Let E be an ellipse centered at the origin on the x, y plane with major radius b and minor radius a. The length of the shortest line segment tangent to E that begins on the x-axis and ends on the y-axis is a+b. This can be shown using Lagrange multipliers. This answer is very simple and leads us to ask the following question:

Can you give a geometric reason for why the length is a+b?

This was originally asked to me by Frank Jones a few years ago.


2 Answers 2


There is a geometric way to show that $n$-gon circumscribed around an ellipse has minimal perimeter if it is inscribed in a confocal ellipse. From Poncelet porism (and generalization of optical property) it follows that we have continuous family of "minimal" polygons.

If we know it, then it is easy to understand that the circumscribed rhomb (from your question) and the circumscribed rectangular (with perimeter $4(a+b)$) are minimal polygons. So, side of the rhomb equals $a+b$.

  • $\begingroup$ That is neat. akopyan, are you the coauthor of Geometry of conics: books.google.com/…? (I deleted a question asking for a reference, because I see that it can be found in that book.) $\endgroup$ Feb 28, 2010 at 22:57
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    $\begingroup$ That is a beautiful answer. Thank you! $\endgroup$ Feb 28, 2010 at 23:18
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    $\begingroup$ Yes, I'm coauthor of this book. Also, look for the famous book "Geometry" by Marcel Berger. In volume 2 you can find many interesting statements similar to the fact in the initial question. – akopyan 53 secs ago $\endgroup$ Mar 1, 2010 at 3:28
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    $\begingroup$ Poncelet's porism is quite elementary to see in this case. Consider the case when $a=b=1$ of a circle. Then the two rectangles are the same, and are inscribed in a circle of radius $\sqrt{2}$. Poncelet's porism just says that there is a 1-parameter family of circumscribing/inscribing squares, in this case related by rotation. The affine transformation $(ax, by)$ gives a projective transformation taking this picture to the ellipse with major and minor axes $a,b$. Then the rotations are sent projectively to a 1-parameter family of projective transformations fixing both ellipses. $\endgroup$
    – Ian Agol
    Mar 1, 2010 at 17:51
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    $\begingroup$ 2 Agol, it is not. In this case you obtain two similar ellipses with different foci. Reason is, under a affine transformation length of segments doesn't preserve. $\endgroup$ Mar 1, 2010 at 19:11

By working simultaneously in the 4 quadrants, this becomes a question of minimizing the perimeter of enclosing rhombi with diagonals on the coordinate axes. Proving the inequality was Problem of the Week No. 13 in Spring 2005 at Purdue. Here is Steven Landy's solution to the problem. The proof is geometric in the sense of Descartes rather than Euclid, and shows that the minimum is at least a+b. There's no calculus, so maybe this is close to what you're looking for. (Edit: I entertained the delusion that this might be close to answering your question only before akopyan's answer was posted.)


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