A quick attempt:

+Matt McIrvin that looks like a version with an equilateral-triangle base and regular-pentagon sides? But then the question becomes whether there is a natural choice for where to put the point. There's also a version that you can obtain by truncating one corner of a cube and pulling the opposite corner farther away, but again, how far?

I don't actually know if those pentagons can be regular and still give you a convex solid.

I made a better version but G+ is unwilling to let me attach two drawings to comments on this post, I guess.

The one in I presume.

...The convex solid definitely exists though: consider the intersection of a regular hexagonal pyramid extending downward from above, and a truncated regular triangular pyramid extending upward from below.

+Matt McIrvin it is easy to visualize. Take three regular pentagons in a plane and use one edge of each to make an equilateral triangle between them. That triangle is the base. Then rotate each pentagon around its base edge towards the center of the base triangle, into the same direction off the plane, until their edges meet. It should be easy to see that this can be done naturally in three dimensions. Or did I miss something?

(I figured out how to install Inkscape so I could draw that correctly)

+Sakari Maaranen The part I'm not sure about is keeping the top part composed of six triangles convex. If everything isn't right, some of the joins between triangles there will go flat or concave. It's possible to keep them all convex but I don't know if it's so for the case where the pentagons are regular.

+Matt McIrvin To make the top convex, you can adjust each pentagon by bringing the vertex opposite to the base edge, closer to the center of each pentagon, so they would no longer be regular.

Draw a circle through the vertices next to the base, so this top circle will be in a plane parallel to the base. Move the top vertex of each (now irregular) pentagon towards its center until a line projecting from the top of the solid through the top vertex of the pentagon meets the circle.

Here's a convex version with regular pentagonal sides:

Here's a version where the top triangular faces are isosceles (as well as the pentagons being regular):

If anyone feels like checking my calculations, the isosceles version has:

r_0 = 1
r_1 = (1+√5)/2
r_2 =1

z_0 = 0
z_1 = (1+√5)/2
z_2 = (3+√5)/2
z_3 = (3+√5+2√2)/2

where r_0, r_1, r_2 are the radial distances of the successively higher triples of points from the central axis, and z_0, z_1, z_2 are the heights of those points, with z_3 the height of the apex.

+Greg Egan I'm curious what the most elegant method of deriving this is? Are you using axial-polar co-ords only to describe it?

My first idea is linear equations in a Cartesian basis but I'm not a real mathematician.

If this is a unit distance graph embedding, why aren't all triangles equilateral +David Eppstein ? ( +Greg Egan , +Matt McIrvin )

+Jon Eckberg This isn’t a unit-distance embedding. This graph has a unit-distance embedding in the plane with edges that cross (as you can see in the Wikipedia article), and also a planar embedding (i.e. with no edges crossing) that is not unit-distance (the image in the body of the post). All we’re doing here is trying to find a “maximally nice” convex polyhedron whose edges and vertices correspond to those of the graph. That doesn’t stretch as far as making all the edges the same length.

To derive this, I took +Matt McIrvin’s shape, assumed it had perfect three-fold symmetry around the axis, and solved for the values of the free parameters that would make the pentagonal faces regular. That left only one free parameter: the height of the apex. For a range of values for that height, the polyhedron is convex ... and that range happened to include a value where the top triangles are isosceles.

I’m not sure that it means much to distinguish between doing this in Cartesian or cylindrical polar coordinates. I wrote down the coordinates of the individual vertices in Cartesian coordinates, but those coordinates were functions of the r_i and z_i, which are cylindrical polar coordinates. Because there were edge lengths, the equations were quadratic in the parameters.

I tried with the kids' mag-tiles but am unable to close the polyhedron. No regular pentagons or obtuse triangles available.

It seems the dual can be done with regular polygons, and it is almost convex.

+Refurio Anachro One hexagon and 7 eq. triangles? Edit 9?

Almost, six eq. triangles, three squares, and a hexagon, +Jon Eckberg .

Neat! It's what you get when you glue a tetrahedron onto a triangular cupola

(maybe ironically, this one actually is a unit distance embedding in 3d).

+David Eppstein +Refurio Anachro Here's the dual rendered in MagGL.

My rendition of the dual (Inkscape drawing, perspective not perfect, but I drew it to correspond more or less to my other drawing's orientation):

Nice, +Matt McIrvin ! For a square cupola (with an octagon as base and a square cap) it's easy to see that the angles of the triangles with respect to the octagon are the same as the ones you'd get for a square pyramid: Just cut it along an edge of the cap, then slice away three squares (two and the cap) by cutting again along the opposite edge. Glue and repeat for the remaining squares. I tried to convince a friend that this would also hold for the triangular cupola, but he wouldn't believe me, as I couldn't show him how to cut that...

So "almost convex" is a cheesy euphemism for the fact that the angle between the squares and the tetrahedral hat's triangles is 180° which makes the edges in between vanish or into virtual ones.