Given a acute triangle $PA_1B_1$ is inscribed in the circle $\Gamma$ with radius $1$. for all integers $n \ge 1$ are defined: $C_n$ the foot of the perpendicular from $P$ to $A_nB_n$ $O_n$ is the center of $\odot (PA_nB_n)$ $A_{n+1}$ is the foot of the perpendicular from $C_n$ to $PA_n$ $B_{n+1} \equiv PB_n \cap O_nA_{n+1}$ If $PC_1 =\sqrt{2}$, find the length of $PO_{2015}$
HIDE: Source Cono Sur Olympiad - 2015 - Day 1 - Problem 3Problem
Source:
Tags: geometry, circumcircle
drmzjoseph
16.05.2015 05:52
Let $D$ be the antipode of $P$ with respect to $\odot (PA_1B_1) \Rightarrow \triangle PDB_1 \sim \triangle PC_1A_2 \Rightarrow \frac{2}{\sqrt{2}}=\frac{PB_1}{PA_2} \Rightarrow \triangle PC_1B_1 \sim \triangle PC_2A_2 \Rightarrow \triangle PA_1B_1 \sim \triangle PB_2A_2$ ratio $\sqrt{2} \Rightarrow \frac{PO_1}{PO_2}=\frac{\sqrt{2}}{1} \Rightarrow PO_{2015}=\frac{1}{(\sqrt{2})^{2014}}=\frac{1}{2^{1007}}$
Luis González
16.05.2015 06:04
Let the perpendicular to $PO_1$ at $O_1$ cut $PA_1,PB_1$ at ${A_2}',{B_2}',$ respectively and $PO_1$ cuts $\odot(PA_1B_1)$ again at $Q_1.$ Since ${|PC_1|}^2=2=|PO_1| \cdot |PQ_1|,$ then $O_1{A_2}'$ is the inverse of $\odot(PA_1B_1)$ under inversion WRT $\odot(P,PC_1).$ ${A_2}',{B_2}'$ are the inverse images of $A_1,B_1$ $\Longrightarrow$ $|PC_1|^2=|P{A_2}'| \cdot |PA_1|=|P{B_2}'| \cdot |PB_1|$ $\Longrightarrow$ ${A_2}',{B_2}'$ coincide with the projections $A_2,B_2$ of $C_1$ on $PA_1,PB_1$ $\Longrightarrow$ $PC_1$ is diameter of the circumcircle of $\triangle PA_2B_2,$ i.e. $O_2$ is midpoint of $PC_1$ $\Longrightarrow$ $O_n$ is the midpoint of $PC_{n-1}$ $\Longrightarrow$ $|PO_n|$ forms a geometric progression with ratio $\tfrac{\sqrt{2}}{2}$ $\Longrightarrow$ $|PO_n|=|PO_1| \cdot \left (\tfrac{\sqrt{2}}{2} \right)^{n-1}$ $\Longrightarrow$ $|PO_{2015}|= \left (\tfrac{\sqrt{2}}{2} \right)^{2014}.$