Let $ABC$ be an acute triangle, let $H$ and $O$ be its orthocentre and circumcentre, respectively, and let $S$ and $T$ be the feet of the altitudes from $B$ to $AC$ and from $C$ to $AB$, respectively. Let $M$ be the midpoint of the segment $ST$, and let $N$ be the midpoint of the segment $AH$. The line through $O$, parallel to $BC$, crosses the sides $AC$ and $AB$ at $F$ and $G$, respectively. The line $NG$ meets the circle $BGO$ again at $K$, and the line $NF$ meets the circle $CFO$ again at $L$. Prove that the triangles $BCM$ and $KLN$ are similar.
Problem
Source: RMM Extralist 2021 G4
Tags: geometry, config geo, RMM Shortlist
19.09.2023 01:25
Let $X = AH \cap (ABC)$. Then $\angle{NKX} = \angle{ABX}$ and $\angle{NLX} = \angle{ACX} \implies NKXL$ is cyclic and its circumcircle is tangent to $(ABC)$ at $X$. Now it suffices to show that $\angle{KXN} = \angle{BCM}$ (and similarly $\angle{LXN} = \angle{CBM}$). Note that $\angle{KXN} = 180^{\circ} - \angle{AXB} - \angle{GNX}$, so it suffices to show that $\angle{GNX} = \angle{ACX} - \angle{BCM}$. Since $\triangle{CST} \sim \triangle{CHA}$, $\triangle{CSM} \sim \triangle{CHN} \implies \angle{BCM} = \angle{ACH} - \angle{CNH} \implies$ it suffices to show that $\angle{GNX} = \angle{NCB}$. If $H_A = AH \cap BC$, then by similar right triangles, this is equivalent to $dist(G, AH) \cdot dist(C, AH) = dist(N, FG) \cdot dist(N, BC)$, which simplifies to $H_AB \cdot \frac{AG}{AB} \cdot H_AC = HH_A \cdot AH_A \cdot \frac{AG}{AB}$, which is true by Power of a Point from $H_A$ to $(ABC)$. $\square$
19.09.2023 04:24
Sadly this took me around an hour since i've never seen this config before. Let $B',C'$ the B,C antipodes in $(ABC)$ respectivily, let $A_1 \in (ABC)$ such that $AA_1 \parallel BC$, let $(CFO) \cap (BGO)=H_A$ and let $AH \cap BC=D$, also let $BF \cap (BGO)=P$ and $CG \cap (CFO)=Q$, let $(BNH_A) \cap BC=R$ and $(CNH_A) \cap BC=S$ and finally let $BF \cap CG=J$. Claim 1: $AD \cap (ABC)=H_A$. Proof: By Miquel point on $\triangle AFG$ we get $ABH_AC$ cyclic, and also since $\angle BH_AO=\angle AGF=\angle ABC$ so $H_A,O,A_1$ are colinear which implies our claim. Claim 2: $J$ lies in $A_1O$ Proof: Fix $(ABC), B, C$ and move $A$ projectivily in $(ABC)$, now note that $\text{deg} A=2$ and $\text{deg} A_1=2$ by reflection, now $O$ is fixed and let $\ell$ the line through $O$ parallel to $BC$, this line is fixed so $\text{deg} F=1$ and $\text{deg}G=1$ as well, now this means $\text{deg} BF=1$ and $\text{deg} CG=1$ so $\text{deg} J=1+1=2$ hence $O,A_1,J$ colinear if it holds for 5 cases. But what if?: If $BC$ is diameter then 4 cases i'll do wont work so we solve this individually, mainly note that $B=G$ and $C=F$ so in fact $O=J$ hence $H_A,O,J,A_1$ are also colinear here, hence we assume $\angle BAC \ne 90$. The cases: My cases are gonna be $A$ lies in $\ell$ (2 cases), $\angle ABC=90$, $\angle ACB=90$ and $\triangle ABC$ isosceles (2 cases), wow i gave 6 cases xD, and all are trivial so we are done by Moving Points. Claim 3: $BNSFH_A$ and $CNGTH_A$ are cyclic. Proof: First note that By Reim's over $(CFO), (ABC)$ we get that $H_A, F, B'$ colinear and in a similar way we get $H_A,G,C'$ colinear, now $\angle BSF=90=\angle BH_AF$ so $BSFH_A$ is cyclic but by PoP we get $BH \cdot HS=AH \cdot HD=NH \cdot HH_A$ so $BNSH_A$ is cyclic hence $BNSFH_A$ is cyclic and in a similar way u prove $CNGTH_A$ cyclic. Claim 4: $M,J$ are isogonal conjugates in $\triangle ABC$ Proof: By Ceva $J$ lies in the A-median in $\triangle ABC$ and by anti-parallels $AM$ is the A-symedian in $\triangle ABC$, now note that by Reim's $NR \parallel AC$ so $\angle FBC=\angle NBS=\angle ABM$ (last holds becuase $NS,NT$ are tangent to $(BSTC)$ by ortogonality of $(AST), (BSTC)$ using 3TL lemma and inversion), both angle equalities finish the claim. Claim 5: $K,Q,P,L$ are colinear. Proof: Proving $K,Q,P$ colinear is enough as the other is analogous, so first by Claim 2 and Radax we get $BQPC$ cyclic, so now $\angle QPB=\angle QCB=\angle GNS=\angle KGB=\angle KPB$ so $K,Q,P$ colinear as desired and $L$ also lies on it by the same process. Finishing: By angle chase $\angle NKL=\angle ABJ=\angle MBC$ and similarily $\angle NLK=\angle MCB$ so $\triangle BCM \sim \triangle KLN$ as desired.
