A math problem I was never able to solve

[NogginBoink]

This one was presented to me in college and I never was able to solve it. In the spirit of the brainteaser threads, maybe someone can make more progress than I was able to:

In two dimensions, you're given the location of point A (Ax, Ay) and point B (Bx, By), and the legths of line segment AC and line segment BC.

Find the location of point C.

Now, before trying to solve algebraically, step back and think about the problem. You can make some general statements of geometry that explain the approach to solve the problem. But I've never been able to arrive at an algebraic solution.

 

[bmacd]

this is saturday...i don't have to do math!

-=bmacd=-

edit: i couldn't solve it if my life depended on it

 

[gopunk]

i really don't know my trig that well any more, but i think you can probably do it by thinking of it as a triangle ABC. you have the length of AB, AC, and BC, and position of A and B, all you need to do is figure out the angles.

 

[zzzz]

Assume C has coordinated Cx Cy

AC= sqrt ( (Cx-Ax)^2 + (Cy-Ay)^2)

simillar equation for BC.
two unknowns, two equations. solve them.

 

[NogginBoink]

Okay. Basically we're looking to draw a triangle with points A, B, and C.

Essentially, the solution is to draw a circle around point B with radius BC, and a circle around point A with radiu AC. Point C is where the circles intersect. There will be either zero, one, or two such points.

If AC + BC < AB, then there are zero solutions.

If AC + BC = AB, then there's one solution and point C is on the line AB.

Assuming two solutions, you get two equations with two unknowns like zzzz says. But when I attempt the math, I find that the two equations lead me in a circle (no pun intended) and can't get a unique solution like I think I should.

 

[Haircut]

I think you're on the right method. What I thought of was to determine Cx and Cy in the following way:

Cx = Ax + AC*Cos(theta) = Bx + BC*Cos(gamma)

Cy = Ay + AC*Sin(theta) = By+BC*Sin(gamma)

where theta and gamma are the angles AC and BC make with the horizontal respectively

If we take Bx and By from the right hand sides and square the two equations and add them up we can cancel the gammas out as we get a multiple of Sin^2(gamma) + Cos^2(gamma) leaving an equation in theta only.
This can be solved to find theta, which can then be put back into the original equations to get gamma.

There will be two answers for theta and gamma as there are two distinct points C.

 

[oLLie]

It's times like these I wish I didn't suck at math .

I wanna look smart too

 

[bizmark]

n/m I didn't read the whole thread. You already know what I said.

 

[silverpig]

Point A = (Ax, Ay)
Point B = (Bx, By)
||AC|| = r
||BC|| = s


Then the point C lies on the circle with equation (x - Ax)^2 + (y - Ay)^2 = r^2
It also lies on the circle with equation (x - Bx)^2 + (y - By)^2 = s^2

Using some manipulation, you can solve the first equation for x, plug it into the x value for the second equation and come up with:

( sqrt [r^2 - (y - Ay) ^2] + Ax - Bx )^2 + (y - By)^2 = s^2

Since you are given Ax, Bx, Ay, By, s, and r, you can easily solve this equation for y. Plug that y value into either one of the original equations, and you'll have your x value.

 

[silverpig]

<-- threadkiller

 

[przero]

To solve it is simple. On the graph (x,y) draw a circle around point A with a radius = to AC. Draw another circle around point B with a radius = to BC. At the point(s) those circles intersect will be point C.