Collinearity of orthocentres

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Source: China TST 2006

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#1 h Jun 18, 2006, 4:39 AM • 2 Y

Y by Adventure10, Mango247

Let $K$ and $M$ be points on the side $AB$ of a triangle $\triangle{ABC}$ , and let $L$ and $N$ be points on the side $AC$ . The point $K$ is between $M$ and $B$ , and the point $L$ is between $N$ and $C$ . If $\frac{BK}{KM}=\frac{CL}{LN}$ , then prove that the orthocentres of the triangles $\triangle{ABC}$ , $\triangle{AKL}$ and $\triangle{AMN}$ lie on one line.

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Amir.S

#2 h Jun 18, 2006, 10:40 AM • 1 Y

Y by Adventure10

let circumcircle of $\triangle AMN,\triangle AKL$ intersect at $A'\not \equiv A$ it's easy to see that a spiral similarity with center $A'$ , $MK$ will be $NL$ and caues $\frac{MK}{NL}=\frac{KB}{LC}$ then $KB$ will be $NC$ .
hence circumcircle of $\triangle ABC$ will pass through $A'$ . it's clear that midpoints of $MN,KL,BC$ are collinear.let $O_1,O_2,O$ be circumcenters of triangle $AMN,AKL,ABC$ respectively.we have in any triangle that $\frac{HG}{OG}=2$ hence it's easy to see (not obvious) that orthocenter of $AMN,ALK,ABC$ are collinear.

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radio

#3 h Dec 5, 2006, 8:34 AM • 2 Y

Y by Adventure10, Mango247

Amir.S wrote:

we have in any triangle that $\frac{HG}{OG}=2$ hence it's easy to see (not obvious) that orthocenter of $AMN,ALK,ABC$ are collinear.

Why is it easy to see? I can't see!

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#4 h Dec 5, 2006, 12:23 PM • 2 Y

Y by Adventure10, Mango247

We can solve this problem, based on the follows simple idea.

LEMMA. – Two lines $(\ell),$ $(\ell'),$ are given and we define three arbitrary points $A,$ $B,$ $C,$ on $(\ell)$ $($ $B,$ between $A,$ $C,$ $)$ and also three points $A',$ $B',$ $C',$ on $(\ell'),$ such that $AB / BC = A'B' / B'C'$ $($ $B',$ between $A',$ $C'$ $).$

Through the points $A,$ $B,$ $C,$ we draw three arbitrary lines $(a),$ $(b),$ $(c),$ parallel each other and also through the points $A',$ $B',$ $C',$ we draw three other ones $(a'),$ $(b'),$ $(c').$ If the directions of these two sets of parallel lines, intersect each other, then the points $K\equiv (a)\cap (a'),$ $L\equiv (b)\cap (b'),$ $M\equiv (c)\cap (c'),$ are collinear.

We can easy to prove this lemma, by Thales theorem.

In your configuration now, we denote as $B',$ $K',$ $M',$ the orthogonal projections of $B,$ $K,$ $M$ respectively, on the sideline $AC$ and also as $C',$ $L',$ $N',$ the ones, of $C,$ $L,$ $N,$ on the sideline $AB.$ So, because of $BB'\parallel KK'\parallel MM'$ and $CC'\parallel LL'\parallel NN',$ we have that the points $H\equiv BB'\cap CC',$ $H'\equiv KK'\cap LL'$ and $H''\equiv MM'\cap NN',$ as the orthocenters of the triangles $\bigtriangleup ABC,$ $\bigtriangleup AKL,$ $\bigtriangleup AMN,$ based on the above lemma, are collinear and the proof is completed.

Kostas vittas.

PS. This proof is dedicated to Silouanos Brazitikos.

This post has been edited 4 times. Last edited by vittasko, Jun 15, 2012, 6:19 PM

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#5 h Dec 5, 2006, 3:49 PM • 1 Y

Y by Adventure10

Amir.S wrote:

that orthocenter of $AMN,ALK,ABC$ are collinear.

I think you haven't been able to claim that yet.

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radio

#6 h Dec 6, 2006, 10:37 AM • 2 Y

Y by Adventure10, Mango247

Okay, vittasko, I agree that your proof is very nice.

However, I still want to see how to finish the solution in Amir.S's way.

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#7 h Dec 6, 2006, 3:07 PM • 2 Y

Y by Adventure10, Mango247

radio wrote:

Okay, vittasko, I agree that your proof is very nice.

However, I still want to see how to finish the solution in Amir.S's way.

I think that the Amir.S’s solution is correct. Although it is a difficult way to solve this problem, however it is based on the below powerful theorem, very useful to all of us in the future.

