## Archive for August, 2014

### Sydney v Richmond

August 27, 2014

You are the captain of Sydney. The phone rings and you pick it up. Trent Cotchin is on the line.

“G’day Trent, what’s up?”

“I’m calling to discuss our game on this weekend. It seems that there’s an opportunity for us both to arrange things to benefit both of us.”

“Um, how?”

“Well what do you guys want to get out of this weekend’s match? The minor premiership, for sure, but even more importantly you want none of your players to get injured or suspended. Agree?”

“Yes”

“Well listen, I have a way to guarantee that you get all of the above. Interested?”

“Um, I’m not sure how you can guarantee that we don’t have any injuries.”

“All I ask for in return is that you let us win. Only by a small enough margin that you still keep top spot from the Hawks on percentage. But you see, we do really need the win more than you guys this time so we can make the eight.”

Trent pauses to take a breath, and continues

“Here’s how I propose we work things out. Come Saturday’s game, when the ball is bounced you allow us with no contest to score a couple of goals. Then both teams agree to sit down for the rest of the game and do nothing. We both get what we want out of the game without having to take any risks. Do you agree to this proposal?

(Also, if you accept, what will be the response of the rest of the world to such a highly contrived outcome?)

### Concyclicity in the Euclidean Plane

August 21, 2014

I want to address the question of how to determine whether or not four points are concyclic. In a previous post, this question reared its head and was solved by an introduction of the circumcentre of three of the points. In this post, I present a direct approach.

We prove the following:

Let $(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4,y_4)$ be four points in $\mathbb{R}^2$. Then these four points are concyclic if and only if

$\displaystyle \left| \begin{array}{cccc} x_1^2+y_1^2 & x_1 & y_1 & 1 \\ x_2^2+y_2^2 & x_2 & y_2 & 1 \\ x_3^2+y_3^2 & x_3 & y_3 & 1 \\ x_4^2+y_4^2 & x_4 & y_4 & 1 \end{array} \right|=0.$

For the proof, first suppose that the points $(x_i,y_i)$ all lie on the curve

$\displaystyle (x-x_0)^2+(y-y_0)^2=r^2.$

The four equations we obtain give a linear dependence of the columns in the $4\times 4$ matrix under consideration, hence its determinant is zero.

Conversely, suppose that the determinant in question is zero. Then there is a nonzero vector in the nullspace of our $4\times 4$ matrix. If its first entry is nonzero, we can normalise it to be of the form $(1,-2x_0,-2y_0,x_0^2+y_0^2-r^2)^T$ and we obtain the circle that the four points lie on. If the first entry of this vector is equal to zero, then the four points must lie on a line, which after is all is just a degenerate circle, QED.