Cono Sur 2020 Problem 3

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Cono Sur 2020 Problem 3

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Source: Cono Sur Math Olympiad 2020 #3

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Vloe

#1 h Dec 3, 2020, 7:27 PM

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Let $ABC$ be an acute triangle such that $AC<BC$ and $\omega$ its circumcircle. $M$ is the midpoint of $BC$ . Points $F$ and $E$ are chosen in $AB$ and $BC$ , respectively, such that $AC=CF$ and $EB=EF$ . The line $AM$ intersects $\omega$ in $D\neq A$ . The line $DE$ intersects the line $FM$ in $G$ . Prove that $G$ lies on $\omega$ .

This post has been edited 1 time. Last edited by Vloe, Jan 4, 2021, 9:38 PM
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#2 h Dec 3, 2020, 9:43 PM • 1 Y

Y by Mango247

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(7cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -18.26753628283823, xmax = 18.278202728019302, ymin = -11.418079024186147, ymax = 11.417036331617656; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); draw((3.5828917555750777,6.736643560989597)--(-8.5,-2.88)--(5.36,-3.46)--cycle, linewidth(2) + zzttqq); /* draw figures */ draw(circle((-1.4116931914993323,0.6129868376194005), 7.902218072112366), linewidth(2.4) + red); draw((3.5828917555750777,6.736643560989597)--(-8.5,-2.88), linewidth(2) + zzttqq); draw((-8.5,-2.88)--(5.36,-3.46), linewidth(2) + zzttqq); draw((5.36,-3.46)--(3.5828917555750777,6.736643560989597), linewidth(2) + zzttqq); draw((3.5828917555750777,6.736643560989597)--(-3.558072464275587,-6.992150010368001), linewidth(2) + blue); draw((-7.626801628273275,5.493302153084237)--(1.037775169046847,-6.900012244631474), linewidth(2) + blue); draw((-7.626801628273275,5.493302153084237)--(-3.558072464275587,-6.992150010368001), linewidth(2) + blue); draw((3.5828917555750777,6.736643560989597)--(1.037775169046847,-6.900012244631474), linewidth(2) + blue); draw((-4.8483321900799465,-3.0328114956532204)--(-4.177775169046848,0.560012244631475), linewidth(2) + blue); draw((5.36,-3.46)--(-0.29744170673588555,3.6483279028105353), linewidth(2) + blue); draw((-4.177775169046848,0.560012244631475)--(5.36,-3.46), linewidth(2) + blue); /* dots and labels */ dot((3.5828917555750777,6.736643560989597),dotstyle); label("$A$", (3.683793227604954,6.974221000885953), NE * labelscalefactor); dot((-8.5,-2.88),dotstyle); label("$B$", (-8.402575360299563,-2.6518788823660677), NE * labelscalefactor); dot((5.36,-3.46),dotstyle); label("$C$", (5.451364918326168,-3.2251453766540292), NE * labelscalefactor); dot((-1.57,-3.17),linewidth(4pt) + dotstyle); label("$M$", (-1.4756052209866972,-2.986284337367379), NE * labelscalefactor); dot((-3.558072464275587,-6.992150010368001),linewidth(4pt) + dotstyle); label("$D$", (-3.4581518470658965,-6.808060965953789), NE * labelscalefactor); dot((-4.177775169046848,0.560012244631475),linewidth(4pt) + dotstyle); label("$F$", (-4.079190549211188,0.7399478755043711), NE * labelscalefactor); dot((1.037775169046847,-6.900012244631474),linewidth(4pt) + dotstyle); label("$T$", (1.1279801072377935,-6.712516550239129), NE * labelscalefactor); dot((-7.626801628273275,5.493302153084237),linewidth(4pt) + dotstyle); label("$G$", (-7.542675618867621,5.68437138873804), NE * labelscalefactor); dot((-4.8483321900799465,-3.0328114956532204),linewidth(4pt) + dotstyle); label("$E$", (-4.74800145921381,-2.8429677137953884), NE * labelscalefactor); dot((1.708332190079945,-3.30718850434678),linewidth(4pt) + dotstyle); label("$V$", (1.7967910172404151,-3.105714857010704), NE * labelscalefactor); dot((-0.29744170673588555,3.6483279028105353),linewidth(4pt) + dotstyle); label("$P$", (-0.20964171276744942,3.8451413862308295), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

This is a fun and challenging problem which I just had to bash

Let $T$ be the second intersection of $\overline{FM}$ with the circle $\omega$ , and let $V$ be the intersection of $\overline{AT}$ with $\overline{BC}$ .
It suffices to show that $ME=MV$ , this stems from the butterfly theorem.

