4
$\begingroup$

There are various well-known ways to construct the common tangents to a pair of circles; this is an easy one.

I also just learned that we can use Pascal's Theorem to construct a tangent to a conic at any desired point. So the question arises: how can we construct the common tangents to a pair of arbitrary conics in the plane? The construction for two circles linked above can't be used (it seems to me) because its key idea is that the circles are images of each other under affine transformations. The conics are images of each other under projective, but generally not affine, maps.

I'm looking for straightedge-and-compass solutions, not analytic ones. Since this is a purely projective problem, straightedge alone might suffice: bonus points for such a solution!

Here's an example setup, with one of the four tangents constructed by eye (orange/green). The only difficulty lies in identifying points $E$ and $E_1$.

enter image description here

[EDIT: Note that this question is a less general version of the present one; unfortunately it's been open for 3+ months without resolution, which doesn't make me very optimistic...]

$\endgroup$
2
$\begingroup$

I got interested in this question through the same route that you mention. Generally the problem is solved analytically using dual conics. But naturally the question of how to do it with just a straightedge is intriguing.

Back in the day, before the fancy mathematics of the 20th century began to push classical geometry out of the mainstream, this sort of subject was all the rage.

The answer is complicated, and there's no way to summarize it here. But it's all spelled out in gory detail (with many special cases) at what should be a stable URL: Chapter XVI of Milne's An elementary treatise on cross-ratio geometry.

The case for two conics that intersect at 4 points is relatively straightforward. These 4 points define 6 lines called common chords. On each common chord you can select a point and draw four tangents to the conics. Then the lines joining the points of contact on the first conic to the points of contact on the second conic will intersect at the points where the shared tangents intersect. You can then join the shared tangent intersection points to get the four shared tangents. This is spelled out in chapter XV, specifically Article 245(1). You may want to start at Chapter XIV and Figure 102 (or even earlier) to get the run-up to the result. There is some very pretty projective geometry there.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.