$ABCDE$ is a regular pentagon. Two circles $C_1$ and $C_2$ are drawn through $B$ with centers $A$ and $C$ respectively. Let the other intersection of $C_1$ and $C_2$ be $P$. The circle with center $P$ which passes through $E$ and $D$ intersects $C_2$ at $X$ and $AE$ at $Y$. Prove that $AX = AY$.
Problem
Source: Canada Repêchage 2021/3 CMOQR
Tags: Canada, repechage, geometry, pentagon
somebodyyouusedtoknow
13.07.2021 19:49
Quite a satisfying problem. This only involves angle chasing. There are no configuration issues so I will just use normal angles instead. For ease of writing, let the circle with center $P$ passing through $E$ be $\omega$. Now we'll proceed to the proof. The crucial claim is Claim 1. $ED = EX = EA$ Its proof: $ED = EA$ is trivial. We'll show $ED = EX$. First let $C_3$ be a circle centred at $E$ with radius $DE$. So let the intersection of $C_2$ and $C_3$ be on $D$ and $X' \neq D$. By SSS Congruency, $\triangle{EDX} \cong \triangle{CDX}$ hence $\angle{XDE} = \angle{XDC} = \frac{1}{2} (108^{\circ}) = 54^{\circ} = \angle{DX'E} = \angle{DX'C}$. Now let's consider that: \[ \angle{DXE} = \frac{1}{2} \angle{DPE} = \frac{1}{2} (360^{\circ} - \angle{DPB} - \angle{EPB}). \]Notice $\angle{DPB} = \angle{EPB} = \frac{1}{2} (360^{\circ} - 108^{\circ}) = 126^{\circ}$ hence $\angle{DXE} = 54^{\circ}$. This concludes that $X = X'$ for clarity's sake I'll include the actual proof for it equalling since it's rather trivial, hidden here.
From above we have $\angle{DXE} = \angle{DX'E} = 54^{\circ}$ so $D, X, X', E$ are concyclic. To show that $X = X'$, notice that $CB = CX = CX'$ however $X' \neq D$ which implies $D, X, X', B$ are concyclic as well. Since $C_2$ and $\omega$ are distinct circles, their radical axis is line $DX$ or $DX'$. Since a line can only have at most 2 intersections with a particular circle, $X = X'$ otherwise $D = X'$ which is a contradiction. Hence proven.
To continue the proof, notice since $DEYX$ is cyclic, $\angle{DXY} = 180^{\circ} - \angle{DEY} = 72^{\circ}$, so $\angle{YXE} = 18^{\circ}$. (1) Corollary of Claim 1. Since $EX = ED$ and $\angle{DXE} = 54^{\circ}$, $\angle{XED} = 72^{\circ}$ which means $\angle{XEA} = 36^{\circ}$. This implies $\angle{AXE} = \angle{XAE} = 72^{\circ}$, and by (1) we have $\angle{AXY} = 54^{\circ}$. This means $\angle{XYA} = 180^{\circ} - 72^{\circ} - 54^{\circ} = 54^{\circ}$ so $AX = AY$, as desired.