Acute scalene triangle $ABC$ has $G$ as its centroid and $O$ as its circumcenter. Let $H_a,\, H_b,\, H_c$ be the projections of $A,\, B,\, C$ on respective opposite sides and $D,\, E,\, F$ be the midpoints of $BC,\, CA,\, AB$ in that order. $\overrightarrow{GH_a},\, \overrightarrow{GH_b},\, \overrightarrow{GH_c}$ intersect $(O)$ at $X,\,Y,\,Z$ respectively. a. Prove that the circle $(XCE)$ pass through the midpoint of $BH_a$ b. Let $M,\, N,\, P$ be the midpoints of $AX,\, BY,\, CZ$ respectively. Prove that $\overleftrightarrow{DM},\, \overleftrightarrow{EN},\,\overleftrightarrow{FP}$ are concurrent.
Problem
Source: Vietnam MO 2nd day 3rd problem (last problem)
Tags: geometry
Duy_Thai2002
12.01.2018 11:16
a. See here:https://diendantoanhoc.net/uploads/monthly_01_2018/post-46921-0-22657500-1515735547.jpg
buratinogigle
12.01.2018 15:06
Lemma. Triangle $ABC$ inscribed in $(O)$ with two point $P$ and $Q.$ $P_aP_bP_c$ and $Q_aQ_bQ_c$ are circumcevian triangle of $P,$ $Q.$ The the lines $P_aQ_a,$ $P_bQ_b,$ $P_cQ_c$ are bound a triangle which is perpsective to $ABC.$ Proof. See https://artofproblemsolving.com/community/c6h420593. Return main problem. From a) then $XG$ meets $(O)$ again at $A_0$ then $AA_0\parallel BC$ so $XA$ and $XA_0$ are isogonal in $\angle BXC.$ Hence $$\frac{UB}{UC}\cdot\frac{H_aB}{H_aC}=\frac{XB^2}{XC^2}\quad (1)$$ From cyclic quadrilateral $ABXC$ then $$\frac{UB}{UC}=\frac{BA}{CA}\cdot\frac{BX}{XC}.$$ Therefore $$\frac{XB^2}{XC^2}=\frac{UB^2}{UC^2}:\frac{AB^2}{AC^2}.$$ So combine with (1) we get $$\frac{UB}{UC}=\frac{UB^2}{UC^2}:\frac{AB^2}{AC^2}:\frac{H_aB}{H_aC}$$ Therfore $$\frac{UB}{UC}=\frac{AB^2}{AC^2}\cdot\frac{H_aB}{H_aC}.$$ Let $BY,$ $CZ$ cut$CA,$ $AB$ at $V,$ $W$ then $$\prod\frac{UB}{UC}=\prod \frac{AB^2}{AC^2}\cdot\frac{H_aB}{H_aC}=1.$$ We get $AX,$ $BY,$ $CZ$ are concurrent $I.$ Let $A_1,$ $B_1,$ $C_1$ be the reflections of $A,$ $B,$ $C$ through midpoints of $BC,$ $CA,$ $AB.$ From midline we see $A_1X\parallel DM$ and $\frac{GA_1}{GD}=3.$ Thus the homothety center $G$ ratio $3$ swap line $DM$ to line $A_1X.$ Similarly with lines $EN$ and $FP.$ so we will prove that $A_1X,$ $B_1Y,$ $C_1Z$ are concurrent. Consider triangle $XYZ$ and two points $I$ and $G.$ Then $ABC$ is circumcevian triangle of $I$ with repsect to triangle$ XYZ.$ Then $A_0B_0C_0$ is circumcevian triangle of $G$ with repsect to triangle$ XYZ.$ Thus from the lemmas triangle $A_1B_1C_1$ is bound by $AA_0,$ $BB_0,$ $CC_0$ which is perspective to triangle $XYZ.$ We are done. Note that. $I$ lies on Euler line but we do not need this condition. But for the general problem, we need this. General problem for b) see https://artofproblemsolving.com/community/c6t48f6h1573769 https://artofproblemsolving.com/community/c6h431409
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jred
22.01.2018 14:27
There is a typo in the statement of part a. It should be $BH_a$ instead of $BH$.
FISHMJ25
17.05.2018 20:31
A) (<XYZ denotes angle )Let L be midpoint of BHa.We shall prove that triangles BXHa XAC sre similar . If HaG cuts (ABC) at T than ABCT is trapezoid(by homotethy) so we see that <BXHa=<B=<AXC and <XBC=<XAC so they are similar.Next BX/BHa=XA/AC from that BX/BL=AX/AE and <XBHa=<XAC so triangles BXL and XAE similar and <BLX=<AEX and from this follows that <XLC+<XEC so XLEC is cyclic.