16

A friend of mine showed me the following problem:

Let $\cal E$ be an ellipse whose semi major axis has length $a$ and semi minor axis has length $b$. Let $\ell_1, \ell_2$ be two parallel lines tangent to $\cal E$. Let $\cal C$ be the circle tangent to $\ell_1$, $\ell_2$, and $\cal E$. Prove that the distance between the centers of $\cal C$ and $\cal E$ is equal to $a+b$.

So far I managed to prove that if we draw the tangent line $k_1$ through $\cal E \cap \cal C$ and the tangent $k_2$ to $\cal E$ parallel to $k_1$ then the circle tangent to $k_1, k_2, \ell_1$ is tangent to $\cal E$ as well.

I'm stuck. I'd like to see some proofs, preferably non-analytic ones.

brainjam
  • 9,172
timon92
  • 12,291
  • 2
    Wild guess: consider the simplest possible case where the ellipse isn't "tilted". If that works, show that rotating the ellipse can't change the distance. –  Jul 15 '17 at 14:50
  • 1
    I found this problem in "sangaku". – Takahiro Waki Jul 15 '17 at 14:53
  • @TakahiroWaki Could you please provide a more precise reference? Thanks. – timon92 Jul 15 '17 at 15:32
  • 2
    (+1) Very nice question. Sangaku problems are well-known to be incredibly deep sometimes. – Jack D'Aurizio Jul 15 '17 at 16:47
  • http://math.a.la9.jp/sangaku.htm. https://www.amazon.com/Japanese-Temple-Geometry-Problems-Sangaku/dp/0919611214/ref=pd_sbs_14_1?_encoding=UTF8&pd_rd_i=0919611214&pd_rd_r=4R9QQTBM2PG7E4J1RGX6&pd_rd_w=qAHyK&pd_rd_wg=ItFyN&psc=1&refRID=4R9QQTBM2PG7E4J1RGX6 – Takahiro Waki Jul 15 '17 at 17:35
  • @TakahiroWaki I don't have access to this book nor I can't afford it. Could you please outline the sketch of proof from this book (if it is essentially different from the one JackD'Aurizio provided)? – timon92 Jul 15 '17 at 17:46
  • @Jack D'Aurizio It would be nice to see an animation of outer rotating tangent circles between parallel tangents of a fixed central ellipse, touching the fixed central ellipse, so that line of centers has a constant radius $(a+b).$ The $OP$ drew two such cases. – Narasimham Jul 16 '17 at 02:18
  • Am sure you would... actually I was scratching my head if parallel conjugate diameters central line makes for a simpler proof ! – Narasimham Jul 16 '17 at 02:27
  • 1
    @Narasimham: animation added to my answer. – Jack D'Aurizio Jul 16 '17 at 02:40
  • Convincing, fine demo! – Narasimham Jul 16 '17 at 03:27
  • Jack D'Aurizio's answer shows rather steiner chain not poncelet's porism. He rejected, though. http://mathworld.wolfram.com/SteinerChain.html – Takahiro Waki Jul 19 '17 at 13:44
  • @TakahiroWaki I'm afraid you are wrong. There is no Steiner chain in Jack's answer. – timon92 Jul 19 '17 at 16:03
  • @timon92 Please search difference between steiner chain and poncelet's porism. – Takahiro Waki Jul 19 '17 at 16:05
  • @TakahiroWaki https://www.youtube.com/watch?v=dqB-EMqpsUA&feature=youtu.be&t=150 – timon92 Jul 19 '17 at 16:31
  • Steiner chain is rotating circle around circle, but poncelet's porism is rotating polygon which tangent to two circle. OK? – Takahiro Waki Jul 19 '17 at 16:37
  • @TakahiroWaki I'm sorry, I don't have time to discuss this. Please read https://en.wikipedia.org/wiki/Poncelet's_closure_theorem and https://en.wikipedia.org/wiki/Steiner_chain – timon92 Jul 19 '17 at 16:39
  • You must read them. – Takahiro Waki Jul 19 '17 at 16:41

