Author |
Topic: How many faces? (Read 1448 times) |
|
rmsgrey
Uberpuzzler
    
Gender: 
Posts: 2873
|
 |
How many faces?
« on: Jan 24th, 2004, 9:40am » |
 Quote  Modify
|
1) Take two regular tetrahedra of unit side length and glue them together so that two faces coincide. How many faces does the resulting solid have?
2) Take a regular tetrahedron and square based pyramid, with all sides unit length and glue two triangular faces together so that they coincide exactly. How many faces does the resulting solid have?
|
|
 IP Logged |
|
|
|
rmsgrey
Uberpuzzler
Gender:
Posts: 2873
|
|
Re: How many faces?
« Reply #2 on: Jan 24th, 2004, 11:07am » |
Quote Modify
|
3) Would I bother asking two almost identical questions if the answers were also almost identical?
|
|
IP Logged |
|
|
|
Sir Col
Uberpuzzler
impudens simia et macrologus profundus fabulae
Gender:
Posts: 1825
|
|
Re: How many faces?
« Reply #3 on: Jan 24th, 2004, 5:35pm » |
Quote Modify
|
on Jan 24th, 2004, 11:07am, rmsgrey wrote:| 3) Would I bother asking two almost identical questions if the answers were also almost identical? |
|
That's because they are almost identical, but not the way you'd expect...
I had to make a paper model to confirm that the square base pyramid and the triangle base pyramid form a pentahedron. Of course, once I did this, the reason is obvious!
Great puzzle, rmsgrey.
|
| « Last Edit: Jan 24th, 2004, 5:40pm by Sir Col » |
IP Logged |
mathschallenge.net / projecteuler.net
|
|
|
rmsgrey
Uberpuzzler
Gender:
Posts: 2873
|
|
Re: How many faces?
« Reply #4 on: Jan 24th, 2004, 7:14pm » |
Quote Modify
|
OK, so the actual numerical answers are pretty similar (both positive integers less than 10) but the solution to 2) has a significant difference.
::If you can't be bothered to actually make a paper model - or consider it insufficiently convincing ("Yeah, sure, it looks like the two faces are coplanar, but they could be slightly off"), an easy way to see what happens is to consider two adjacent square based pyramids with unit edges resting on their square faces on a plane, with a common edge. If you then add a straight line connected the peaks of the pyramids, it obviously has length 1 so completes the skeleton of a regular tetrahedron sharing two faces and five edges with the two pyramids. The two new faces, each sharing three vertices with one of the co-planar pairs of triangles mentioned earlier are obviously also co-planar with the pair. Removing one of the pyramids then gives the exact situation of 2)::
I encountered this puzzle in Arthur C Clarke's "The Ghost from the Grand Banks", in which he credits it to an actual real life US school math paper, in which the examiners got the obvious wrong answer (so towr is in good company)
|
|
IP Logged |
|
|
|
Sir Col
Uberpuzzler
impudens simia et macrologus profundus fabulae
Gender:
Posts: 1825
|
|
Re: How many faces?
« Reply #5 on: Jan 25th, 2004, 2:23am » |
Quote Modify
|
Sorry, when I said the reason is obvious, I didn't mean because the faces looked coplanar, I had verified it; although I did it differently. I started by calculating the angles between the faces in each solid, the tetrahedron was straight-forward, but I was battling with the angles between the triangle faces on the square base pyramid, then I realised... duh! The square base pyramid is one half of a regular octahedron. To save time and effort, I consulted my copy of Mathematical Models, by Cundy and Rollet, and I was able to obtain the dihedral angles between the triangle faces on the square base pyramid (109o 28'), and the dihedral angle for the tetrahedron (70o 32'). As these add to 180o, the faces are coplanar.
I prefer your visual method, though. Nice!
[e]Posted edited to correct dihedral angle for tetrahedron[/e]
|
| « Last Edit: Jan 25th, 2004, 4:15am by Sir Col » |
IP Logged |
mathschallenge.net / projecteuler.net
|
|
|
rmsgrey
Uberpuzzler
Gender:
Posts: 2873
|
|
Re: How many faces?
