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I have two convex quadrilaterals (ABCD and WXYZ). Their sides and their interior angles are known. I also know that WXYZ fits perfectly inside ABCD with each corner point touching a different side.

Is there any way to figure out analytically where the touches occur?
For example, if I knew AX or angle BXY, everything else would be obvious because it is all triangles. But how to find out where X is on line AB?

... Of course, I can always draw the things on different pieces of paper, cut WXYZ out, fit it by hand and then measure the result. But that isn't terribly precise. (!!)

I've shown WXYZ inside ABCD in the linked image

SJ.Storms
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  • How do you know wxyz fits perfectly inside abcd? There are at most 4 ways to do this. (four possible lines for the first point). Start be assuming A goes on line WX. Label one of the resulting angles theta and figure the remaining 7 angles in terms of theta. Use law of cosines and law of sines to calculate what the new lengths must be and if they have a solution. Try this four times until you get something that works. – fleablood Apr 27 '16 at 15:57
  • So in that image. Let the measure of angle AXW = theta. sin A/WX = sin theta/WA = sin(180-theta-A)/AX, sin B/XY = sin(180-theta-X)/BY = sin(theta+ X - B)/BX, etc. Use these to figure out the lengths and angles. – fleablood Apr 27 '16 at 16:04

1 Answers1

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Assume a solution.

Let a = m angle A; b= m angle B; and so on.

Let $\theta = $ m angle AXW.

By law of sins:

$\sin a/WX = \sin \theta/AW = \sin(180 - \theta - a)/AX$

$\sin b/XY = \sin(180 - \theta - x)/BY = \sin(\theta + x - b)/XB$

$\sin c/YZ = \sin(180-y - \theta - x + b)/ZC = \sin(y + \theta + x - b - c)/YC$

and

$\sin d/WZ = \sin (\theta + a - w)/DZ = \sin (180 - d - \theta - a +w)/DW$.

(And we can also note: $180 - d -\theta - a + w = 180 - z - y - \theta -x + b + c$)

Using these was can solve AW,AX,BY,XB,ZC,YC,DZ, DW.

fleablood
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  • Mind you, the fact that 180−d−θ−a+w=180−z−y−θ−x+b+c doesn't help because it just resolves into a+b+c+d = w+x+y+z. – SJ.Storms Apr 27 '16 at 20:58
  • Thank you! Mind you, I'm baffled as to solving for theta and AW, etc with just the above. But if I add the fact that AB, BC, CD and DA are known, then with (1)DW+AW=DA and (2)sin(d)/ WZ = sin(180−d−θ−a+w)/DW and (3)sin(a)/WX = sinθ/AW, then, substituting for DW and AW in (1) I get WZsin(180−d−θ−a+w)/sin(d) + WXsinθ/sin(a) = DA. And the only unknown in that is θ. Yay! .... But that isn't an equation I know how to solve.... maybe there's an easier equation still to be found! – SJ.Storms Apr 27 '16 at 21:33
  • Well, sin a/wx = sin theta/aw and sin d/wz = sin(180−d−θ−a+w)/dw. And aw + wd = ad is known. That should make theta solvable. ad = aw + wd = sin thetawz/sin a + sin(180 - d - thteta - a + w)wz/sin d. So ad/wz = sin theta/sin a + sin(180 - d - theta - a + 2)/sin d. It's a single trig equation with only theta unknown. – fleablood Apr 27 '16 at 23:33
  • Yes, that's sorted, thank you very much! I don't actually know how to solve for theta when the equation looks like 0 = sin(θ +c) + sin(θ) + g, but I can always plug different values in and find one that works. :) Thanks again; I was 100% stuck without your help! – SJ.Storms Apr 28 '16 at 01:57