Given a convex pentagon $ABCDE$, points $A_1, B_1, C_1, D_1, E_1$ are such that $$AA_1 \perp BE, BB_1 \perp AC, CC_1 \perp BD, DD_1 \perp CE, EE_1 \perp DA.$$In addition, $AE_1 = AB_1, BC_1 = BA_1, CB_1 = CD_1$ and $DC_1 = DE_1$. Prove that $ED_1 = EA_1$
Problem
Source: St Petersburg 2021 10.3
Tags: geometry, pentagon
cadaeibf
25.12.2021 00:49
Let $A'$ be the projection of $A$ on $EB$, and cyclically $B',C',D',E'$. Also let $A''$ be the intersection of $EE'$ and $BB'$ and cyclically $B'',...,E''$. Let $s_A=A'A_1^2$ (note that therefore it doesn't actually matter where $A_1$ is wrt $A'$), and cyclically $s_B,...,s_E$. Note that $s_B-s_E=(AB_1^2-AB'^2)-(AE_1^2-AE'^2)=AE'^2-AB'^2$. Similar relationships hold because of the equal distances (except for the relation $s_A-s_D=ED'^2-EA'^2$ which is equivalent to the thesis).
Since $\sum_{cyc}s_B-s_E=0$ and $$\sum_{cyc}AE'^2-AB'^2=\sum_{cyc}AE'^2-AB'^2+\sum_{cyc}EE'^2-\sum_{cyc}BB'^2=\sum_{cyc}AE^2-AB^2=0$$it follows that we in fact have also the fifth relationship $s_A-s_D=ED'^2-EA'^2$, which then implies $EA_1^2-ED_1^2=(s_A+EA'^2)-(s_D+ED'^2)=(ED'^2-EA'^2)+(EA'^2-ED'^2)=0$
Note that in fact the same thing holds for all polygons.