Same symmedian point. [In memory of Vladimir Zajic (yetti)]

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Same symmedian point. [In memory of Vladimir Zajic (yetti)]

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#1 h Mar 12, 2019, 1:23 AM • 64 Y

Y by Amir Hossein, no_u, 62861, AlastorMoody, buratinogigle, MathStudent2002, khan.academy, Kayak, centos6, Pluto1708, MNJ2357, Arhaan, livetolove212, RMO-prep, rmtf1111, AnArtist, RC., tenplusten, Kezer, MathPassionForever, franchester, mathman3880, yayups, rkm0959, enhanced, e_plus_pi, khanhnx, EinsteinXXI, tworigami, InCtrl, Mathlete99s, bluephoenix, winnertakeover, yshk, RAMUGAUSS, ValidName, WizardMath, Centralorbit, math_pi_rate, A-B-C, foxtrot3, anantmudgal09, WeakMathemetician, Kagebaka, MathGenius_, fibonaccilegend, Vasu090, Awesome555, amar_04, AmirKhusrau, GEO_ADDICT, Functional_equation, Aryan-23, Kanep, OlympusHero, DepressedSucculent, Siddharth03, Mathbee2020, lneis1, Aimingformygoal, Quidditch, LoloChen, nguyenducmanh2705, Adventure10

Let $H$ and $G$ be the orthocenter and centroid of $\triangle ABC.$ Let $X,Y,Z$ be the projections of $G$ on $HA,HB,HC.$ Show synthetically that $\triangle ABC$ and $\triangle XYZ$ have the same symmedian point.

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#2 h Mar 12, 2019, 2:54 PM • 7 Y

Y by Pluto1708, yumeidesu, nguyenhaan2209, ValidName, amar_04, Functional_equation, Adventure10

Let $K$ be symmedian point of triangle $ABC$ , $D$ be the projection of $A$ onto $BC$ , $R$ be the intersection of the tangents through $B,C$ of $(O)$ , $N_a$ be the projection of $H$ onto median $AM_a,$ similarly define $N_b, N_c$ .
Line through $N_a$ and perpendicular to $BC$ intersects $BC$ at $T$ , $AR$ at $S$ , $DR$ at $Q$ , $DK$ at $P$ .
Since $(AZ,KR)=-1,$ we have $D(AZ,KR)=-1$ . Project onto $PQ$ we get $TP=TQ.$
On the other side, $(AS,ZR)=-1$ then $D(AS,ZR)=-1$ , we get $ST=SQ$ . But $TN_a=TS$ then $N_aS=2PN_a.$
We get $\frac{AX}{XD}=2=\frac{SN_a}{N_aP}$ . Then $X,K,N_a$ are collinear. Similarly we get $XN_a, YN_b, ZN_c$ concur at $K$ .
Now we need to show that $XN_a, YN_b, ZN_c$ are symmedians of triangle $XYZ$ .
Since $H(BC,DN_a)=-1$ and $GX\perp AH, GY\perp BH, CZ\perp CH, GN_a\perp HN_a$ then $G(YZ,XN_a)=-1$ . Then $XYN_aZ$ is a harmonic quadrilateral. We are done.

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#3 h Mar 12, 2019, 11:21 PM • 4 Y

Y by yumeidesu, amar_04, Functional_equation, Adventure10

Let $D$ , $E$ , $F$ be the foot of the altitudes from $A$ , $B$ , $C$ . We have

$D=\frac{(a^2+b^2-c^2)B+(a^2+c^2-b^2)C}{2a^2}$
and

$3X=2D+A=\frac{(a^2+b^2-c^2)B+(a^2+c^2-b^2)C+a^2A}{a^2}.$
Thus

$3a^2X=(a^2+b^2-c^2)B+(a^2+c^2-b^2)C+a^2A.$
Similarly

$3b^2Y=(b^2+c^2-a^2)C+(b^2+a^2-c^2)A+b^2B$
and

$3c^2Z=(c^2+a^2-b^2)A+(c^2+b^2-a^2)B+c^2C.$
Let $S'$ and $S$ be symmedian points of $XYZ$ and $ABC$ . Note that $\triangle XYZ\sim\triangle ABC$ , we have

$S'=\frac{YZ^2\cdot X+ZX^2\cdot Y+XY^2\cdot Z}{YZ^2+ZX^2+XY^2}=\frac{a^2X+b^2Y+c^2Z}{a^2+b^2+c^2}=\frac{a^2A+b^2B+c^2C}{a^2+b^2+c^2}=S.$
We are done.

