Unique questions about pentagon Consider the (non-regular) pentagon with consecutive vertices at (-1,-1), (-1,1), (0,2), (1,1), and (1,-1).
a) Prove that there is no circle that is tangent to all 5 sides of the pentagon
b) Is there an ellipse that is tangent to all five sides of the pentagon? If so, find its equation.
Here is what I have done so far:
I have drawn out the pentagon. For part a) I am very bad at proving. For part b) I have some idea. I drew the ellipse and tried to match the ellipse inside of the pentagon and tried to see its tangency. My best idea is to take the derivative. I may be completely off from the true solution, I am not sure. Please click the solution button below and tell me the correct solution. This is a question that is not for homework, so if you posted the solution there would be no issue.
 A: Hint: consider ellipses with centre on the $y$ axis and axes parallel to the coordinate axes, tangent to the vertical and horizontal sides.  If the major axis of the ellipse is too small, it will look like this:

If too large, like this:

You want it just right so it is tangent to the slanted sides.

A: If a polygon is circumscribed to a circle, then the bisectors of its internal angles all meet at a same point. Hence, for part a) just show that this is not the case.
For b) it is a general theorem that there always exists a conic tangent to five given lines. In particular, if the lines form a convex pentagon then the conic is an ellipse.
There is a simple geometric construction for the ellipse inscribed in a given convex pentagon $ABCDE$. One can, first of all, find tangency points $PQRST$ (see figure below). Draw, for instance, diagonals $AC$ and $BD$, meeting at $F=(1/2,1/2)$. Line $EF$ meets then side $BC$ at tangency point $P=(1,1/3)$.
This construction depends on Brianchon's theorem:

The three opposite diagonals of every hexagon circumscribing a conic
are concurrent.

In fact, if $P$ is the tangency point on $BC$, we can view $ABPCDE$ as a limiting case of a hexagon circumscribed to the ellipse. Hence diagonals $AC$, $BD$ and $EP$ must intersect at the same point $F$.
For a generic pentagon one would go on finding the other tangency points and remembering then that the line, passing through the intersection point of two tangents to an ellipse and through the midpoint of their tangency points, also passes through the center of the ellipse. Once the center is found one can construct a pair of conjugate diameters and finish then the construction by standard techniques.
In this particular case the construction is simpler, as by symmetry we know that major axis of the ellipse lies on $y$-axis and $PR$, parallel to $x$-axis, is minor axis. Hence semi-minor axis $PO$ has a length $b=1$ and semi-major axis $QO$ has a length $a=4/3$. Focal distance is then $c=\sqrt{a^2-b^2}=\sqrt7/3$ and from that we can find foci $L$ and $M$.

A: For a), note that the diameter formed by the points of tangency of the vertical sides, must be horizontal, and therefore must be 2. In other words, the radius must be 1. This implies that the center of the circumference must have an X coordinate equal to 0. But then for it to be tangent to the horizontal side, it must have Y coordinate equal to 0. So the center is (0,0). Prove that that is impossible so that it is tangent to the crooked sides.
For b) use a general formula and try to clear the parameters, knowing that the foci must have an x ​​coordinate equal to 0.
