VMO 2020

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Source: VMO 2020 D2 P6

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199 posts

#1 h Dec 28, 2019, 5:40 AM • 2 Y

Y by Adventure10, Mango247

Let a non-isosceles acute triangle ABC with tha attitude AD, BE, CF and the orthocenter H. DE, DF intersect (AD) at M, N respectively. $P\in AB,Q\in AC$ satisfy $NP\perp AB,MQ\perp AC$
a) Prove that EF is the tangent line of (APQ)
b) Let T be the tangency point of (APQ) with EF,.DT $\cap$ MN={K}. L is the reflection of A in MN. Prove that MN, EF ,(DLK) pass through a piont

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2125 posts

#2 h Dec 28, 2019, 6:05 AM • 3 Y

Y by lachimolala, GeoMetrix, Adventure10

Solution for Part a)
Solution for Part b)

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924 posts

#3 h Dec 28, 2019, 6:41 AM • 4 Y

Y by amar_04, mueller.25, Adventure10, Mango247

For part a)
Perform an inversion centred at $A$ with radius $\sqrt{AH*AD}$ . You get the following equivalent problem.

Inverted problem wrote:

Let $\Delta ABC$ be a triangle and let a line parallel to $BC$ through $H$ intersect $\odot(AHB)$ and $\odot(AHC)$ at $G,I$ . Let the tangent to $\odot(AHB)$ at $G$ meet $AB$ at $J$ and let the tangent to $\odot(AHC)$ at $I$ meet $AC$ at $K$ . Prove that $JK$ is tangent to $\odot(ABC)$ .

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Proof: Let $A'$ be the $A$ -antipode w.r.t $\odot(ABC)$ . Notice that $G-B-A'$ collinear . Similiarly $I-C-A'$ collinear. Now note that $AI$ is a diameter of $\odot(AHC)$ . Well known fact that radius of $\odot(ABC)$ is equal to radius of $\odot(AHC) \implies \overline{AI}=\overline{AA'}$ . Similiarly $\overline{AG}=\overline{AA'}$ . Now note that $\angle JA'G=\angle JGA'=90^\circ -\angle AGB =90^\circ-\angle AA'B \implies JA'$ is tangent to $\odot(ABC)$ . Similiarly $KA'$ is tangent to $\odot(ABC)$ . But note that since $\angle JA'A+\angle KA'A=90^\circ+90^\circ=180^\circ \implies J-K-A'$ collinear which implies the result. $\blacksquare$ .

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237 posts

ABCCBA

#4 h Dec 28, 2019, 9:18 AM • 1 Y

Y by Adventure10

For a) draw $AT \perp BC$ first
For b)
Let $EF \cap MN =X,$ then we need to prove $(X,K;M,N)=-1$
It reduces to prove the following problem :
$\triangle ABC,$ orthic triangle $DEF.$ $M, N$ is the reflection of $D$ with $AB, AC.$ $MN \cap BC =X$ then $(X,D;B,C)=-1$
It suffices to prove $M$ and $N$ lies on $EF$ which is simple

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2327 posts

#6 h Dec 29, 2019, 1:08 AM • 2 Y

Y by qua96, Adventure10

General problem. Let $DEF$ be pedal triangle of any point $J$ with respect to triangle $ABC$ . $A^*$ is isogonal conjugate of $A$ with respect to triangle $DEF$ . $M$ , $N$ are projections of $A^*$ on the lines $DE$ , $DF$ respectively. $EF$ cuts $MN$ at $P$ . $PA^*$ cuts $DA$ at $X$ . $K$ is projection of $A$ on the line $EF$ . $DK$ cuts $MN$ at $Q$ . Prove that reflection of $X$ in line $MN$ lies on circle $(DPQ)$ .

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225 posts

jbaca

#7 h Dec 30, 2019, 12:13 AM • 2 Y

Y by mijail, Adventure10

Solution for part a). Let $T$ be the projection of $A$ on $EF$ . Clearly, $ATFN$ is cyclic. Since $AB$ is the external bisector of $\angle DFE$ we easily get that $ATFN$ is a cyclic kite (in fact, $AMDN$ is a cyclic kite as well), hence $N,\ P$ and $T$ are collinear. Similarly, $ATEM$ is a cyclic kite and $M,\ Q$ and $T$ lie on the same line. We deduce that
$\angle PAT=\angle NAF=\angle PTF$ so $EF$ touches $(PAQ)$ at $T$ .

