- #1
Robin04
- 260
- 16
- 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...
##\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...