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Source: 7th Iranian Geometry Olympiad (Advanced) P1

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162 posts

#1 h Nov 4, 2020, 2:05 AM • 5 Y

Y by Eliot, itslumi, TheDarkLord1, Mango247, Rounak_iitr

Let $M,N,P$ be midpoints of $BC,AC$ and $AB$ of triangle $\triangle ABC$ respectively. $E$ and $F$ are two points on the segment $\overline{BC}$ so that $\angle NEC = \frac{1}{2} \angle AMB$ and $\angle PFB = \frac{1}{2} \angle AMC$ . Prove that $AE=AF$ .
Proposed by Alireza Dadgarnia

This post has been edited 3 times. Last edited by Gaussian_cyber, Nov 4, 2020, 6:34 PM

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3787 posts

vsamc

#2 h Nov 4, 2020, 2:08 AM

Y by

nevermind

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435 posts

#3 h Nov 4, 2020, 2:31 AM

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Where can we see the whole igo test?

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142 posts

#4 h Nov 4, 2020, 4:21 AM • 4 Y

Y by GEO_ADDICT, Mango247, Mango247, Mango247

Let $AD$ be altitude of $\triangle ABC$ . Let $Z$ be midpoint of $PN$ and let $Y$ and $Z$ be intersections of angle bisectors of $\angle AMB$ and $AMC$ with $PN$ respectively.
Note that $Z$ lies on $AM$ and that $PN\parallel BC$ .

Now $\angle ZMX=\angle XMC=\angle ZXM$ . So $MZ=ZX$ . Similarly $MZ=ZY$ . So $YZ=ZX$ . Also $PN=NZ$ . So $NZ=PY$ .
Now $PDMN$ is isoceles trapezium with $PN\parallel DM$ . $PY=NZ\implies ZDMY$ is also isoceles trapezium with $DM\parallel YZ$ . So $\angle ZDM=\angle YMD=\frac{\angle AMB}{2}=\angle NEC$ . So $NE\parallel DZ$ . Also $NZ\parallel DE$ . So $NZDE$ is parallelogram. So $DE=NZ$ . Similarly $DF=PY$ . But we know that $PY=NZ$ . So $DE=DF$ . But $AD$ is altitude. So $AE=AF$ as desired.

Q.E.D.

This post has been edited 1 time. Last edited by Euler365, Nov 4, 2020, 4:21 AM

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#6 h Nov 4, 2020, 12:36 PM • 1 Y

Y by Mango247

hydo2332 wrote:

Where can we see the whole igo test?

https://igo-official.ir/events/7/

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162 posts

#7 h Nov 4, 2020, 12:42 PM

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hydo2332 wrote:

Where can we see the whole igo test?

There is an IGO 2020 in contest collection, now.

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r_ef

#8 h Nov 4, 2020, 2:03 PM • 1 Y

Y by GEO_ADDICT

Nice problem! here is sketch of my solution.
Let $T$ be intersections of $AM,PN$ and $S$ be intersection of $PF,NE$ .Let $K$ be midpoint of $EF$ .
Case-1: $FSE$ is right angled triangle.
Case-2: $K,S,T$ are conlinear.
Case-3: $MT=TA=KT$
The final Case is enough to prove that $AK$ is perpendicular bisector of $EF$ and we are done.

This post has been edited 4 times. Last edited by r_ef, Nov 6, 2020, 7:55 PM
Reason: sorry for Bad english

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222 posts

#9 h Nov 5, 2020, 10:46 AM • 1 Y

Y by jelena_ivanchic

Let $D$ be the foot of perpendicular from $A$ to $BC$
Let $E'$ be the reflection of $C$ over $E$ and $F'$ be the reflection of $B$ over $F$ . Now, we have $\angle AE'M=\dfrac{1}{2}\angle AME'$ . Now, construct a tangent from $A$ to $(AME')$ , let it intersect $BC$ at $X$ . Define $F'$ and $Y$ similarly.
$\angle XAM=\angle AE'M=\frac{1}{2}AME'=\angle AME'-\angle XAM=\angle AXM\implies MX=AM$ . Similarly, $MY=AM$ . Thus, $CX=BY$ .
Also, we get that $E'AX$ is isosceles, thus midpoint of $E'X$ is $D$ , thus, $ED=\frac{1}{2}(E'X-E'C)=\frac{1}{2}CX$ . Similarly, $FD=\frac{1}{2}BY$ . Thus, $FD=ED$ . Thus, both are equidistant from $A$ .