THEOREM. – A quadrilateral $ABCD$ is given and let $E,$ $F,$ two points on $AB,$ $CD$ respectively $($ $E,$ between $A,$ $B$ and $F,$ between $C,$ $D$ $)$ , such that $AE / EB = DF / FC = p.$

We define three points $K,$ $L,$ $M,$ on the segments $AD,$ $EF,$ $BC$ respectively, such that $AK / KD = EL / LF = BM / MC = q.$ Prove that the points $K,$ $L,$ $M$ are collinear, such that $KL / LM = p.$

REMARK. – This theorem is true, when the given quadrilateral $ABCD$ is convex, or non-convex and also when it's vertices $A,$ $B,$ $C,$ $D,$ are not in the same plane.

Kostas Vittas.

PS. I have been used the above theorem, for the solution of a problem presented by Tiks, in this forum. You can see at http://www.Mathlinks.ro/Forum/viewtopic.php?t=107395

As I have been promised, I will post here next time, the two elementary proofs of this theorem, I have in mind.

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#8 h Dec 6, 2006, 3:09 PM • 2 Y

Y by Adventure10, Mango247

Extremely nice solution Mr Vittas .Thank you for the dedication .

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radio

#9 h Dec 8, 2006, 2:42 PM • 2 Y

Y by Adventure10, Mango247

vittasko wrote:

THEOREM. – A quadrilateral $ABCD$ is given and let $E,$ $F,$ two points on $AB,$ $CD$ respectively $($ $E,$ between $A,$ $B$ and $F,$ between $C,$ $D$ $)$ , such that $AE / EB = DF / FC = p.$

Okay. Please prove it.

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#10 h Dec 8, 2006, 10:42 PM • 2 Y

Y by Adventure10, Mango247

THEOREM. – A quadrilateral $ABCD$ is given and let $E,$ $F$ be, two points on $AB,$ $CD$ respectively $($ $E,$ between $A,$ $B$ and $F,$ between $C,$ $D$ $)$ , such that $AE / EB = DF / FC = p.$
We define three points $K,$ $L,$ $M,$ on the segments $AD,$ $EF,$ $BC$ respectively, such that $AK / KD = EL / LF = BM / MC = q.$ Prove that the points $K,$ $L,$ $M$ are collinear, such that $KL / LM = p.$

PROOF 1. – $($ In my drawing $AB = 5.5,$ $BC = 8.0,$ $AC = 6.3,$ $CD = 2.8,$ $AD = 4.5$ $).$

Through vertices $A,$ $C$ of $ABCD,$ we draw two lines parallel to $CD,$ $AD$ respectively and we denote as $A',$ their intersection point. Through $E,$ we draw two lines parallel to $BA',$ $AA',$ which intersect the segment lines $AA',$ $BA',$ at points $F',$ $E',$ respectively.

Because of $EF'\parallel BA'$ $\Longrightarrow$ $\frac{AE}{EB}= \frac{AF'}{F'A'}$ $\Longrightarrow$ $\frac{AF'}{F'A'}= \frac{DF}{FC}$ $($ because of $\frac{AE}{EB}= \frac{DF}{FC}$ $)$ $\Longrightarrow$ $AD\parallel F'F\parallel A'C$ $,(1)$

Through the point $K,$ we draw a line parallel to $AA'\parallel CD,$ which intersects the segment line $A'C,$ at a point so be it $K'$ and let $L'$ be, the intersection point of $E'C,$ $MK'.$

$\bullet$ From the parallelograms $EF'A'E',$ $F'FCA'$ $\Longrightarrow$ $EE\parallel = F'A'\parallel FC$ and $E'C\parallel = EF$

From the parallelogram $AA'CD$ $\Longrightarrow$ $\frac{AK}{KD}= \frac{A'K'}{K'C}$ $\Longrightarrow$ $\frac{A'K'}{K'C}= \frac{BM}{MC}$ $,(2)$ $($ because of $\frac{AK}{KD}= \frac{EL}{LF}= \frac{BM}{MC}$ $)$

From $(2)$ $\Longrightarrow$ $MK'\parallel BA'$ $\Longrightarrow$ $\frac{E'L'}{L'C}= \frac{BM}{MC}= \frac{EL}{LF}$ $\Longrightarrow$ $LL'\parallel = EE'$ $,(3)$

From $(3)$ and because of $KK'\parallel = AA'$ $,(4)$ and $EE'\parallel AA'$ $\Longrightarrow$ $LL'\parallel KK'$ $,(5)$

From $MK'\parallel BA'$ and $EE'\parallel AA',$ we have also $\frac{ML'}{MK'}= \frac{BE'}{BA'}= \frac{EE'}{AA'}$ $,(6)$

From $(3),$ $(4),$ $(5),$ $(6)$ $\Longrightarrow$ $\frac{ML'}{MK'}= \frac{LL'}{KK'}$ $,(7)$

From $(7),$ we conclude that the points $K,$ $L,$ $M,$ are collinear, such that $\frac{KL}{LM}= \frac{K'L'}{L'M}= \frac{A'E'}{E'B}= \frac{AE}{EB}= p$ and the proof is completed.