« Reply #6 on: Jan 25th, 2004, 3:24am » |
Quote Modify
|
Since I'm not entirely sure how to calculate dihedral angles in the first place, and probably wouldn't want to bother even if I did, I spent quite some time playing with mental visualisations, trying to find a sufficiently intuitively obvious proof.
By the way, I think there's a typo in your figure for the dihedral angle of a tetrahedron - spot research gives a value of about 70.53o - which brings the sum of the two angles back down from 182o...
|
|
IP Logged |
|
|
|
Sir Col
Uberpuzzler
impudens simia et macrologus profundus fabulae
Gender:
Posts: 1825
|
Oops! Good spot. It should have been 70o 32' (in degrees and seconds).
The dihedral angles are quite easy. Consider a regular tetrahdron (see diagram below):
BC = sqrt(3)/2.
AB = (1/3)BC = 1/(2sqrt(3))
cos(B) = (1/(2sqrt(3)))/(sqrt(3)/2) = 1/3 [bigto] B ~= 70o 32'.
The octahedron is found in the following way. The dihedral angle, Y, is twice the angle between the apex of the square base pyramid, the centre of one edge and the centre of the square base. Let this angle be X.
cos(X) = (1/2)/(sqrt(3)/2) = 1/sqrt(3)
cos(Y) = cos(2X) = 2cos2(X)–1 = -1/3 [bigto] Y ~= 109o 28'.
Now we hope to prove that B+Y = 180o.
sin(B) = sqrt(1–cos2(B)) = 2sqrt(2)/3
Considering angles carefully, sin(Y) = 2sqrt(2)/3
sin(B+Y) = sin(B)cos(Y)+cos(B)sin(Y) = (2sqrt(2)/3)(-1/3)+(2sqrt(2)/3)(1/3) = 0
cos(B+Y) = cos(B)cos(Y)–sin(B)sin(Y) = (1/3)(-1/3)–(2sqrt(2)/3)(2sqrt(2)/3) = -1
Hence, B+Y = 180o.
|
| « Last Edit: Jan 25th, 2004, 4:19am by Sir Col » |
IP Logged |
mathschallenge.net / projecteuler.net
|
|
|
towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730
|
|
Re: How many faces?
« Reply #8 on: Jan 25th, 2004, 7:59am » |
Quote Modify
|
hmm.. I should have realized I made an error after I completed the other puzzle you posted (with the two halves of a regular tetrahedron)
But I only tried to visualize it and I based on that thought they weren't coplanar..
I should know better than to trust my visualizations.. And also I should have picked up on 3) ...
|
|
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
rmsgrey
Uberpuzzler
Gender:
Posts: 2873
|
|
Re: How many faces?
« Reply #9 on: Jan 25th, 2004, 10:21am » |
Quote Modify
|
A much simpler description for the half-tetrahedra, which I didn't use for obvious reasons, is that each is formed of a square pyramid with two regular tetrahedra joined to non-adjacent triangular faces.
|
|
IP Logged |
|
|
|
SWF
Uberpuzzler
Posts: 879
|
|
Re: How many faces?
« Reply #10 on: Jan 25th, 2004, 6:11pm » |
Quote Modify
|
Another easy way to see this (Edit: I see towr refers to another riddle that relates to this): start with a regular tetrahedron. Cut off the 4 corners to remove 4 smaller tetrahedra with half the side length of the orginal tetrahedron. What remains is a regular octahedron (two square base pyramids). Four of the octahedron's faces were originally flush with the faces of the large tetrahedron.
I think the math exam with this question which rmsgrey mentioned was the SAT test. Back when they first started revealing the "correct" solutions to the SAT, this was the first instance of a question that had been graded incorrectly.
|
| « Last Edit: Jan 25th, 2004, 6:13pm by SWF » |
IP Logged |
|
|
|
Barukh
Uberpuzzler
Gender:
Posts: 2276
|
I thought some visualization wouldn't hurt, ah? 
Very nice puzzle, rmsgrey!
|
|
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print