# Simple geometry. Or is it?

I've got a regular tetrahedron and a square pyramid. Every edge of the two solids has the same length. If I perfectly attach one face of the tetrahedron to one of the triangular faces of the square pyramid (I.E. every point of one face overlaps a point of the other face, edges and vertexes included) how many faces will the new solid have?

Edit, let me clarify: this only has to do with geometry. No lateral-thinking, no word puns, no silly explanations (otherwise I would have added some of these tags), just pure and simple geometry. Yes, it may look stupid, but it isn't.

Inspired by "How many faces does the resulting polyhedron have?" on Math SE.

• I downvoted this at first because it didn't seem to be much of a puzzle, rather just an exercise in visualisation. But what happens with the faces labelled 1 and 4 in Rubio's answer is interesting enough that I've reverted my downvote. Nov 15, 2016 at 12:46
• There's an implicit assumption here that if two faces share a common edge and lie in a common plane, we should regard them as the same face. But this ought to be specified rather than left implicit. It is by no means clear that the number arrived at after deleting "removable" edges is the universal proper answer to the question of "how many faces" there are. Nov 15, 2016 at 14:53
• +1 for a nice puzzle. This was in the Arthur C. Clarke book Ghost from the Grand Banks, used as a plot device for the protagonists to recognise their daughter's mathematical brilliance. Nov 15, 2016 at 16:53
• I'm pretty sure this was an SAT question that got reported lol Nov 15, 2016 at 17:50
• Carl, the issue was obvious to me and so it didn't occur to me that I might be spoiling anything. To me the work in figuring out the answer to the puzzle is in establishing that the angle at the join is in fact 180 degrees. If there were a way to spoiler the content of my comment in the same way parts of answers may be spoilered, then I would do it, but it doesn't appear that this can be done. I stand by the content of my comment and do not propose to delete it. Nov 15, 2016 at 22:55

5

I can't draw this, but

if the square pyramid sits on its base
and has its four triangular faces oriented N S E W,
and you set the tetrahedron to the east of the pyramid on its base,
with one side facing full west - call this face 1,
leaving face 2 pointing roughly NE and face 3 facing roughly SE,
and face 4 the face it sits on the ground on...

now tip the tetrahedron over to rest its face 1 against the E face of the pyramid, and glue it like that. you'll find that the edge between its faces 2 and 3 is now parallel to the ground, extending due east from the point of the pyramid. face 2 is perfectly aligned with pyramid's N face, and face 3 with the pyramid's S face. face 4 is now off the ground and doesn't align with anything in particular.

faces of the conjoined figure are now:
1: pyramid W
2: pyramid N + tetrahedron 2
3: pyramid S + tetrahedron 3
4: pyramid base
5: tetrahedron 4

with pyramid E and tetrahedron 1 glued and no longer externally visible as faces at all.

(Removed the OP-suggested image I had here. It wasn't accurate, and one commenter plus two other answers in the thread have posted much better images.)

• I thougt of it for a moment but as @matsmath, I couldn't imagine it and thougt it wasnt the same plane Nov 15, 2016 at 9:43
• The image that the OP provided really does not make a case for the two adjacent faces being coplanar. As is shown here, the tet faces should have an X component, where X is an axis parallel to the front and rear sides of the pyramid's base. This component valishes ahen the ridge is parallel to X, which is what actually happens and what is described in the answer. See here for an alternative representation. Nov 15, 2016 at 10:40
• The image is inaccurate, in that faces 2 and 4 of the compound solid are actually rhombuses, so the top edge should be parallel to (and equally long as) the front and back edges of the square base. Nov 15, 2016 at 13:15
• I googled "tetrahedron and square pyramid" and found this image... if you can provide a better visualization then you're welcome, that would be great Nov 15, 2016 at 16:44
• For those who are still having trouble visualizing this, this answer on Mathematics might help. Nov 15, 2016 at 19:14

I know the answer is already given but I'd like to show an easy explanation of why the 2 planes are coplanar. Take this image:

Consider two pyramids sitting side by side, and draw a line between their tops. This line must be of unit length, because it is the same length as the line joining the midpoints of the bases of the pyramids. It can now be seen that the space between the pyramids, which is bounded by four equilateral triangles, is identical to the unit tetrahedron. The tetrahedron will therefore fit perfectly into this space, making a five-sided prism. Each exposed face of the tetrahedron must therefore lie in the same plane as the two adjacent faces of the pyramids, which are obviously coplanar with each other by symmetry.

• Mmm... makes me think of Toblerone in the box and out of it :-) Nov 15, 2016 at 13:02
• I think people are upvoting Toblerone :9
– Rubio
Nov 15, 2016 at 17:03
• You're looking at a piece of the tetrahedral-octahedral tiling: en.wikipedia.org/wiki/Tetrahedral-octahedral_honeycomb Nov 15, 2016 at 18:27
• yay, got my first gold badge with this :) Populist: Highest scoring answer that outscored an accepted answer with score of more than 10 by more than 2x.
– Ivo
Nov 16, 2016 at 7:00
• The new (UK) Toblerone... naj, surely not. Too few gaps... @TheVee Nov 16, 2016 at 7:06

When I started this problem, this is how I first assumed you meant to answer it. When you lay the two faces together, there are two options for which direction we can orient the tetrahedron, keeping the square based pyramid fixed. I saw it as letting the tetrahedron be on the inside of the pyramid and let it poke out of the bottom, which will give us all 4 of the triangle faces of the pyramid and one weird face on the bottom with 3 more from the tetrahedron. That is 8 faces. Just thought I would give a weird other option that fits what you asked for, even though the solids have intersection.

And the view from the bottom:

And the one the OP was thinking of:

• I didn't actually ever thought of the first one... cool. I was assuming the solids were.. you know... solid (i.e. you cannot intersect them). The second image displays the solution really well. Nov 15, 2016 at 17:27
• Gee -- I assumed that this was the answer OP had in mind. This is a famous problem from the USA's SAT (Scholastic Aptitude Test). After the problem had been on the advanced math exam for a few years, one student found the alternate solution, challenged the question, and received credit for the question after the fact. It's one of the very few cases where a "wrong" answer has been globally corrected in the history of the Educational Testing Service. Nov 15, 2016 at 19:13
• @Prune, which was the alternate solution? (Which solution did the SAT test expect?) Nov 16, 2016 at 5:29
• The SAT listed answers 5 through 8 and something else. They had 5 as the only acceptable answer. Nov 16, 2016 at 6:52

5

Because

The answer really should be five, because the two pairs of adjacent triangular faces in the combined solid are actually coplanar, and therefore make two rhombus-shaped faces rather than four triangular faces.

• I beat you there by under 30 seconds. :) Great minds think alike!
– Rubio
Nov 15, 2016 at 9:54
• This is a very nice written explanation. I specifically like the use of the word "coplanar".
– josh
Nov 15, 2016 at 10:38