nevermind

Where can we see the whole igo test?

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.

hydo2332 wrote: Where can we see the whole igo test? https://igo-official.ir/events/7/

hydo2332 wrote: Where can we see the whole igo test? There is an IGO 2020 in contest collection, now.

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.

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 $A$.

One solution for this problem.

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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][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][/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, $AE=AF$.$\square$

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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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 $AD\perp EF\implies AE=AF$.

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, $AE=AF$.

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.

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 $AE=AF$