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105 posts

#14 h Nov 5, 2020, 7:51 PM

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One solution for this problem.

Attachments:

This post has been edited 1 time. Last edited by dungnguyentien, Nov 5, 2020, 7:52 PM
Reason: .

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#18 h Dec 7, 2020, 10:05 PM

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Fairly tricky for a problem 1.

Let the altitude from $A$ cut $PN$ and $BC$ at $D'$ and $D$ , respectively. Let $M'$ be the midpoint of $PN$ . Now, consider the points $E'$ and $F'$ on $BC$ such that $PE'\parallel NE$ and $NF'\parallel PF$ . Let $PE'$ and $NF'$ cut at $R$ .

$[asy] defaultpen(fontsize(10pt)); size(280); pair A=dir(70); pair B=dir(200); pair C= dir(-20); pair M=midpoint(B--C); pair N=midpoint(A--C); pair P=midpoint(A--B); pair Mp=midpoint(N--P); pair D=foot(A,B,C); pair Dp= extension (A,D,N,P); pair Dr=2*Mp-D; pair R=IP(CP(Mp,P), Mp--Dr); pair Ep= extension (R,P,B,C); pair Fp= extension (R,N,B,C); pair E=Ep+N-P; pair F=Fp+P-N; dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$M$", M, dir(M)); dot("$N$", N, dir(N)); dot("$P$", P, dir(P)); dot("$A$", A, dir(A)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$D'$", Dp, dir(Dp)); dot("$E'$", Ep, dir(Ep)); dot("$F'$", Fp, dir(Fp)); dot("$R$", R, dir(R)); dot("$M'$", Mp, dir(Mp)); draw(A--B--C--cycle); draw(P--N); draw(P--F, blue); draw(N--E, red); draw(R--Ep, red+dashed); draw(R--Fp, blue+dashed); draw(R--D, green); draw(A--M); draw(C--Fp); draw(A--D); [/asy]$

Note that

$\begin{align*} \angle PRN &=180^{\circ}-\angle NPR-\angle RNP\\ &=180^{\circ}-\angle EE'P-\angle NF'F\\ &=180^{\circ}-\angle CEN-\angle PFB\\ &=180^{\circ}-\dfrac{\angle AMB+\angle CMA}{2}\\ &=180^{\circ}-90^{\circ}\\ &=90^{\circ}\\ \end{align*}$
Then, $\angle NM'R=2\angle NPR=2\angle ME'P=2\angle CEN=180^{\circ}-\angle CMA=180^{\circ}-\angle D'M'A=180^{\circ}- \angle DM'N$ so $R-M'-D$ are collinear. Finally, homothety from $R$ implies that $D$ is the midpoint of $E'F'$ , so $DE=E'D-E'E=DF'-PN=DF'-FF'=DF$ Hence, $AE=AF$ . $\square$

This post has been edited 1 time. Last edited by Modesti, Dec 7, 2020, 10:10 PM
Reason: Typo

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116 posts

#19 h Jul 30, 2021, 5:01 AM

Y by

My solution on the test:

Let $X$ $=$ $PF \cap EN$ and $H$ $=$ $PN \cap AM$ .

We see,

$\angle EXF = 180^{\circ} - \angle XEF - \angle XFE = 180^{\circ} - \angle NEC - \angle PFB = 180^{\circ} - \frac{1}{2} (\angle AMB + \angle AMC) = 180^{\circ} - \frac{1}{2} (180^{\circ}) = 90^{\circ}$ .

Let $D$ be midpoint of $EF$ . It will suffice to show tat $AD \perp EF$ .

Clearly $PN \parallel BC$ , then $H$ is midpoint of $PN$ .

Now $\triangle PXN \sim \triangle FXD$ and $PN \parallel EF$ , then $D, X, H$ are collinear.

Finally,

$\angle HDM = \angle XDM = \angle XDF = 2 \angle XED = \angle AMB$

thus $HD = HM = HA$ , therefore $H$ is circumcenter of $\triangle ADM$ , then $ADM$ is right triangle with $AD \perp EF$ , as needed $\blacksquare$

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This post has been edited 1 time. Last edited by UI_MathZ_25, Jul 30, 2021, 5:02 AM
Reason: language

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309 posts

#20 h Aug 15, 2022, 7:20 AM

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Let $G=NE\cap PF$ . Clearly $\angle EGF=90^\circ$ from the angle condition. Let $K=AM\cap PN$ . So $K$ is the midpoint of $PN$ . Let $D=KG\cap BC$ . So $D$ is the midpoint of $EF$ and center of $(EFG)$ because $\triangle EFG$ is right and $K$ is the midpoint of $PN$ . So $\angle KDM=\angle DEG+\angle DGE=2\angle DEG=\angle KMD$ Means $KM=KD$ . Let $L$ be the midpoint of $DM$ . Notice $KLMN\cong KLDP\implies PD=MN\implies PNMD$ is isosceles trapezoid. So it is cyclic. Since $M,N,P$ are the midpoint so $D$ has to be the foot from $A$ to $BC$ . Since $DE=DF$ and $AD\perp EF\implies AE=AF$ .

