On triangle $ABC$ the $AM$ ($M\in BC$) is median and $BB_1$ and $CC_1$ ($B_1 \in AC,C_1 \in AB$) are altitudes. The stright line $d$ is perpendicular to $AM$ at the point $A$ and intersect the lines $BB_1$ and $CC_1$ at the points $E$ and $F$ respectively. Let denoted with $\omega$ the circle passing through the points $E, M$ and $F$ and with $\omega_1$ and with $\omega_2$ the circles that are tangent to segment $EF$ and with $\omega$ at the arc $EF$ which is not contain the point $M$. If the points $P$ and $Q$ are intersections points for $\omega_1$ and $\omega_2$ then prove that the points $P, Q$ and $M$ are collinear.
Problem
Source: Balkan MO Shortlist 2008 G6
Tags: geometry, collinear, circles, tangent, altitudes, median
amar_04
06.04.2020 13:42
BMOSL 2008 G6 wrote: On triangle $ABC$ the $AM$ ($M\in BC$) is median and $BB_1$ and $CC_1$ ($B_1 \in AC,C_1 \in AB$) are altitudes. The stright line $d$ is perpendicular to $AM$ at the point $A$ and intersect the lines $BB_1$ and $CC_1$ at the points $E$ and $F$ respectively. Let denoted with $\omega$ the circle passing through the points $E, M$ and $F$ and with $\omega_1$ and with $\omega_2$ the circles that are tangent to segment $EF$ and with $\omega$ at the arc $EF$ which is not contain the point $M$. If the points $P$ and $Q$ are intersections points for $\omega_1$ and $\omega_2$ then prove that the points $P, Q$ and $M$ are collinear. Let $H$ be the Orthocentre of $\triangle ABC$ and let $T\in AM$ such that $HT\|FE\implies HT\perp AM$. So, by Brocard's Theorem on Cyclic Quadrilateral $BC_1B_1C$ we get that $\{B_1C_1,HT,BC\}$ concurs at a point $X_A$. Let $A_1$ be the foot of altitude from $A$ to $BC$. Then $-1=(X_A,A_1;B,C)\overset{H}{=}(\infty_{EF},A;E,F)\implies M$ is the midpoint of Minor Arc $EF$ of $\omega$. Let the tangency points made by $\{\omega_1,\omega_2\}$ with $EF$ be $\{K_1,K_2\}$ and the tangency points made by $\{\omega_1,\omega_2\}$ with $\omega$ be $\{Z_1,Z_2\}$ respectively. Now by Shooting lemma we get that $\overline{Z_1-K_1-M}$ and $\overline{Z_2-K_2-M}$ and Inversion around $M$ with radius $ME=MF$ swaps $\{K_1,Z_1\}$ and $\{K_2,Z_2\}$ respectively. Hence, $MK_1\cdot MZ_1=MK_2\cdot MZ_2\implies M$ has Equal Powers WRT $\{\omega_1,\omega_2\}$. Hence $M$ lies on the radical axis of $\{\omega_1,\omega_2\}$ which is $PQ$. Hence $\overline{M-P-Q}$. $\blacksquare$
Suntafayato
10.05.2020 00:26
It seems like there's something missing about circles $\omega_1, \omega_2$. There are infinitely many circles that are tangent to minor arc $EF$ and segment $EF$.
parmenides51
10.05.2020 01:16
parmenides51 wrote: On triangle $ABC$ the $AM$ ($M\in BC$) is median and $BB_1$ and $CC_1$ ($B_1 \in AC,C_1 \in AB$) are altitudes. The stright line $d$ is perpendicular to $AM$ at the point $A$ and intersect the lines $BB_1$ and $CC_1$ at the points $E$ and $F$ respectively. Let denoted with $\omega$ the circle passing through the points $E, M$ and $F$ and with $\omega_1$ and with $\omega_2$ the circles that are tangent to segment $EF$ and with $\omega$ at the arc $EF$ which is not contain the point $M$. If the points $P$ and $Q$ are intersections points for $\omega_1$ and $\omega_2$ then prove that the points $P, Q$ and $M$ are collinear. that was the wording
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