE = 39
is a partition of an integer-sided triangle into 3 mutually dissimilar
integer-sided triangles. It is not the smallest, see below.
The [7,7,13] triangle is not the smallest (in perimeter) which can
be
partitioned into 2 dissimilar integer-sided triangles. In a [6,8,10]
right triangle, the line from the right angle to the midpoint of
the
hypotenuse partitions it into [5,5,6] and [5,5,8], which are dissimilar
as required. In general, a [2x,2y,2z] right triangle can be partitioned
into triangles [z,z,2x] and [z,z,2y].
The general problem of finding integer-sided 2-partitions goes like
this:
C.
/ \ .
/ \ .
d/ \
.e
/ a\
.
/
\ .
/
\ .
A-------------D-------B
b
c
One requires angle CDA to be complementary to CDB. This means that
the
sum of their cosines is 0. Using the law of cosines and eliminating
the
cosine term, one arrives at
f(a,b,c,d,e) = ba^2 + ca^2 + cb^2 + bc^2 - cd^2 - be^2 = 0
This defines a surface in 5 dimensions. Suppose that (A,B,C,D,E)
is any
known solution. If we substitute (a,b,c,d,e) = (A,B,C,D,E) +
(r,s,t,u,v)*x, with rational parameters r,s,t,u,v to be determined
and x
variable, we get a cubic in x with the known root x = 0, thus of
the
form:
K*x^3 + L*x^2 + M*x = 0
Here, K, L and M are rational functions of (A,B,C,D,E) and (r,s,t,u,v).
M in particular is linear in (r,s,t,u,v), thus we can impose the
extra
condition M = 0 by setting r equal to some appropriate rational
linear
function of s,t,u,v. This leaves
K*x^3 + L*x^2 = 0
which has a double root x = 0 and a single root x = -L/K, with K
and L
rational. If we then substitute x = -L/K, we get (after clearing
denominators)
(a,b,c,d,e) = (A,B,C,D,E)*K - (r,s,t,u,v)*L
which gives another generally distinct rational solution. Any rational
solution (a,b,c,d,e) corresponds to an integer solution
(a,b,c,d,e)/gcd(a,b,c,d,e).
The condition M = 0 means that the vector (r,s,t,u,v) lies in the
hyperplane tangent to the surface f at (A,B,C,D,E). This hyperplane
is
orthogonal to the vector
W = (2AB+2AC,A^2+2BC+C^2-E^2,A^2+B^2+2BC-D^2,-2CD,-2BE)
thus we will have M = 0 if (r,s,t,u,v).dot.W = 0. The components
of W
are the partial derivatives of f with respect to each variable,
evaluated at (A,B,C,D,E). It is easily verified that (r,s,t,u,v)
=
(-A,-B,-C,-D,-E) satisfies this condition, thus the point (0,0,0,0,0)
=
(A,B,C,D,E)+(r,s,t,u,v) always lies in the tangent hyperplane.
If the required calculations are carried out, the result is a parametric
solution in 4 rational parameters s,t,u,v which gives all rational
solutions in the tangent hyperplane at (A,B,C,D,E). If we choose
any
other solution in the hyperplane and repeat the procedure, we will
obtain another parametric family of all rational solutions in a
distinct
tangent hyperplane. Note that a given hyperplane cannot be tangent
to f
at two distinct points.
Another way of obtaining new solutions is, given any two distinct
solutions (A,B,C,D,E) and (A',B',C',D',E'), where neither point
lies in
the tangent hyperplane of the other point, if we write
(a,b,c,d,e) = (A,B,C,D,E) + (A'-A,B'-B,C'-C,D'-D,E'-E)*x
we get a cubic in x with known roots x = 0 and x = 1. The third
root
will perforce be rational, and will generate a new solution.
Most of the above is analogous to the well-known chord-tangent procedure
for finding rational points on an elliptic curve. An elliptic curve
is
equivalent to a homogeneous cubic in three variables, while the
surface
f is a homogeneous cubic in five variables. I don't know if anyone
has
systematically investigated such curves, but it seems likely to
me that
there is a group law for f which gives all rational solutions of
f.
