How many isosceles triangles on grid?

If you want to avoid brute force, you can say that if OBC is isoscles in O and A is the middle of BC. Then OAB is a right triangle of area 4.5, and if you expand it by a factor 2, you are looking for a right triangle OHK with integer coordinates, OH is the hypotnuse, H has even coordinates and area of OHK is 18.

Call OK =sqrt(p) and HK = sqrt(q). Then sqrt(pq) = 36 so pq = 2^43^4 and p + q = 4n , where n = OB^2. So p and q have to be multiples of 4, namely (p,q) = (4,324),(12,108)(36,36),(108,12),(324,4)

The decomposition as a sum of squares of 4 offers few possiblities, same for 36 and 12 is out. So that gives you only triangles having one of the coordinates axes as axis of symmetry.

You get back Slowfinger's list and take care of Rita the dog's objection.

Tough one. I've starred it so hopefully others see it. I'll keep working on it, but this is as far as I got:

First consider the cases where the vertex at the origin is where the two congruent sides meet. Call the other two points (a,b) and (c,d). Then the distance formula says:

√[ (a-0)^2 + (b-0)^2 ] = √[ (c-0)^2 + (d-0)^2 ]

a^2 + b^2 = c^2 + d^2

The area is 9, so:

(1/2)base * height = 9

base * height = 18

[distance between (a,b) and (c,d) ] * [distance between (0,0) and midpoint of base] = 18

√[ (a-c)^2 + (b-d)^2 ] * √[ ((a+c)/2 - 0)^2 + ((b+d)/2 - 0)^2 ] = 18

[ (a-c)^2 + (b-d)^2 ] [ ((a+c)/2)^2 + ((b+d)/2)^2 ] = 18^2

[ (a-c)^2 + (b-d)^2 ] [ (a+c)^2 + (b+d)^2 ] = 4*18^2

[ a^2 - 2ac + c^2 + b^2 - 2bd + d^2 ] [ a^2 + 2ac + c^2 + b^2 + 2bd + d^2 ] = 4*18^2

[ (a^2 + b^2) - 2ac - 2bd + (c^2+d^2) ] [ (a^2+b^2)+ 2ac + 2bd + (c^2+d^2) ] = 4*18^2

Since a^2 + b^2 = c^2 + d^2 , we have

[ 2(a^2 + b^2) - 2ac - 2bd ] [ 2(a^2+b^2)+ 2ac + 2bd ] = 4*18^2

[ 2(a^2 + b^2) - 2(ac+bd) ] [ 2(a^2+b^2) + 2(ac+bd) ] = 4*18^2

[ (a^2 + b^2) - (ac+bd) ] [ (a^2+b^2) + (ac+bd) ] = 18^2

(a^2 + b^2)^2 - (ac+bd)^2 = 18^2

(ac + bd)^2 + 18^2 = (a^2 + b^2)^2

I suppose this at least tells you that a^2 + b^2 > 18, that a^2 + b^2 = even, 0 mod 3, 0 mod 9, etc.

The way I read the question any of the 3 vertices can be at the origin. There seems to be no requirement that the equal edges meet at the origin. Therefore Slowfinger's triangles can have (0,0) at any of the 3 vertices, giving 3*12 = 36 triangles in all. Ohterwise, I agree with Slowfinger, although proof that there are no more, e.g. very long skinny triangles with some vertices outside the present search region, is lacking.

Let's represent two (equal) sides of an isosceles triangle as vectors:

a = x₁ i + y₁ j

b = x₂ i + y₂ j

Their lengths are equal, so

|a| = |b|

this can be translated as

x₁² + y₁² = x₂² + y₂²

Area of the triangle can be written via cross product:

|a × b| / 2 = 9

a × b = (x₁ i + y₁ j) × (x₂ i + y₂ j) = (x₁y₂ - x₂y₁) k

|a × b| = x₁y₂ - x₂y₁

|a × b| / 2 = (x₁y₂ - x₂y₁)/2 = 9

So we have two equations:

x₁² + y₁² = x₂² + y₂²

x₁y₂ - x₂y₁ = 18

By brute force method (I wrote a simple Fortran program to check ALL combinations of integers x₁ , y₁, x₂ and y₂ from -20 to 20) I found these coordinates for points A and B of a triangle (third point is the origin, of course):

...A............B

(-9,-1)......(9,-1)

(-9,1)......(-9,-1)

(-3,-3)......(3,-3)

(-3,3)......(-3,-3)

(-1,-9)......(1,-9)

(-1,9)......(-1,-9)

(1,-9)......(1,9)

(1,9)......(-1,9)

(3,-3)......(3,3)

(3,3)......(-3,3)

(9,-1)......(9,1)

(9,1)......(-9,1)

So there are 12 such triangles.