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515 posts

#21 h Mar 29, 2023, 7:49 AM • 1 Y

Y by anurag27826

Solved with AaravGP a.k.a everythingpi3141592
Sol:- . Let $A'$ be the reflection of $A$ across perpendicular bisector of $BC$ .Let the line through $P \parallel$ to $A'M$ meet $BC$ at $H$ and the line through $N \parallel$ to $A'M$ meet $BC$ at $G$ . $D$ be the foot of $A$ on $BC$ . Note that $D$ is midpoint of $HG$ . The angle condition implies $HP=HF$ and $GN=GE$ . Since $HP=GN$ we get that $HG$ and $EF$ share same midpoint i.e. $D$ . So, $AE=AF$ .

https://cdn.discordapp.com/attachments/961202142809563140/1090542777827328011/image.png

This post has been edited 1 time. Last edited by GeoKing, Mar 29, 2023, 8:03 AM
Reason: sorry

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101 posts

#23 h Aug 14, 2023, 7:28 PM

Y by

Nice problem, for P1. Let $NE$ and $PF$ meet at point $Z$ . Then $\angle EZF=90^\circ$ , let R be the midpoint of $EF$ , then $\angle ZRM=\angle AMB$ (1). Obviously $RZ$ passes through midpoint of $PN$ (because $PN$ // $EF$ ), and $AM$ also passes through midpoint of $PN$ , which means that $AM$ and $RZ$ at midpoint of $PN$ , let $M'$ be the midpoint of $PN$ . From (1) $M'M$ = $M'R$ , and from $PN$ // $BC$ and from $P$ is midpoint of $AB$ , we have $M'A$ = $M'M$ = $M'R$ , which implies that $\angle ARM=90^\circ$ , which gives that $AEF$ is isosceles, as desired.

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127 posts

KAME06

#24 h Oct 18, 2024, 11:46 PM

Y by

Let $O = AM \cap PN$ , $X$ and $Y$ points on $PN$ such that $\angle AMY = \angle YMB$ and $\angle AMX = \angle XMC$
(O between Y and N, O between P and X) Consider $\angle AMB < \angle AMC$ wlog.
Note that it's well known that $PN \parallel BC$ so $\angle YMB = \angle YME =\angle MYN = \angle NEC = \angle ENY$ So $MEYN$ must be an isosceles trapezoid
and $\angle PFB = \angle FPX =\angle XMC$ so $MFPX$ must be an isosceles trapezoid.
Also, $\angle AMY = \angle YME = \angle MYX = \angle MYO$ and $\angle XMC = \angle OXM =\angle AMX = \angle OMX$ , so $OY=OM=OX$ and $O$ is the circumcenter of $\triangle YXM$ . $O$ is on $YX$ so $\angle YMX = 90$ .
Note that $O$ is also the midpoint of $AM$ ( $PO \parallel BM$ and $P$ midpoint of $AB$ ). So $OA=OM=OY=OX$ and $A, Y, X, M$ are concyclic.
1) $\angle AYX = \angle AMX = \angle XMC = \angle PXM = \angle FPX$
$AM$ and $YX$ are diameters, so $\angle YAX = \angle AXM = \angle XMY = \angle MYA = 90$ and $AXMY$ is a parallelogram such that $AY=MX$
Now, using the isosceles trapezoid, $YE=MN=AB/2 = AP=BP$ and using the parallelogram and the isosceles trapezoid, $AY=MX=FP$
2) $YE=BP$ and $YP \parallel BE$ so $YPEB$ must be an isosceles trapezoid and $\angle EYP = \angle YPB = \angle APN$
Using 1 and 2 $\angle AYX + \angle EYP = \angle AYE = \angle FPX + \angle APN = \angle APF$
For LAL ( $YE=AP , \angle AYE = \angle APF, AY=FP$ ) , $\triangle EYA \cong \triangle APF$ so $AE=AF$

This post has been edited 3 times. Last edited by KAME06, Oct 18, 2024, 11:50 PM

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