Geometry 3D- number of faces

[Annelle]

In a certain convex polyhedron, each face is a tetrahedron or hexagon and three edges meet at each vertex. How many tetrahedral faces does the polyhedron have?
I'd be grateful for help

 

[Dr.Peterson]

In a certain convex polyhedron, each face is a tetrahedron or hexagon and three edges meet at each vertex. How many tetrahedral faces does the polyhedron have?
I'd be grateful for help

I think you miscopied or mistranslated. Faces of a polyhedron are polygons, not polyhedra (such as a tetrahedron).

My guess is that you meant either triangle or quadrilateral.

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[Annelle]

I mistranslated indeed. I mean each face is quadrilateral or hexagon
I thought about doing it in 4 cases- with all hexagons in each corner (which is impossible so i skip it), with all quadrilaterals in each corner, with 2 quadrilaterals and one hexagon and with one quadrilateral and 2 hexagons, but actually i dont know what to do next and how to do that. I can't use euler formula for polyhedra ( V + F - E = 2 ), because i dont know any value. I only drew the case with 2 hexagons and one quadrilateral in each corner

 

[Dr.Peterson]

Use the Euler formula, together with what you were told: "each face is a quadrilateral or hexagon and three edges meet at each vertex". That tells you a second relationship between V and E in the formula.

I might define two variables, Q and H, where Q + H = F in the formula. What else can you say about them?

 

[Annelle]

three edges meet at each vertex tells that 3E=2V
There can be 3 options:
- F=Q
- Q=2H, so F=3H
- H=2Q, so F=3Q
but it is still not enough

 

[Dr.Peterson]

three edges meet at each vertex tells that 3E=2V
There can be 3 options:
- F=Q
- Q=2H, so F=3H
- H=2Q, so F=3Q
but it is still not enough

So look for more! You haven't used the fact that Q is the number 4-sided faces and H is the number of 6-sided faces. Write an equation for that.

Also, are you sure about 3E = 2V?

I get four equations in 5 variables (V, E, F, Q, H), and end up with a family of answers, one of which is the cube (Q = 6, H = 0).

 

[Annelle]

Also, are you sure about 3E = 2V?

I think it's true because 3 edges meet at each corner, but every edge velongs to 2 faces, so it was counted two times. Am I wrong?

Is it good to consider 4 cases: only Q faces, two Q faces and one H face in each corner, 2 H faces and one Q at each corner and only H faces?

I thought that the number of edges would be like that:
(4Q+6H)/2=E
because we have 4 edges for every Q face and 6 for every H face and we count every edge 2 times, am I right?

There would be:
V+F-E=2
2Q+3H=E
3E=2V
Q+H=F

so V=3*(2Q+3H)/2

I think I do something wrong, because I have an equation with 2 unknown elements:Q and H: 2Q+2,5H=2 and it's wrong because a cube has 6 faces, not 10...

 

[Dr.Peterson]

I think it's true because 3 edges meet at each corner, but every edge velongs to 2 faces, so it was counted two times. Am I wrong?

I mentioned one example, the cube. Three edges meet at every vertex. Is it true that 3E = 2V? Since E = 12 and V = 8, 3E = 36 while 2V = 16.

Maybe you got something backward?

Is it good to consider 4 cases: only Q faces, two Q faces and one H face in each corner, 2 H faces and one Q at each corner and only H faces?

That's far more work, and more risky, than what you will be able to do once you have the right equations.

I thought that the number of edges would be like that:
(4Q+6H)/2=E
because we have 4 edges for every Q face and 6 for every H face and we count every edge 2 times, am I right?

Yes, that will be one of your equations. (And you can apply the same sort of thinking to correct 3E = 2V.)

Fix the one equation and try again! You'll get there.

 

[Annelle]

noo, i know where I made a mistake, it should be 2E=3V and then we have Q=6 and that's the correct answer

 

[Dr.Peterson]

noo, i know where I made a mistake, it should be 2E=3V and then we have Q=6 and that's the correct answer

Good. You caught it.

Now, it's interesting that you were only asked about the number of quadrilaterals. How many hexagons can it have?

 

[Annelle]

I think that the number of hexagon can be different
Thank you a lot for help