By centroid of a quadrilateral $PQRS$ we call a common point of two lines through the midpoints of its opposite sides. Suppose that $ABCDEF$ is a hexagon inscribed into the circle $\Omega$ centered at $O$. Let $AB=DE$, and $BC=EF$. Let $X$, $Y$, and $Z$ be centroids of $ABDE$, $BCEF$; and $CDFA$, respectively. Prove that $O$ is the orthocenter of triangle $XYZ$.
Problem
Source: III Caucasus Mathematical Olympiad
Tags: geometry
Gryphos
18.03.2018 00:51
Nice problem indeed
Using vectors, it is easy to see that $x=\frac{a+b+d+e}{4}$, analogous formulas hold for $y, z$. This implies that
$$z-y = \frac{a+c+d+f}{4}-\frac{b+c+e+f}{4} = \frac{a-e}{4} + \frac{d-b}{4}.$$Since $ABDE$ is an isosceles trapezoid, $AE \parallel BD$, hence $YZ \parallel AE \parallel BD$.
On the other hand, since $ABDE$ is an isosceles trapezoid, $X$ lies on the common perpendicular bisector of $AE,BD$. Since $O$ also lies on this bisector, we get $OX \perp AE$, thus $OX \perp YZ$.
Analogously, one shows that $OY \perp XZ$. Consequently, $O$ is the orthocenter of $\triangle XYZ$.
rafaello
25.01.2021 23:26
First, let us note since $AB=CE$ and $BC=EF$, $AFCD$ is isosceles trapezoid.
We use complex numbers.
We have $x=\frac{a+b+d+e}{4}$ and $y=\frac{b+c+e+f}{4}$ and $z=\frac{c+d+f+a}{4}$
$$(a+b)^2de=(d+e)^2ab$$
$$\frac{z}{\overline{z}}=-\frac{x-y}{\overline{x-y}}$$$$\overline{z}=\frac{acd+afd+cdf+acf}{4acdf}$$$$\frac{z}{\overline{z}}=\frac{(a+c+d+f)acdf}{acd+afd+cdf+acf}$$
$$x-y=\frac{a+b+d+e}{4}-\frac{b+c+e+f}{4}=\frac{a+d-c-f}{4}$$
$$\overline{x}-\overline{y}=\frac{abd+abe+bde+ade}{4abde}-\frac{bce+bcf+bef+cef}{4bcef}=\frac{(abd+abe+bde+ade)cf-(bce+bcf+bef+cef)ad}{4abcdef}=\frac{abcef+bcdef-abcde-abdef}{4abcdef}$$We need
$$\frac{(a+c+d+f)acdf}{acd+afd+cdf+acf}=-\frac{(a+d-c-f)abcdef}{abcef+bcdef-abcde-abdef}$$
$$\frac{(a+c+d+f)}{acd+afd+cdf+acf}=-\frac{(a+d-c-f)be}{abcef+bcdef-abcde-abdef}=-\frac{(a+d-c-f)be}{be(acf+cdf-acd-adf)}=-\frac{a+d-c-f}{acf+cdf-acd-adf}$$
$$(a+c+d+f)(acf+cdf-acd-adf)=a^2cf+acdf-a^2cd-a^2df+ac^2f+c^2df-ac^2d-acdf+acdf+cd^2f-acd^2-ad^2f+acf^2+cdf^2-acdf-adf^2$$$$=a^2cf-a^2cd-a^2df+ac^2f+c^2df-ac^2d+cd^2f-acd^2-ad^2f+acf^2+cdf^2-adf^2$$
$$(c+f-a-d)(acd+afd+cdf+acf)=ac^2d+acfd+c^2df+ac^2f+acdf+adf^2+cdf^2+acf^2-a^2cd-a^2df-acdf-a^2cf-acd^2-ad^2f-cd^2f-acdf$$$$=ac^2d+c^2df+ac^2f+adf^2+cdf^2+acf^2-a^2cd-a^2df-a^2cf-acd^2-ad^2f-cd^2f$$
We need $$a^2cf-a^2cd-a^2df+ac^2f+c^2df-ac^2d+cd^2f-acd^2-ad^2f+acf^2+cdf^2-adf^2=ac^2d+c^2df+ac^2f+adf^2+cdf^2+acf^2-a^2cd-a^2df-a^2cf-acd^2-ad^2f-cd^2f$$That simplifies to the following to show:
$$2(a^2cf-ac^2d+cd^2f-adf^2)=2a^2cf-2ac^2d+2cd^2f-2adf^2=0$$
We also have $$(a-d)\left(\frac{1}{a}-\frac{1}{d}\right)=(c-f)\left(\frac{1}{c}-\frac{1}{f}\right)
\implies (a-d)^2cf=(c-f)^2ad\implies a^2cf+cd^2f=ac^2d+adf^2$$Hence, we just shown that $XY\perp OZ$, we work similarly and conclude that $XY\perp OZ$, $XZ\perp OY$ and $YZ\perp OX$, hence $O$ is the orthocenter of triangle $XYZ$.
First, let us note since $AB=CE$ and $BC=EF$, $AFCD$ is isosceles trapezoid. Also, $ABDE$ and $BCEF$ are isosceles trapezoids.
Let $M_1$ be the midpoint of $AB$, define $M_2$ as the midpoint of $CD$, $M_3$ as the midpoint of $DE$ and $M_4$ as the midpoint of $FA$.
Notice that $M_2M_3\parallel CE \parallel BF\parallel M_1M_4$.
Let us consider quadrilateral $M_1M_2M_3M_4$, notice that $X$ is the midpoint of $M_1M_3$ and $Z$ is the midpoint of $M_2M_4$.
Let $N$ be the midpoint of $M_1M_2$, thus considering midlines we obtain $NX\parallel M_2M_3\parallel M_1M_4\parallel NZ\implies XZ\parallel M_1M_4$.
Also, since $Y$ is by definition the centroid of $BCEF$, we have $XZ\parallel BF\perp OY\perp CE$, thus $XZ\perp OY$. Similarly, we have $XY\perp OZ$ and $YZ\perp OX$, therefore we conclude that $O$ is the orthocenter of triangle $XYZ$.