2015 Taiwan TST Round 1 Mock IMO Day 2 Problem 1

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2015 Taiwan TST Round 1 Mock IMO Day 2 Problem 1

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Source: 2015 Taiwan TST Round 1 Mock IMO Day 2 Problem 1

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102 posts

#1 h Jul 12, 2015, 10:51 AM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be a triangle and $M$ be the midpoint of $BC$ , and let $AM$ meet the circumcircle of $ABC$ again at $R$ . A line passing through $R$ and parallel to $BC$ meet the circumcircle of $ABC$ again at $S$ . Let $U$ be the foot from $R$ to $BC$ , and $T$ be the reflection of $U$ in $R$ . $D$ lies in $BC$ such that $AD$ is an altitude. $N$ is the midpoint of $AD$ . Finally let $AS$ and $MN$ meets at $K$ . Prove that $AT$ bisector $MK$ .

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#2 h Jul 12, 2015, 9:32 PM • 3 Y

Y by thczarif, Adventure10, Mango247

Let the tangents of the circumcircle $\odot(ABC)$ at $B,C$ meet at $Y.$ $AS \equiv AY$ is the A-symmedian of $\triangle ABC.$ If $W \equiv AY \cap BC,$ then $NW$ cuts $MY$ at its midpoint $M',$ due to $AD \parallel MY.$ Since $M'(D,N,A,\infty)=-1=M'(S,W,A,Y)$ $\Longrightarrow$ $M',S,D$ are collinear. If $J \equiv MS \cap DY,$ then $JW \parallel YM$ as $S$ is on the D-median $DM'$ of $\triangle DMY.$ Thus from the complete $MWJY,$ it follows that $(M,W,D,L)=-1.$

Let $F$ be the reflection of $D$ on $M.$ Since $M',S,D$ are collinear, then by symmetry $M',R,F$ are collinear and since $UR \parallel MY,$ then $T \in YF.$ By symmetry, $Z \equiv MR \cap YF$ is the reflection of $J$ across $YM$ and therefore the projection $V$ of $Z$ on $BC$ is the reflection of $W$ on $M$ $\Longrightarrow$ $(M,V,F,U)=(M,W,D,L)=-1$ $\Longrightarrow$ $(M,V,F,U)=(Y,Z,F,T)=A(K,M,F,T)=-1.$ Since $AF \parallel MN,$ it follows that $AT$ is the A-median of $\triangle AKM,$ i.e. $AT$ bisects $\overline{MK}.$

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2312 posts

#3 h Aug 23, 2015, 6:24 PM • 7 Y

Y by Aiscrim, thczarif, enhanced, Siddharth03, Hasin_Ahmad, Adventure10, Mango247

Let $V$ be the projection of $S$ on $BC$ and $J$ $\equiv$ $MN$ $\cap$ $UR,$ $P$ $\equiv$ $AT$ $\cap$ $MN$ . From $VS$ $\stackrel{\parallel}{=}$ $UR$ $\stackrel{\parallel}{=}$ $RT$ we know $VRTS$ is a parallelogram. From $AD$ $\parallel$ $UR$ $\Longrightarrow$ $J$ is the midpoint of $UR$ , so $2MJ$ $\stackrel{\parallel}{=}$ $VR$ $\stackrel{\parallel}{=}$ $ST$ $\Longrightarrow$ $KP$ $\parallel$ $ST$ . Let $X$ be the point on $AM$ s.t. $XP$ $\perp$ $BC$ and $Y$ $\equiv$ $XP$ $\cap$ $BC$ . Since $\triangle KPX$ and $\triangle STR$ are homothetic (with center $A$ ), so $XK$ $\parallel$ $RS$ $\Longrightarrow$ $XK$ $\parallel$ $BC$ $\equiv$ $MY$ , hence combine $XP$ $=$ $YP$ ( $\because$ $XY$ $\parallel$ $AD$ ) we get $XKYM$ is a parallelogram $\Longrightarrow$ $KP$ $=$ $MP$ .

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#4 h Aug 24, 2015, 11:09 PM • 3 Y

Y by govind7701, Adventure10, Mango247

I think we can prove a more general statement:

Let $\triangle ABC$ be a triangle and let $M$ be a variable point on $BC.$ Let $AM$ meet $\odot(ABC)$ for a second time at $R.$ Let $U$ be the projection of $R$ onto $BC$ and let $T$ be the reflection of $U$ in $R.$ Let $D$ be the projection of $A$ onto $BC$ and let $N$ be the midpoint of $\overline{AD}.$ Let $\tau$ be the line passing through $R$ parallel to $BC.$ Point $S$ lies on $\tau$ and satisfies $MR = MS.$ Prove that if $K \equiv AS \cap MN$ , then $AT$ bisects $\overline{MK}.$

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695 posts

#5 h Aug 24, 2015, 11:10 PM • 2 Y

Y by Adventure10, Mango247

My solution to the generalization:

Let $R'$ be the reflection of $R$ in $M$ and let $H$ be the projection of $AD$ onto $\tau.$ Denote $V \equiv AT \cap MN, X \equiv AT \cap BC, Y \equiv AS \cap BC$ , and let $D'$ be the reflection of $D'$ in $M.$

Note that $-1 = A(U, T; R, \infty) = (U, X; M, D).$ Then from $\triangle MUR \sim \triangle MDA$ , we infer $XM : XD = UM : UD = RM : RA.$ Note that $M$ is the circumcenter of $\triangle RR'S$ , which implies that $\angle R'SR = 90^{\circ}.$ Therefore, $\triangle RR'S \sim \triangle RAH$ , implying that $\tfrac{RM}{RA} = \tfrac{RR'}{2RA} = \tfrac{RS}{2RH} = \tfrac{MY}{2MD}.$ Hence, $\tfrac{XM}{XD} = \tfrac{MY}{2MD}.$ It follows that $\frac{XY}{XM} = \frac{YM}{XM} - 1 = 1 + 2\left(\frac{MD}{XD} - 1\right) = 1 + \frac{2XM}{XD} = 1 + \frac{MY}{MD} = \frac{D'Y}{D'M}.$ Because $AD' \parallel MN$ , we obtain $-1 = A(Y, M; X, D') = (K, M; V, \infty).$ $\square$

This post has been edited 1 time. Last edited by Dukejukem, Aug 24, 2015, 11:10 PM

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#9 h Sep 5, 2015, 10:53 AM • 1 Y

Y by Adventure10

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#10 h Nov 4, 2018, 1:50 PM • 2 Y

Y by Adventure10, Mango247

Define $E\equiv\overline{AS}\cap\overline{BC}$ , $P\equiv\overline{AT}\cap\overline{MN}$ , $P'$ as the midpoint of $\overline{KM}$ , and $F$ as the reflection of $M$ over $\overline{AD}$ . To begin, observe that $[AMT]=[DMR]$ and $[ANT]=[NDU]$ , so
$\frac{[AMT]}{[ANT]}=\frac{[DMR]}{[NDU]}=\frac{UR}{DN}\cdot\frac{DM}{DU}.$ Furthermore, because $\triangle ADM\sim\triangle RUM$ and $\overline{AF}\parallel\overline{MS}$ ,
$\frac{UR}{DN}\cdot\frac{DM}{DU}=\frac{2RM}{AR}=\frac{2SE}{AS}=\frac{2EM}{FM}=\frac{EM}{DM}.$ $[asy] size(9cm); defaultpen(fontsize(9pt)); pair A=(-56.8,69.1), B=(-80,-40), C=(80,-40), D=(-56.8,-40), E=(-33.8,-40), M=(0,-40), N=(-56.8,14.5), K=(-42.4,0.7), U=(24.1,-40), R=(24.1,-86.1), S=(-24.1,-86.1), T=(24.1,-132.3), F=(-113.8,-40); draw(A--B--C--cycle, linewidth(0.5)); draw(circumcircle(A,B,C), linewidth(0.5)); draw(A--R--U, linewidth(0.5)); draw(S--R--T, linewidth(0.5)); draw(A--D, linewidth(0.5)); draw(A--S, linewidth(0.5)); draw(M--N, linewidth(0.5)); draw(A--T, linewidth(0.5)); draw(D--R, linewidth(0.4)+grey); draw(M--T, linewidth(0.4)+grey); draw(N--T, linewidth(0.4)+grey); draw(N--U, linewidth(0.4)+grey); draw(A--F--B, linewidth(0.4)+grey); draw(S--M, linewidth(0.4)+grey); dot("$A$", A, NW); dot("$B$", B, SW); dot("$C$", C, SE); dot("$D$", D, SW); dot("$E$", E, SW); dot("$M$", M, (-0.6,-1)); dot("$R$", R, SE); dot("$S$", S, SW); dot("$T$", T, SE); dot("$U$", U, SE); dot("$N$", N, SW); dot("$K$", K, SW); dot("$F$", F, SW); [/asy]$
Now, we apply Menelaus on $\triangle MND$ to obtain
$\frac{MK}{KN}\cdot\frac{NA}{AD}\cdot\frac{DE}{EM}=1\implies \frac{MK}{KN}=\frac{2EM}{DE};$ thus,
$\frac{MP'}{P'N}=\frac{EM}{DM}=\frac{[AMT]}{[ANT]}=\frac{MP}{PN},$ which immediately implies the desired conclusion.

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1900 posts

#11 h Dec 1, 2018, 5:05 PM • 2 Y

Y by Adventure10, Mango247

Why $[AMT]=[DMR]$ ?

Blast_S1 wrote:

To begin, observe that $[AMT]=[DMR]$

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#12 h Dec 3, 2018, 2:25 AM • 2 Y

Y by WolfusA, Adventure10

@above: Sorry for the confusion, $[AMT]=[ART]-[MRT]=[DUR]-[MUR]=[DMR]$ . Hope that helps

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yayups

#13 h Apr 9, 2019, 10:40 PM • 1 Y

Y by Adventure10

https://lh3.googleusercontent.com/-MIjPKqHFuAU/XK0fKcXTweI/AAAAAAAAFAM/Woh3R42D2ssRiRmjWtiGJu3BITDIZezYACK8BGAs/s0/2019-04-09.png

This is quite a terrible solution that I somehow hacked up. Let $H$ be the reflection of $D$ about $M$ , let $V$ be the reflection of $U$ about $M$ , let $I=AS\cap BC$ , let $G=AT\cap BC$ , and let $F=AT\cap MK$ . We wish to show that $F$ is the midpoint of $MK$ .

We see that $\widehat{SB}=\widehat{RC}$ , so $AS$ is the $A$ -symmedian. Quiz 2 problem 2 actually implies then that $K$ is the symmedian point of $ABC$ , but we won't be needing this fact. We need to first prove a lemma and a claim.

Lemma: $(DV;IM)=-1$ .

Proof of Lemma: By $\sqrt{bc}$ inversion, we have that $AM\cdot AS=AI\cdot AR$ , so
$\frac{SA}{AI}=\frac{RA}{MA}.$ This actually implies that
$\frac{SI}{AI}=\frac{MR}{MA}.$ But from similar triangles we have that $SI/AI=IV/ID$ , and $MR/MA=MU/MD=MV/MD$ , so we have
$\frac{IV}{ID}=\frac{MV}{MD},$ or $(DV;IM)=-1$ . $\blacksquare$

Claim: $(GH;IM)=-1$ .

Proof of Claim: Note that $(UT;R\infty)=-1$ , so porjection through $A$ implies that $(UG;MD)=-1$ . The key to crack this claim is to invert about $M$ with radius $MB=MC$ . For a point $P$ , we will denote its image under this map by $P'$ . We have $(UG;MD)=-1$ , so $(U'G';\infty D')=-1$ , so
$2D'=U'+G'.$ From the lemma, we have $(DV;IM)=-1$ , so $(D'V';I'\infty)=-1$ , so
$2I'=D'+V'.$ Adding these two equations, we learn that
$2D'+2I'=D'+G'+2M,$ or
$2I'=(2M-D')+G',$ or $2I'=H'+G'$ . Thus, $(G'H';I'\infty)=-1$ , so $(GH;IM)=-1$ , as desired. $\blacksquare$

Note that $\triangle NDM\sim\triangle ADH$ by SAS, so $AH\parallel MK$ . Thus, projecting $(GH;IM)$ through $A$ , we get that
$(F\infty;KM)=-1.$ Thus, $F$ is the midpoint of $KM$ , as desired. $\blacksquare$

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