nice analog of Euler line (Kazakhstan NMO 2009 10 grade P 2)

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nice analog of Euler line (Kazakhstan NMO 2009 10 grade P 2)

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Ovchinnikov Denis

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Ovchinnikov Denis

#1 h Sep 6, 2010, 3:54 PM • 3 Y

Y by amatysten, Adventure10, Mango247

Let in-circle of $ABC$ touch $AB$ , $BC$ , $AC$ in $C_1$ , $A_1$ , $B_1$ respectively.
Let $H$ - intersection point of altitudes in $A_1B_1C_1$ , $I$ and $O$ -be in-center and circumcenter of $ABC$ respectively.
Prove, that $I, O, H$ lies on one line.

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skytin

#2 h Sep 6, 2010, 4:14 PM • 1 Y

Y by Adventure10

it is famous lemma.
let take midpoints of arc AB BC and AC points P Q R. easy tosee that PQR and A_1B_1C_1 is homotety to each other and I is orhocenter of PQR so center of homotety is on line IH and I is circumcenter of A_1B_1C_1 so I H O is on same line

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#3 h Sep 6, 2010, 4:58 PM • 3 Y

Y by opptoinfinity, Adventure10, Mango247

I will use complex numbers.Let us asume that $A',B'$ and $C'$ are on the unit cercle.Because $A$ is the intersection of 2 tangents we get $a=\frac{2c'b'}{c'+b'}$ and analogues the other 2 points. $h=a'+b'+c'$ .As $O$ is the intersection of perpendicular bisectors from the middle points of segment $AC$ and $BC$ we get ,afther a few calculations ,that $o=\frac{2a'b'c'(a'+b'+c')}{(a'+b')(b'+c')(c'+a')}$ .We now the complex numbers of all 3 points,we now just need to show that they are collinear,because $i=0$ we need to prove the following statement $\frac{h}{\overline h}=\frac{o}{\overline o}$ afther taking every complex number in form $x=y+z\sqrt {-1}$ we easily prove the statement.
So we are done.

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jayme

#4 h Sep 7, 2010, 12:13 PM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,
1. HI is the Euler's line of A1B1C1.
2. It is known that this line passes through the center of the circumcircle of the tangential triangle ABC of A1B1C1 (Gob result).
Sincerely
Jean-Louis

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#5 h Sep 14, 2010, 1:22 AM • 2 Y

Y by Adventure10, Mango247

Ovchinnikov Denis wrote:

Let in-circle of $ABC$ touch $AB$ , $BC$ , $AC$ in $C_1$ , $A_1$ , $B_1$ respectively.
Let $H$ - intersection point of altitudes in $A_1B_1C_1$ , $I$ and $O$ -be in-center and circumcenter of $ABC$ respectively.
Prove, that $I, O, H$ lies on one line.

I use different notation.
Problem. The incircle $(I)$ of the triangle $\triangle{ABC}$ touche sides $BC$ , $AC$ and $AB$ at $A_1$ , $B_1$ and $C_1$ respectively. let $H_1$ be the orthocenter of $\triangle{A_1B_1C_1}$ . Prove that the circumcenter of $\triangle{ABC}$ lies on $IH_1$ .

My solution

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