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Circular Reasoning first shows an interesting property
of circles: if you draw a line from a point P and it cuts
a circle in points Q and R, the product of the distances PQ
and PR is equal to the value of the point with respect to
the equation of the circle. We then continue the discussion
from January by talking about Ptolemy's Theorem, and showing
how it can be used to link Snell's Law and Fermat's Principle
of Least Time for refraction.
Marco Corvi has observed that this linkage seems to have
a flaw. The basic problem is that when we choose a new point
on the surface that is to the left or right of the point dictated
by Snell's Law, the condition that four of the points remain
on a circle can be violated. This is correct, but not a fatal
flaw in the argument. Rather, it was just a flaw in my presentation.
Here's the way to fix things up.
Ptolemy's Theorem can be considered a generalization of the
triangle inequality. This states that for any triangle ABC,
|AC|>=|AB|+|BC|. The two sides are equal if the triangle
is degenerate (that is, B lies on the line segment AC). Otherwise
the triangle actually encloses some area, and the relation
is strictly greater-than.
Similarly, Ptolemy's Theorem states (take a deep breath)
that the product of the lengths of the diagonals of a quadrilateral
is greater than or equal to the sum of the products of the
lengths of opposite sides.
If the four points lie on a circle, then equality holds.
If it's a more general quadrilateral, then like the triangle
inequality, the relation is strictly greater-than.
In the final step of the column, I use Ptolemy's Theorem
to substitute for four points on a circle, and four points
off the circle, to show that the distance covered by the four
non-cyclic points is longer. In this step, when I substitute
for the cyclic points the replacement is exact. When I substitute
for the non-cyclic points, the value given by Ptolemy's Theorem
is larger than the value we're replacing. But since we're
only showing that this value is bigger than the one for the
cyclic points, it's okay.
In other words, if lengths of the non-cyclic points is N
and the length of the cyclic points is C, and their substituted
values from Ptolemy's Theorem are N' and C', then N'>N
and C'=C. We start with N>C, so the new version N'>C'
still holds, since N'>N, and we've returned to Snell's
Law.
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