We shall assume that $V'$ is the point such that $MV'=ME$ , and we shall use $T'$ as the intersection of $AV'$ and $\omega$ .

We throw the configuration onto the complex plane and we denote with the lowercase letters the respected points with the same uppercase letters.
Now let $\omega$ be the unit circle, and let $c=\frac{1}{b}$ , by this way we have that $m=\overline{m}$ .

We define point $P$ to be the foot of the perpendicular line from $C$ onto $AB$ , then we have that $p=\frac{1}{2}\left(a+b+\frac{1}{b}-ab^2\right)$ .
Since we have that $CF=CA$ we have that:
$f+a=2p$ this implies that:
$f=b+\frac{1}{b}-ab^2$
Now we calculate point $E$ and $V'$ .
Since we must have that $E$ lies on $BC$ then we must have that:
$\frac{b-e}{\overline{b-e}}=\frac{b-c}{\overline{b-c}}=-1$ this gives us that $\overline{e}=2m-e$ .
But since we have that $BE=BF$ , we must have the following:
$\frac{e-\frac{b+f}{2}}{\overline{e-\frac{b+f}{2}}}=-\frac{b-f}{\overline{b-f}}=-\frac{b-a}{\overline{b-a}}=ab$ from here we get that:
$e=m+\frac{(b^2-2)(1-ab)}{2b(ab+1)}$ since we have that $e+v=2m$ , then we have that:
$v=m+\frac{(ab-1)(b^2-2)}{2b(ab+1)}$
Now finally we get $T'$ .
Since $T$ is on $\omega$ we get the following:
$\frac{a-v}{\overline{a-v}}=\frac{a-t}{\overline{a-t}}=-at$ this easily simplifies to the following:
$t=\frac{v-a}{1-ae}$
But when we plug in all those definitions we get that:
$\frac{f-m}{\overline{f-m}}=\frac{f-t}{\overline{f-t}}$ this implies that the points $F,M$ and $T'$ are all colinear this implies that $T'\equiv T$ and this implies that $V'\equiv V$ .

Now by the butterfly theorem we get that $G$ lies on $\omega$ .

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1079 posts

#10 h Dec 3, 2020, 10:43 PM • 3 Y

Y by hectorraul, Mathematicsislovely, myh2910

This might seem challenging, but this is actually a very easy one.
Draw parallel line to $AB$ through $C$ and let that intersect $\omega$ at $H$ .
Let $F'$ be the reflection of $H$ over $M$ . Obviously, $F'$ lies on $AB$ , since $M$ is the midpoint of $BC$ and $F'H$ and since $AB\parallel CH$ . We show that $F\equiv F'$ . $\angle BAC=180^\circ-\angle BHC=180^\circ-\angle BF'C=\angle AF'C,$ thus $AF'=AC=AF$ and since line intersects circle at most $2$ points, we have that $F\equiv F'$ . Let $J$ be the intersection of $CH$ and line through $B$ which is parallel to $AC$ , then $\angle BJC+\angle BFC=\angle BAC+180^\circ-\angle AFC=180^\circ,$ hence $BFCJ$ is cyclic and from here we have that $\angle BFE=\angle FBE=\angle FBC=\angle BCJ=\angle BFJ,$ which implies that $E$ lies on $JF$ . Let $I=AH\cap BC$ . Since $BF=CH$ we have that $AFJH$ is a parallelogram, hence $EM=MI$ . Let $G'=MH\cap \omega$ . Thus by Butterfly theorem, we have that $E'M=MI$ , where $E'=G'D\cap BC$ . Hence, we have that $E\equiv E'$ (notice that $E'$ cannot be $I$ ) and thus $G\equiv G'$ , we are done.

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508 posts

Jjesus

#11 h Dec 4, 2020, 3:04 PM • 2 Y

Y by myh2910, mijail

Problem proposed by Jefferson Lopez, Peru https://m.facebook.com/story.php?story_fbid=3419851701461711&id=100003107432841&sfnsn=mo

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#12 h Dec 7, 2020, 5:07 AM

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Let $G' = DE\cap\omega$ .

