In the plane, three distinct non-collinear points $A,B,C$ are marked. In each step, Ege can do one of the following: For marked points $X,Y$, mark the reflection of $X$ across $Y$. For distinct marked points $X,Y,Z,T$ which do not form a parallelogram, mark the center of spiral similarity which takes segment $XY$ to $ZT$. For distinct marked points $X,Y,Z,T$, mark the intersection of lines $XY$ and $ZT$. No matter how the points $A,B,C$ are marked in the beginning, can Ege always mark, after finitely many moves, a) The circumcenter of $\triangle ABC$. b) The incenter of $\triangle ABC$. Proposed by Deniz Can Karaçelebi
Problem
Source: Turkey Olympic Revenge 2024 P2
Tags: geometry, complex numbers, Spiral Similarity
sami1618
06.08.2024 17:10
see below
Consider $A=(1,0)$, $B=(0,0)$, and $C=(0,1)$. We claim that any new constructed points have both rational coordinates. This is obviously true for reflection and intersection lines. Notice that the circumcenters of three rational points will itself be rational. We can think of the second property as reflecting the intersection of two lines over the line connecting the circumcircles of two rational triangles, which itself must be rational. As the incenter does not have rational coordinates, we are done.
i3435
06.08.2024 21:00
I think you're not allowed to reflect $A$ over $BC$ in one step, so the above proof for circumcenter is incorrect. Here is another one: Let $O$ be the circumcenter of $ABC$. Reflect $B$ over $A$ to $B'$ and $A$ over $C$ to $A'$. Then the spiral center from $AB'$ to $CA'$ is the spiral center from $BA$ to $AC$, the 'Dumpty' point $D_A$, which is the foot from the circumcenter to the $A$-symmedian. Let $\overline{AD_A}\cap\overline{BC}=P_A$ and define $P_B,P_C$ similarly. Then if $T=\overline{P_BP_C}\cap\overline{BC}$, $(T,P_A;B,C)=-1$. If $K=\overline{AD_A}\cap (ABC)$, then $T=\overline{AA}\cap\overline{KK}$, so $\overline{OT}\perp\overline{AK}$ and $T-D_A-O$. Thus drawing the lines in $B$ and $C$ corresponding to $\overline{TD_A}$ and intersecting them will give $O$.