Let $ABCDEF$ a convex hexagon, and let $A_1,B_1,C_1,D_1,E_1$ y $F_1$ be the midpoints of $AB,BC,CD,$ $DE,EF$ and $FA$, respectively. Construct points $A_2,B_2,C_2,D_2,E_2$ and $F_2$ in the interior of $A_1B_1C_1D_1E_1F_1$ such that both 1. The sides of the dodecagon $A_2A_1B_2B_1C_2C_1D_2D_1E_2E_1F_2F_1$ are all equal and 2. $\angle A_1B_2B_1+\angle C_1D_2D_1+\angle E_1F_2F_1=\angle B_1C_2C_1+\angle D_1E_2E_1+\angle F_1A_2A_1=360^\circ$, where all these angles are less than $180 ^\circ$, Prove that $A_2B_2C_2D_2E_2F_2$ is cyclic. Note: Dodecagon $A_2A_1B_2B_1C_2C_1D_2D_1E_2E_1F_2F_1$ is shaped like a 6-pointed star, where the points are $A_1,B_1,C_1,D_1,E_1$ y $F_1$.
Problem
Source: 2024 Mexican Mathematical Olympiad, Problem 3
Tags: geometry, hexagon, dodecagon, Mexico
plagueis
07.11.2024 00:34
Any solutions?
a_507_bc
07.11.2024 00:57
Just an idea. Seems like the classical idea of constructing parallelograms in hexagon problems is relevant here (like in ISL 2013 G5 for example). The center $O$ of the desired circle is such that $OA_2A_1B_2, OB_2C_1C_2, \ldots $ are parallelograms (actually, rhombuses). The details seem nontrivial though.
soryn
07.11.2024 07:15
Very good idea
Geolympics
08.11.2024 01:22
Very tricky problem. Here's my solution (in both Spanish and English). The diagram is in the attachments.
Hint1Prove that triangles $\triangle ACE$ and $\triangle BDF$ have the same circunradius.
Hint2Let $O_1, O_2$ be the cricumcenters of $\triangle ACE$ and $\triangle BDF$, respectively, and let $O$ be the
midpoint of $O_1O_2$. Prove that $O$ is the circumcenter of $A_2B_2C_2D_2E_2F_2$.
SolutionWe'll begin by proving that triangles $\triangle ACE$ and $\triangle BDF$ have the same circunradius. To do so,
let $O_1, O_2$ and $R_1,R_2$ be the circumcenters and circucircumradius of $\triangle ACE$ and $\triangle BDF$, respectively, and let $r$
be the side length of the given dodecagon (that is, $r=A_2A_1=A_1B_2= \cdots = F_1A_2$).
Consider a point $Q$ such that $\angle QA_2A_1 = \angle C_1C_2B_1$, $A_2Q=A_2A_1=r$, and $Q$ and $A_2$ lie on the same side
of $A_1F_1$. From the condition $\angle A_1A_2F_1+\angle C_1C_2B_1 + \angle E_1E_2D_1 = 360$, it follows that
\[
\angle QA_2F_1= \angle E_1E_2D_1.
\]Notice also that, since $A_2Q=A_2A_1=A_2F_1=r$, then $\triangle A_1F_1Q$ has circumradius $r$ and
circumcenter $A_2$. Also, using $SAS$,
\[
\triangle QA_2A_1 \cong \triangle C_1C_2B_1, \triangle QA_2F_1 \cong \triangle D_1E_2E_1.
\]Hence, $QA_1=C_1B_1$ and $QF=D_1E_1$. However, we see from the midpoint condition that
\[
A_1F_1=\frac{BF}{2}, B_1C_1=\frac{BD}{2}, D_1E_1 = \frac{DF}{2}.
\]Therefore,
\[
A_1F_1=\frac{BF}{2}, A_1Q=\frac{BD}{2}, QF_1 = \frac{DF}{2}.
\]Using $SSS$, we get that $\triangle A_1F_1Q \sim \triangle BFD$ with ratio 1:2. Nevertheless, we know that $r$ and $R_2$
are the circumradius of $\triangle A_1F_1Q$ and $\triangle BFD$ respectively. Whence,
\[
r = \frac{R_2}{2}.
\]Similarly,
\[
r = \frac{R_1}{2},
\]implying that $R_1 = R_2 = R = 2r$.
On the other hand, we see that $A_1F_1 || BF$, $BF = 2A_1F_1$, and that $\triangle A_1A_2F_1$ and $\triangle BO_2F$ are similar
isosceles triangles. Therefore, triangles $\triangle A_1A_2F_1$ and $\triangle BO_2F$ are homothetic with ratio 1:2 and
homothety center $A$. In consequence, $A, A_2, O_2$ are collinear and $A_2$ is the midpoint of $AO_2$. This argument
works in a similar way with $B_2, C_2, D_2, E_2, F_2$.
