How can I calculate if a given point is wrapped inside a pentagon? If I have a pentagon and I know the coordinates of it's nodes, how do I calculate if a point is wrapped inside it?
An example to clarify what I mean: 
Assume that we know the coordinates of the points a,b,c,d,e in the pentagon below. How can I calculate if the point α is actually inside the shape?

I can calculate if a point is wrapped inside a square (nodes: a,b,c,d) by creating a boolean expression:

(α.x >= a.x AND α.x <= b.x AND α.y >= a.y AND α.y <= c.y)
 A: Let $\alpha$ be located at $(x_\alpha,y_\alpha)$. Let $AB(x)$ be a function denoting the line connecting the points $A$ and $B$. Then we know that $\alpha$ is below the line if $AB(x_\alpha)>y_\alpha$. Use this for the other $4$ lines and see if you can create another boolean expression. 
A: Choose an arbitrary point in the pentagon (say point $a$) and an arbitrary direction $\hat{u}$ not parallel to any of the five sides nor to any of the five lines from $\alpha$ to a vertex. Calculate the distances from A to each of the other points in the pentagon; let the maximum of those four distances be $\mu$. Choose point $\Omega$ be starting at $a$ and moving by $2 \times 5 \mu \hat{u}$; it is easy to show that 
$\Omega$ must lie outside the pentagon.  
Now draw line the segment from $\alpha$ to $\Omega$ and test it against each line segment making up a side of the pentagon:  The test, involving two line segments whose extended lines meet in a point $P$, is to determine if $P$ lies on both line segments (pass) or outside at least one of the segments (fail). Having tested against each of the 5 sides, there are five results.  $\alpha$ is in the interior of the pentagon if and only if an odd number of the tests have passed.
This method works for any polygon. 
