?ÿ
OK, here's where they came from.?ÿ They are a mixture of Ptolemy's Theorem and the Law of Cosines while trying to solve for the radius of the inscribed circle on the left.?ÿ The circumscribed triangle ABC was to be shown by a minimal about of information, but can be solved easily.?ÿ?ÿ The line DE is parallel with the base of the triangle.?ÿ The inscribed circle is tangent to the triangle, the circumscribed circle and the parallel line.?ÿ I thought they might degenerate into identities.?ÿ Negative values are not valid.
From the drawing, which I had difficulty plotting because I couldn't get the inscribed circle to be tangent to the lines I wanted, gives V=275.7?ñ, W=506.0?ñ, and X=610?ñ.?ÿ I'm betting a better CAD user can plot it perfectly.?ÿ I try not to use problems that can be solved by AutoCAD.
I may even be on the wrong track to a solution!
They both subtend chord/arc CB.
Angle EAC and EBC are equal for the same reason: they both subtend chord/arc EC.
Angle CEA is the supplement of angle CBA, or 103°32'22".
And angle ECA =angle EBA by subtending chord/arc EA.
You might want to look at the book "The Secrets of Triangles" by Alfred S. Posamentier and Ingmar Lehmann
on page 136-138. See Simson's Theorem (Robert Simson (1687-1768).
JOHN NOLTON
Yes, the opposite triangles are similar. It made sense since the opposite angles in the quadrilateral are supplementary and the sum of the angles of the triangles is 180 degrees.
Now the question of the algebra remains
You are TOUGHT. But I go by what a very good Professor friend of mine told me once;
"Remember never teach your students every thing you know"
JOHN NOLTON
LOL! Maybe he did it for you and didn't tell you how! 😉
I put Dave's drawing in Geogebra with some of the givens and some eyeballing and got v = 276.33, w = 505.47, and x = 610.33. Here's the Geogebra drawing, incliuding some of the doodles that always find their way into such stuff. Note that the point names are different from Dave's.
If you plug your numbers back into the equations, the fit is exactly the same as for two of the solutions that we got above, so solutions are not unique.
Perhaps the computer algebra systems need one or more starting values in order to converge to the correct solution.
OK this is how I did it. First lets change one letter in Dave's Diagram (he did the diagram fast just for
us and had 2 points labeled the same). From point C to point D there is a point E. Also from point A to
point B there is a point E. I changed the point from C to D to "F".
Now we can start; I solved the triangle ABC for its sides and for angle at A ,B-C. I solved for other information
but did not use it (only for later work). I then put coordinates on points on the BIG circle; starting at point
A (and going clockwise around the circle) E,B,C,F(new lettered point), and D. Point F is the point we need
and its just a GUESS on its coord. I then put it into STAR-NET( I know, I am getting lazy but I am old) and let
it iterate point F. That's how I got the value for V,W X.
JOHN NOLTON
So it's a least squares solution and the values are not zeroes of the four equations. We have two adjacent sides of the quadrilateral and one of the diagonals. We're trying to make the other diagonal a function of the other two sides, and it isn't. It's actually a function of the other diagonal and all four of the sides.
?ÿI thought so, too, but I don't see the line in any of the four equations. I think that the line constrains the location and size of the small circle but doesn't have anything to do with the quadrilateral.
Perhaps it needs another given, maybe a piece of the angle at B, one of the missing sides, or the missing diagonal.
It's an intriguing concept and you've taught me a good bit by sharing it. But, then, all of your problems are intriguing and you teach more people than you'll ever know by sharing them.
Point E, the point where the two circles are tangent, can be moved randomly without changing the given information. Here are two cases from Geogebra:
The givens stay the same, but the non-given sides and non-given diagonal change.?ÿ
Something has to fix the location of Point E.
This is a problem that Dave Lindell is working UP. (A problem in its infancy.)
Remember that we had to go back to Dave and ask what he was trying to do because when we worked
the equations they did not come out looking good.
Dave said "they are a mixture of Ptolemy's theorem and the Law of Cosines while trying to solve
for the radius of the inscribed circle on the left".
In another post Dave (to you) said "it is constrained by the line parallel with the triangle base. I think
With the "I think" part I think he means if the inscribed circle is tangent at point "G" and point "H" both on a fixed
line(point H is on fixed line AC and point G is on fixed line DE) then there should be only one point
F.
Is that point F unique and can it be solved for and can the radius be solved for?
John Nolton
MT; how many people let alone surveyors know who Ptolemy was and what he did? Without looking him UP.
This is a problem that Dave Lindell is working UP.?ÿ (A problem in its infancy.)
Gee, aren't there already enough problems in the world without creating more? ?????ÿ