I think it is likely that any rational-sided triangle can be partitioned
into two other rational-sided triangles in an infinite number of
ways.
Let's see for example whether an equilateral triangle can be thus
partitioned. In order for the outer triangle to be equilateral,
we
require
(a,b,c,d,e) = (x,y,z-y,z,z)
for some unknowns x,y,z. This gives
f = z*(x^2 - y^2 + yz - z^2) = 0
and since we want z <> 0, we need
x^2 = y^2 - yz + z^2
i.e.
(2x)^2 = (2y-z)^2 + 3z^2
This has the parametric solution
2x = u^2 + 3v^2
2y-z = u^2 - 3v^2
z = 2uv
With (u,v) = (5,3) we get (x,y,z) = (26,14,30), which leads to the
integer solution (a,b,c,d,e) = (13,7,8,15,15). In general, to obtain
distinct integer solutions, we want u and v to be coprime and we
need u
and v to have the same parity, thus both odd, and to ensure 0 <
y < z we
need v < u < 3v. The resulting values (x,y,z) will always
be even, so we
can reduce them to (x/2,y/2,z/2). This leads to an infinite number
of
distinct solutions for the equilateral triangle.
This gives another general procedure for finding solutions. Suppose
that
the sides of the outer triangle d:e:(b+c) are in the ratio R:S:T.
Choose
odd coprime integers u,v so that
v*(S^2-(R-T)^2) < u < v*((S+T)^2-R^2)
Let
D = (R+S+T)*(R+S-T)*(S+T-R)*(T+R-S)
x = u^2 + D*v^2
y = (u + v*((R+T)^2-S^2)) * (u + v*((R-T)^2-S^2))
z = 4*T*u*v
(At this point, any common divisor of x,y,z can be factored out.)
(a,b,c,d,e) = (x,y,T*z-y,R*z,S*z)
This gives an integer solution whose perimeter is (R+S+T)*z.
Note that the discriminant D is 16 times the square of the area
of a
triangle with sides R,S,T.
I used this to search for a small perimeter solution without any
isosceles triangles and came up with (11,2,6,12,10) with perimeter
30. I
haven't tried an exhaustive search, so I don't know if this is
minimal.
Working from the (11,2,6,12,10) solution, I found the following
solution
to the 3-partition problem:
C
. .
. .
.
.
E
.
.
.
.
.
A---------D---------------B
with lengths
AD = 4
DB = 12
AE = 6
EC = 18
CB = 20
CD = 22
DE = 5
This looks pretty good, with perimeter only 60. I wouldn't be surprised
if it is the smallest possible. This is spoiled somewhat by the
fact
that ADE is similar to ABC, but the three small triangles are pairwise
dissimilar.
Somewhat surprisingly, it is also possible to add a point F to this
along the CB edge, with
CF = 5
FB = 15
DF = 18
This gives a 4-partition of the triangle with the same small perimeter.
But wait, there is more. Point G can be added to the BD edge with
DG = 7
GB = 5
CG = 20
Using points A,B,C,D,G we have a 3-partition in which no two triangles,
including outer triangles, are similar. It's too bad that BCG is
isosceles.
Unfortunately, DF intersects CG, so you can't use both. Wait, there's
still more. Point H can be added to the DF edge, with
DH = 10
HF = 8
BH = 10
With points A,B,C,D,E,F,H this gives a 5-partition of the triangle.
I'm not even sure that there aren't more points that can be added.
I've
found at least several points that can be added if all the lengths
are
doubled.
Of course, this is not quite as magic as it might appear, since
DE is
parallel to BC and DF is parallel to AC.
