Let $ABCD$ be a convex quadrilateral with $\angle B < \angle A < 90^{o}$. Let $I$ be the midpoint of $AB$ and $S$ the intersection of $AD$ and $BC$. Let $R$ be a variable point inside the triangle $SAB$ such that $\angle ASR = \angle BSR$. On the straight lines $AR, BR$ , take the points $E, F$, respectively so that $BE , AF$ are parallel to $RS$. Suppose that $EF$ intersects the circumcircle of triangle $SAB$ at points $H, K$. On the segment $AB$, take points $M , N$ such that $\angle AHM =\angle BHI$ , $\angle BKN = \angle AKI$. a) Prove that the center $J$ of the circumcircle of triangle $SMN$ lies on a fixed line. b) On $BE, AF$ , take the points $P, Q$ respectively so that $CP$ is parallel to $SE$ and $DQ$ is parallel to $SF$. The lines $SE, SF$ intersect the circle $(SAB)$, respectively, at $U, V$. Let $G$ be the intersection of $AU$ and $BV$. Prove that the median of vertex $G$ of the triangle $GPQ$ always passes through a fixed point .
Problem
Source: Vietnam TST 2023 P5
Tags: geometry
14.04.2023 19:01
Bài này nhìn khủng bố quá ))). This problem has too many points and contestants in the test room may even find it difficult to draw the diagram ,not talking about solving alone yet . A real challenge... My sol is pretty standard ,I think the idea may be same to official sol. $\textbf{Solution}$ a) Not hard so I will just say the idea: Use ratio of symmedian line, then use cyclic quad ratio lemma, Steiner ratio and it becomes a well known problem ,proof by angle chasing, SM,SN isogonal so center of (SMN) collinear with center (SAB),S. b) We use a well known lemma $\textbf{Lemma}$ Triangle $ABC$ has bisector $d$. $D,E$ lie on the line from $B,C$ parallel to $d$ and $AD,AE$ isogonal in angle $BAC$. Then the line connecting the midpoints of $DE,BC$ parallel to $d$. Applying the lemma, let $N$ be the midpoint of $EF$ ,$H$ midpoint $AB$ so $NH//SR$. We now will prove $\overline{G,N,S}$. We let $GS$ cut $UV,EF$ at $Z,N'$. $SG$ cut $BA$ at $A_1$. $\frac{N'E}{N'F}=\frac{sin(N'SE)}{sin(N'SF)}\frac{SF}{SE}=\frac{sin(ZSU)}{sin(ZSV)}\frac{SF}{SE} = \frac{ZU}{ZV}\frac{SV}{SU}\frac{SA}{SB}$ ($\triangle{SEB} \sim \triangle{SFA}$) = $\frac{A_1A}{A_1B}\frac{SV}{SU}\frac{SA}{SB} \overset{well-known}{=} 1$. So $N \equiv N'$.done. Let $L$ be midpoint of minor arc $AB$ in $(SAB)$. Now we will prove $\overline{G,W,J}$ with $J$ is the midpoint of minor arc $CD$ of $(SCD)$ or $\frac{HN}{HW}=\frac{LS}{LJ}$ We have $\frac{HN}{HW}=\frac{\frac{BE+FA}{2}}{\frac{BP+AQ}{2}} \overset{Thales}{=} \frac{\frac{BE+FA}{2}}{\frac{\frac{BE}{BS}BC+\frac{AF}{AS}AD}{2}}=\frac{BE+FA}{\frac{AF}{AS}BC+\frac{AF}{AS}AD}=\frac{BE+AF}{BC+AD}\frac{AS}{AF}$ By similar we have $\frac{BE+AF}{SB+SA}=\frac{AF}{AS}$ So $\frac{HN}{HW}=\frac{BE+AF}{BC+AD}\frac{AS}{AF}=\frac{AF}{AF}\frac{(SB+SA)}{BC+AD}\frac{AS}{AF}=\frac{SB+SA}{BC+AD} \overset{(*)}{=} \frac{LS}{LJ}$. $(*)$ $\frac{BC+AD}{SB+SA}+1=\frac{SC+SD}{SB+SA}$ $\frac{LJ}{LS}+1=\frac{SJ}{SL}$ But it is known that (use Ptoleme) $\frac{SJ}{SL}=\frac{SC+SD}{SB+SA}$ since $L,J$ are midpoints of arcs $AB,CD$ So the fixed point it passes through is $J$,which is fixed. $Q.E.D.$
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14.04.2023 19:56
$\textbf{Generalization}$ (My teacher) Let $ABCD$ be a quadrilateral with $\angle{B} < \angle{A} < 90$. Suppose $AD$ cut $BC$ at $S$. $Y,Z$ lie on $AB$ such that $SY,SZ$ isogonal in angle $\angle{ASB}$. $V,U$ lie on $(SAB)$ such that $UV//AB$.The line from $A//SY$ intersects $SV$ at $F$,from $B//SZ$ intersects $SU$ at $E$,from $D//SV$ intersects $AF$ at $P$,from $C//SE$ intersects $BE$ at $Q$. Let $AU$ intersects $BV$ at $G$. $X$ is the midpoint of $PQ$.Prove that: $GX$ passes through a fixed point called $T$ and $TC=TD$. https://www.facebook.com/photo?fbid=6842270659121383&set=a.167495923265590