How many triangles?
Solution by Bill Daly

I think there is a straightforward way to calculate the number of
triangles in the figure, or for that matter to come up with a general
formula for similar figures. Suppose that S is the length of a side of
the figure; the figure as shown has S = 10.

There are six sets of parallel lines in the figure:

S lines A[1]..A[S] parallel to the base of the figure (indexed by
length).
S lines B[1]..B[S] parallel to the right side of the figure (indexed
similarly).
S lines C[1]..C[S] parallel to the left side of the figure (indexed
similarly).
2S-1 lines D[1]..D[2S-1] perpendicular to the A lines (indexed from left
to right along the base).
2S-1 lines E[1]..E[2S-1] perpendicular to the B lines (indexed
similarly).
2S-1 lines F[1]..F[2S-1] perpendicular to the C lines (indexed
similarly).

The sides of any triangle must consist of one line from each of three
distinct parallel sets, thus there are triangles ABC, ABD, etc. If we
define #ABC as the number of ABC triangles, etc., then the total number
of triangles will be #ABC + #ABD + ... + #DEF, a total of twenty subsets
to be considered. The symmetries of the figure imply #ABF = #ACE = #BCD,
#AEF = #BDF = #CDE, #ABD = #ABE = #ACD = #ACF = #BCD = #BCE, #ADE = #BDE
= #ADF = #CDF = #BDE = #BEF, thus we can reduce the problem to computing

  #ABC + #DEF + 3*#ABF + 3*#BDF + 6*#ABD + 6*#ADF

Note that the triangles in #ABC and #DEF are equilateral, those in #ABF
and #BDF are isosceles with angles 120-30-30, and those in #ABD and #ADF
are right with angles 90-60-30.

To compute (for example) #ABC, we want to know how many of the sets of
lines (A[a],B[b],C[c]) are such that the three line intersections are
distinct and all lie within the figure. The constraints are

  a+b >= S
  a+c >= S
  b+c >= S
  a+b+c <> 2S

This can be solved by brute force algebra to give

  #ABC = (2S^3+5S^2+2S-(S mod 2)) / 8

Here as usual, (S mod 2) is 1 if S is odd and 0 if S is even. For S =
10, we get #ABC = 315.

Here follow the constraints and the derived formulas for the remaining
sets of triangles.

#DEF: (D[d],E[e],F[f])

  d+2e >= 2S
  2d+e <= 4S
  e-d <= S
  e+2f >= 2S
  2e+f <= 4S
  f-e <= S
  f+2d >= 2S
  2f+d <= 4S
  d-f <= S
  d+e+f <> 3S

  #DEF = (3S^3 - S - 2(S mod 3)) / 3

  for S = 10, #DEF = 996

#ABF: (A[a],B[b],F[f])

  a+b >= S
  2b+f >= 2S
  b+f <= 2S
  f-a >= 0
  f-2a <= 0
  -a+b+f <> S

  #ABF = (2S^3 + 3S^2 - 2S - 3(S mod 2)) / 12

  for S = 10, #ABF = 190

#BDF: (B[b],D[d],F[f])

  d-b >= 0
  d-2b <= 0
  2b+f >= 2S
  b+f <= 2S
  f+2d >= 2S
  2f+d <= 4S
  d-f <= S
  3b-d+f <> 2S

  #BDF = (14S^3 + 33S^2 + 4S - 6(S mod 4) + 3(S mod 2)) / 48

  for S = 10, #BDF = 361

#ABD: (A[a],B[b],D[d])

  a+b >= S
  a+d >= S
  d-a <= S
  d-b >= 0
  d-2b <= 0
  a+2b-d <> S

  #ABD = (5S^3 + 12S^2 + 3S - 2(S mod 3)) / 18

  for S = 10, #ABD = 346

#ADF: (A[a],D[d],F[f])

  a+d >= S
  d-a <= S
  f-a >= 0
  f-2a <= 0
  f+2d >= 2S
  2f+d <= 4S
  d-f <= S
  -3a+d+2f <> S

  #ADF = (18S^3 + 37S^2 - 2S - 5(S mod 2) - 8(S^3-S^2+S mod 5)) / 40

  for S = 10, #ADF = 542

We can now calculate the total number of triangles for S = 10:

  #ABC + #DEF + 3*#ABF + 3*#BDF + 6*#ABD + 6*#ADF =
  315 + 996 + 3*190 + 3*361 + 6*346 + 6*542 =
  8292

The formula for general S is

  (1678*S^3 + 3117*S^2 + 88*S - 345*(S mod 2) - 320*(S mod 3) - 90*(S
mod 4)
  - 288*((S^3-S^2+S) mod 5)) / 240

This can probably be simplified.

