A proof of Poncelet Porism with two circles

Persiflage 2024-07-03

Brian Lawrence showed me the following conceptual proof of Poncelet porism in the case of two circles, which I thought was neat and wanted to sketch here. (This is only a sketch, since I’m not really defining the integration.)

Let ${P}$ be a point on the outer circle, and let ${Q}$ be the point you get when you take the counterclockwise tangent from ${P}$ to the inner circle. Consider what happens if we nudge the point ${P}$ by a small increment ${dP}$ .

poncelet-1

The similar triangles in power of a point then give us the approximation

$\displaystyle \frac{dP}{t(P)} = \frac{dQ}{t(Q)}$

where ${t(X)}$ is the length of the tangent from ${X}$ on the large circle to the smaller one. (Note that because we’re working with circles, the definition of ${t}$ doesn’t care about clockwise vs counterclockwise tangent).

Now, suppose ${(P_0, P_1, P_2, \dots, P_n)}$ be a sequence of points on the large circle such that ${P_i P_{i+1}}$ is the counterclockwise tangent to the inner circle for all ${i}$ . Now, suppose ${(P_0', P_1', P_2', \dots, P_n')}$ be another such sequence where ${P_0'}$ is slightly counterclockwise of ${P_0}$ . Then we have the integral relations

$\displaystyle \int_{P_0}^{P'_0} \frac{dX}{t(X)} = \int_{P_1}^{P'_1} \frac{dX}{t(X)} = \int_{P_2}^{P'_2} \frac{dX}{t(X)} = \dots = \int_{P_n}^{P'_n} \frac{dX}{t(X)}.$

So if ${P_0 = P_n}$ , it follows ${P_0' = P_n'}$ as well. Hence Poncelet’s closure theorem is proved.