Visualising regular pyramids All but the last question in this are rhetorical and not the main question I'm asking
I recently had a high school maths lesson about frustums and it got me thinking about ways of viewing pyramids. And, while imagining up various views of square based pyramids, I found something odd. Which them seemed to work for pentagonal based pyramids. And so I would like a bit of confirmation about this theory.
The theory
Imagine a regular, square based pyramid in such a way that you can rotate it in every $3$-dimensional plane of rotation.
Now, rotate it so that you can only see $3$ of the triangular faces. The base doesn't count. Can you?


*

*It is fairly easy to see $1$ face; look at it face on.

*$2$ faces is also quite basic; look at a corner.

*For $4$ faces, simply look at a plan view of the pyramid.
But how do you see $3$ faces?
Now picture a pentagonal based pyramid in the same position. Can you rotate it in any way to see $4$ of the triangular faces?
My theory is that given a regular polygonal based pyramid where the number of sides of the base is $n$, it is impossible to see $n-1$ triangular faces as long as $n > 3$
(Actual question)
Is this true? I'm looking for a mathematical proof to this as I can't simply imagine up a $200000$ faced pyramid and test this but I can apply a mathematical formula to it.
However! As you may have noticed above, I am in high school and so please keep your proofs fairly basic. If this is impossible, please offer a detailed explanation of how your proof works. I don't think that it is too unreasonable to say that if I can't understand your answer, I can't accept it.
 A: This is not true - you can see only three of the four triangular faces of a regular square pyramid. For example:

I got that image as a screenshot from this video since I didn't have a regular square pyramid nearby. 

As Ross Millikan pointed out in the comments, it is also not true for a pyramid with  $n$ triangular faces and a regular $n$-gon as the base. 
The method of starting with a point of view from which you can see all $n$ triangular faces, then tilting the pyramid until the furthest face is no longer visible, will still allow you to see exactly $n-1$ triangular faces. 

This is not a rigorous proof - in order to write on you would probably need to formally section $\mathbb{R}^3$ into $n$ different subsets based on which points can see each of the $n$ faces, then show that there exists a collection of $n-1$ of these "visibility" sets whose intersection is nonempty. 
For me, the non-rigorous reasoning is enough to convince me that the statement isn't true, so I am not going to spend the time to try to write a rigorous disproof. 