Solution by James Boyce

I will attempt to count the triangles.

As a warm-up. I will count the points in the diagram.  There are 55
little right-side-up equilateral triangles, each of which has a topost
vertex, 3 midpoints of sides, and a centroid.  That accounts for 275
points.  In addition, there are 45 upside-down triangles of that same
size, each of which has a centroid.  Finally, there are 11 points
along the bottom edge that I have not already counts.  That gives a
total of 331 points.

The diagram has lines in 6 different directions.  I will name the
directions parallel to the side of the diagram A, B, and C, and denote
their corresponding normals a, b, and c.  A triangle in the diagram
will have sides in three of those directions.  There are 20 triples of
directions, and each triple defines triangles of two orientations.
There are many symmetries in the diagram, which reduces the number of
different cases I need to count carefully.

The cases to considers are these:
ABC: triangles with 2 edges parallel to the edges of the diagram.
AaB: triangles with 2 edges parallel to the edges of the diagram, and
1 edge normal to one of those edges.  (There are 6 cases like this,
all the same.)
ABc: triangles with 2 edges parallel to the edges of the diagram and 1
edge normal to the other edge of the diagram.  (There are 3 cases like
this.)
Abc: triangles with 1 edge parallel to an edge of the diagram and with
2 edges normal to the other 2 edges of the diagram.  (There are 3
cases like this.)
Aab: right triangles with 1 edge parallel to and edge of the diagram
and with 2 edges normal to edges of the diagram.  (There are 6 cases
like this.)
abc: triangles with edges normal to all three sides of the diagram.

Remember that there are two orientations for each of those cases, so I
might have as many as 12 counting problems.  I will add up the answers
to those problems (with suitable multipliers) to get the actual
answer.

case ABC: Each right-side-up ABC triangle has a topmost point.  Each
'topmost-point' may be the topmost point of more than one triangle.
There are 55 topmost points in the diagram.  10 of them have only 1
triangle.  9 of them have 2, ... and 1 of them has 10.  That is a
total of 220.

Each upside-down ABC triangle has a bottom-most point.  There are 45
bottom-most points in the diagram.  17 of them have only one triangle;
13 have 2; 9 have 3; 5 have 4; and 1 has 5.  That is a total of 95.

case AaB: The smallest AaB triangle is composed of 3 of the little
triangles in the diagram.  An AaB triangle is half of an equilateral
triangle.  For right-side-up equilateral triangles, all of the
eqilater triangle is presents, so the answer is the same as the ABC
case: 220 triangles.

The upside-down case requires more counting.  Each AaB triangle has a
60 degree angle, with wlog is its bottom-most point.  There are still
45 bottom-most points to consider.  There are 9 with only 1 triangle;
15 have 2; 6 have 3; 9 have 4; 3 have 5; and 3 have 6.  This total is
126.

case ABc: The smallest ABc triangle is composed of 6 of the little
triangles in the diagram.  It is 1/4 of an equilateral triangle.  In
the case of right-side-up equilateral triangles, the entire triangle
will be there.  So we need to count right-side up equilateral
triangles with even edge length.  There are 45 triangles of side 2; 28
of side 4; 15 of side 6; 4 of side 8; and 1 of side 10.  That makes 95.

The upside-down case is also easy.  These are the same in number as
the upside-down equilateral triangles.  The bottom-most point of the
equilateral triangle is the obtuse vertex of the ABc triangle.  So
there are 95 of these, too.

case Abc: The smallest Abc triangle is composed of 2 of the little
triangles in the diagram.  It is 1/3 of an equilateral triangle.  In
the case of right-side-up equilateral triangles, the entire
equilateral triangle is in the diagram, so the answer is the same as
the ABD case: 220.

