## Always or Never

Take two ellipses, one within the other. Take a point on the outer ellipse, and draw one of the two tangents to the inner ellipse, and find its second intersection with the outer ellipse. Use this point to start the process again, and again. You will get a polygonal path in the ellipse that will most likely not close up. But in case you are lucky, something miraculous happens: If you pick any other point and repeat the game, the polygon will again close up.

This is the content of a famous theorem by Jean-Victor Poncelet.

In spirit, it is similar to a theorem of Jakob Steiner that asserts that a chain of circles in an annulus bounded by two circles either always or never closes up. While Steiner’s theorem follows immediately by inverting the circles into a pair of concentric circles, such a simple proof is not available for Poncelet’s theorem. Until recently, all proofs I know of were, let’s say, advanced.

At the core of a new proof by Lorenz Halbeisen and Norbert Hungerbühler are some fundamental theorems from projective geometry.

Let’s first recall that five points, no three collinear, determine a unique conic.

This is because through four points, you can find two different degenerate conics consisting each of a pair of lines, and by forming linear combinations, accommodate a fifth point. Below we will need the dual theorem: Given five lines, no three concurrent, there is a unique conic tangent to them.

Pascal’s theorem is a condition for six points to lie on a conic: They do if and only if opposite sides intersect in collinear points. Above you see this for six points on the two branches of a hyperbola.

Dual to this is Brianchon’s theorem (illustrated above): The sides of a hexagons are tangent to a conic if and only of its diagonals are concurrent.

As an application, Halbeisen and Hungerbühler show: If the six vertices of two triangles a1,a2,a3 and b1,b2,b3 lie on a conic, than there is a conic tangent to the six sides of the triangles. The proof is easy: Applying Pascal to the hexagon a1,b2,a3,b1,a2,b3 gives us three collinear points c12,c13,c23.

Then applying Brianchon to the hexagon a1,c12,b1,b3,c23,a3 shows that it is tangent to a conic. But the sides of this hexagon are the same as the sides of the two triangles, so we are done

From here, we obtain Poncelet’s theorem for triangles: Suppose you have two ellipses inside each other, and a triangle whose vertices lie on the outer ellipse and whose sides are tangent to the inner. Take another point on the outer ellipse, and form a second triangle by drawing the tangents to the inner ellipse. We have to show that the third side of the triangle is also tangent to the inner ellipse.

By the theorem by Halbeisen and Hungerbühler, the two triangles have an inscribed common ellipse. The given inner ellipse touches five of the same six lines by construction. But a conic is uniquely determined by five tangent lines.

The general case follows of n-gons the same idea, but requires more bookkeeping.

## Brianchon’s Theorem

When you rotate a straight line about the vertical axis, you will generally get a hyperboloid of revolution. By construction, this is a ruled surface, and by symmetry, there is a second set of lines on the surface. We call these two sets of lines the A-lines and B-lines.

These lines dissect the hyperboloid into lots of skew quadrilaterals, reminding us that any quadrilateral can be doubly ruled, and opening up more possibilities for our previously discussed bent rhombi.

Let’s form a hexagon, following the A- and B-lines alternatingly once around the hyperboloid. Then a theorem by Charles Julien Brianchon states that the three main diagonals (i.e. those connecting opposite vertices) of this hexagon will meet in one point.

One reason why this is curious is that it quite unexpected: In space, we don’t even expect two lines to meet in a point, let alone three. The other reason is that it has such a simple proof, due to Germinal Pierre Dandelin: Any pair of A- and B-lines will lie in a common plane, because they either intersect or are parallel. So the pairs of opposite edges give us three planes, which will meet in a common point. Because the diagonals of the hexagon are also the intersections of any pair of the three planes, we are done.

If we project the hyperboloid with all its decorations into the plane, like done so in the images above, the outline of the hyperboloid becomes a common hyperbola, and the six lines of the hexagon tangential to it. This leads to Brianchon’s theorem in the plane: The main diagonals of a hexagon circumscribed in a conic section meet in a point.

This theorem becomes easier to parse if the conic is just an ellipse:

We also have enough room here to see that there is a second dual conic on which the A-lines and B-lines, respectively, meet.