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I'm trying to prove the claim:

Two triangles inscribed in a conic section, there must be another conic tangent to the six edges of the two triangles.

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The problem looks simpler than this question: A conic inside a hexagon , but I haven't found a way to use Pascal + Desargues + Brianchon directly.

Here is my approach:

  1. Pascal's theorem tells us that AD∩BE lies on line GJ;
  2. Pappus's theorem applying on ALE and BHD tells us that G, AD∩BE and LD∩HE are collinear, so LD∩HE lies on GJ;
  3. Pappus's theorem applying on LKE and HID tells us that LI∩HK, LD∩HE and J are collinear, so LI∩HK lies on GJ, i.e. GJ, HK and LI are concurrent;
  4. The converse of Brianchon's theorem tells us that there must be a conic tangent to the hexagon GHIJKL.

Are there any simple ways to prove it?

brainjam
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auntyellow
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1 Answers1

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Your proof is already pretty short, but it can be shorter. For the hexagon $\pmb{H}=AGBDJE$ you've shown that the diagonals $AD,GJ,BE$ are concurrent. By the converse of Brianchon, $\pmb{H}$ has an inscribed conic. But the sides of $\pmb{H}$ are the same as the sides of the triangles, so we're done.

brainjam
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