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If the coordinates of A are (xa,ya) and the coordinates of B are (xb,yb) then the length AB is sqr((xa-xb)^2 + (ya-yb)^2)
The length CB is sqr(AB^2 - AC^2) (Length AC is known)
Thus (xc-xb)^2 + (yc-yb)^2 = CD^2
and (xc-xa)^2 + (yc-ya)^2 = AC^2
As xa, ya and xb,yb and AC are known just expand the two equations, subtract one from the other and the answer should drop out.
Haven't tried it, but the maths is OK!
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but (xc-xb)^2 + (yc-yb)^2 = BC^2
It needs to solve 2nd order equations Ж8-О I hed the same before but if there is any easier method?
chesnokov
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No, if you reeduce the equations you are left with a simple quadratic.
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I think you need to provide more constraints as each side of the hypotenuse segment will result in 2 unique solutions.
A clue to get you going, the distance from the mid of the segment AB to C is equal to the half of |AB|
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Hi arash. Your the expert in geometry, I've seen some of your projects.
There are no more constraints. That ones should be sufficient.
Would you help? to solve the problem?
Is it always half AB if ACB is 90'?
chesnokov
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There are two ways of solving this problem, one is to do it algebraically by solving a set quartics based around the circumcircle, the other is simple geometric constructions.
I'll explain the 2nd one as its easier to understand and much easier to program a solution for.
1. Extend perpendiculars to the segment AB, 2 segments one each centered at A and B.
2. The end points for these 2 segments will be called A_0,A_1 and B_0,B_1
3. Extend diagonals from A_0 to B_1 and B_0 to A_1
4. The mid points between A and A_0 will be A_2, similar for the other side B_2
5. Extend diagonals from A_2 to B_0 and from B_2 to A_0
6. The intersection point between the segments A_2B_0 and A_0B_1 will be x_0, similar for the other two diagonals
7. Extend perpendiculars from x_0 and x_1 upon A_0B_0, the points of intersection will be p_0 and p_1
8. Extend segments from A to p_0 and B to p_1, called s_0 and s_1
9. The intersection point between s_0 and A_0B_1 will be C_0
10. The intersection point between s_1 and B_0A_1 will be C_1
11. Symmetric solution can be found by mirroring C_0 and C_1 about AB
A diagram of the solution can be found here:
http://www.partow.net/images/90degreetriangle.png[^]
Various routines in Wykobi can be used in conjunction with each other to solve this construction.
(perpendicular,mirror,project,closest_point_on_segment_from_point etc...)
have fun 
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That sounds like a whole bunch of work. A simpler solution is to find the mid-point of AB as D, and construct a circle of radius AD centered at D. Since it is given that angle ACB is 90 degrees, point C must also lie on this circle. Since it is given that the distance AC is known, construct a circle at A with a radius AC. Solve the two equations for the two circles for their intersection points. You are correct that there are two possible solutions, one solution on each side of AB.
Dave Augustine
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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????
Start with simpler problem, A at origin, B at (0,c) (i.e. AB = c) and we know AC = b and BC = a. (In your problem you can easily find BC using pythagoras).
The two intersecting circles can be solved almost trivially giving the co-ords of C (choosing y>0 option) as
x = (b^2 + c^2 - a^2)/(2a) y = sqrt(b^2 - x^2)
now rotate co-ords until AB has correct slope (cos = (xb - xa)/c, sin = -(yb - ya)/c) and translate A into position.
So the general solution is:
xc = xa + (b * (xb-xa) - a*(yb-ya))*b/c^2
yc = ya + (b * (yb-ya) + a*(xb-xa))*b/c^2
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."
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I don't know why this thread has attracted the anonymous univoter. As far as I can see this is the simplest solution so far - can the univoter improve on 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."
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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(A-B) * |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 non-trivial solutions are more interesting.
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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(A-B) * |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;}
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 x-axis. 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((xb-xa)^2+(yb-ya)^2)
b = sqrt(c^2-a^2)
xc = xa + (b * (xb-xa) - a*(yb-ya))*b/c^2
yc = ya + (b * (yb-ya) + a*(xb-xa))*b/c^2
for the other solution change the sign of 'a' in the last two equations
BTW - It's a pity people 1-vote 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."
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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]
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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 non-trivial 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. 
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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).
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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 x-axis, 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."
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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!
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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."
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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.
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The cross-posting 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."
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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.
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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."
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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:
(ys-y)^2 = (x1-x)^2 + (y1-y)^2 = (x2-x)^2 + (y2-y)^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
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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
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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."
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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.
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