Edit:

It was only after I had posted the answer that I read the link you provided and it appears now that we independently followed the same logic. Btw here's my fortran code:

integer a, b, c, d, e, f, g

do 10 a=-20,20

do 20 b=-20,20

do 30 c=-20,20

do 40 d=-20,20

e=a**2+b**2

f=c**2+d**2

if(e.ne.f) goto 40

g=a*d-b*c

if(g.ne.18) goto 40

write (*,*) a, b, c, d

40 continue

30 continue

20 continue

10 continue

end

Edit 2:

@Rita the dog

"proof that there are no more, e.g. very long skinny triangles with some vertices outside the present search region, is lacking."

Sorry, I thought it was almost obvious. Let's stick to my initial reading of this question (I'll comment this later) where vertex in the origin is the one opposite to base of triangle. Because height has direction that passes through the origin and base is perpendicular to that direction, we'll distinguish between two possibilities:

1) height lays on one of coordinate axes

2) height is inclined at 45° to axes.

In the first case the axis that height lays on bisects the base, and as vertices at both ends of base can't be closer to axis than 1 unit, two halves of the base must be at least 1 unit in length each, or in another words base can't be shorter than 2 units.

Consequently, if area is exactly 9 sq.units, and base is at least 2 min. then height is 9 units at most.

Examples with base 2 units long:

http://www.wolframalpha.com/input/?i=triangle+%280...

http://www.wolframalpha.com/input/?i=triangle+%280...

(there are two more)

As long as height is horizontal or vertical we can't get two ends of base closer to each other than 2 units and that prevents us from making "very long skinny triangles"

Second case - height is inclined at 45° to axes

Now length of base is in form

m√2

where m is an odd integer, and height is in form

n√2/2

where n is another odd integer

Area is

A= (1/2) m√2 n√2/2 = mn/2

product of two odd numbers is odd number, so area will always be odd number of halves and we need round number, 9.

Examples:

http://www.wolframalpha.com/input/?i=triangle+%280...

(area=17/2)

http://www.wolframalpha.com/input/?i=triangle+%280...

(area=19/2)

http://www.wolframalpha.com/input/?i=triangle+%280...

(area=21/2)

Area is always (odd number)/2 and we can't get 9 square units in case (2) so we're left with case (1) that is horizontal or vertical height.

"The way I read the question any of the 3 vertices can be at the origin. There seems to be no requirement that the equal edges meet at the origin."

The way I initially read the question it seemed there is requirement that the equal edges meet at the origin. But after your remark I read it again and I think you're right.

I'm glad actually because it allows for the fun part.

Assuming there is exactly 12 triangles such that equal edges meet at the origin, then for any point on the grid we can construct exactly 12 triangles such that equal edges meet in that point and any of these triangles will be translated image of one of triangles we've already constructed in the origin. We can always choose point where equal edges meet such that one of two remaining vertices is at the origin.

For example, one of 12 triangles which equal edges meet at (9,1) is this one:

http://www.wolframalpha.com/input/?i=triangle+%289...

if we compare it to triangle No1 on my list:

http://www.wolframalpha.com/input/?i=triangle+%280...

we can come to alternative formulation that new triangle is formed by sliding triangle No1 along one of its (equal) edges.

By sliding triangle No1 along other equal edge we get this one:

http://www.wolframalpha.com/input/?i=triangle+%28-...

And from triangle No2

http://www.wolframalpha.com/input/?i=triangle+%280...

we derive these two:

http://www.wolframalpha.com/input/?i=triangle+%289...

http://www.wolframalpha.com/input/?i=triangle+%289...

For all 12 original triangles we can derive two new triangles that comply to Rita's demand that is ANY vertex at the origin, so there will be 36 triangles in total.