Figure of midpoints and feet of altitude triangle

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Figure of midpoints and feet of altitude triangle

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Source: VMO 2019

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#1 h Apr 15, 2019, 9:32 PM • 3 Y

Y by nguyendangkhoa17112003, Adventure10, Mango247

Let $ABC$ be an acute, nonisosceles triangle with inscribe in a circle $(O)$ and has orthocenter $H$ . Denote $M,N,P$ as the midpoints of sides $BC,CA,AB$ and $D,E,F$ as the feet of the altitudes from vertices $A,B,C$ of triangle $ABC$ . Let $K$ as the reflection of $H$ through $BC$ . Two lines $DE,MP$ meet at $X$ ; two lines $DF,MN$ meet at $Y$ .
a) The line $XY$ cut the minor arc $BC$ of $(O)$ at $Z$ . Prove that $K,Z,E,F$ are concyclic.
b) Two lines $KE,KF$ cuts $(O)$ second time at $S,T$ . Prove that $BS,CT,XY$ are concurrent.

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#2 h Apr 15, 2019, 10:21 PM • 2 Y

Y by Adventure10, Mango247

Assume $AB < AC$ .

Let $Z' = (EFK)\cap (ABC)$ , then we want to show $Z'\in XY$ . In fact if $J=EF\cap BC$ , then $Z' = JK\cap (ABC)$ , so projecting $(D,J;B,C) = -1$ through $K$ we have $(A,Z;B,C) = -1$ or that $AZ'$ is the $A$ -symmedian. So, it suffices to show that $XY$ is also the $A$ -symmedian, which is true by HMMT 2018 Team 4. This proves a).

Now by Pascal $BS\cap CT = Q$ lies on $EF$ by Pascal on $BACTKS$ . Now move $K$ on $(ABC)$ , we have that for any point $K\in (ABC)$ , projecting through $E$ to $S\in (ABC)$ , then projecting through $B$ to $Q\in EF$ is a projective map. We have $A\to J$ , $B\to E$ , and $C \to F$ . Therefore, $(A,K;B,C) = (J,Q;E,F)$ . To show that $Q\in XY$ it suffices to show that $QE = QF$ , for which it suffices to show that $\frac{JE}{JF} = \frac{AB}{AC}\cdot \frac{KC}{KB}$ . But $\frac{AB}{AC}\cdot \frac{KC}{KB} = \frac{AB}{AC}\cdot \frac{HC}{HB} = \frac{AB}{AC}\cdot \frac{CE}{BF} = \frac{AB}{BF} \cdot \frac{EC}{CA},$ which equals $\frac{FJ}{EJ}$ by Menelaus on $\triangle AEF$ and line $BCJ$ , which solves b). $\blacksquare$

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#3 h Apr 15, 2019, 10:24 PM • 2 Y

Y by Adventure10, amirhsz

Let $Q = EF \cap BC.$

(a) It suffices to prove that $Q \in KZ$ , since that would imply that $QK \cdot QZ = QB \cdot QC = QF \cdot QE$ by PoP, and hence PoP would finish. Let $Z'$ be the $A-$ HM point of $\triangle ABC.$ It's well-known/easy to see that $Q, H, Z'$ are collinear. Hence, since $K, H$ are symmetric w.r.t $BC$ , in order to prove that $Q \in KZ$ , we just need to show that $Z, Z'$ are symmetric w.r.t $BC.$ However, it's well-known that the point which is the reflection of the $A-$ HM point over $BC$ is just the point where the $A-$ symmedian meets $(\triangle ABC)$ for a second time, and so it would therefore suffice to show that $XY$ is the $A-$ symmedian of $\triangle ABC$ . By Pascal on $EDFPMN$ (cyclic hexagon on the nine-point circle of $\triangle ABC$ ), we see that $A, X, Y$ are collinear. Hence, we just need to show that $AX$ is the $A-$ symmedian.