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#4 h Mar 13, 2019, 1:09 AM • 4 Y

Y by brokendiamond, amar_04, Functional_equation, Adventure10

Let $K$ be the symmedian point of $\triangle ABC,$ then by the properties of Hagge circle $\color{blue} ^{[1]}$ we know that $\odot (HG)$ is the K-Hagge circle of $\triangle ABC$ and $\triangle ABC \cup K \stackrel{-}{\sim} \triangle XYZ \cup K,$ so $K$ is the symmedian point of $\triangle XYZ.$

[1] Blog-Properties of Hagge circle (Corollary 1.2 and Property 3)

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#5 h Mar 13, 2019, 3:45 PM • 3 Y

Y by amar_04, Functional_equation, Adventure10

Here is my solution for the problem
Solution
Firstly, we need to prove the lemmas:
Lemma 1: Let $\triangle$ $ABC$ be a given triangle, symmedian point $L$ , altitude $AD$ . Let $M$ , $N$ be midpoint of $BC$ , $AD$ . Prove that: $M$ , $L$ , $N$ are collinear
Proof: Let $S$ be intersection of tangents at $B$ , $C$ of ( $ABC$ ); $T$ $\equiv$ $AL$ $\cap$ $BC$
We have: $B(CALS)$ = ( $TALS$ ) = $-$ 1
So: ( $ASTL$ ) = 2 or $\dfrac{\overline{TA}}{\overline{TS}}$ : $\dfrac{\overline{LA}}{\overline{LS}}$ = 2
Then: $\dfrac{\overline{TA}}{\overline{TS}}$ = 2 $\dfrac{\overline{LA}}{\overline{LS}}$
But: $\dfrac{\overline{TA}}{\overline{TS}}$ = $\dfrac{\overline{AD}}{\overline{SM}}$ = 2 $\dfrac{\overline{AN}}{\overline{SM}}$ so: $\dfrac{\overline{LA}}{\overline{LS}}$ = $\dfrac{\overline{AN}}{\overline{SM}}$ or $M$ , $L$ , $N$ are collinear
Lemma 2: Let $\triangle$ $ABC$ be a given triangle, $G$ , $L$ be centroid, symmedian point of $\triangle$ $ABC$ , altitude $AD$ . Let $Q$ be orthogonal projection of $G$ on $AD$ , $T$ be second intersection of $AL$ with ( $ABC$ ), $P$ be reflection of $T$ through $BC$ . Prove that: $Q$ , $L$ , $P$ are collinear
Proof: Let $U$ be midpoint of $AD$ ; $M$ be midpoint of $BC$ ; $S$ be intersection of tangents at $B$ , $C$ of ( $ABC$ ), $N$ be a point lie on $MS$ which satisfies $\dfrac{\overline{NM}}{\overline{NS}}$ = $\dfrac{1}{4}$
From the proof of lemma 1, we have: $\dfrac{\overline{LA}}{\overline{LS}}$ = $\dfrac{\overline{AU}}{\overline{SM}}$
But: $\dfrac{\overline{UA}}{\overline{QA}}$ = $\dfrac{4}{3}$ = $\dfrac{\overline{NS}}{\overline{MS}}$ then: $\dfrac{\overline{LA}}{\overline{LS}}$ = $\dfrac{\overline{QA}}{\overline{NS}}$ or $Q$ , $L$ , $N$ are collinear
Let $I$ $\equiv$ $AL$ $\cap$ $BC$ ; $J$ $\equiv$ $PT$ $\cap$ $BC$
It's easy to see that: $P$ is $A$ - Humpty point of $\triangle$ $ABC$
We have: $\dfrac{\overline{IJ}}{\overline{ID}}$ = $\dfrac{\overline{TJ}}{\overline{AD}}$ = $\dfrac{- \overline{PJ}}{\overline{AD}}$ = $\dfrac{- \overline{MJ}}{\overline{MD}}$ or ( $DJIM$ ) = $-$ 1
So: ( $MDIJ$ ) = 2 or $\dfrac{\overline{IM}}{\overline{ID}}$ : $\dfrac{\overline{JM}}{\overline{JD}}$ = 2