The upside-down case requires more counting.  The designated point in
these triangles is the Ab intersection.  There are 45 such points in
the diagram.  There are 9 points with just 1 triangle each; 8 have 2,
13 have 3; 5 have 4; 4 have 5; 5 have 6; and 1 has 7.  This total is
141.

case Aab: The smallest Abc triangle  is just 1 of the little triangles
in the diagram.  It is 1/6 of an equilateral triangle.  First, I count
the  ones  that  are  easily  contained in  right-side  up  equilteral
triangles.   Note that the  entire right-side-up  equilateral triangle
need  not be  in the  diagram, so  there is  something to  count.  The
designated point  is the Ab  intersection.  There are 55  such points.
There are 10 points  with just 1 triangle; 9 have 3;  8 have 4; 7 have
6; 6 have 7; 5 have 9; 4 have  10; 3 have 12; 2 have 13; and 1 has 15.
The  total is  315.  (This  is  220 +  95, the  number of  equilateral
triangles oriented with  the edges of the diagram, but  I don't see an
obvious relationship that would predict this.)

The Ab intersection is the designated point for the upside-down case,
too.  There are 45 relevant vertices.  There are 9 points with just 2
triangles; 8 have 3; 7 have 4; 11 have 6; 4 have 8; 3 have 9; 2 have
10; and 1 has 12.  That is a total of 227.

case abc: The smallest abc triangle is composed of just 2 of the
little triangles in the diagram.  An abc triangle has a vertical edge
either on the left or on the right.  By symmetry, the counts are the
same for both cases.  The topmost point in an abc triangle can be in 3
types of places: the centroid of a right-side-up triangle, the
centroid of an upside-down triangle, or at the vertex of a 6-triangle
equilateral triangle.  There are 45 places to count in each of the 3
cases.  Let's start with upside-down centroids.  There are 9 along the
bottom with just 1 (oriented) abc triangle; 15 have 2; 6 have 4; 9
have 5; 3 have 7; and 3 have 8.  That is a total of 153.

Now let's count for the centoids of the right-side-up triangles.
There are 9 with just 1 triangle each; 8 have 2; 7 have 3; 6 have 4; 5
have 5; 4 have 6; 3 have 7; 2 have 8; and 1 has 9.  That is a total of
165.

Finally, let's count the ones with topmost point a vertex of the
obvious equilateral triangles.  There are 45 interesting vertices.
There are 9 points with just 1 triangle each; 15 have 3; 6 have 4; 9
have 6; 3 have 7; and 3 have 9.  That is a total of 180.  That make a
total of 498 triangles for a particular orientation of abc.

Now let's add everything up.
ABC: 1 * (220 +  95) =  315
AaB: 6 * (220 + 126) = 2076
ABc: 3 * ( 95 +  95) =  570
Abc: 3 * (220 + 141) = 1083
Aab: 6 * (315 + 227) = 3252
acb: 1 * (498 + 498) =  996
total =                8292

I have also written a program to count the triangles.  I was gratified
to see that it got the same answer.  It is possible that I should have
counted triangles of a particular shape and size, rather than
triangles of a particular shape with a given vertex.  That amounts to
interchanging the order of a summation, but it might have made some
things clearer.

-----

When I wrote my program, I used the following technique for
identifying co-linear points.  I embedded the diagram in the plane
x+y+z=10.  I identified the extreme points of the diagram with
(10,0,0), (0,10,0), and (0,0,10).

The points of intersection are
(i,j,k) i,j,k are integers in [0,10] and i+j+k=10

(i+1/2, j+1/2, k) i,j,k are integers in [0,9] and i+j+k=9
(i+1/2, j, k+1/2)
(i, j+1/2, k+1/2)
(i+1/3, j+1/3, k+1/3)

(i+2/3, j+2/3, k+2/3) i,j,k are integers in [0-8] and i+j+k=8.

The lines in the diagram are
x=integer
y=integer
z=integer
x-y=integer
x-z=integer
y-z=integer

Solution by Wei-Hwa Huang

here is his PDF file.

Torsten Sillke also solved this problem... and was also asked by Singmaster! http://www.mathematik.uni-bielefeld.de/~sillke/SEQUENCES/grid-triangles