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Source: Own , Mock Thailand Mathematic Olympiad P9

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151 posts

#1 h Apr 30, 2025, 9:30 AM • 1 Y

Y by Dasfailure

Let $ABC$ be triangle with point $D,E$ and $F$ on $BC,AB,CA$
such that $BE=CF$ and $E,F$ are on the same side of $BC$
Let $M$ be midpoint of segment $BC$ and $N$ be midpoint of segment $EF$
Let $G$ be intersection of $BF$ with $CE$ and $\dfrac{BD}{DC}=\dfrac{AC}{AB}$
Prove that $MN\parallel DG$

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22 posts

moony_

#2 h Apr 30, 2025, 11:56 AM

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ddit for AEGF and point D: pairs (DE, DF), (DA, DG), (DB, DC) are in involution. Now move this lines to point A in parallel: (AK, AI), (AL, AD), (Ainf, Ainf) and we wanna proof that AL is bissector of angle BAC. Now we move this involution on line BC: (K, I), (L, D), (inf, inf).
CI/CD = CA/CF => CI/BK = BE/CF * CA/BA * CD/BD = 1 => CI = BK => this involution is symmetry by point M => D and L are symmetrical by M => BL/CL = BA/CA => BL - bissector of angle BAC => DG || AL || MN => DG || MN

soooo nice geo ^_^

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moony_

#3 h Apr 30, 2025, 12:18 PM

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ItzsleepyXD wrote:

Let $ABC$ be triangle with point $D,E$ and $F$ on $BC,AB,CA$
such that $BE=CF$ and $E,F$ are on the same side of $BC$
Let $M$ be midpoint of segment $BC$ and $N$ be midpoint of segment $EF$
Let $G$ be intersection of $BF$ with $CE$ and $\dfrac{BD}{DC}=\dfrac{AC}{AB}$
Prove that $MN\parallel DG$

another solution:
Let ABA'C - parallelogram
BD/DC = CA/BA = BA'/CA' => A'D - bissector of angle BA'C.
G, D, A' collinear cuz parallelogram bissector lemma
GD || bissector of angle BAC || MN => GD || MN

you can proove parallelogram bissector lemma by using pappus theorem for lines AB and AC and points E, B, inf and F, C, inf on them

it may be easier than first one... hahaha

upd: point names on pic are wrong, sry

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#4 h Apr 30, 2025, 7:58 PM

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Is $BE=CF$ true for D E F symmetrically ?

This post has been edited 1 time. Last edited by FarrukhBurzu, Apr 30, 2025, 7:59 PM

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Tkn

#6 h May 1, 2025, 1:52 PM

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$[asy] size(9cm); defaultpen(fontsize(11pt)); pair A = (-0.25,2); pair B = (-1,0); pair C = (2,0); path C1 = circle(B,1); path C2 = circle(C,1); pair E1 = intersectionpoint(C1, A--B); pair F = intersectionpoint(C2, A--C); pair M = (E1+F)/2; pair N1 = (B+C)/2; pair P = 2N1-A; pair Q = 2M-A; pair R = extension(Q,F,B,P); pair S1 = extension(E1,Q,C,P); pair T = extension(B,F,C,E1); pair D = extension(T,P,B,C); pair U = extension(F,Q,C,E1); draw(A--B--C--cycle, black); draw(E1--Q--F, black); draw(B--P--C, black); draw(R--Q--S1, black+dashed); draw(E1--F, black); draw(B--F,blue); draw(C--E1,blue); draw(T--P, blue+dashed); draw(M--N1, red+dashed); dot(A); dot(B); dot(C); dot(E1); dot(F); dot(P); dot(Q); dot(R); dot(S1); dot(M,red); dot(N1,red); dot(T,blue); dot(D); label("$A$", A, N, black); label("$B$", B, SW, black); label("$C$", C, E, black); label("$E$", E1, NW, black); label("$F$", F, NE, black); label("$Q$", Q, 1.3W+0.5S, black); label("$P$", P, S, black); label("$R$", R, SW, black); label("$S$", S1, SE, black); label("$N$", N1, SW, black); label("$M$", M, N, black); label("$G$", T, S+0.5W, black); label("$D$", D, NE, black); [/asy]$

