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Difference between revisions of "2021 USAMO Problems/Problem 6"
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==Solution==
 
==Solution==
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[[File:2021 USAMO 6b.png|430px|right]]
 
We construct two equal triangles, prove that triangle <math>XYZ</math> is the medial triangle of both this triangles, use property of medial triangle and prove that circumcenters of constructed triangles coincide with given circumcenters.
 
We construct two equal triangles, prove that triangle <math>XYZ</math> is the medial triangle of both this triangles, use property of medial triangle and prove that circumcenters of constructed triangles coincide with given circumcenters.
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Denote <math>A' =  C + E – D, B' = D + F – E, C' =  A+ E – F,</math>
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<math>D' =  F+ B – A, E' =  A + C – B, F' =  B+ D – C.</math>
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Then <math>A' – D'  =  C + E – D –  ( F+ B – A) = (A + C + E ) – (B+ D + F).</math>
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Denote  <math>D' – A' = 2\vec V.</math>
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Symilarly we get <math>B' – E' = F' – C' =  D' – A'  \implies</math>
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<math>\triangle ACE = \triangle BDF,</math> and the translation vector is <math>2\vec {V.}</math>
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<math>X = \frac {A+D}{2} =  \frac { (A+ E – F) + (D + F – E)}{2} =  \frac {C' + B'}{2} = \frac {E' + F'}{2},</math>
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so <math>X</math> is midpoint of <math>AD, B'C',</math> and <math>E'F'.</math> Symilarly <math>Y</math> is the midpoint of <math>BE, A'F',</math> and <math>C'D', Z</math> is the midpoint of <math>CF, A'B',</math> and <math>D'E'.</math>
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<math>Z + V =  \frac {A' + B'}{2}+ \frac {D' – A'}{2} =  \frac {B' + D'}{2} = Z'</math> is the midpoint of <math>B'D'.</math>
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Symilarly <math>X' = X + V</math>  is the midpoint of <math>B'F',Y'= Y + V</math>  is the midpoint of <math>D'F',</math> so <math>X'Y'Z'</math> is the medial triangle of <math>\triangle B'D'F',</math> translated on <math>– \vec {V}.</math>
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It is known (see diagram) that circumcenter of triangle coincite with orthocenter of the medial triangle. Therefore  orthocenter <math>H</math> of <math>\triangle XYZ</math> is circumcenter of <math>\triangle B'D'F'</math> translated on <math>– \vec {V}.</math> It is the midpoint of segment <math>OO'</math> connected circumcenters of <math>\triangle B'D'F'</math> and <math>\triangle A'C'E'.</math>

Revision as of 07:52, 15 September 2022

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

2021 USAMO 6b.png

We construct two equal triangles, prove that triangle $XYZ$ is the medial triangle of both this triangles, 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 $D' – A' = 2\vec V.$

Symilarly we get $B' – E' = F' – C' = D' – A' \implies$ $\triangle ACE = \triangle BDF,$ and the translation vector 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 $B'D'.$ Symilarly $X' = X + V$ is the midpoint of $B'F',Y'= Y + V$ is the midpoint of $D'F',$ so $X'Y'Z'$ is the medial triangle of $\triangle B'D'F',$ translated on $– \vec {V}.$ It is known (see diagram) that circumcenter of triangle coincite 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 $\triangle A'C'E'.$

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