But: $\dfrac{\overline{IM}}{\overline{ID}}$ = $\dfrac{\overline{SM}}{\overline{AD}}$ , $\dfrac{\overline{JM}}{\overline{JD}}$ = $\dfrac{\overline{PM}}{\overline{PA}}$ then $\dfrac{\overline{SM}}{\overline{AD}}$ = 2 $\dfrac{\overline{PM}}{\overline{PA}}$ = 2 $\dfrac{\overline{MN}}{\overline{AQ}}$ or $Q$ ; $P$ ; $N$ are collinear
Hence: $Q$ , $L$ , $P$ , $N$ are collinear
Back to the main problem
Let $L$ be symmedian point of $\triangle$ $ABC$ ; $P$ is $A$ - Humpty point of $\triangle$ $ABC$
We have: ( $XZ$ ; $XY$ ) $\equiv$ ( $HZ$ ; $HY$ ) $\equiv$ ( $AB$ ; $AC$ ) $\equiv$ $-$ ( $AC$ ; $AB$ ) (mod $\pi$ ) and ( $YX$ ; $YZ$ ) $\equiv$ ( $HX$ ; $HZ$ ) $\equiv$ ( $BC$ ; $BA$ ) $\equiv$ $-$ ( $BA$ ; $BC$ ) (mod $\pi$ )
So: $\triangle$ $XYZ$ $\stackrel{-}{\sim}$ $\triangle$ $ABC$ or $\dfrac{XY}{XZ}$ = $\dfrac{AB}{AC}$
But: ( $PY$ ; $PZ$ ) $\equiv$ ( $XY$ ; $XZ$ ) $\equiv$ $-$ ( $AB$ ; $AC$ ) $\equiv$ ( $HB$ ; $HC$ ) $\equiv$ ( $PB$ ; $PC$ ) (mod $\pi$ ) and ( $ZP$ ; $ZY$ ) $\equiv$ ( $HP$ ; $HY$ ) $\equiv$ ( $HP$ ; $HB$ ) $\equiv$ ( $CP$ ; $CB$ ) (mod $\pi$ ) then: $\triangle$ $PYZ$ $\stackrel{+}{\sim}$ $\triangle$ $PBC$
Hence: $\dfrac{PY}{PZ}$ = $\dfrac{PB}{PC}$ = $\dfrac{AB}{AC}$ = $\dfrac{XY}{XZ}$ or $XP$ is $X$ - symmedian line of $\triangle$ $XYZ$
But from lemma 2, we have: $X$ , $L$ , $P$ are collinear so: $XL$ is $X$ - symmedian line of $\triangle$ $XYZ$
Similarly: $YL$ , $ZL$ are $Y$ - symmedian line, $Z$ - symmedian line of $\triangle$ $XYZ$
Therefore: $L$ is symmedian point of $\triangle$ $XYZ$ or $\triangle$ $ABC$ , $\triangle$ $XYZ$ have common symmedian point

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#6 h Mar 19, 2019, 2:55 AM • 12 Y

Y by buratinogigle, livetolove212, sunken rock, MissionModi2019, GEO_ADDICT, amar_04, Functional_equation, Kanep, OlympusHero, brokendiamond, Adventure10, Mango247

Thank you my friends for your solutions. This is what I did:

Let $D,E,F$ be the projections of $A,B,C$ on $BC,CA,AB$ and let $K$ be the symnmedian point of $\triangle ABC.$ $M$ is the midpoint of $\overline{BC}$ and $AM,AK$ cut $\odot(ABC)$ again at $U,V,$ resp. It's well-known that the 2nd intersection $S$ of $\odot(HBC)$ and $\odot(AEHF)$ lies on $AM$ being the center of the spiral that swaps $\overline{CYF} \sim \overline{BZE}$ and the center of the spiral similarity that swaps $\overline{EF}$ and $\overline{BC}.$ This yields, keeping in mind that $AS$ is the A-symmedian of $\triangle AEF,$ $\triangle SYZ \sim \triangle SCB \sim \triangle SFE \sim \triangle VCB$ $\Longrightarrow$ $\triangle ABC \cup V \sim \triangle XYZ \cup S$ $\Longrightarrow$ $XS$ is the X-simmedian of $\triangle XYZ$ and $S,V$ are symmetric across $BC,$ i.e. $SV \perp BC.$

