The Dayan Gem VI is a twisty puzzle, and there are some pictures of it here: https://www.amazon.com/DaYan-Gem-Cube-VI-black/dp/B00PDVZ6YQ

The polyhedron consists of 6 octagons arranged like cube faces, with 24 pentagons surrounding the octagons, with 8 triplet of pentagons arranged like cube vertices.

This is not a Johnson solid, so pentagons and/or octagons are not regular polygons. Is there a name for this polyhedron?

  • $\begingroup$ As can be seen at this other view of it images-na.ssl-images-amazon.com/images/I/518sEw-4N9L.jpg (from the very website), the bit of the pentagons at the 3-pentagon-vertices is smaller than the other corner bits. $\endgroup$ May 14, 2018 at 18:41
  • 1
    $\begingroup$ Btw. a cube has 6 faces; accordingly your polyhedron would have just 6 octagons, not 8. $\endgroup$ May 14, 2018 at 18:55
  • $\begingroup$ You are right of course about the number of octagons. $\endgroup$
    – Nishant
    May 14, 2018 at 19:52

1 Answer 1


If you would apiculate the octagons, that is, each pentagon expands into a very flat triangle, then you would recognize this thingy as the dual of the truncated cube. Thus, if $C$ means cube, $t$ means truncation, $d$ means dual, then this depicted polyhedron would be $tdtC$, i.e. the truncation of the dual of the truncated cube. For sure, the second truncation here applies to the 8-fold vertices only.

This then proves that the octagons are - or at least can be chosen to be - regular. The angle of the pentagons at the 3-pentagons-vertices will be defined by the obtuse angle of the triangles of the dual of the truncated cube.

The final truncation depth than could be accommodated such that the 3 other edges of the pentagon, not incident to these 3-pentagons-vertices, have the same size.

--- rk

  • 1
    $\begingroup$ You can visualize "$tdtC$" here: levskaya.github.io/polyhedronisme (The second truncation includes the three-fold vertices, but it's pretty clear what the figure would look like if that didn't happen.) $\endgroup$
    – Blue
    May 14, 2018 at 19:17
  • 2
    $\begingroup$ replace t by t8, i.e. t8dtC, will only truncate the 8-fold vertices. $\endgroup$ May 14, 2018 at 19:24
  • $\begingroup$ @achillehui: I should've scrolled down! :) $\endgroup$
    – Blue
    May 14, 2018 at 19:28
  • $\begingroup$ Ah, so it's a truncated triakis octahedron. Thanks! $\endgroup$
    – Nishant
    May 14, 2018 at 19:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .