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Source: Turkey Olympic Revenge 2024 P4

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210 posts

egxa

#1 h Aug 6, 2024, 1:11 PM • 2 Y

Y by SerdarBozdag, Rounak_iitr

Let the circumcircle of a triangle $ABC$ be $\Gamma$ . The tangents to $\Gamma$ at $B,C$ meet at point $E$ . For a point $F$ on line $BC$ which is not on the segment $BC$ , let the midpoint of $EF$ be $G$ . Lines $GB,GC$ meet $\Gamma$ again at points $I,H$ respectively. Let $M$ be the midpoint of $BC$ . Prove that the points $F,I,H,M$ lie on a circle.

Proposed by Mehmet Can Baştemir

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904 posts

#3 h Aug 6, 2024, 1:44 PM

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Consider the linear function $f(X)=\mathbb{P}(X,\Gamma)-\mathbb{P}(X,G)$ . Then $\mathbb{P}(G,\Gamma)=f(G)=\frac{f(E)+f(F)}{2}=\frac{BE^2+BF \cdot CF -EG^2-FG^2}{2}=$ $\frac{(BM^2+BF\cdot CF+EM^2)-2EG^2}{2}=\frac{(FM^2+EM^2)-2EG^2}{2}=\frac{EF^2-2EG^2}{2}=EG$ Thus inverting about $G$ with radius $EG$ maps $F$ , $I$ , $H$ , and $M$ all to line $BC$ , finishing the problem.

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This post has been edited 1 time. Last edited by sami1618, Aug 6, 2024, 1:45 PM

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#4 h Aug 6, 2024, 2:21 PM • 4 Y

Y by alexanderhamilton124, bin_sherlo, egxa, ehuseyinyigit

$\angle FME = 90, GF = GE$ implies $GF = GM = GE$ . Note that $G$ is on the radical axis of the point circle $(E)$ and $(ABC)$ because it lies on the $E$ -midline of the triangle $EBC$ . This gives
$GF^2 = GM^2 = GB \cdot GI = GH \cdot GC$ and because $B,C \in FM$ ; $F,I,M,H,G$ are cyclic.

This post has been edited 2 times. Last edited by SerdarBozdag, Aug 6, 2024, 2:23 PM

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#5 h Aug 6, 2024, 4:15 PM • 1 Y

Y by ehuseyinyigit

We complex bash. Let $\Gamma$ be the unit circle (and let $j$ , rather than $i$ , denote the coordinate of $I$ ), so that
$|a|=|b|=|c|=1$ $e=\frac{2bc}{b+c}$ $m=\frac{b+c}2$ $\overline{f} = \frac{b+c-f}{bc}$ $g = \frac{e+f}2 = \frac{2bc+bf+cf}{2(b+c)}$ $\overline{g} = \frac{\frac2{bc}+\frac{b+c-f}{b^2c}+\frac{b+c-f}{bc^2}}{2\left(\frac1b+\frac1c\right)} = \frac{b^2+4bc+c^2-bf-cf}{2bc(b+c)}$ $j = \frac{b-g}{b\overline{g}-1} = \frac{\frac{2b^2-bf-cf}{2(b+c)}}{\frac{b^2+2bc-c^2-bf-cf}{2c(b+c)}} = \frac{c(2b^2-bf-cf)}{b^2+2bc-c^2-bf-cf}$ $h = \frac{c-g}{c\overline{g}-1} = \frac{\frac{2c^2-bf-cf}{2(b+c)}}{\frac{-b^2+2bc+c^2-bf-cf}{2b(b+c)}} = \frac{b(2c^2-bf-cf)}{-b^2+2bc+c^2-bf-cf}$ $j-f = \frac{bf^2+cf^2-b^2f-3bcf+2b^2c}{b^2+2bc-c^2-bf-cf} = \frac{(b-f)(2bc-bf-cf)}{b^2+2bc-c^2-bf-cf}$ $h-f = \frac{bf^2+cf^2-c^2f-3bcf+2bc^2}{-b^2+2bc+c^2-bf-cf} = \frac{(c-f)(2bc-bf-cf)}{-b^2+2bc+c^2-bf-cf}$ $j-m = \frac{-b^3+b^2c-bc^2+c^3+b^2f-c^2f}{2(b^2+2bc-c^2-bf-cf)} = \frac{(c-b)(b^2+c^2-bf-cf)}{2(b^2+2bc-c^2-bf-cf)}$ $h-m = \frac{b^3-b^2c+bc^2-c^3-b^2f+c^2f}{2(-b^2+2bc+c^2-bf-cf)} = \frac{(b-c)(b^2+c^2-bf-cf)}{2(-b^2+2bc+c^2-bf-cf)}$ Thus
$\frac{(j-f)(h-m)}{(h-f)(j-m)} = -\frac{b-f}{c-f}$ which is real since $F$ lies on line $BC$ . $\blacksquare$

