TITLE: Ten and eleven sided polygons with circles.
QUESTION [6 upvotes]: I am having trouble with the question below: 
Given a not necessarily convex 10 sided polygon, draw circles with its sides as diameters. Is it possible that all these circles pass through a point which is not a vertex of this 10 sided polygon?
Do the same for an 11 sided polygon. 
I've tried simply drawing sketches of various situations and am currently thinking that it is not possible

REPLY [7 votes]: Given points $A$ and $B$, it is a well-known fact that the points $P$ for which $\angle APB$ is $90$ degrees form a circle with $AB$ as its diameter.
The question asks for a decagon such that the circles drawn around each edge (using that edge as its diameter) all intersect in one point. If that point is called $P$, then we know that $\angle A_iPA_{i+1}$ is $90$ degrees for any pair of successive vertices $A_i$ and $A_{i+1}$.
Lets put the origin at $P$, and have the x-axis going through the first vertex $A_1$. The segment $PA_1$ therefore goes along the x-axis. The next segment, $PA_2$, is perpendicular to $PA_1$ and goes through the origin so must go along the y-axis. The next, $PA_3$, is perpendicular to $PA_2$ so goes along the x-axis again, and so on. Therefore, the vertices $A_1$, $A_3$, $A_5$, $A_7$, and $A_9$ lie on the x-axis and the vertices with the even indices lie on the y-axis.
Here is one example:

The question then asks about an 11-gon. It would seem impossible to make the construction work for any polygon with an odd number of vertices, because the vertices must alternately lie on the x and the y axis. The last vertex, $A_{11}$, therefore lies on the x-axis just like $A_1$, and so does the edge connecting them. The only way for the circle around that edge to go through the origin is if $A_{11}$ actually lies at the origin.
And this actually works. By putting $A_{11}$ at the origin, it lies on both axes, and because of this ambiguity it can be considered to be on a different axis to either of its neighbours, even though one neighbour is on the x axis and the other on the y axis. It is impossible however if we require the intersection point of the circles to be distinct from the vertices of the polygon.