Post by Blair Dowden on Dec 18, 2019 18:22:27 GMT
First, I want to thank you so much for this very helpful website. It is what makes this excellent textbook really work for me.
I would like to comment on your solution to Exercise 2 in Chapter 3, the Peaucellier Linkage. I add to your diagram by labeling the other two points on the rhombus as C and D.
First, your proof assumes points p and p ̃ will always be aligned with the origin. Here is a justification: CpDp ̃ is a rhombus, so its diagonals form a right angle. Triangle ODC is bisected by line Opp ̃ therefore points p and p ̃ will remain on a straight line originating at point O.
Second, there is no need to invoke the Tangent-Secant theorem. Because the inversion circle K cuts the circle centered at C, if follows that any straight line starting from the origin (P) will cut the circle at its two inverse points relative to circle K. Therefore points p and p ̃ are the inverse of each other. Done.
The Tangent-Secant theorem itself is a direct consequence of inversion: Draw a circle, and a point P outside it. Join P to the two tangent points, and draw an inversion circle through those points entered at P. Any line drawn from P intersecting the circle will cut it at the two inversion points, p1 and p2. Multiplying length p1 by p2 gives the square of the radius, which is the distance to the tangent points. Done.
Unlike the standard proof, which relies on the Pythagorean theorem and is only valid in a flat space, this proof should work in any space that has a meaningful inverse, such as the surface of a sphere.
The author likes elegant solutions to otherwise complex problems, so I am surprised that he did not include this proof of such a fundamental theorem.
I would like to comment on your solution to Exercise 2 in Chapter 3, the Peaucellier Linkage. I add to your diagram by labeling the other two points on the rhombus as C and D.
First, your proof assumes points p and p ̃ will always be aligned with the origin. Here is a justification: CpDp ̃ is a rhombus, so its diagonals form a right angle. Triangle ODC is bisected by line Opp ̃ therefore points p and p ̃ will remain on a straight line originating at point O.
Second, there is no need to invoke the Tangent-Secant theorem. Because the inversion circle K cuts the circle centered at C, if follows that any straight line starting from the origin (P) will cut the circle at its two inverse points relative to circle K. Therefore points p and p ̃ are the inverse of each other. Done.
The Tangent-Secant theorem itself is a direct consequence of inversion: Draw a circle, and a point P outside it. Join P to the two tangent points, and draw an inversion circle through those points entered at P. Any line drawn from P intersecting the circle will cut it at the two inversion points, p1 and p2. Multiplying length p1 by p2 gives the square of the radius, which is the distance to the tangent points. Done.
Unlike the standard proof, which relies on the Pythagorean theorem and is only valid in a flat space, this proof should work in any space that has a meaningful inverse, such as the surface of a sphere.
The author likes elegant solutions to otherwise complex problems, so I am surprised that he did not include this proof of such a fundamental theorem.
