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Let $P_0, P_1, P_2$ be three points on the circumference of a circle with radius $1$, where $P_1P_2 = t < 2$. For each $i \ge 3$, define $P_i$ to be the centre of the circumcircle of $\triangle P_{i−1}P_{i−2}P_{i−3}$.

  1. Prove that the points $P_1, P_5, P_9, P_13, \ldots$ are collinear.
  2. Let $x$ be the distance from $P_1$ to $P_{1001}$, and let $y$ be the distance from $P_{1001}$ to $P_{2001}$. Determine all values of $t$ for which $\sqrt[500]{\frac{x}{y}}$ is an integer.

This question is from the 2001 Canada National Olympiad.

Is there a cleaner way of solving the problem, using e.g. vectors, complex numbers, or some theorem, so that one does not need to take cues from such an accurate diagram?


The diagram below (from GeoGebra) shows an example of a possible (converging) sequence of points. (It is possible for the points to diverge if $t$ is nearly zero or nearly diametral.)

IteratedCircumcenters

Note that every triangle other than possibly the initial one is isosceles, i.e. $$P_{i-2}P_{i}=P_{i-1}P_{i},\qquad i\ge 3 \tag{1}$$

Since $P_i$ (for $i\ge3$) is equidistant from all of $P_{i−1},P_{i−2},P_{i−3}$, then $P_i$ must be on the perpendicular bisector of $P_{i−1}P_{i−2}$ and $P_{i−2}P_{i−3}$.
So, if $\angle{P_{i-2}P_{i-1}P_{i}}=\alpha,\angle{P_{i-1}P_{i}P_{i+1}}=\beta$ (as labeled for the $i=3$ example), then $$\beta=\frac{\pi}{2}-\alpha \tag{2}$$ This relationship can be applied inductively to all larger $i$ to show that: $$\triangle{P_{i-2}P_{i-1}P_{i}} \sim \triangle{P_{i}P_{i+1}P_{i+2}},\qquad i\ge 3 \tag{3}$$ and that $$\triangle{P_{i-2}P_{i-1}P_{i}} \perp \triangle{P_{i}P_{i+1}P_{i+2}},\qquad i\ge 3 \tag{4}$$ Also because triangles turn through $90^\circ$ every two point, then side $P_{i-2}P_{i-1}$ is anti-parallel to side $P_{i+2}P_{i+3}$, i.e. $$P_{i-2}P_{i-1} \parallel P_{i+3}P_{i+2},\qquad \tag{5}$$

As before, let $\angle{P_{i-2}P_{i-1}P_{i}}=\alpha,\angle{P_{i-1}P_{i}P_{i+1}}=\beta$, with $i\ge3$ odd, and let distance $P_{i-2}P_{i-1}=x$. Then $$\begin{align} P_{i-1}P_{i}&=\frac{1}{2}x\sec\alpha \\[1em] P_{i}P_{i+1}=\frac{1}{2}\left(\frac{1}{2}x\sec\alpha\right)\sec{\beta}&=\frac{1}{4}x\,\sec\alpha\,\mathrm{cosec}\,\alpha \\ \end{align}$$ so $$\frac{P_{i-2}P_{i-1}}{P_{i}P_{i+1}}=4\sin\alpha\cos\alpha=2\sin{2\alpha}=2\sin{2\beta}=4\sin\beta\cos\beta$$

From the problem definition, assuming WLOG that $\beta<\frac{\pi}{2}$, we have $\sin\beta=\frac{t}{2}$ and $\cos\beta=\frac{\sqrt{4-t^2}}{2}$ so that

$$\frac{P_{i-2}P_{i-1}}{P_{i}P_{i+1}}=t\sqrt{4-t^2},\qquad i\ge3 \tag{6}$$

Applying (6) repeatedly:

$$\frac{P_{i}P_{i+1}}{P_{i+2k}P_{i+2k+1}}=\left(t\sqrt{4-t^2}\right)^k,\qquad i\ge1,k\ge0 \tag{7}$$

Part 1

I'm not quite sure how to prove this rigorously.

It's obvious that the next point in the sequence depends only on the previous three and nothing beyond this.
So if triangles turn through $90^\circ$ every two points (and hence $180^\circ$ every four points) and are similar, then any four consecutive segments in the path are a translated, mirrored and scaled image of the next four segments in the path. Then $P_1,P_5,P_9,\ldots$ must be collinear. And $P_2,P_6,P_{10},\ldots$, etc.

Should I formalise this e.g. by using vectors?

Part 2

The path is a regular spiral (claimed), so by similarity arguments

$$\frac{x}{y}=\frac{P_{1}P_{2}}{P_{1001}P_{1002}}=\left(t\sqrt{4-t^2}\right)^{500}$$ Therefore, for the ratio as a function of $t$: $$f(t)=\sqrt[500]{\frac{x}{y}}=t\sqrt{4-t^2}$$ Since $f(t)$ has a maximum when $t=\sqrt2$ and $f(\sqrt2)=\sqrt2\sqrt{4-2}=2$, we immediately have one integer value at: $$\boxed{t=\sqrt2}$$ So it remains to find $t$ at which $f(t)=1$. $$\begin{align} t\sqrt{4-t^2}=1 \implies t^2(4-t^2)=1 &\implies t^4-4t^2+1=0 \implies t^2=2\pm\sqrt3 \end{align}$$

From $t^2=2+\sqrt3$ we get $\boxed{t=\dfrac{1+\sqrt3}{\sqrt2}}$

From $t^2=2-\sqrt3$ we get $\boxed{t=\dfrac{-1+\sqrt3}{\sqrt2}}$

Remark: If $\frac{1+\sqrt3}{\sqrt2}<t<2$ or $t<\frac{-1+\sqrt3}{\sqrt2}$, the sequence of points diverges ($f(t)<1$).

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