1. If any of the points is the origin (0,0), done — TRUE.

2. Check quadrants: if either all x or all y values are all either positive or negative, then done — FALSE.

3. At this point one point must have an x value with a different sign than the other 2 points, and it’s the lines between this point and each of the other 2 points that must have y intercepts on either side of the origin. Let the single point p0 (x0, y0) and the other points p1 (x1, y1) and p2 (x2, y2). The y intercepts between p0 and each of p1 and p2 are given by

(y0-x0*(y1-y0)/(x1-x0))

and

(y0-x0*(y2-y0)/(x2-x0))

and these must have different signs (one or the other could be exactly 0).

Now I need to figure out the supporting maths…

Here is the main function as well as the CalcDirection helper function:

Public Function InTheTriangle(Pt As Coordinates) As Boolean
Dim InOut As Directioning
InTheTriangle = False
m = CalcSlope(Vertices1, Vertices2): b = CalcIntercept(Vertices1, m)
InOut = CalcDirection(m, b, Pt)
If InOut = Equal Then
InTheTriangle = True
Else
d = CalcDirection(m, b, Vertices3)
If d = InOut Then
m = CalcSlope(Vertices2, Vertices3): b = CalcIntercept(Vertices2, m)
InOut = CalcDirection(m, b, Pt)
If InOut = Equal Then
InTheTriangle = True
Else
d = CalcDirection(m, b, Vertices1)
If d = InOut Then
m = CalcSlope(Vertices3, Vertices1): b = CalcIntercept(Vertices3, m)
InOut = CalcDirection(m, b, Pt)
If InOut = Equal Then
InTheTriangle = True
Else
d = CalcDirection(m, b, Vertices2)
InTheTriangle = (d = InOut)
End If
End If
End If
End If
End If
End Function

Private Function CalcDirection(Slope As Variant, Intercept As Double, Vertex3 As Coordinates) As Directioning
Dim Result As Double
If CStr(Slope) = Infinity Then
Result = Vertex3.X + Intercept
Else
Result = Vertex3.Y – (Slope * Vertex3.X) – Intercept
End If

If Abs(Result) 0 Then
CalcDirection = Greater
Else
CalcDirection = Less
End If
End Function

http://newtonexcelbach.wordpress.com/2008/03/09/section-properties-from-coordinates/

shows how the area calculation works for an irregular quadrilateral, but the same principle applies to a triangle (or any other polygon). It’s actually based on the trapezoid rule (area = base x average height) rather than the 1/2(base * height).

The other method was to ensure that the angles formed around the origin with each pair of vertices must equal 360 degrees.

But, I rejected both as being sufficiently compled to compute given just the vertex coordinates that there had to be a better approach.

A Google search pointed to an approach using vector products. This would be easy to compute.

In fact, the computations are sufficiently straightforward that the problem can be solved in Excel without any code support.

http://www.tushar-mehta.com/misc_tutorials/project_euler/euler102.html

I like it when you’re workin’. I take it you used 1/2(base*height), and while I think I see 1/2 in there, I can’t vision the area computation loop. Could I impose upon you to talk thru one of the XDiff and YSum/2 loops? I do see how you employed the mindiff concept. Thanks for that.

…mrt

The function below uses the area method, but finds the areas without finding the side lengths, thus saving a lot of squaring and square rooting. On my machine it runs in 7.8 milliseconds, compared with about 15 milliseconds for the routine potsed by Michael.

Replace .LT. with the Less Than symbol, and enter as an array function in two adjacent cells to get the solution time as well as the result.

Function Triangles2(Coords As Variant) As Variant
Dim NumTri As Long, i As Long, j As Long, Area1 As Double, Area2 As Double
Dim XDiff(1 To 3) As Double, Ysum(1 To 3) As Double, Area2inc As Double
Dim count As Long, ResA(1 To 1, 1 To 2) As Double

Const MinDiff As Double = 0.0000000001

ResA(1, 2) = Timer
Coords = Coords.Value2

NumTri = UBound(Coords)

For i = 1 To NumTri
Area1 = 0
Area2 = 0
Area2inc = 0

XDiff(1) = Coords(i, 3) – Coords(i, 1)
XDiff(2) = Coords(i, 5) – Coords(i, 3)
XDiff(3) = Coords(i, 1) – Coords(i, 5)

Ysum(1) = (Coords(i, 2) + Coords(i, 4)) / 2
Ysum(2) = (Coords(i, 4) + Coords(i, 6)) / 2
Ysum(3) = (Coords(i, 6) + Coords(i, 2)) / 2

For j = 1 To 3
Area1 = Area1 + XDiff(j) * Ysum(j)
Next j
Area1 = Abs(Area1)

XDiff(2) = -Coords(i, 3)
XDiff(3) = Coords(i, 1)

Ysum(2) = (Coords(i, 4)) / 2
Ysum(3) = (Coords(i, 2)) / 2

For j = 1 To 3
Area2inc = Area2inc + XDiff(j) * Ysum(j)
Next j

Area2 = Abs(Area2inc)

XDiff(1) = Coords(i, 3)
XDiff(2) = Coords(i, 5) – Coords(i, 3)
XDiff(3) = -Coords(i, 5)

Ysum(1) = (Coords(i, 4)) / 2
Ysum(2) = (Coords(i, 4) + Coords(i, 6)) / 2
Ysum(3) = (Coords(i, 6)) / 2

Area2inc = 0
For j = 1 To 3
Area2inc = Area2inc + XDiff(j) * Ysum(j)
Next j

Area2 = Area2 + Abs(Area2inc)

XDiff(1) = -Coords(i, 1)
XDiff(2) = Coords(i, 5)
XDiff(3) = Coords(i, 1) – Coords(i, 5)

Ysum(1) = (Coords(i, 2)) / 2
Ysum(2) = (Coords(i, 6)) / 2
Ysum(3) = (Coords(i, 6) + Coords(i, 2)) / 2

Area2inc = 0
For j = 1 To 3
Area2inc = Area2inc + XDiff(j) * Ysum(j)
Next j

Area2 = Area2 + Abs(Area2inc)

If Abs(Area2 – Area1) .LT. MinDiff Then count = count + 1

Next i

ResA(1, 1) = count
ResA(1, 2) = Timer – ResA(1, 2)

Triangles2 = ResA

End Function

It’s one of those things that is harder than it looks.

Probably the easiest way (which I used) is to project a horizontal line from the point you are checking to infinity and count how many times it crosses the boundary of the shape.

An odd number of crossings indicates inside, and zero or an even number indicates outside.

But you get situations like intersecting a corner, or a segment of the shape exactly overlying the projected line from the point, so you have to have a consistent way of dealing with those things.

Thank you two times. I’ll go get your routine. I found out I didn’t know how to write one.

…mrt