Let $ABC$ be an acute-angled triangle. Let $A_1$ be the midpoint of the arc $BC$ which contain the point $A$. Let $A_2$ and $A_3$ be the foot(s) of the perpendicular(s) of the point $A_1$ to the lines $AB$ and $AC$, respectively. Define $B_2,B_3,C_2,C_3$ analogously. a) Prove that the line $A_2A_3$ cuts $BC$ in the midpoint. b) Prove that the lines $A_2A_3,B_2B_3$ and $C_2C_3$ are concurrents.
Problem
Source: Brazil National Olympiad Junior 2021 #5
Tags: geometry
mathisreal
15.02.2022 01:38
Define $M,N,P$ the midpoints of $BC,AC,AB$.
a) Note that $A_1M$ is perpendicular to $BC$, because $A_1B=A_1C$, so $A_2,A_3,M$ are in the simson line of the point $A_1$(then these points are collinear). Analogously $B_2,B_3,N$ is a simson line and $C_2,C_3,P$ is a simson line.
b) Note that $\angle A_1A_2A=\angle A_1MB=90^{\circ}$, so $\angle AA_2M=\angle BA_1M=\frac{\angle BAC}{2}$, then $A_2M$ is parallel to the angle bisector of $\angle BAC$. As the triangles $MNP$ and $ABC$ are homothetics, we have $MN$ is parallel to $AB$ then $\angle A_3MN=\frac{\angle BAC}{2}=\frac{\angle NMP}{2}$. Finally the three lines are concurrents in the incenter of $\triangle MNP$.
guptaamitu1
15.02.2022 02:28
This problem was also Delta 8.3 of the book Lemmas in Olympiad Geometry.
Humberto_Filho
10.05.2022 02:22
I found a solution using Simson line and complex bash. a) $A_1$ lies on the circumcircle of ABC and $A_2$ and $A_3$ are the projections of this point to the sides of the triangle, $A_2 A_3$ is the simson line of the triangle ABC relative to verex A. So, It suffices to show that the projection of $A_1$ to the side BC meets the midpoint of BC. But this is clear, because we can work with complex numbers. Let A = A = $a^2$, B = $b^2$ , C = $c^2$ and $A_1$ is the well-known point $bc$. Lemma : the complex foot of a point z into a chord AB is given by the formula $\frac{1}{2} * (z + a^2 + b^2 + ab \overline{z})$ So, lets compute using the complex numbers : $\frac{1}{2} * (bc + b^2 + c^2 - \frac{b^2 c^2}{bc}$ but this simplifies to $\frac{1}{2} * ( bc + b^2 + c^2 - bc)$, that is $\frac{1}{2}*(b^2 + c^2)$. But this point is well-known, the MIDPOINT OF BC. Doing similarly to the other points in the other sides, we can check this property. b) Now, the first thing you need to prove is that $\overline{A_2 A_3}$ is parallel to the internal bissector going trought A. This is proven checking if the angular coefficients of the lines are equal. The angular coefficient of the line ab is $\frac{a-b}{\overline{a} - \overline{b}}$ First doing the line $\overline{A_2 A3}$ $\frac{1}{2}*(bc + a^2 + c^2 - b^2 - a^2 c^2 \frac{1}{bc})$ - $\frac{1}{2}*(bc + a^2 + b^2 - a^2 b ^2 * \frac{1}{bc})$ Superior part of the fractions $\frac{1}{2}*(\frac{1}{bc} + \frac{1}{a^2} + \frac{1}{c^2} - \frac{bc}{a^2c^2}$) - $ \frac{1}{2}*(\frac{1}{bc} + \frac{1}{a^2} + \frac{1}{b^2)} - \frac{bc}{a^2 b^2}$ Inferior part of the fraction = eliminate the $\frac{1}{2}$ and do some simplifications. = $(c^2 - \frac{a^2 c}{b}) - (b^2 - \frac{a^2 b}{c})$ superior part of the fraction $(\frac{1}{c^2} - \frac{b}{a^2 c}) - (\frac{1}{b^2} - \frac{c}{a^2 b})$ inferior part of the fraction And this translates to : $\frac{\frac{b c ^3 - a ^2 c ^2 - b ^3 c + a ^2 b^2}{bc}}{\frac{a^2 b^2 - b^3 c - a^2 c^2 + b c^3}{a^2 b^2 c^2}}$ . And this let us left with $a^2 b c$. Now, we do the internal bissector of A, that gives us : $\frac{a^2 + bc}{\frac{1}{a^2} + \frac{1}{bc}}$ that is $a^2 b c$ . So, the lines are indeed parallel. Since the triangles $M_a M_b M_c$ are homothetic to the triangle ABC and their respective lines are parallel, the Simson lines meet themselves at the incenter of $M_a M_b M_c$, so the lines concurr.
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