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ohhh

#6 h Aug 7, 2024, 12:39 AM

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Fix triangle $ABC$ and move the point $F$ along the line $BC$ (Yes, ignore the "which is not on the segment" condition).
$degF = 1 \implies degG = 1 \implies degH = degI =1$
Notice that the condition $(IMGF)$ cyclic has degree at most $degGI + degMI + degMF +deg GF= 1 + 1 + 0 + 1= 3$ .
So $4$ points are enough!
$1: F = B$
$2: F = M$
$3: F$ such that $G = M_{bc}$
$4: F$ such that $G = M_a$
Okay! Similarly, we can prove $(HMGF)$ and consequently $(IMHGF)$ , as desired.

This post has been edited 3 times. Last edited by ohhh, Aug 7, 2024, 12:41 AM
Reason: Typos

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#7 h Aug 21, 2024, 2:00 PM • 2 Y

Y by ehuseyinyigit, Ege_Saribass

Let $T$ be the point that satisfies $FBET$ parallelogram. $\angle{TFB}=\angle{EBC}=\angle{BCE},FC||TE\Rightarrow \hspace{1mm}\text{FTEC is cyclic}$ $\angle{TEC}+\angle{FCE}=180^\circ=\angle{TEC}+\angle{TIC}\Rightarrow \text{I lies on FTEC circle}$ . So $FB\cdot BC=2FB\cdot BM=IB\cdot BT=2BG\cdot IB\Rightarrow \text{FGMI is cyclic}$ . We know that $FG=GM$ so $\angle{FIG}=\angle {GIM}=\angle{FMG}$ so $\triangle {GMB}\sim \triangle {GIM}\Rightarrow GM^2=GB\cdot GI=GH\cdot GC$ so $\angle{GMF}=\angle{MHC}$ and we get $\angle{GIM}+\angle{GHM}=180^\circ$ so $H$ lies on $FGMI$ circle. We're done.

This post has been edited 3 times. Last edited by Anancibedih, Aug 21, 2024, 2:01 PM

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#8 h Sep 14, 2024, 2:53 PM • 1 Y

Y by MS_asdfgzxcvb

We have $BC$ is polar of $E$ WRT $\Gamma$ so $\bigodot(EF)$ is orthogonal to $\Gamma$ . From this, we have $\mathcal{I}^{k = GM^2}_G: \Gamma \longleftrightarrow \Gamma$ . Hence $\mathcal{I}^{k = GM^2}_G: F \longleftrightarrow F, M \longleftrightarrow M, B \longleftrightarrow I, C \longleftrightarrow H$ . But $B, C, M, F$ are collinear then $G, I, H, F, M$ lie on a circle

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#10 h Apr 4, 2025, 8:19 PM

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The most natural way is constructing parallelograms. Let $C'$ and $B'$ be reflection of the points $C$ and $B$ wrt to $G$ respectively. Then $\angle BHC=\angle BAC=\angle BCE=\angle BFC'$ implies $BHC'F$ is cyclic and thus $FHMG$ (Since $BC.CF=2MC.CF=C'C.CH=2GC.CH$ ). On the other hand, $BE=EC$ implies $CEB'F$ is cyclic-isosceles trapezoid. Furthermore $\angle BHC=\angle CIB'=\angle CEB'$ gives $B'EICF$ is cyclic. Then PoP gives $BC.BF=2BM.BF=BI.BB'=2BI.BG$ implies $FGMI$ is cyclic. The result follows from $(FGMI)$ and $(FHMG)$ as desired.

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#11 h Apr 4, 2025, 10:31 PM

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$G$ is on the radax of $(ABC)$ and pointcircle $E$ , since that radax is the midparallel in triangle $EBC$ . So inversion with radius $GE$ around $G$ sends $H$ to $C$ and $I$ to $B$ . Now since $G$ is midpoint of $EF$ and $\angle EMF=90^\circ$ we see that $GE=GM=GF$ so $M$ and $F$ remain fixed. So the images of $F, I, H, M$ are collinear so $F, I, H, M$ are concyclic.

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