To show this, let $W = AA \cap MP,$ where $AA$ denotes the tangent to $(\triangle ABC)$ at $A.$ It's easy to see by angle-chasing that $AEXW$ is an isosceles trapezoid with $AE || XW$ , and so $PA = PE \Rightarrow P$ is the midpoint of $WX.$ Hence, if we let $P_{\infty}$ denote the point at infinity along $WX$ , projecting the harmonic bundle $(W, P, X, P_{\infty})$ from $A$ onto $(\triangle ABC)$ implies that $ABZC$ is harmonic, where we used that $Z = AX \cap (\triangle ABC).$ This clearly implies the desired, and so we're done.

$\square$

(b) By Pascal on $BSKTCA$ we know that $BS \cap CT \in EF.$ Since $EF$ is anti-parallel to $BC$ , it suffices to show that $BS \cap CT$ is the midpoint of $EF$ , since the median in $\triangle AEF$ is the $A-$ symmedian of $\triangle ABC$ , which we know from part (a) is $XY$ .

So let's show this. If we let $M$ be the midpoint of $EF$ , it suffices by symmetry to show that $CM \cap KF \in (\triangle ABC).$ Notice that $\triangle FHD \sim \triangle FEC$ , and so hence it's clear that $\triangle FHK \sim \triangle MEC$ . This implies that $\angle ECM = \angle HKF$ and so now the result is clear

.

$\square$

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#4 h Apr 16, 2019, 3:20 AM • 1 Y

Y by Adventure10

It was posted here.

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#6 h May 16, 2019, 10:08 PM • 1 Y

Y by Adventure10

For part (a), by Sharygin 2016/12, we know that it suffices to show $XY$ is the $A$ -symmedian. This is just standard projective stuff.

For part (b), we make the following claim: if $L$ is the midpoint of $EF$ then lines $BL$ and $KE$ meet on $\odot(O)$ . Proving this will imply that $BS \cap CT=L$ , which lies on the $A$ -symmedian, so we would be done.

Let $X$ be the spiral center for $EF \mapsto CB$ . Then $\triangle XEL \sim \triangle XCM \sim \triangle XKB$ since $KXBC$ is harmonic. Hence $X$ is the spiral center for $EL \mapsto KB$ , proving the claim.

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#7 h May 27, 2019, 6:10 PM • 2 Y

Y by Adventure10, Mango247

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(a) We make a very important structural claim first.

Claim: $XY$ is the $A$ -symmedian.

Proof of Claim 1: It suffices to show that $AX$ is the isogonal of $AM$ in $\angle A$ . To do this, let $U$ be on $MP$ such that $AU$ tangent to $(ABC)$ . Then, we have
$\angle AUX=\angle(AU,AC)=\pi-\angle CAU=\angle ABC$ and
$\angle EXU = \angle(ED,MP)=\angle(ED,AC)=\angle CED=\angle ABC.$ Thus, $\angle AUX=\angle EXU$ , so $AUXE$ is an iscoleces trapezoid. But we have that $AEDB$ cyclic with center $P$ , so $PA=PE$ , implying that $P$ is the midpoinit of $UX$ . Thus, $(UX;P\infty_{AC})=-1$ . Projecting through $A$ , we get that
$(AU,AX;AB,AC)=-1,$ which is certainly enough to imply that $AX$ is the $\angle A$ -symmedian. It similarly follows that $AY$ is the $\angle A$ -symmedian, so we are done. $\blacksquare$

Now, we have that $Z$ is the harmonic conjugate of $A$ wrt $BC$ in $(ABC)$ . Let $T=EF\cap BC$ and $G=AT\cap (ABC)$ . It is well known that if we define an inversion $\phi$ at $T$ with power $TB\cdot TC$ , then $\phi(G)=A$ . Also, we have that $\phi$ swaps $(E,F)$ .

Now, we know $(BC;TD)=-1$ , so projecting from $A$ , we get $(GK;BC)=-1$ . Inverting this statement, we get $(A\phi(K);CB)=-1$ , which readiliy implies $\phi(K)=Z$ . Thus, $\phi(K)=Z$ and $\phi(E)=F$ , so $EFKZ$ cyclic, as desired.

