

I'm not sure what you're on about, but the simplest solution to the problem is to assume a square,
from that you get:
c_1,2 = mid(AB) +/ (perp(AB) * AB/ 2)
code:
template<typename T>
void generate_right_triangle(const wykobi::point2d<T>& a, const wykobi::point2d<T>& b,
wykobi::point2d<T>& c1, wykobi::point2d<T>& c2)
{
T distance = wykobi::distance(a,b);
wykobi::point2d<T> mid = wykobi::segment_mid_point(a,b);
wykobi::vector2d<T> v = wykobi::normalize(wykobi::perpendicular(a  b)) * (distance / T(2.0));
c1 = mid + v;
c2 = mid  v;
}
I went for the construction route as most times the nontrivial solutions are more interesting.





Arash Partow wrote: I'm not sure what you're on about, but the simplest solution to the problem is to assume a square,
from that you get:
c_1,2 = mid(AB) +/ (perp(AB) * AB/ 2)
code:
templatevoid generate_right_triangle(const wykobi::point2d& a, const wykobi::point2d& b, wykobi::point2d& c1, wykobi::point2d& c2){ T distance = wykobi::distance(a,b); wykobi::point2d mid = wykobi::segment_mid_point(a,b); wykobi::vector2d v = wykobi::normalize(wykobi::perpendicular(a  b)) * (distance / T(2.0)); c1 = mid + v; c2 = mid  v;}
But what if AC != BC? My simple solution solves this case and does not need wykobi.
All I did was solve for the triangle with A at (0,0) and with AB along the xaxis. The equations solve easily for C, then all you ahve to do is a rotation (using the vector AB) then translate to place A in the correct position. The general problem is done with 4 lines of code and no wykobi:
Input: (xa, ya), (xb, yb) and length of BC = a, to find (xc,yc) st angle(acb) = 90
c = sqrt((xbxa)^2+(ybya)^2)
b = sqrt(c^2a^2)
xc = xa + (b * (xbxa)  a*(ybya))*b/c^2
yc = ya + (b * (ybya) + a*(xbxa))*b/c^2
for the other solution change the sign of 'a' in the last two equations
BTW  It's a pity people 1vote solutions they don't understand.
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."





I doubt he was the univoter here.
Unfortunately Mr.Univoter appears quite often in replies to Mr. Chesnokov Yuriy's posts.
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
 Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
 Iain Clarke
[My articles]





1. It wasn't me, I don't vote on forum articles
2. The question didn't place any requirements on edge lengths, and if it did my original solution would provide all 4 nontrivial solutions.
3. Can you provide some code to your solution, it would be interesting to see.
4. You provided a solution, others read it and decided to give a rating, no need to get your knickers in a knot.





I didn't look carefully over the other answers but glancing at them, I didn't see anybody mention the obvious: C is on the circle whose diameter is AB. So it amounts to finding the intersection of two circles  one centered at the midpoint of AB with a radius of AB/2, the other with a center at A with a radius of AC. Pretty simple algebra. There will be two answers (one on each side of AB) unless AC = 0 or AC = AB in which case the unique answer is B or A respectively (in which case the "angle" ACB is really not well defined so my claim would be that this case wouldn't fit the requirements of the original question where it was claimed that this angle was unambiguously a right angle  so I would claim there are always two answers).





darrellp wrote: didn't see anybody mention the obvious: C is on the circle whose diameter is AB. So it amounts to finding the intersection of two circles  one centered at the midpoint of AB with a radius of AB/2, the other with a center at A with a radius of AC
This was cross posted, and there was a link on the other page that explained it, and Member 4194593 also explained this.
darrellp wrote: Pretty simple algebra.
Try it  you end up with a quartic for the general case.
If, however, you rotate the triangle so that AB lies on the xaxis, and move A to the origin, the algebra becomes much simpler, you end up with a quadratic that is easily solved. All you have to do then is rotate the result and move it back into position and you have the general solution as I explained. There may be other ways to solve it, but this is the simplest I have found. My solution is correct, as are others (except for those claiming 4 solutions), for some reason the univoters didn't like it.
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."





I don't understand the crossposting stuff  I just did a response to the original query. If I screwed up, I apologize but I'm not sure what I need to do differently. If you would enlighten me, that would be great!
I'm glad to hear I just didn't look carefully enough. I'll take your word on complexity. I just saw two second degree polynomials and two solutions and assumed it all boiled down to a simple quadratic. That'll teach me to assume in the future!





Sorry, didn't mean to offend.
Like many things  the devil is in the detail  and solving those intersecting circles is tricky.
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."





No offense taken. I was serious  if I did some stupid crossposting thing, I really didn't mean to and I'd love to learn what I did so I could avoid it in the future.





The crossposting reference was about the original post (not you)  Cpallini's reply linked to the an identical post on another forum that had some useful discussion.
What i'd like to know is why folk want to solve this particular problem?
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."





Ah  okay. Sorry to be so dense. I don't often interact with these forums (fora?) so assumed I'd made some sort of boneheaded blunder.
Don't know about a motive. Could just be "recreational math". I think the original poster was under the impression that there was a single unique solution which suggests to me that he/she really was trying to solve a "real world" problem, but who knows?
I do know that my estimation of the simplicity of finding intersections of quadrics hasn't been good. There's a similar problem in the Fortune algorithm for Voronoi diagrams where you have to find the intersection of a couple of parabolas and I went in with the assumption that it would be a trivial problem. I was way off. Eventually, I just dumped everything into Mathematica and let it solve the problem. Every description I've ever seen of this algorithm passes this problem off as "find the interesection of the parabolas" as if it was some of throw away detail. You'd think somebody would stop and note that this is a damnedly difficult problem and here is the (rather ugly)solution. Oh well  if only the world were run the way I think it ought to be...
Just FYI, the problem is essentially that you're given two focii of parabolas with a common horizontal line as the directrix for both parabolas. Find the (two) intersections between the parabolas. Seemed simple. Wasn't. I'm still not sure of any terribly "intuitive" way of solving this. No obvious transform to do on this one since the directrix is already horizontal.





