, 2017

Geometry Difficulty 7.5 National olympiad, round 2 Prove it Balkan Mathematical Olympiad

Given an acute triangle $\triangle ABC$ ( $AC \neq AB$ ) and let $(C)$ be its circumcircle. The excircle $C_1$ corresponding to the vertex $A$ , of center $I_a$ , tangents to the side $BC$ at the point $D$ and to the extensions of the sides $AB$ , $AC$ at the points $E$ , $Z$ respectively. Let $I$ and $L$ be the intersection points of the circles $(C)$ and $(C_1)$ , $H$ the orthocenter of the triangle $EDZ$ and $N$ the midpoint of segment $EZ$ . The parallel line through the point $I_a$ to the line $HL$ meets the line $HI$ at the point $G$ . Prove that the perpendicular line $(e)$ through the point $N$ to the line $BC$ and the parallel line $(\delta)$ through the point $G$ to the line $IL$ meet each other on the line $HI_a$ .

Solution

We have $(e) \perp BC$ and $I_aD \perp BC$ , so $(e) \parallel I_aD$ . Let $T$ , $S$ be the midpoints of the segments $HI_a$ , $HD$ respectively and $Y$ the point of intersection of the lines $HD$ , $EZ$ . Then, $TS \parallel I_aD$ , $TS \perp BC$ and $SY \perp EZ$ .
The Euler circle $(\omega)$ of the triangle $EDZ$ passes through the points $N$ , $Y$ , $S$ . Therefore, the segment $SN$ is a diameter of the circle $(\omega)$ . Thus, the center of $(\omega)$ , let $T'$ , is the midpoint of the segment $SN$ .

On the other hand, we know that the center of Euler circle $(\omega)$ is the midpoint $T$ of $HI_a$ . So $T = T'$ . Therefore, the line $(e)$ passes through the points $T$ , $S$ .
Therefore, we get that the quadrilateral $HSI_aN$ is parallelogram and its diagonals meet each other at the point $T$ .
We consider the inversion $I(I_a, I_aZ^2)$ . As $I_aZ^2 = I_aA \cdot I_aN$ we have $I(N) = A$ . Similarly, if $M_1$ , $M_2$ the midpoints of the segments $DE$ , $DZ$ respectively, we get, $I(M_1) = B$ and $I(M_2) = C$ .
Therefore, the circumcircle $(C)$ of the triangle $ABC$ is the image of the circle $(\omega)$ under the inversion $I$ and the points of the intersection of the circles and $(\omega)$ are invariant under this inversion. But it is well known that the circle of inversion passes through the points of the intersection of the circles $(C)$ and $(\omega)$ . Thus, the Euler circle $(\omega)$ passes through the points $I$ , $L$ .
Also, we consider the inversion $J(H, r^2)$ with
$r^2 = HX \cdot HZ = HD \cdot HY = HW \cdot HE$
where $X$ , $W$ , $Y$ the traces of the altitudes of the triangle $EDZ$ on its sides. Then, $J(Z) = X$ , $J(D) = Y$ and $J(E) = W$ . Therefore, the circumcircle $(C_1)$ of the triangle $ABC$ is the image of the circle $(\omega)$ under the inversion $J$ . Thus, the circle of inversion $J$ passes through the points $I$ , $L$ .
We conclude that $HI = HL$ and $HI_a \perp IL$ and since $(\delta) \parallel IL$ , we have $HI_a \perp (\delta)$ .
If $R$ is the point of intersection of the lines $(\delta)$ , $HL$ , we get that quadrilateral $HRI_aG$ is parallelogram and its diagonals meet each other at the point $T$ . So, the perpendicular line $(e)$ through the point $N$ to the line $BC$ and the parallel line $(\delta)$ through the point $G$ to the line $IL$ , meet each other on the line $HI_a$ .

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.