Given are two points on a line with coordinates. How do we calculate the third forming a perfect 60 degree triangle? So we have X,Y, but need Z...

X: 0,0    ( 0,0 i.e. horizontal, vertical )
Y: 50, 0
Z: 25, ??

How to calculate the missing horizontal coordinate for Z? Forming a perfect 60 degree triangle?

EDIT: I'm trying to figure out a formula that by providing only 50 as the length of one side of the triangle all the point coordinates can be figured out. The first 2 and half of them are easy (0,0 50, 0, ??, 25)... but how to calculate the number at '??' is what I'm trying to figure out.

EDIT2 (@Dennis): It would probably help if I explained that my coordinates are not of the actual point but a 100px wide circle that the point is in the middle of. I made a small video to show you guys exactly what I'm doing and will post is just as soon as it's done uploading.

EDIT3: Here is a video showing exactly what I mean and why I came up with the question in the first place: http://archebian.org/videos/math/triangle-question.mp4


closed as off-topic by barak manos, Behrouz Maleki, Rohan, Claude Leibovici, E. Joseph Dec 22 '16 at 13:34

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  • 3
    $\begingroup$ Show some effort please. Add an illustration of the coordinates on the plane. Describe your thoughts on the problem and where you got stuck. This is not a 'do my homework for free' service! $\endgroup$ – barak manos Dec 21 '16 at 11:14
  • 1
    $\begingroup$ Lol I'm not asking for home work help smart guy. $\endgroup$ – OB7DEV Dec 21 '16 at 11:18
  • $\begingroup$ Take the midpoint of the line joining the two given points, draw a perpendicular to this line and intersect it with $\;y=25\;$ . Why this works? $\endgroup$ – DonAntonio Dec 21 '16 at 11:24
  • $\begingroup$ @DonAntonio - You are correct, but what I need to do is make this into a formula. Provide the length of one side and calculate the rest from only that. $\endgroup$ – OB7DEV Dec 21 '16 at 11:28
  • $\begingroup$ @OB7DEV "Rofl" But it really looks like one... Nevertheless it is really hard to help you, when you don't tell, what you have already tried and where you got stuck. See: math.stackexchange.com/help/how-to-ask $\endgroup$ – ctst Dec 21 '16 at 11:29

With the video explanation

You are asking for the intersection of two circles (this is not creating an equilateral triangle). You can take the equations of your circles $c_1: x^2+y^2=50^2$ and $c_2: (x-50)^2+y^2=50^2$.

$$x = \frac{d^2-r^2+R^2}{2d}$$ For your question the distance between the center points is $d=50$ and $r=R=50$. And surprisingly you'll find it at 25. This you can put into the circle equation again and find $$y=\sqrt{50^2 - 25^2}\approx 43,30127$$ Just as in your picture.

You find the solution here: http://mathworld.wolfram.com/Circle-CircleIntersection.html.

For the equilateral part

What about the height of an equilateral triangle being $\frac{\sqrt{3}}{2}a$ where $a$ is the length of your side. With this you should see that 25 isn't suitable for your chosen height of 25.

e.g. (0,0); (50,0); (25,21.65) is close to perfect. In your post I see the 25 in the y coordinate, which doesn't make sense to me.

  • $\begingroup$ Thanks, I fixed the coordinate mistake in the question. $\endgroup$ – OB7DEV Dec 21 '16 at 12:42

I gave your question to Pythagoras (a friend of mine): He said: the missing horizontal coordinate is given by

$$ \sqrt{50^2+25^2}.$$

OOps !

Edit: coordinate is given by

$$ \sqrt{50^2-25^2}.$$

  • $\begingroup$ Are you sure that number is the missing abscissa of the third vertex with ordinate $\;y=25\;$ ? We have that $\;\sqrt{50^2+25^2}>50\;$ and thus the resulting triangle would have an angle greater than $\;90^\circ\;$, and of course won't be equilateral. Am I missing something? $\endgroup$ – DonAntonio Dec 21 '16 at 11:38
  • $\begingroup$ Thanks ! I meant $ \sqrt{50^2-25^2}$ and not $ \sqrt{50^2+25^2}.$ $\endgroup$ – Fred Dec 21 '16 at 11:43
  • $\begingroup$ @Fr That could be, unless the OP meant equilateral...but right now I've no real idea what the OP meant. Read last comments below question. $\endgroup$ – DonAntonio Dec 21 '16 at 11:44
  • $\begingroup$ @Fred - You got answers and humor! $\endgroup$ – OB7DEV Dec 21 '16 at 11:48

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