Difference between revisions of "2021 USAMO Problems/Problem 6"
(→Solution) |
|||
Line 3: | Line 3: | ||
==Solution== | ==Solution== | ||
+ | [[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. | ||
+ | |||
+ | 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> | ||
+ | Then <math>A' – D' = C + E – D – ( F+ B – A) = (A + C + E ) – (B+ D + F).</math> | ||
+ | Denote <math>D' – A' = 2\vec V.</math> | ||
+ | |||
+ | Symilarly we get <math>B' – E' = F' – C' = D' – A' \implies</math> | ||
+ | <math>\triangle ACE = \triangle BDF,</math> and the translation vector 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> | ||
+ | |||
+ | 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> | ||
+ | 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> | ||
+ | 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 be a convex hexagon satisfying
,
,
, and
Let
,
, and
be the midpoints of
,
, and
. Prove that the circumcenter of
, the circumcenter of
, and the orthocenter of
are collinear.
Solution
We construct two equal triangles, prove that triangle 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
Then
Denote
Symilarly we get
and the translation vector is
so is midpoint of
and
Symilarly
is the midpoint of
and
is the midpoint of
and
is the midpoint of
Symilarly
is the midpoint of
is the midpoint of
so
is the medial triangle of
translated on
It is known (see diagram) that circumcenter of triangle coincite with orthocenter of the medial triangle. Therefore orthocenter
of
is circumcenter of
translated on
It is the midpoint of segment
connected circumcenters of
and