Now here's my chance to overestimate the simplicity of a problem, surely the equation of a parabola with horizontal directrix is y = P(x) where P(x) is a quadratic, so for two such parabolas you need to solve
P1(x) = P2(x)
which is a 'simple' quadratic. I can see that the coefficients may be messy, but straightforward (?).
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."





Yes, that's the kind of thinking I had for both of these questions  equating two quadratics shouldn't be that tough. It's the fact that you've got two quadratic equations in two unknowns in both cases that complicate matters. In the Fortune case, the equations are:
(ysy)^2 = (x1x)^2 + (y1y)^2 = (x2x)^2 + (y2y)^2
where ys is the y coordinate of the sweepline and (x1,y1), (x2,y2) are the two points. Using one equation, you can get y as a quadratic function of x. Substituting that into the other equation gives a quartic equation in x and you have similar problems to the ones you deal with in the original problem of this thread. I'm glad I'm not the only one who can be misled by these problems! What's more, like I said, I don't see any nice intuitive transforms or any other easy way, geometric or otherwise, of coming up with the solution for this one. Certainly doesn't mean there isn't one  in fact I suspect that there is such a thing  I just haven't seen it yet. When I got the solution from Mathematica, I was able to break it into lots of redundant pieces to improve performance which suggests to me that there ought to be a way. Perhaps if I meditated on the solution long enough, I could see it, but I haven't spent the time.
Minus some exceptional cases, here's the code I ended up with (directly from my code):
double a1 = 1 / (2 * (pt1.Y  ys));
double a2 = 1 / (2 * (pt2.Y  ys));
double da = a1  a2;
double s1 = 4 * a1 * pt1.X  4 * a2 * pt2.X;
double dx = pt1.X  pt2.X;
double s2 = 2 * Math.Sqrt(2 * (2 * a1 * a2 * dx * dx  da * (pt1.Y  pt2.Y)));
double m = 0.25 / da;
double xs1 = m * (s1 + s2);
double xs2 = m * (s1  s2);
It's been a long time since I did this. I think the rounding was done to avoid some numerical instability problems.
xs1 and xs2 are the x coordinates of the two solutions. For my purposes, I didn't really care about the y coordinates.
Darrell





Oops  okay  I see what you mean. Doh! I think you're right. I'm sorry to be so dim witted! Since you end up with the y^2 terms dropping out, you can solve for y, set the resulting two quadratics in x equal to each other for one quadratic in x and solve. Man, I better just stop before I stick my foot in my mouth again!
Darrell





Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."





Peter,
cp9876 wrote: as are others (except for those claiming 4 solutions)
Unless the triangle has the same length legs (ac=bc), there will be 4 solutions. Nothing in the stated problem indicated any orientation. For example, line ab can be taken as a vertical line starting with a at the bottom and b at the top, or, flip that upside down, then point c can be on either side of that vertical line, in either orientation of ab. Essentially, you need to draw two circles of radius ac, one at one end of the diameter ab and one at the other end of the diameter ab. There will be 4 points of intersection of these two circles with the circle drawn from the midpoint of ab with radius of ab/2. There is also no constraint on whether ac is the longest or shortest leg.
Dave.





Chesnokov Yuriy wrote: There is ABC triangle randomly rotated on XY plane. A,B coords are known, ACB angle is 90'. AC length is known. How to find C coords????
Member 4194593 wrote: Nothing in the stated problem indicated any orientation.
We seem to be solving different problems, I assumed that as the poster says that the coordinates of A and B are known that he knows which are which, in particular which coordinates are the end of the side AC. i.e. we are given (xa, ya), (xb, yb) and the length AC.
If he really doesn't know which is which, he can easily get the other solutions by swapping the A and B coords.
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."





Peter,
You are correct as usual. You cannot draw the circle of length AC at B, it has to be drawn at A, thus only two solutions, one on either side of line AB.
One solution could be to start at the origin and call the point A, drop down distance AC on the y axis and call the point C, move to the right (the mirrored solution would be obtained by moving to the left) a distance sqrt(AB^2ac^2) and call that point B. Now rotate ABC about A until the line AB matches the original slope of the given line AB, then translate ABC to position A at the origin to the original given position of A.
Dave.





Dave,
Neat constructional idea, but the algebra looks nasty to work out the sin and cosine of that angle for the rotation,
Peter.
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."





I would like to check if a segment and a polygon have some intersections. Could you plz indicate me some algorithms to solve it. I knew Ray Casting algorithm and I wonder if there are other algorithms ?






The above algorithm won't detect the case where the segment is entirely contained in the polygon, without intersecting any edge. Let me know if you need to detect this case too.





Plz show me if that is the most optimal algorithm ? I really want to know the most optimal algorithm to solve this problem. So, would you mind helping me ?





If it isn't optimal it's pretty close. The calculation to determine if two segments intersect isn't time consuming. The only case it doesn't handle, as I've mentioned, is when the segment is entirely contained within the polygon.
modified on Friday, October 17, 2008 9:55 AM





OK. Thanks so much for replying




