Let $A_1A_2A_3 \cdots A_{2013}$ be a cyclic $2013$-gon. Prove that for every point $P$ not the circumcenter of the $2013$-gon, there exists a point $Q\neq P$ such that $\frac{A_iP}{A_iQ}$ is constant for $i \in \{1, 2, 3, \cdots, 2013\}$. Proposed by Robin Park
Problem
Source: ELMO 2014 Shortlist G3, by Robin Park
Tags: geometry, circumcircle, ratio, projective geometry, geometry proposed
leminscate
25.07.2014 08:56
I don't think this works if $P = A_1$. Because then we would require $\frac{A_i P}{A_i Q} = 0$ for $i \in \{1, 2, 3, \cdots , 2013 \}$, which is not possible. So maybe the problem should say: for every point $P$ not the circumcentre and not on the circumcircle of the $2013$ - gon.... In this case if the circumcircle is the unit circle, we can simply let $q = \frac{1}{\overline{p}}$, which is no equal to $p$ and is defined due to the additional conditions stated above. Or you can think of this as an inversion about the circumcircle which kind of flips ratios and stuff.
AnonymousBunny
25.07.2014 16:04
The title gives most of the problem away. Let $AB$ be the diameter of $(A_1A_2 \cdots A_{2013})$ containing $P$ ($A$ and $B$ are distinct points and lie on $(A_1A_2\cdots A_{2013})$). Let $Q$ be the harmonic conjugate of $A,P,B.$ It's easy to see that this selection for $Q$ works (note that the circumcircle of $A_1A_2 \cdots A_{2013}$ is the Appolonius circle of $P$ and $Q$ with ratio $\dfrac{AP}{AQ},$ so $\dfrac{A_iP}{AQ} = \dfrac{AP}{AQ} \quad \forall \ i$).
thecmd999
23.10.2014 10:27
Set $Q$ as the inverse of $P$ with respect to $(A_1A_2A_3...A_{2013})$. Then $(A_1A_2A_3...A_{2013})$ is the $A_i$-Apollonius circle of triangle $A_iPQ$, and the result follows.
guptaamitu1
08.06.2021 18:32
For any $i \in \{1,2,\dots,2013\}$, let $\omega_i$ be the locus of all points $X$ such that $\frac{A_iP}{A_{i+1}P} = \frac{A_iX}{A_{i+1}X}$ (indices modulo $2013$). Then all the $\omega_i \text{'s}$ are circles. Note that it suffices to show all these circles intersect at a point not equal to $P$. Since these circles clearly cannot be tangent to one another at a point $P$, so it enough to show that there is a point (not equal to $P$) which has same power WRT all these circles. But we note that the circumcircle of the $2013$-gon has this property, and we are done!
Can someone please check my solution and tell if it correct or not.
john0512
28.01.2023 07:48
Let Q be such that the Apollonian circle of PQ is the circumcircle of the 2013-gon and we are done.