$\angle{CFE} = 180 - \angle{AFC}- \angle{EFB} = 180 - \angle{A} - \angle{B} = \angle C$

$\frac{BE}{EC} = \frac{BG'}{CG'}\cdot\frac{BD}{CD}\rightarrow \frac{G'B}{G'C} = \frac{BE}{EC}\cdot \frac{CD}{BD} = \frac{BE}{EC}\cdot \frac{AB}{AC}$

$\frac{BE}{EC} = \frac{BF}{FC}\cdot \frac{\sin{\angle{EFB}}}{\sin{\angle{CFE}}} = \frac{BF}{FC}\cdot \frac{\sin{\angle B}}{\sin{\angle C}} = \frac{BF}{FC}\cdot\frac{AC}{AB}$

$\frac{G'B}{G'C} = \frac{BF}{FC} = \frac{BF}{AC}$

$\triangle G'FB \sim \triangle G'CA$
$\angle FG'B = \angle CG'A = \angle B$
$BC$ is tangent to $(G'FB)$

$\triangle G'CB \sim \triangle G'AF$
$\angle FCB = \angle A - \angle B$
$\angle CG'F = \angle AG'F - \angle AG'C = \angle CG'B - \angle B = \angle A - \angle B$
$\angle CG'F = \angle FCB$
$BC$ is tangent to $(G'FC)$

Since $BC$ is the common tangent of $(G'FB)$ and $(G'FC)$ and $G'F$ is their radical axis we conclude that $G'$ , $F$ and $M$ are collinear, finishing the proof.

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222 posts

#13 h Apr 26, 2021, 12:26 PM • 2 Y

Y by I_am_human, ike.chen

Very cute and short problem!

Let the reflection of $F$ over $M$ be $F'$ . Now, $F'\in (ABC)$ . Also, let $U$ be the foot of perpendicular from $C$ to $AB$ . Also, let $AF'\cap BC=X$ . Now, $\angle XAB=\angle F'AB=\angle F'CB=\angle ABC$ . Now, observe that $FE\parallel MU\parallel AX$ . Thus, since $U$ is midpoint of $AF$ , we have that $M$ is midpoint of $XE$ . Thus, we are done by Butterfly theorem!

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hakN

#14 h Apr 26, 2021, 1:31 PM

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Mine is quite different.
Let the reflection of $F$ across $M$ be $H$ . Because $FBHC$ is a parallelogram we have $180 - \angle A = \angle BFC = \angle BHC \implies H\in (ABC)$ .
Let $MF\cap (ABC) = G$ and let $DE\cap (ABC) = G'$ .
We have $\frac{EB}{2\cdot EM} = (B,M;E,C) \stackrel{D} = (B,A;G',C)$ .
Also $(B,A;G,C) \stackrel{H} = (B,A;F,AB_{\infty}) = \frac{FB}{FA}$ .
So it suffices to show that $\frac{EB}{2\cdot EM} = \frac{FB}{FA}$ .
Let $N$ be the reflection of $E$ across $M$ . It is clear that $FEHN$ is a parallelogram and so
$\angle BNA = 180 - \angle ENH = 180 - \angle FEM = \angle FEB \implies AN\parallel FE$ .
So the desired result follows: $\frac{FB}{FA} = \frac{EB}{EN} = \frac{EB}{2\cdot EM}$ . $\blacksquare$

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#15 h May 6, 2021, 7:23 PM

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Sketch

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#16 h Aug 15, 2022, 3:35 AM • 3 Y

Y by Mango247, Mango247, Mango247

Let $A'$ be the point on $\omega$ such that $AA' \parallel BC$ , the reflection of $A$ in $M$ be $A_1$ , the reflection of $F$ over $M$ be $F_1$ , and the projection of $C$ onto $AB$ be $P$ . In addition, we redefine $G$ as the second intersection between $\omega$ and $\overline{FMF_1}$ .

Because $BFCF_1$ is a parallelogram, $\measuredangle BF_1C = \measuredangle CFB = \measuredangle CFA = - \measuredangle CAF = \measuredangle BAC$ so $F_1$ lies on $\omega$ . Now, since $MB = MP$ follows from Thales', $\measuredangle BPM = \measuredangle MBP = \measuredangle EBF = \measuredangle BFE$ which gives $EF \parallel MP$ .