Finally, let $O$ be the midpoint of $O_1O_2$. We will now prove that $O$ is the circumcenter of
$A_2B_2C_2D_2E_2F_2$. To do so, we see that since $O$ is the midpoint of $O_1O_2$ and $C_2$ is the midpoint
of $O_2C$, then
\[
OC_2 = \frac{O_1C}{2} = \frac{R}{2} = r.
\]Analogously for the other five points, we conclude that $A_2B_2C_2D_2E_2F_2$ has circumcenter
$O$ and circumradius $r$.
Hint1Demuestra que los triángulos $\triangle ACE$ y $\triangle BDF$ tienen el mismo circunradio.
Hint2Sean $O_1, O_2$ los circuncentros de $\triangle ACE$ y $\triangle BDF$, respectivamente, y sea $O$ el punto
medio de $O_1O_2$. Demuestra que $O$ es el circuncentro de $A_2B_2C_2D_2E_2F_2$.
SoluciónComenzaremos por demostrar que los triángulos $\triangle ACE$ y $\triangle BDF$ tienen el mismo circunradio. Para ello,
sean $O_1, O_2$ y $R_1,R_2$ los circuncentros y circunradios de $\triangle ACE$ y $\triangle BDF$, respectivamente, y sea $r$
la longitud de los lados del dodecágono dado (esto es, $r=A_2A_1=A_1B_2= \cdots = F_1A_2$).
Considera un punto $Q$ tal que $\angle QA_2A_1 = \angle C_1C_2B_1$, $A_2Q=A_2A_1=r$, y $Q$ y $A_2$ están en el mismo lado de $A_1F_1$.
De la condición $\angle A_1A_2F_1+\angle C_1C_2B_1 + \angle E_1E_2D_1 = 360$, se sigue que
\[
\angle QA_2F_1= \angle E_1E_2D_1.
\]Notemos también que, ya que $A_2Q=A_2A_1=A_2F_1=r$, entonces $\triangle A_1F_1Q$ tiene circunradio $r$ y
circuncentro $A_2$. Además, por $LAL$,
\[
\triangle QA_2A_1 \cong \triangle C_1C_2B_1, \triangle QA_2F_1 \cong \triangle D_1E_2E_1.
\]En consecuencia, $QA_1=C_1B_1$ y $QF=D_1E_1$. No obstante, observamos de la condición de ser puntos medios que
\[
A_1F_1=\frac{BF}{2}, B_1C_1=\frac{BD}{2}, D_1E_1 = \frac{DF}{2}.
\]Por tanto,
\[
A_1F_1=\frac{BF}{2}, A_1Q=\frac{BD}{2}, QF_1 = \frac{DF}{2}.
\]Usando $LLL$, obtenemos que $\triangle A_1F_1Q \sim \triangle BFD$ con razón 1:2. No obstante, sabemos que $r$ y $R_2$
son los circunradios de $\triangle A_1F_1Q$ y $\triangle BFD$ respectivamente. Luego,
\[
r = \frac{R_2}{2}.
\]Similarmente,
\[
r = \frac{R_1}{2},
\]implicando que $R_1 = R_2 = R = 2r$.
Por otro lado, vemos que $A_1F_1 || BF$, $BF = 2A_1F_1$, y que $\triangle A_1A_2F_1$ y $\triangle BO_2F$ son triángulos isósceles semejantes.
Por ende, los triángulos $\triangle A_1A_2F_1$ y $\triangle BO_2F$ son homotéticos con razón 1:2 y centro de homotecia
$A$. En consecuencia, $A, A_2, O_2$ son colineales y $A_2$ es el punto medio de $AO_2$. Este argumento funciona
de manera similar para $B_2, C_2, D_2, E_2, F_2$.
Finalmente, sea $O$ el punto medio de $O_1O_2$. Procederemos a demostrar que $O$ es el circuncentro de
$A_2B_2C_2D_2E_2F_2$. Para ello, observamos que al ser $O$ el punto medio de $O_1O_2$ y $C_2$ el punto medio de
$O_2C$, entonces
\[
OC_2 = \frac{O_1C}{2} = \frac{R}{2} = r.
\]Análogamente para los otros cinco puntos, concluimos que $A_2B_2C_2D_2E_2F_2$ tiene circuncentro
$O$ y circunradio $r$.
Attachments:
