Let $ABC$ be a triangle with $\angle BAC=60^\circ$. Let $D$ be a point on the line $AC$ such that $AB = AD$ and $A$ lies between $C$ and $D$. Suppose that there are two points $E \ne F$ on the circumcircle of the triangle $DBC$ such that $AE = AF = BC$. Prove that the line $EF$ passes through the circumcenter of $ABC$.
Problem
Source: MEMO 2024 T5
Tags: geometry, circumcircle, geometry proposed
VicKmath7
28.08.2024 00:22
Nice and easy. Let $A'$ be the circumcenter of $\triangle DBC$. Then $\triangle A'BC$ is equilateral. Hence, $EF$ is the perpendicular bisector of $AA'$ as $EAFA'$ is a rhombus (all its sides are equal to $BC$). But the circumcenter of $ABC$ lies on this perpendicular bisector as $BAA'C$ is cyclic - done.
VedViesKto
28.08.2024 00:23
Hint:
Using the condition $\angle BAC=60^\circ$ find another points, at which we have length $BC$ .
Solution:
Let $N$ be the midpoint of the longer arc $BAC$. Since $N$ is on the perpendicular bisector of $BC$, the triangle $BNC$ is equilateral. Moreover, it is seen, that $N$ is the circumcentre of the circumcircle of the triangle $DBC$ ($N$ lies on the perpendicular bisector of $BC$ and also perpendicular bisector of $DB$, which is seen to be the outer angle bisector at the point $A$).
Therefore, the radius of the circumcircle of the $DBC$ is $BC =: r$. This gives that for the points $E$ and $F$ we have $AE = AF = r = NE = NF$, meaning that the quadrilateral $AFNE$ is a rhobus. This means that the points $E$ and $F$ both lie on the perpendicular bisector of the chord $AN$ of the circumcircle of $ABC$.
From this we get that the line $EF$ passes through the circumcentre of the triangle $ABC$ as desired.
Cali.Math
17.11.2024 22:41
We uploaded our solution https://calimath.org/pdf/MEMO2024-T5.pdf on youtube https://youtu.be/boQGJcd9tQo.