15
$\begingroup$

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.

$\endgroup$
17
  • 2
    $\begingroup$ 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. $\endgroup$
    – user2469
    Jul 15, 2017 at 14:50
  • 1
    $\begingroup$ I found this problem in "sangaku". $\endgroup$ Jul 15, 2017 at 14:53
  • $\begingroup$ @TakahiroWaki Could you please provide a more precise reference? Thanks. $\endgroup$
    – timon92
    Jul 15, 2017 at 15:32
  • 2
    $\begingroup$ (+1) Very nice question. Sangaku problems are well-known to be incredibly deep sometimes. $\endgroup$ Jul 15, 2017 at 16:47
  • 1
    $\begingroup$ @Narasimham: animation added to my answer. $\endgroup$ Jul 16, 2017 at 2:40

1 Answer 1

18
$\begingroup$

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

$\endgroup$
10
  • 2
    $\begingroup$ 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? $\endgroup$
    – timon92
    Jul 15, 2017 at 17:25
  • 1
    $\begingroup$ @timon92: you are not missing anything, you are completely right. $\endgroup$ Jul 15, 2017 at 17:37
  • 2
    $\begingroup$ Thank you very much. This was incredible. $\endgroup$
    – timon92
    Jul 15, 2017 at 17:42
  • 1
    $\begingroup$ @hypergeometric: Geogebra allows that, too, one needs just a bit of patience. $\endgroup$ Jul 16, 2017 at 14:40
  • 1
    $\begingroup$ @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. $\endgroup$ Jul 19, 2017 at 8:02

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.