Let $T$ be the projection of $K$ on $BC$ and let $L \in AS$ be the symmedian point of $\triangle AEF.$ Thus $\tfrac{AL}{AS}=\tfrac{AK}{AV}$ $\Longrightarrow$ $KL \perp BC,$ i.e. $L \in KT.$ If $J$ is the midpoint of $AD,$ we know that $M,K,J$ are collinear (Schawtt line). Hence from $TL \parallel AD,$ it follows that $K$ is the midpoint of $\overline{LT}$ $\Longrightarrow$ $\tfrac{LK}{AX}=\tfrac{\frac{1}{2}LT}{\frac{2}{3}AD}=\tfrac{3}{4} \cdot \tfrac{LM}{AM}.$ But since $ME,MF$ are tangents of $\odot(AEF),$ we have $\tfrac{LS}{AS}=\tfrac{3}{4} \cdot \tfrac{LM}{AM}$ $\Longrightarrow$ $\tfrac{LS}{AS}=\tfrac{LK}{AX}$ $\Longrightarrow$ $K \in XS,$ i.e $XK$ is the X-symmedian of $\triangle XYZ$ and the same for the other symmedians of $\triangle XYZ.$

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#7 h Jun 25, 2022, 4:09 PM

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My solution.
Let $D,E,F$ be the projections of $A,B,C$ on $BC,CA,AB$ and let $L$ be the symmedian point of $\triangle ABC.$ $M$ is the midpoint of $\overline{BC}$ and $AL$ cuts $BC, \odot(ABC)$ again at $T,Q,$ respectively.
Let $P$ be the projection of $H$ on $AM$ . It's well-known that $P$ and $Q$ are symmetrical wrt $BC$ . $PQ$ cuts $BC$ at $R$ .
Easily we have $H,X,Y,G,P$ are concyclic and $XP$ is the symmedian line of $\triangle YXZ$ since $H(AP,EC)=-1$ .
It suffices to prove that $X, L, P$ are collinear, or:
$\frac{LA}{LQ}=\frac{AX}{PQ}\iff \frac{\sin ABL}{\sin LBQ}\cdot\frac{AB}{BQ}=\frac{AD}{3RQ}\iff \frac{\sin GBM}{\sin BGM}\cdot\frac{AM}{MC}=\frac{AT}{3TQ}\iff \frac{AM^2}{BM^2}=\frac{AC}{CQ}$ because it's clear that $\angle LBQ=\angle BGM$ . But the last ratio is always true since $CT$ is also the symmedian line of $\triangle ACQ$ , so $X, L, P$ are collinear.
Let $S$ be the intersection of $XP$ and $YZ$ . Easily we have $\triangle XYZ \sim \triangle ABC$ , so $\frac{XS}{SP}=\frac{AT}{TQ}=\frac{DT}{TR}\implies ST\parallel AD\parallel PR$ or $L$ is also the same symmedian point of $\triangle XYZ$ .

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#8 h Jun 26, 2022, 10:26 AM • 1 Y

Y by Mango247

A similar problem.
Let $H$ and $N_a$ be the orthocenter and Nagel's point of $ABC$ . Let $X,Y,Z$ be the projections of $N_a$ on $HA, HB, HC$ .
* $ABC$ and $XYZ$ have same incenter.

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#10 h Jun 26, 2022, 6:43 PM

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Let $K$ be the symmedian point of $ABC$ . Let $X,Y,Z$ be the projections of $K$ on perpendicular bisectors of $BC, CA, AB$
* $ABC$ and $XYZ$ have same centroid.

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#11 h Jul 1, 2022, 2:14 PM

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archimedes26 wrote:

Let $K$ be the symmedian point of $ABC$ . Let $X,Y,Z$ be the projections of $K$ on perpendicular bisectors of $BC, CA, AB$
* $ABC$ and $XYZ$ have same centroid.