(b) We claim that they are concurrent on the midpoint of $EF$ which we'll denote $L$ . Note that $AL$ is the $A$ symmedian of $ABC$ since $AEF$ and $ABC$ are inversely similar, so this in fact suffices. So it suffices to show that $T,L,C$ collinear in the following diagram:
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We have $\angle FHD=\angle FEC=\pi-B$ and $\angle DFH=\angle CFE$ , so $\triangle DHF\sim \triangle CEF$ . But then by SAS, we have $\triangle KHF\sim\triangle CEL$ , so $\angle FKH=\angle LCE$ , so $\angle TKA=\angle LCA$ , implying that $LC$ passes through $T$ , as desired.

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#8 h Apr 12, 2020, 3:35 PM

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Solution

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#9 h Jun 5, 2020, 7:48 PM • 4 Y

Y by a_simple_guy, GeoMetrix, mueller.25, Bumblebee60

VMO 2019 P6 wrote:

Given an acute triangle $ABC$ and $(O)$ be its circumcircle, and $H$ is its orthocenter. Let $M, N, P$ be midpoints of $BC, CA, AB$ , respectively. $D, E, F$ are the feet of the altitudes from $A, B$ and $C$ , respectively. Let $K$ symmetry with $H$ through $BC$ . $DE$ intersects $MP$ at $X$ , $DF$ intersects $MN$ at $Y$ .
a) $XY$ intersects smaller arc $BC$ of $(O)$ at $Z$ . Prove that $K, Z, E, F$ are concyclic.
b) $KE, KF$ intersect $(O)$ at $S, T$ ( $S, T\ne K$ ), respectively. Prove that $BS, CT, XY$ are concurent.

$(a)$ By Pascal's Theorem on $DFPMNE$ we get that $\overline{A-X-Y}$ . So, redifine $Z$ as $AX\cap\odot(ABC)$ .

Claim:- $AZ$ is the $A-\text{Symmedian}$ of $\triangle ABC$ .
By Three Tangents Lemma we get that $MF=ME$ . Hence, $MM\|EF$ WRT $\odot(X_5)$ . So, by Pascal's Theorem on $DEFPMM$ we get that $BX\|FE$ . Let $BX\cap AC=L$ . Now as $MP\|AC$ we get that $BX=XL$ . Hence, $AZ$ bisects $FE$ . Now as $FE$ and $BC$ are antiparallel, we get that $AZ$ is the $A-\text{Symmedian}$ of $\triangle ABC$ .

Let $KZ\cap BC=X_A'$ also it's well known that $K\in\odot(ABC)$ . Then $-1=(AZ;BC)\overset{K}{=}(DX_A';BC)$ Now if $EF\cap BC=X_A$ then $(X_AD;BC)=-1\implies\boxed{X_A'\equiv X_A}$ . Hence, $EF,BC,KZ$ are concurrent at a point $X_A$ .So, $X_AF\cdot X_AE=X_AB\cdot X_AC=X_AK\cdot X_AZ\implies E,F,K,X\text{ are concyclic}.\blacksquare$
$(b)$ From ELMO Shortlist 2019 G1 we get that $CT$ bisects $EF$ and $BS$ bisects $EF$ . Hence, $CT,BS,XY$ are concurrent at the midpoint of $XY$ . $\blacksquare$

This post has been edited 1 time. Last edited by amar_04, Jun 5, 2020, 7:48 PM

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#10 h Jul 3, 2023, 6:14 AM

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$\text{I think this one is just about angle chasing and some basic ratios; the problem would be interesting if it didn't have part a.}$

https://i.ibb.co/0XmRdJV/2019-VMO-P6.png

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#11 h Dec 24, 2024, 5:58 PM • 1 Y

Y by ehuseyinyigit

as neither a geo main nor exactly an anti-geo main, this would have been impossible without peeking at the "Muricaaaa" handout
solution

This post has been edited 2 times. Last edited by onyqz, Dec 24, 2024, 6:00 PM

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