Probability - Camera in a parallelogram shaped room

[Robin04]

Homework Statement

We place a camera on the floor of a parallelogram shaped two dimensional room. What is the probability that the camera sees at least three sides of the room at an angle less or equal to 90 degrees? The camera has an equal chance to be placed anywhere in the room. How does this depend on the sides of the room and the angles between them? Can we place the camera so that it sees every sides at an angle less or equal to 90 degrees? What is its probability?

Relevant Equations

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I tried to mess around with some equations in Mathematica with more or less success. Let's call $a,b$ the two sides and $\delta$ the angle between them. I contructed vectors that point to the corners of the room.
$\vec{a} = (0, 0)$
$\vec{b} =(a, 0)$
$\vec{c}=(a+b \cos{\delta}, b \sin{\delta})$
$\vec{d}=(b \cos{\delta}, b \sin{\delta})$
To express the angle the camera sees between the corners I can say (between corners a and b):
$(\vec{r}-\vec{a})\cdot (\vec{r}-\vec{b})=x^2-ax+y^2$ This has to be $\geq 0$
I can do this for all sides but the equations get more complicated. I don't think this is the right way to go...

 

[phinds]

So, you don't think the camera's field of focus matters?

 

[Robin04]

So, you don't think the camera's field of focus matters?

What do you mean by that? Angle of sight? It's not given so I think it doesn't matter. It's just a "simple" probability problem. Determining the area where the sides can and can't be seen from the given angles would solve the problem I think. It should be just pure geometry.

 

[phinds]

What do you mean by that? Angle of sight? It's not given so I think it doesn't matter. It's just a "simple" probability problem. Determining the area where the sides can and can't be seen from the given angles would solve the problem I think. It should be just pure geometry.

So the answer will be the same whether the camera has a very narrow field of focus or a very wide field of focus? I don't think so, unless you don't mean all three walls at the same time but rather just having the camera point at one wall at a time, in which case my caveat does not apply.

I think you need to draw a diagram of just what it is you mean.

 

[Robin04]

So the answer will be the same whether the camera has a very narrow field of focus or a very wide field of focus? I don't think so, unless you don't mean all three walls at the same time but rather just having the camera point at one wall at a time, in which case my caveat does not apply.

I think you need to draw a diagram of just what it is you mean.

The way I interpreted the problem is that the camera's properties do not play any role. It could be regarded as just a random point inside the parallelogram from which we have to draw lines to the corners and measure the angles between those lines. Calling it a camera is only for making it more illustrative.

 

[Robin04]

Maybe my bad English made the problem a bit confusing, hope this makes it clear.
In this draw scenario only two of the angles that the camera sees are acute.

 

[SammyS]

Problem Statement: We place a camera on the floor of a parallelogram shaped two dimensional room. What is the probability that the camera sees at least three sides of the room at an angle less or equal to 90 degrees? The camera has an equal chance to be placed anywhere in the room. How does this depend on the sides of the room and the angles between them? Can we place the camera so that it sees every sides at an angle less or equal to 90 degrees? What is its probability?

What is a 2 dimensional room? What is the floor of such a room? (OK. Maybe post #6 helps clarify those issues.)

Added in Edit:
(This does remind me of the book Flatland. That leads to the observation: There appears to be no door.)

 

[Robin04]

What is a 2 dimensional room? What is the floor of such a room? (OK. Maybe post #6 helps clarify those issues.)

Yes, my teacher doesn't like to be precise... I already got used to it

 

[jim mcnamara]

Trying to get this one into my head... Parallelograms have the following properties of interest:
The sum of the interior angles = 360°. Acute angles are defined to have less than 90 ° So, there can be at most 2 acute interior angles, or zero acute angles. No other possibilities. Right? So what angle are we actually talking about? Corners (the interior angles)? The camera's field of vision (undefined as far as I can tell) ?

If you copied the question from a book or worksheet, then the problem needs a tune up. Or maybe the drawing should show a line from the camera to the midpoint of each side.

I am confused - as you can tell. Maybe someone else will see more than I do. @fresh_42 ?

 

[LCKurtz]

Here's my take. Using the figure in post #6, look at the 4 central angles from the camera to the corners. If any of those are acute, you can expand the angle to 90 degrees in such a way that the angle includes parts of the neighboring two sides. That means three sides will be in the 90 degree field of view if the camera is aimed properly. All four angles can't be obtuse. The extreme case is where all four angles are 90 degrees, like at the center of the square. Then it can't see 3 walls. So I figure the probability is 1 that it can see 3 walls.
Edit: There is a non-zero probability that I am mis-interpreting the problem.

 

[Robin04]

I updated the figure. $\alpha_i$ are the angles I'm talking about. Trivially $\sum{\alpha_i} = 360^\circ$, but some of them are acute, some of them are obtuse angles depending on 1) where the point is inside the parallelogram, 2) what the parameters of the parallalogram are $(a, b, \delta)$.

The first part of the problem is to find the set of points inside the parallelogram where at least 3 $\alpha_i$ are acute or right angles (where $\alpha_i \leq 90^\circ$). The only case when there are four such $\alpha_i$ is when $a=b$.

The second part is just simply about calculating the probability of finding such a point if we randomly choose a point from inside the parallelogram.

Forget all the camera, 2d room stuff, my teacher has some weirdly phrased exercises, but this is the mathematical problem I'm dealing with.