Let $O, G$ be circumcenter, centroid of $\triangle ABC;$ $AK$ intersects $BC, (O)$ at $D, S;$ $M, Q$ be midpoint of $BC, AS;$ $R$ be intersection of tangents at $B, C$ of $(O)$ . Since $\dfrac{\overline{QD}}{\overline{QS}} = \dfrac{\overline{QS}}{\overline{QR}} = \dfrac{\overline{AQ}}{\overline{QR}} = \dfrac{\overline{AQ} + \overline{QD}}{\overline{QR} + \overline{AQ}} = \dfrac{\overline{AD}}{\overline{AR}}$ we have $\dfrac{\overline{GM}}{\overline{GA}} \cdot \dfrac{\overline{QA}}{\overline{QR}} \cdot \dfrac{\overline{XR}}{\overline{XM}} = \dfrac{- 1}{2} \cdot \dfrac{- \overline{QS}}{\overline{QR}} \cdot \dfrac{\overline{KR}}{\overline{KD}} = \dfrac{\overline{QS}}{\overline{QR}} \cdot \dfrac{\overline{AR}}{\overline{AD}} = 1$ Then $X, G, Q$ are collinear. It's easy to prove that $\triangle ABC$ $\stackrel{-}{\sim}$ $\triangle XYZ$ . Suppose that $N, P$ are midpoint of $CA, AB$ . We have $(YZ, YQ) \equiv (OZ, OQ) \equiv (AB, AQ) \equiv (NP, NQ) \pmod \pi$ $(ZQ, ZY) \equiv (OQ, OY) \equiv (AQ, AC) \equiv (PQ, PN) \pmod \pi$ So $\triangle QYZ$ $\stackrel{+}{\sim}$ $\triangle QNP$ . Hence $\dfrac{QY}{QZ} = \dfrac{QN}{QP} = \dfrac{AN}{AP} = \dfrac{CA}{AB} = \dfrac{ZX}{XY}$ or $XQ$ is $X$ - median of $\triangle XYZ$ . But $X, G, Q$ are collinear then $XG$ is $X$ - median of $\triangle XYZ$ . Similarly, we have $G$ is centroid of $\triangle XYZ$

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#12 h Jul 2, 2022, 1:33 PM

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archimedes26 wrote:

A similar problem.
Let $H$ and $N_a$ be the orthocenter and Nagel's point of $ABC$ . Let $X,Y,Z$ be the projections of $N_a$ on $HA, HB, HC$ .
* $ABC$ and $XYZ$ have same incenter.

It's easy to see that $AX = BY = CZ$ and $\triangle ABC$ $\sim$ $\triangle XYZ$ . Suppose that $I$ is incenter of $\triangle ABC;$ $AI$ intersects $(ABC)$ again at $P,$ $P'$ be reflection of $P$ in $BC$ . We have $\angle{P'BY} = \angle{HBC} - \angle{P'BC} = 90^{\circ} - \angle{ACB} - \angle{PBC} = 90^{\circ} - \angle{ACB} - \angle{IAC}$ $= \dfrac{\angle{ABC} - \angle{ACB}}{2} = \angle{ABC} + \angle{IAC} - 90^{\circ} = \angle{PCB} - \angle{HCB} = \angle{P'CZ}$ Then $\triangle P'BY = \triangle P'CZ$ . So we have $P'Y = P'Z$ and $\angle{YP'Z} = \angle{YP'B} + \angle{BP'Z} = \angle{ZP'C} + \angle{BP'Z} = \angle{BP'C} = \angle{BPC}$ Hence $P'$ is midpoint of arc $YZ$ not containing $X$ of $(XYZ)$ . Let $M$ be midpoint of $BC$ and $U$ be orthogonal projection of $I$ on $CA$ then it's easy to see that $\triangle AIU$ $\sim$ $\triangle BPM$ . So we have $\dfrac{AX}{AI} = \dfrac{2 IU}{AI} = \dfrac{2 PM}{BP} = \dfrac{PP'}{IP}$ Hence $X, I, P'$ are collinear or $XI$ is internal bisectorof $\angle{YXZ}$ . Similarly, we have $I$ is incenter of $\triangle XYZ$

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