incenters of XYZ and ABC are collinear - ISL 1986

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incenters of XYZ and ABC are collinear - ISL 1986

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#1 h Aug 31, 2010, 10:20 AM • 2 Y

Y by Adventure10 and 1 other user

The circle inscribed in a triangle $ABC$ touches the sides $BC,CA,AB$ in $D,E, F$ , respectively, and $X, Y,Z$ are the midpoints of $EF, FD,DE$ , respectively. Prove that the centers of the inscribed circle and of the circles around $XYZ$ and $ABC$ are collinear.

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#2 h Aug 31, 2010, 2:38 PM • 4 Y

Y by Amir Hossein, AlastorMoody, Adventure10, Mango247

+Let $I_1;I_2;I_3$ be the excenter of $\triangle ABC$ respectively. $G_1;G_2$ be the centroid of $\triangle DEF;\triangle I_1I_2I_3$ . $O_1;O_2;O_2$ be the circumcenter of $\triangle DEF ;\triangle ABC ; \triangle I_1I_2I_3$
+We'll have $\triangle DEF;\triangle I_1I_2I_3$ is homothetic.Let $G_3$ be the center of a homothety takes $\triangle DEF$ to $\triangle I_1I_2I_3$ .We'll get $\overline {G_1;G_2;G_3} ;\overline {G_3;O_1;O_3} ; \overline {O_1;O_2;O_3}$ .So We'll conclude $\overline{G_1;O_1;O_2}$ Our proof is completed
Our proof is completed

*Remark : Notation $\overline {A;B;C}$ mean $A;B;C$ are collinear

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#3 h Aug 31, 2010, 4:09 PM • 3 Y

Y by Amir Hossein, Adventure10, Mango247

Let $O_1,I,O_2$ be circumcenter of $\Delta ABC,\Delta DEF,\Delta XYZ$ respectively.
Let $A'=AI\cap (O_1),B'=BI\cap (O_1),C'=CI\cap (O_1)$
It's easy to prove that $XZ\parallel FD\parallel C'A'\Rightarrow \frac{IX}{IA'}=\frac{IY}{IB'}$
Smilarity we have $\frac{IY}{IB'}=\frac{IZ}{IC'}$ so $I$ is the homothetic center of two triangles $XYZ$ and $A'B'C'$
Thus $O_1,I,O_2$ are collinear.

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#4 h Sep 25, 2018, 3:57 PM • 4 Y

Y by Amir Hossein, AlastorMoody, Adventure10, Mango247

Let $I$ and $\triangle I_AI_BI_C$ be the incenter and excentral triangle of $\triangle ABC$ . Also, Let $V$ be the Bevan point of $\triangle ABC$ . Then $V$ lies on line $OI$ , where $O$ is the circumcenter of $\triangle ABC$ .

Now, $\angle FDB=\angle FIB=90^{\circ}-\frac{B}{2}=\angle I_ABD \Rightarrow DF \parallel I_AI_C$ . Thus, we get that $\triangle DEF$ and $\triangle I_AI_BI_C$ are homothetic, and their homothety center lies on the line joining their centers, i.e. line $IV$ . This gives that line $OI$ is the Euler line of $\triangle DEF$ , proving that the nine point center of $\triangle DEF$ (i.e. the circumcenter of $\triangle XYZ$ ) lies on line $OI$ .

This post has been edited 1 time. Last edited by math_pi_rate, Sep 25, 2018, 3:58 PM

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#5 h Sep 25, 2018, 6:22 PM • 3 Y

Y by AlastorMoody, Adventure10, Mango247

Amir Hossein wrote:

The circle inscribed in a triangle $ABC$ touches the sides $BC,CA,AB$ in $D,E, F$ , respectively, and $X, Y,Z$ are the midpoints of $EF, FD,DE$ , respectively. Prove that the centers of the inscribed circle and of the circles around $XYZ$ and $ABC$ are collinear.

It is very easy with inversion around the incircle of the triangle $ABC$ . Then the circumcircle of the triangle $ABC$ will be the nine points circle of $DEF$ (the nine points circle of $DEF$ is the circumcircle of the triangle $XYZ$ ).
So we have $O$ , $I$ and the circumcenter of $XYZ$ are collinear.
So we also have $OI$ is the euler line of the contact triangle of the triangle $ABC$ .

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