----------------------------------------------------------------------
I have a fairly simple algorithm to search for small integer-sided
triangles which can be partitioned into two smaller integer-sided
triangles. To recap a part of my earlier message:
Given a triangle with sides in the proportion R:S:T, we want to
divide
the T edge at an integer point so that the distance from this point
to
the opposite vertex is also an integer. The parametric solution
which I
derived is (written somewhat differently)
D1 = (R+S+T)*(R-S+T)
D2 = (S-T+R)*(T-R+S)
D3 = (R+S+T)*(T-R+S)
D4 = (S-T+R)*(R-S+T)
D2*v < u < D3*v
x = u^2 + D1*D2*v^2
y = (u + D1*v)*(u - D2*v)
z = 4T*u*v
T*z-y = (D4*v + u)*(D3*v - u)
corresponding to the partition
^.
/ \ .
/ \ .
R*z/ \ .S*z
/ x\
.
/
\ .
/
\ .
+-------------+-------+
y
T*z-y
As I noted before, any common divisor of x,y,z can be factored out.
The
perimeter of this triangle is (R+S+T)*z, thus for any given values
of
R,S,T, the solution with the smallest perimeter will be the one
with the
smallest value of z. Apart from the issue of common factors, this
implies that u and v should be chosen to be as small as possible.
Now
suppose that p is a common prime divisor of y and z. Then p also
divides
T*z-y. Thus, p must divide one of the factors u + D1*v or u - D2*v
of y,
and also one of the factors D4*v + u and D3*v - u of T*z-y. If
p divides
u + D1*v and D4*v + u, then it also divides (D1-D4)*v. If it divides
v,
then it also divides u, thus assuming u and v are coprime, p must
divide
D1-D4 = 2T*(R-S+T). Using the other possible pairs of factors,
we can
conclude that p must divide one of
D1-D4 = 2T*(R-S+T)
D1+D3 = 2T*(R+S+T)
D3-D2 = 2T*(T-R+S)
D4+D2 = 2T*(S-T+R)
If p divides D1*D2 = D3*D4, then it must also divide u. If p does
not
divide u, then it must divide 2T and cannot divide R^2-S^2.
I have chosen (for the moment) to ignore the primes dividing 2T.
My
algorithm works as follows. For each divisor g of D1*D2, let u
be the
smallest multiple of g which is larger than D1. If g is odd and
u does
not have the same parity as D1*D2, then replace u by u+g; this
ensures
that a factor of 4 can always be removed. If u is not less than
D3, go
to the next iteration. Then let v = 1 and calculate x,y,z. Factor
out
the gcd of x,y,z (which will always be a multiple of g). Find the
result
for which the final value of z is smallest.
I have found several new results with this algorithm. This includes
a
2-partition smaller than the [7,7,13] or [6,8,10], namely that
the
[6,4,5] triangle can be partitioned into a [6,4,4] and a [4,4,1].
This
is obtained from the above using R:S:T = 6:4:5 and (u,v) = (35,1).
There
is also a 4-partition constructible as follows. Start with triangle
ABC
with
AB = 15
AC = 18
BC = 12
Add D on AB so that
AD = 12
DB = 3
CD = 12
Add E on AC so that
AE = 10
EC = 8
DE = 8
Add F on AD so that
AF = 3
FD = 9
EF = 8
The resulting triangles are pairwise dissimilar, albeit three of
them
are isosceles and ADE is similar to ACB.
It is also possible to add another point G on AE with
AG = 2
GE = 8
FG = 2
but there are two pairs of similar triangles here.
All of these results have been derived from triangles similar to
[6,5,4], which appears to be an amazingly good source of solutions.
***
Of course, there is one other aspect of the 4-partition problem
that I
haven't attempted to approach yet. A triangle can also be 4-partitioned
by choosing one point on each edge and drawing the three
edges
connecting these points. I'm not sure even how to begin analyzing
this
type of 4-partition.
This raises an interesting question. The 4-partitions that I have
analyzed are all created by applying the following operation three
times:
O1. Draw a line from a vertex of a triangle to a point on
the opposite
edge.
As noted above, this does not generate all possible partitions of
a
triangle into smaller triangles. We need at least one other operation:
O2. Choose one point on each edge of a triangle and draw
the three
edges connecting these points.
Are these two operations sufficient to construct all possible partitions
of a triangle into smaller triangles, or is there a partition that
cannot be constructed in this way?