Let $P$ and $Q$ denote the reflection of $A$ across $N$ and $M$ , respectively.
Let $R=\overleftrightarrow{FQ}\cap\overleftrightarrow{BP}$ , and $S=\overleftrightarrow{EQ}\cap \overleftrightarrow{CP}$
Note that $AM=MQ$ , and $AN=NP$ . It implies that $AEQF$ , and $ABPC$ are parallelograms.
Since $BE=CF$ , we have $QS=RP=RQ=PS$ . This indicates that $PRQS$ is a rhombus. Furthermore, the line $\overleftrightarrow{PQ}$ must bisect $\angle{BPC}$ .
By the given ratio condition of the point $D$ :
$\frac{BD}{CD}=\frac{AC}{AB}=\frac{BP}{CP},$ so, $D$ lies on the angle bisector of $\angle{BPC}$ ; that is $D\in \overleftrightarrow{PQ}$ .
Next, it suffices to show that $G$ also lies on $\overleftrightarrow{PQ}$ , and it could be proceed using Menelaus' theorem on $\triangle{ABF}$ and the line $\overleftrightarrow{EGC}$ :
$\begin{align*} 1&=\frac{BE}{AE}\cdot \frac{AC}{CF}\cdot \frac{FG}{GB}\\ &=\frac{AE}{BE}\cdot \frac{CF}{AC}\cdot \frac{GB}{FG}\\ &=\frac{FQ}{QR}\cdot \frac{RP}{BP}\cdot \frac{GB}{GF}. \end{align*}$ The converse of Menelaus' theorem also holds for $\triangle{FBR}$ , so $G\in \overleftrightarrow{PQ}$ .
The last step is to observe that $A$ is the homothetic center sending $\overline{MN}\mapsto \overline{QP}$ . Therefore, $\overleftrightarrow{PQ}\parallel\overleftrightarrow{MN}$ .
In conclusion $\overleftrightarrow{MN}\parallel\overleftrightarrow{GD}$ as desired.

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1924 posts

#7 h May 1, 2025, 3:49 PM

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Swap the names of $E$ and $F$ .

Move $E$ linearly along $AC$ , then $F$ moves linearly along $AB$ and $N$ moves linearly. Assume WLOG $AB<AC$ , then let the value of $E$ when $F=A$ be $P$ . Then when $F=B$ and $F=A$ we get that $N$ lies on the line through $M$ and the midpoint of $AP$ . A homothety of scale factor $2$ from $A$ takes this line through $D$ parallel to the $A$ -angle bisector. A simple bary (or MMP) argument shows that $G$ moves on a line (since it's degree $1+1-1=1$ ). When $E=P$ we get $G=P$ which does lie on the desired line. When $E=A$ we get $G=DP\cap AB$ which does lie on the desired line. $\blacksquare$

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DottedCaculator

7357 posts

DottedCaculator

#8 h May 1, 2025, 4:24 PM

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Beautiful problem with a beautiful solution!

Let $D=(0:c:b)$ , $E=(t:c-t:0)$ , and $F=(t:0:b-t)$ . Then, $M=(0:1:1)$ and $N=((b+c)t:b(c-t):c(b-t))$ , so $MN$ has direction $(b+c:-b:-c)$ . In addition, $G=(t:c-t:b-t)$ , so $DG$ also has direction $(t(b+c):(c-t)(b+c)-c(b+c-t):(b-t)(b+c)-b(b+c-t))=(t(b+c):-bt:-ct)=(b+c:-b:-c)$ . Therefore, $MN$ and $DG$ are parallel.

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#9 h May 1, 2025, 11:37 PM

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In this config it is also well known (follows through the midpoint of $BF$ or $CE$ ) that $MN$ is parallel to the angle bisector of $\angle BAC$ .
In fact, the foot of this bisector is isotomic with $D$ by the angle bisector property.

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Ianis

#10 h May 2, 2025, 12:08 AM

Y by

Assassino9931 wrote:

In this config it is also well known (follows through the midpoint of $BF$ or $CE$ ) that $MN$ is parallel to the angle bisector of $\angle BAC$ .
In fact, the foot of this bisector is isotomic with $D$ by the angle bisector property.

Indeed! And hence it suffices to prove that if $G'$ is the reflection of $G$ wart to $M$ then $G'$ lies on the angle bisector of $\angle BAC$ , for which it suffices to prove that the distances from $G'$ to $CA$ and $AB$ are equal. Since $CF=BE$ it suffices to prove that $[G'CF]=[G'EB]$ . Now, observe that $[G'CF]=[G'CB]=[G'EB],$ because $BF\parallel BG\parallel CG'$ and $CE\parallel CG\parallel BF'$ . Done.

Credits go to @magnusarg for this solution.

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#11 h May 2, 2025, 3:42 AM

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Let $P$ be the arc midpoint of $BAC$
Let an arbitrary circle from $P$ and $A$ intersect $AB$ and $AC$ at $E'$ and $F'$ respectively.
As $P$ will be the center of spiral similarity that sends $E'F'\rightarrow BC$
$\implies \angle E'PB=\angle F'PC$ and $E'P=F'P, BP=PC \implies E'B=F'C$
$\therefore E'\equiv E$ and $F'\equiv F$
If $AB<AC$ then let $H=MN\cap AC$
see that as $P$ is the center of spiral similarity that sends $NF\rightarrow MC$ and a fixed point $\implies PHMC$ is cyclic $\implies H$ is a fixed point
and $\angle NHF=\angle NPF=\angle \frac{A}{2} \implies MN \parallel A\text{ angle bisector}$
Now after the introduction of point $A'$ such that $ABA'C$ is a parallelogram the problem reduces to-

Quote:

Let $G$ be a point on the angle bisector of $A$ and let $l_1$ and $l_2$ be lines from $B$ and $C$ such that $l_1$ and $l_2$ is parallel to $AC$ and $AB$ . Let $E=CG\cap l_1$ and $F=BG\cap l_2$ . Prove $BE=CF$ .

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