# 2021 USAMO Problems/Problem 6

## Problem 6

Let $ABCDEF$ be a convex hexagon satisfying $\overline{AB} \parallel \overline{DE}$, $\overline{BC} \parallel \overline{EF}$, $\overline{CD} \parallel \overline{FA}$, and$$AB \cdot DE = BC \cdot EF = CD \cdot FA.$$Let $X$, $Y$, and $Z$ be the midpoints of $\overline{AD}$, $\overline{BE}$, and $\overline{CF}$. Prove that the circumcenter of $\triangle ACE$, the circumcenter of $\triangle BDF$, and the orthocenter of $\triangle XYZ$ are collinear.

## Solution

We construct two equal triangles, prove that triangle $XYZ$ is the same as medial triangle of both this triangles. We use property of medial triangle and prove that circumcenters of constructed triangles coincide with given circumcenters.

Denote $A' = C + E – D, B' = D + F – E, C' = A+ E – F,$ $D' = F+ B – A, E' = A + C – B, F' = B+ D – C.$ Then $A' – D' = C + E – D – ( F+ B – A) = (A + C + E ) – (B+ D + F).$

Denote $D' – A' = 2\vec V.$

Similarly we get $B' – E' = F' – C' = D' – A' \implies$ $\triangle A'C'E' = \triangle D'F'B'.$

The translation vector maps $\triangle A'C'E'$ into $\triangle D'F'B'$ is $2\vec {V.}$ $X = \frac {A+D}{2} = \frac { (A+ E – F) + (D + F – E)}{2} = \frac {C' + B'}{2} = \frac {E' + F'}{2},$

so $X$ is midpoint of $AD, B'C',$ and $E'F'.$ Symilarly $Y$ is the midpoint of $BE, A'F',$ and $C'D', Z$ is the midpoint of $CF, A'B',$ and $D'E'.$ $Z + V = \frac {A' + B'}{2}+ \frac {D' – A'}{2} = \frac {B' + D'}{2} = Z'$ is the midpoint of $B'D'.$

Similarly $X' = X + V$ is the midpoint of $B'F',Y'= Y + V$ is the midpoint of $D'F'.$

Therefore $\triangle X'Y'Z'$ is the medial triangle of $\triangle B'D'F'.$

$\triangle XYZ$ is $\triangle X'Y'Z'$ translated on $– \vec {V}.$

It is known (see diagram) that circumcenter of triangle coincide with orthocenter of the medial triangle. Therefore orthocenter $H$ of $\triangle XYZ$ is circumcenter of $\triangle B'D'F'$ translated on $– \vec {V}.$

It is the midpoint of segment $OO'$ connected circumcenters of $\triangle B'D'F'$ and $\triangle A'C'E'.$

According to the definition of points $A', C', E',$ quadrangles $ABCE', CDEA',$ and $AFEC'$ are parallelograms. Hence $$AC' = FE, AE' = BC, CE' = AB, CA' = DE, EA' = CD, EC' = AF \implies$$ $$AC' \cdot AE' = CE' \cdot CA' = EA' \cdot EC' = AB \cdot DE \implies$$ Power of points A,C, and E with respect circumcircle $\triangle A'C'E'$ is equal, hence distances between these points and circumcenter of $\triangle A'C'E'$ are the same. Therefore circumcenter $\triangle ACE$ coincide with circumcenter $\triangle A'C'E'.$

Similarly circumcenter of $\triangle BDF$ coincide with circumcenter of $\triangle B'D'F'.$