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

  • 2
    $\begingroup$ If A and B are opposite on a circle, so that AB is the circle's diameter, and C is another point on that circle, what do you know of the angle ACB? $\endgroup$ – Jaap Scherphuis May 4 '17 at 11:32
  • $\begingroup$ ohhh 90 degrees, I am bit confused as to how that helps though $\endgroup$ – hyul May 4 '17 at 11:37
  • $\begingroup$ So now try to construct such a polygon. Put this supposed intersection point of the circles at the origin, put the first vertex somewhere on the x axis for example, and then figure out where you could put each of the other vertices in turn. $\endgroup$ – Jaap Scherphuis May 4 '17 at 11:41
  • $\begingroup$ Ok I have tried some more sketches but I'm lost $\endgroup$ – hyul May 4 '17 at 12:00
  • $\begingroup$ If you draw the first vertex A on the x-axis, where can you put the second vertex B, such that the angle AOB is 90 degrees? $\endgroup$ – Jaap Scherphuis May 4 '17 at 12:04

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: enter image description here

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.

  • $\begingroup$ Oh that's nice. +1 I wonder if the circles can be so arranged to make a prettier, more symmetric figure. $\endgroup$ – Brian Tung May 5 '17 at 18:45

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