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Exercise 10.3.14
As explained in [21](C.R.Videla, On points constructible from conics), Pappus used intersections of conics to trisect angles as follows. Consider the unit circle centered at the origin, and let . Then is the corresponding point on the unit circle. We assume that is known. Also let and set . Thus .
- (a)
- Consider the curve consisting of all points such that the distance from to is twice the distance from to the -axis. The curve intersects the unit circle at a point lying in the interior of . Prove that .
- (b)
- Show that the curve is a hyperbola. It follows that we have trisected an angle using the intersection of a hyperbola and a circle, i.e., an intersection of conics.
Answers
Proof.
- (a)
-
Let
be the orthogonal projection of
on the
-axis, and
such that
is the midpoint of
, so
is the reflection of
with regard to the
-axis. Since
is on the curve
,
therefore the measures of and are equal, and is twice the measure of , so trisects the angle and a measure of is .
- (b)
-
Let
The discriminant of the quadratic form is , so is not empty and is a hyperbola.