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Track with rectangle, the coordinates of which I want

Consider the image above. DEGP is a rectangle that I'm trying to draw programmatically so I need the coordinates of the points.

What's known is this:

  • The coordinates of D
  • The coordinates of the inside path that P and G are on (semi-circles with straight segments, nothing too weird).
  • DE is 20 (as is PG, therefor)
  • D is such that a rectangle like this will exist, D will never be lower than C or such that PG is either entirely on the straight or entirely on the semi-circle.

I can't seem to figure out how to get the coordinates of E or G or P because they're all dependent on each other. I tried to draw this by saying that a line from D straight down to AB should be equal to EA, but that gives a wonky result shown below:

Attempt to define rectangle with wonky result

Below is how I would do it if PG was entirely on the semi-circle:

If PG was entirely on the semi-circle

So I'd use the symmetry that the rectangle now has over line AQ to find E and draw DP and EG perpendicular until they meet the semi-circle.

EDIT: I originally tagged this algebraic-geometry and I'm coming back to that with a possible approach to solve this:

The semi circle over CJ can be described as $x=-\sqrt{12.5^2-y^2}+\Delta AD$

What formula would describe the path of G as we rotate the rectangle around D and stretch DP (and thus EG) to have P be on the path. If I have that formula, I could equate it to the formula for the semi-circle to determine for which y point G is equal to a point on the semi-circle.

Is this a feasible approach?

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This does not seem a trivial problem at all. Are there any restrictions on the choice of $D$, or do you know a priori that $D$ is chosen such that a rectangle of the type you have shown exists? How far have you gotten with any calculations? The way I would try and tackle this is to find the points $G$ and $P$ such that the line $GP$ is orthogonal to $DP$. Once you have found these points then $E$ is automatically defined. Have you tried something like this already? –  Keeran Brabazon Sep 21 '13 at 18:00
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You know a priori that such a rectangle exists. If D is such that PG would be entirely on the straight or entirely on the semi-circle, the problem is much easier and would be solved by a different calculation. Furthermore, the ovals actually represent a race-track, so if D is inside the inner oval or outside the outer oval, the calculation wouldn't be carried out. So far I've tried to split the rectangle into 2 triangles with line DG so I could maybe use trigonometry. But I'm unsure how to get the angle EDG from the DeltaY and DeltaX for D and C as those points are known. –  funkwurm Sep 21 '13 at 22:52
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(Edited the question with more info and example if PG is entirely on the semi-circle) –  funkwurm Sep 21 '13 at 23:48
    
There is only one degree of freedom here, the position of $P$ on the inner path. This determines $G$ as the point on the path counterclockwise from $P$ with $|PG|=20$ (this is unique as long as the diameter $|CJ|\ge20$). Then the only thing left to solve for is the condition that $PD\perp PG$. –  Rahul Sep 22 '13 at 1:06
    
I don't see how the position of P is a degree of freedom as it is constraint by the fact that PD has to be equal to EG for it to be a rectangle. For every valid position of D (which is known) there really is only one possible rectangle. CJ is and will always be 25, coordinates and measurements of the inner path are fully known. –  funkwurm Sep 23 '13 at 23:59
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