1 Answers1

20

I have a roadmap for a simple solution through analytic geometry / trigonometry:

  1. Let us consider an ellipse $\mathcal{E}$ with equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, i.e. $(x,y)=(a\cos\theta,b\sin\theta)$;
  2. Let us consider a generic point $P\in\mathcal{E}$, i.e. some $\theta\in[0,2\pi)$;
  3. Let us compute the slope of the tangent $\tau_{P}$ through $P$ and consider that the parallel tangent is $\tau_{-P}$, since in a conic the midpoints of parallel chords are aligned along a line through the center of the conic;
  4. Let $R_\theta$ be half the distance between $\tau_P$ and $\tau_{-P}$ and $\ell_\theta$ the line through the original parallel to $\tau_P$. Let $C_\theta$ be an intersection between $\ell_\theta$ and the circle $x^2+y^2=(a+b)^2$;
  5. In order to check that $\mathcal{E}$ and the circle with radius $R_\theta$ centered at $C_\theta$ are tangent it is enough to check that a discriminant equals zero.

enter image description here

Here there are some ideas for a purely Euclidean solution:

  1. Let $Q=\mathcal{C}\cap\mathcal{E}$. First Claim: the parallelogram having sides $\tau_P,\tau_{-P},\tau_Q,\tau_{-Q}$ has its vertices on a fixed ellipse $\mathcal{E}'$, having the same foci as $\mathcal{E}$;
  2. Let $U,V$ the vertices of the previous parallelogram on $\tau_Q$. Second Claim: $C_\theta U$ and $C_\theta V$ are orthogonal tangents to $\mathcal{E}'$, so $C_\theta$ lies on the orthoptic of $\mathcal{E}'$ and has a constant distance from $O$.

enter image description here

Indeed, the first claim follows from Poncelet's porism and the second claim is just a matter of angle chasing.

Here it comes an animation, too:

enter image description here

Jack D'Aurizio
  • 361,689
  • 2
    I think these two claims lead to a solution. You have proved that $OC_\theta$ does not depend on $\theta$, so it is enough to check what happens in one particular case - e.g. the case when $C_\theta$ lies on the major axis of $\cal E$, but in this case the problem is trivial. Am I missing something? – timon92 Jul 15 '17 at 17:25
  • 1
    @timon92: you are not missing anything, you are completely right. – Jack D'Aurizio Jul 15 '17 at 17:37
  • 2
    Thank you very much. This was incredible. – timon92 Jul 15 '17 at 17:42
  • Amazing solution (+1), and beautiful animation! How did you produce the animation? – Hypergeometricx Jul 16 '17 at 14:35
  • 1
    @hypergeometric: Geogebra allows that, too, one needs just a bit of patience. – Jack D'Aurizio Jul 16 '17 at 14:40
  • But affine transformation is instant solution. – Takahiro Waki Jul 19 '17 at 07:46
  • 1
    @TakahiroWaki: no, it is not. I do not see how you can prove that $OC_\theta$ is constant with an affine transformation. If you are able to do that you are probably able to prove Poncelet's porism with affine transformations only: I do not believe that's possible. – Jack D'Aurizio Jul 19 '17 at 08:02
  • https://books.google.com/books/about/Mathematical_Omnibus.html?hl=ja&id=IiG9AwAAQBAJ&redir_esc=y and search affine transformation, poncelet. By the way I think ellipse version of poncelet's porism must need outside circumscribed circle. – Takahiro Waki Jul 19 '17 at 08:23
  • 1
    @TakahiroWaki: do what you want 'cause a pirate is free, if you have a simpler solution, just post it. – Jack D'Aurizio Jul 19 '17 at 08:47
  • @JackD'Aurizio: your proof if very beautiful, but I'm a little confused about the angle chasing of 2nd claim. Can you give some hint for me? Thank you very much. – Ichungchen Mar 29 '21 at 11:06