Thread: triangle problem concept help

Contains? Or intersects?
Hmmmm.
Contains: check that each of the three vertices is inside other triangle. Then do similar check to see if triangle B may be in A instead.
Intersects: check if any of the three lines intersects with any of the other triangle's three lines.

ya, contains. but what can be the logic behind that check. give some hint.

wow.. found it.. salem lots of thanks...
this is the page if in any case someone needs Determining Whether A Point Is Inside A Complex Polygon
solves much more complex problem than i require....

A lot of complicate solutions on-line. What if you calculate the angle from the questionable point to each of the three vertices. If the angles are all within a 180 degree swath then the point was outside. This should be all simplifiable by eliminating any arctan after optimization.

I AM SORRY nonoob but i didnt get your point. can you please explain your point w.r.t to the picture attached. please mention the name of the angle when you are explaining.

Ok. In the first diagram, I would calculate angles from "o" to A, B, C. They would be something like 80 degrees, 220 degrees, and 300 degrees.
If you subtract the lowest value 80 from each, you'd get 0, 140, and 220 respectively. The points are NOT all <= 180.

Think of how you would tell if you stood in a large field. There are three posts around you. Are you inside or outside the triangle? If you are outside the triangle then all posts lie on one side of a straight line that intersects with your origin.

This algorithm is just one I came up with. In its initial non-optimized phase you'd be calling atan2(). I haven't tried to clean it up by eliminating all trig and just ending up with vector math.

i still didnt get your point
1.

Ok. In the first diagram, I would calculate angles from "o" to A, B, C. They would be something like 80 degrees, 220 degrees, and 300 degrees.
If you subtract the lowest value 80 from each, you'd get 0, 140, and 220 respectively. The points are NOT all <= 180.

WHAN TWO ANGLES ARE 80,220 THE OTHER MUST BE 60 (SINCE SUM OF ALL ANGLES IS 360).
IN THAT CASE IF WE SUBTRACT 60 FROM EACH THEY BECOME 0,20,160 AND NONE OF THEM ARE GREATER THAN 180 ie.they are all <=180.
if u meant it for the picture in which the point is outside the triangle, look at fig.2 of the attachment. your theory is not always true. HERE ANGLE AOC IS SLIGHTLY GREATER THAN 180 AND SUBTRACTING THE LEAST ANGLE WILL MAKE IT <180.
2.

Think of how you would tell if you stood in a large field. There are three posts around you. Are you inside or outside the triangle? If you are outside the triangle then all posts lie on one side of a straight line that intersects with your origin.

THIS IS ALSO NOT TRUE. SEE THE ATTACHMENT.FIG.1

I'm not talking about sum of all angles of the triangle. I am talking about azimuth angle from "o" to each of the vertices.
In Fig.1, from "o" angle B is approx. 20 degrees, angle C is 70 degrees, A is 80. All three points lie in a sector that's less that 180. That is, I don't have to turn myself any more than within a slice of 60 to view all points.

o thats quiet clear now.

I just had a brainwave. Find area of triangles OAB, OBC, OCA. Is the sum of those triangles equal to area ABC or greater? If equal then the point O is inside.

To follow that up nonoob, I found a cool method for calculating area of a triangle. I think it is some sort of variant on Heron's formula when you have vertex points instead of side lengths:

Area= abs(x1*y2+x2*y3+x3*y1-x1*y3-x3*y2-x2*y1)/2 (from Point in triangle algorithm)

Should be nice and speedy.

Yeah, that's the way to calculate area of any polygon: Polygon Area -- from Wolfram MathWorld
Now to optimize using my untested brainwave algorithm...

I'm no mathematician, don't even play one on TV... but...

Why would you not calculate the center point of each triangle and see if the center of one is inside the other?

(Hope that's not too confusing...)