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 1

Let $M_1$ , $M_2$ , and $M_3$ be the midpoints of $CE$ , $AE$ , $AC$ and $N_1$ , $N_2$ , and $N_3$ be the midpoints of $DF$ , $BF$ , and $BD$ . Also, let $H$ be the orthocenter of $XYZ$ . Note that we can use parallel sides to see that $X$ , $Z$ , and $M_3$ are collinear. Thus we have $\text{Pow}(M_3,(XYZ)) = M_3Z \cdot M_3X = \frac 14 AB \cdot DE$ by midlines. Applying this argument cyclically, and noting the condition $AB \cdot DE = BC \cdot EF = CD \cdot FA$ , $M_1$ , $M_2$ , $M_3$ , $N_1$ , $N_2$ , $N_3$ all lie on a circle concentric with $(XYZ)$ .

Next, realize that basic orthocenter properties imply that the circumcenter $O_1$ of $(ACE)$ is the orthocenter of $\triangle M_1M_2M_3$ , and likewise the circumcenter $O_2$ of $(BDF)$ is the orthocenter of $\triangle N_1N_2N_3$ .

The rest is just complex numbers; toss on the complex plane so that the circumcenter of $\triangle XYZ$ is the origin. Then we have $o_1 = m_1+m_2+m_3 = (c+e)/2+(a+e)/2+(a+c)/2=a+c+e$ $o_2 = n_1+n_2+n_3 = (b+d)/2+(d+f)/2+(b+f)/2=b+d+f$ $h = x+y+z = (a+d)/2+(b+e)/2+(c+f)/2=(a+b+c+d+e+f)/2.$ Note that from the above we have $h=\frac{o_1+o_2}{2}$ , so $H$ is the midpoint of segment $O_1O_2$ . In particular, $H$ , $O_1$ , and $O_2$ are collinear, as required.

~ Leo.Euler

Solution 2

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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 $\triangle B'D'F'.$

vladimir.shelomovskii@gmail.com, vvsss

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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2021 USAMO Problems/Problem 6

Problem 6

Solution 1

Solution 2