I think you miscopied or mistranslated. Faces of a polyhedron are polygons, not polyhedra (such as a tetrahedron).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
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.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
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?Also, are you sure about 3E = 2V?
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.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?
That's far more work, and more risky, than what you will be able to do once you have the right equations.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?
Yes, that will be one of your equations. (And you can apply the same sort of thinking to correct 3E = 2V.)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?
Good. You caught it.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