Solution for part b). Let $S$ be the second intersection point of $(APQ)$ and $(MDN)$ . Because $\angle TSA=90^\circ=\angle DSA$ we infer that $D,\ T$ and $S$ are collinear. Let $R=\overline{AS}\cap \overline{MN}$ . Together with $RK\perp AD$ we conclude that $K$ is the orthocenter of $\bigtriangleup RAD$ , therefore
$\angle DKR=180^\circ-\angle RAD=180^\circ-\angle RLA=\angle RLD$ thus $DLKR$ is cyclic. It suffices to show that $EF$ passes through $R$ , but we know that
$\angle ASN=\angle AMN=\angle ANM$ so $AN$ is tangent to $(NSR)$ , thus
$AT^2=AN^2=AS\cdot AR$ from which $\angle ATR=90^\circ$ and the desired collinearity follows. $\blacksquare$

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phuonghoang2130

90 posts

phuonghoang2130

#8 h Jan 2, 2020, 3:36 PM • 1 Y

Y by Adventure10

Yahhh! A solution by Barycentric Coordinates

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635 posts

#9 h Jan 9, 2020, 7:20 PM • 1 Y

Y by Adventure10

Solution for part a) Obviously, $\widehat {ABN}=\widehat {AHN}=90^{\circ}, \widehat {ANB}=180^{\circ}-\widehat {AHB}=\widehat {ATB},\widehat {ABT}=90^{\circ}$ . This $\implies$ $N$ is the reflection of $T$ by $B\implies \widehat {PTB}=\widehat {PNB}=\widehat {PAN}=\widehat {TAB}$ . Similarly, $\widehat {QTC}=\widehat {TAC}\implies\widehat {PTQ}=180^{\circ}$

Solution for part b)
Applying homothety of center $T$ and ratio $\tfrac 12$ it is sufficient to prove this lemma.
Lemma- Let $\triangle IBC$ be inscribed in $(L)$ has $(T)$ at its incircle. Let $H$ be the midpoint of $BC$ and $O\neq I$ be the intersection of $IT$ and $L$ . Assume that $\odot (OHT)$ cuts $L$ at $K\neq 0$ , then $\odot (HLK)$ , $(OT)$ , $(L )$ all pass through $S$ . Note, $(OT)$ is the circle of diameter $OT$ .
Proof of the Lemma- Let $G$ be $\mathrm {I-mixtilinear incircle}$ of $\triangle IBC$ , then $(G)$ touches $(L)$ , $AB$ , $AC$ at $S, X, Y$ .
Property of Mixtilinear
1. $ST$ passes through $R$ , the reflection of point $O$ through $L \implies S\in\odot (OT)$ .
2. $SO, BC, XY$ are concurrent at $J$ .
Now back to the main proof.
Let $L'$ be the reflection of $O$ by $H$ and $U$ is the center of $\odot (JHO)$ . Let $SR$ cuts $BC$ at $V$ . $SR$ is the bisector of $\widehat {BSC}$ , and $SJ\perp SR$ , $(JBVC)=-1$ . Then $HJ\cdot HV=HB^2=HO\cdot HR=HL'\cdot HR\implies V\in\odot (JLR)$ . Therefore $SR$ , $BC$ , $(JLR)$ all pass through $V$ . The inversion of center $O$ and $\mathrm {power}(OB^2)$ shows that $(SLH)$ , $(L)$ , $(OHT)$ all pass through $K$ . This implies that $S\in\odot (HLK)$ . $(HLK)$ , $(L)$ $(OT)$ all pass through $S$ .
$\blacksquare$

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1618 posts

#10 h Feb 12, 2020, 4:11 PM • 2 Y

Y by Adventure10, Mango247

Here is my solution for this problem
Solution
a) Let $O$ be circumcenter of $\triangle ABC$ ; $T$ $\equiv$ $AO$ $\cap$ $EF$
Since: $AO$ $\perp$ $EF$ and $A$ is $D$ - excenter of $\triangle DEF$ ; we have: $FT = FN$ , $ET = EN$
Then: $AE$ , $AF$ are perpendicular bisectors of $NT$ , $TM$ or $N$ , $P$ , $T$ are collinear and $T$ , $Q$ , $M$ are collinear
So it's easy to see that: $(APQ)$ tangents $EF$ at $T$
b) Let $U$ be midpoint of $MN$ ; $W$ $\equiv$ $EF$ $\cap$ $MN$
Since: $(MNKW) = (EFTW) = - 1$ , we have: $\overline{UK} . \overline{UW} = UN^2 = UM^2 = - \overline{UA} . \overline{UD} = \overline{UL} . \overline{UD}$ or $D$ , $L$ , $K$ , $W$ lie on a circle

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