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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]]
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[[File:2021 USAMO 6b.png|350px|right]]
 
[[File:2021 USAMO 6c.png|300px|right]]
 
[[File:2021 USAMO 6c.png|300px|right]]
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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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[[File:2021 USAMO 6a.png|300px|right]]
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We construct two equal triangles, prove that triangle <math>XYZ</math> 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 <math>A' =  C + E – D, B' = D + F – E, C' =  A+ E – F,</math>
 
Denote <math>A' =  C + E – D, B' = D + F – E, C' =  A+ E – F,</math>
 
<math>D' =  F+ B – A, E' =  A + C – B, F' =  B+ D – C.</math>
 
<math>D' =  F+ B – A, E' =  A + C – B, F' =  B+ D – C.</math>
 
Then <math>A' – D'  =  C + E – D –  ( F+ B – A) = (A + C + E ) – (B+ D + F).</math>
 
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>  
 
Denote  <math>D' – A' = 2\vec V.</math>  
  
 
Symilarly we get <math>B' – E' = F' – C' =  D' – A'  \implies</math>
 
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>\triangle A'C'E' = \triangle D'F'B'.</math>
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The translation vector maps <math>\triangle A'C'E'</math> into <math>\triangle D'F'B'</math> is <math>2\vec {V.}</math>
 
<math>X = \frac {A+D}{2} =  \frac { (A+ E – F) + (D + F – E)}{2} =  \frac {C' + B'}{2} = \frac {E' + F'}{2},</math>
 
<math>X = \frac {A+D}{2} =  \frac { (A+ E – F) + (D + F – E)}{2} =  \frac {C' + B'}{2} = \frac {E' + F'}{2},</math>
  
 
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>
 
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>
 
<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>
 
<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>
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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>
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Therefore <math>\triangle X'Y'Z'</math> is the medial triangle of <math>\triangle B'D'F'.</math>
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<math>\triangle XYZ</math> is <math>\triangle X'Y'Z'</math> translated on <math>– \vec {V}.</math>
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It is known (see diagram) that circumcenter of triangle coincide 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>
  
 
According to the definition of points <math>A', C', E', ABCE', CDEA',</math> and <math>AFEC'</math> are parallelograms <math>\implies</math>
 
According to the definition of points <math>A', C', E', ABCE', CDEA',</math> and <math>AFEC'</math> are parallelograms <math>\implies</math>
 
<cmath>AC' = FE, AE' = BC, CE' = AB, CA' = DE, EA' = CD, EC' = AF \implies</cmath>
 
<cmath>AC' = FE, AE' = BC, CE' = AB, CA' = DE, EA' = CD, EC' = AF \implies</cmath>
 
<cmath>AC' \cdot AE' = CE' \cdot CA' = EA' \cdot EC' \implies</cmath> 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 \implies circumcenters of constructed triangles coincide with given circumcenters.
 
<cmath>AC' \cdot AE' = CE' \cdot CA' = EA' \cdot EC' \implies</cmath> 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 \implies circumcenters of constructed triangles coincide with given circumcenters.

Revision as of 08:57, 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
2021 USAMO 6c.png
2021 USAMO 6a.png

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 $D' – A' = 2\vec V.$

Symilarly 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 $B'D'.$

Symilarly $X' = X + V$ 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 $A', C', E', ABCE', CDEA',$ and $AFEC'$ are parallelograms $\implies$ \[AC' = FE, AE' = BC, CE' = AB, CA' = DE, EA' = CD, EC' = AF \implies\] \[AC' \cdot AE' = CE' \cdot CA' = EA' \cdot EC' \implies\] 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 \implies circumcenters of constructed triangles coincide with given circumcenters.

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