Consider the inversion about $(BCP)$ , which has center $M$ . It's clear that $A_1 \leftrightarrow D$ and $F \leftrightarrow G$ , so we know $MBGP$ , $A_1DEE^*$ , and $FGE^*E$ are cyclic. Now, inversion angle properties yield $\measuredangle BGE^* = \measuredangle MGE^* - \measuredangle MGB = \measuredangle FEM - \measuredangle FBM$ $= \measuredangle FEB + \measuredangle EBF = \measuredangle EFB = - \measuredangle EBF = \measuredangle ABC$ $= \measuredangle BAA' = \measuredangle BGA'$ implying $A, E^*, G$ are collinear.

It's easy to see that $A'$ and $A_1$ are symmetric about $\overline{BCE^*}$ . Thus, $\measuredangle ADE = \measuredangle A_1DE = \measuredangle A_1E^*E = - \measuredangle A'E^*E = \measuredangle EE^*G$ $= \measuredangle EFG = \measuredangle EFM = \measuredangle PMF = \measuredangle PMG = \measuredangle PBG = \measuredangle ABG = \measuredangle ADG$ so $D, E, G$ are collinear, which finishes. $\blacksquare$

Remarks: We can use inversion to deduce that there is something linking $A_1 \in EF$ and $G \in DE$ . In particular, the angle chases which I used to prove these two collinearities are not disjoint.

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#17 h Jan 22, 2023, 6:03 AM

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Let $K$ be a point such that $ABKC$ is a parallelogram. We have that $CF = AC = BK$ then $FBKC$ is a isosceles trapezium. Easy to see that $F$ , $E$ and $K$ are collinear. Apply the inversion about the circle of centre $M$ and radius $BM$ . Since $\ BM^{2} = MA \cdot MD = MK \cdot MD$ therefore $D$ and $K$ swap. Let $F'$ be the image of $F$ , then $MD \cdot MK = MF \cdot MF'$ so $F'$ lies on $\odot (\triangle DFK)$ .

Claim: $L = KC \cap MF$ lies on $\omega$
Proof: It's clear that $\frac{MB}{MC} = \frac{MF}{ML}$ then $MF = ML$ and $AFKL$ is a parallelogram. Hence $\angle LAB = \angle LAF = \angle FKL = \angle EKC = \angle ECK = \angle BCL \square$
Notice that $ML \cdot MF' = MF \cdot MF' = MB \cdot MC$ and as $L$ lies on $\omega$ , thus $F'$ lies on $\omega$ too. Redefine $G = DE \cap \omega \neq D$ then $ED \cdot EG = EB \cdot EC = EF \cdot EK$ thereby $G \equiv \omega \cap \odot(\triangle DFK) \equiv F'$ , which concludes the problem $\blacksquare$

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This post has been edited 6 times. Last edited by UI_MathZ_25, Jan 22, 2023, 6:28 AM
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v4913

#18 h Jan 22, 2023, 5:12 PM

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Let FM intersect major and minor arcs BC at K, K’ respectively. Then note that CFBK is a parallelogram since CM = MD, so CF || K’D meaning the angle between CK and FM is equal to the angle between CF and CB. Since <CFE = <ACB, this implies <CFM / <EFM = <BCK / <ACK, and thus (ME/FE) / (MC/CF) = AK/BK, so (ME/BE) / (MD/DB) = AK/BK, and by the Ratio Lemma that implies D, E, K collinear, done.

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#19 h Dec 18, 2024, 10:41 PM

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Using the converse of Butterfly's, we can reduce the problem to the following:

Claim: If $Y = FM \cap \omega$ , $Z = AY \cap BC$ , then $M$ is the midpoint of $EZ$ .

The key observation is to notice that $(BCF)$ and $(BCA)$ are congruent by Law of Sines, and thus are reflections over $M$ . In particular, $M$ bisects both $BC$ and $FY$ , so $BFCY$ is a parallelogram. Angle chasing then gives
$\angle BFE = \angle B = \angle BCY = \angle BAY \implies EF \parallel AY,$
which also makes $FEYZ$ a parallelogram, from which the claim follows. $\blacksquare$

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