20.09.2023 09:12
G3 is definitely harder than this. Also, I believe this solution is different from the above ones.
04.07.2024 21:13
Cute! Let $H_A$ denote the reflection of orthocenter in $BC$, then we have $H_ALKN$ cyclic and tangnet to $(ABC)$ at $H_A$ due to angles, so it suffices to show $\measuredangle GNH_A = \measuredangle ACH_A - \measuredangle MCB$, it is well known in this config that $90 = \measuredangle BNG = \measuredangle CNF$ and therefore $BNH_AG$ is cyclic, and so it suffices to show $\measuredangle GNH_A = \measuredangle XBC$ but this follows from $BNH_AG$ cyclic. Remark: This config also appears in InfinityDots 2017
08.07.2024 02:13
Complex bash with $(ABC)$ unit circle. Since the line through $O$ parallel to $BC$ intersects the unit circle at two points with product $bc$ and sum $0$, we have that $f=\frac{bc(a+c)-ac(0)}{bc-ac}=\frac{b(a+c)}{b-a}$ and similarly $g=\frac{c(a+b)}{c-a}$. We have $n=\frac{a+h}2=\frac{2a+b+c}2$ and $m=\frac{s+t}2=\frac{(b+a+c-ac/b)+(c+a+b-ab/c)}4=\frac{-ab^2-ac^2+2abc+2b^2c+2bc^2}{4bc}$, so $\frac{m-b}{m-c}=\frac{-ab^2-ac^2+2abc-2b^2c+2bc^2}{-ab^2-ac^2+2abc+2b^2c-2bc^2}=\frac{-ab+ac-2bc}{-ab+ac+2bc}$. If $(BGO)$ has center $x$ and squared radius $R$, then $(n-k)(\overline{n-g})=\text{Pow}_{(BGO)}(N)=(x-n)(\overline{x-n})-R=(x-n)(\overline{x-n})-x\overline x=n\overline n-n\overline x-\overline nx$. We compute \[x=\frac{gb(\overline b-\overline g)}{g\overline b-\overline gb}=\frac{\frac{bc(a+b)}{c-a}(\frac1b+\frac{a+b}{b(c-a)})}{\frac{c(a+b)}{b(c-a)}+\frac{a+b}{c-a}}=\frac{bc}{c-a}\] Now, \[n\overline n-n\overline x-\overline nx=\frac{2a+b+c}2\cdot\frac{ab+ac+2bc}{2abc}-\frac{2a+b+c}2\cdot\frac{-a}{b(c-a)}-\frac{ab+ac+2bc}{2abc}\cdot\frac{bc}{c-a}=\frac{(a+c)(b-c)(2a^2+ab+ac+2bc)}{4abc(a-c)}\] So \[\frac{n-k}{n-l}=\frac{\frac{(a+c)(b-c)(2a^2+ab+ac+2bc)}{4abc(a-c)}/(\overline{n-g})}{\frac{(a+b)(c-b)(2a^2+ab+ac+2bc)}{4abc(a-c)}/(\overline{n-f})}=\frac{\frac{(a+c)(b-c)}{4abc(a-c)}/(\frac{(a+c)(ab-ac-2bc)}{2abc(a-c)})}{\frac{(a+b)(c-b)}{4abc(a-b)}/(\frac{(a+b)(-ab+ac-2bc)}{2abc(a-b)})}=-\frac{-ab+ac-2bc}{ab-ac-2bc}=\frac{m-b}{m-c}\]as desired.
06.10.2024 06:36
I have a purely synthetic solution
06.10.2024 06:54
Ritwin wrote: I have a purely synthetic solution
splendid solution!