Kostas Vittas.

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#11 h Dec 11, 2006, 9:00 PM • 1 Y

Y by Adventure10

PROOF 2. – $($ In my drawing $AB = 3.5,$ $BC = 5.2,$ $AC = 8.2,$ $CD = 9.0,$ $AD = 6.7$ $).$

Through $K,$ we draw two lines $(\ell),$ $(\ell'),$ parallel to $AB,$ $DC,$ respectively.Through the points $E,$ $B,$ we draw two lines parallel to $AD,$ which intersect the line $(\ell),$ at points $E',$ $B',$ respectively. Through the points $F,$ $C,$ we draw two lines also parallel to $AD,$ which intersect the line $(\ell'),$ at points $F',$ $C',$ respectively.

$\bullet$ From the parallelograms $AKB'B,$ $DKC'C$ $\Longrightarrow$ $BB'\parallel = EE'\parallel = AK$ $,(1)$ and $CC'\parallel = FF'\parallel = DK$ $,(2)$

From $(1),$ $(2)$ $\Longrightarrow$ $BB'\parallel CC'$ $,(3)$ and $EE'\parallel FF'$ $,(4)$ and let $L\equiv EF\cap E'F'$ be and $M\equiv BC\cap B'C'$

We will prove that $\frac{EL}{LF}= \frac{BM}{MC}= \frac{AK}{KD}= q$ and that the points $K,$ $L,$ $M,$ are collinear, such that $\frac{KL}{LM}= p$

From $(3),$ $(4)$ $\Longrightarrow$ $\frac{BM}{MC}= \frac{BB'}{CC'}= \frac{EE'}{FF'}= \frac{AK}{KD}= q$ $,(5)$

Similarly we have that $\frac{B'M}{MC'}= \frac{BM}{MC}= \frac{E'L}{LF'}= \frac{EL}{LF}= q$ $,(6)$

From $(1),$ $(2),$ we have also that $\frac{KE'}{E'B'}= \frac{AE}{EB}= \frac{DF}{FC}= \frac{KF'}{F'C'}= p$ $\Longrightarrow$ $E'F'\parallel B'C'$ $,(7)$

From $(6,)$ $(7)$ $\Longrightarrow$ the points $K,$ $L,$ $M,$ are collinear such that $\frac{KL}{LM}= p$ and the proof is completed.

Kostas Vittas.

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#12 h Dec 15, 2006, 2:48 PM • 2 Y

Y by Adventure10, Mango247

Well, say the circumcenters of $AMN,AKL,ABC$ are $O_{1},O_{2},O_{3}$ . To use your lemma, I think we need to prove $\frac{O_{1}O_{2}}{O_{2}O_{3}}=\frac{MK}{KB}$ , which I can't.

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#13 h Dec 15, 2006, 5:10 PM • 2 Y

Y by Adventure10, Mango247

barasawala wrote:

Well, say the circumcenters of $AMN,AKL,ABC$ are $O_{1},O_{2},O_{3}$ . To use your lemma, I think we need to prove $\frac{O_{1}O_{2}}{O_{2}O_{3}}=\frac{MK}{KB}$ , which I can't.

$($ In my drawing $AB = 11.8,$ $BC = 6.0,$ $AC = 12.3,$ $BK = 1.2,$ $KM = 1.7,$ $CL = 1.3,$ where $K,$ between $B,$ $M$ and $M,$ in the extension of $AB.$ Also $N,$ between $A,$ $C$ and $L,$ between $N,$ $C$ $).$

If $O_{1},$ $O_{2},$ $O_{3},$ are the circumcenters of the triangles $\bigtriangleup ABC,$ $\bigtriangleup AKL,$ $\bigtriangleup AMN$ respectively, we consider their orthogonal projections $O'_{1},$ $O'_{2},$ $O'_{3},$ on the sideline $AB,$ of $ABC.$ So, we have that $\frac{O_{1}O_{2}}{O_{2}O_{3}}= \frac{O'_{1}O'_{2}}{O'_{2}O'_{3}}$ and it is enough to prove that $\frac{O'_{1}O'_{2}}{O'_{2}O'_{3}}= \frac{BK}{KM}$

But this result is true because of $O'_{1}O'_{2}= \frac{BK}{2}$ and $O'_{2}O'_{3}= \frac{KM}{2}$ $($ easy to prove because of the points $O'_{1},$ $O'_{2},$ $O'_{3},$ are the midpoints of the segments $AB,$ $AK,$ $AM,$ respectively $).$

Kostas Vittas.

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#14 h Jan 1, 2007, 1:07 PM • 1 Y

Y by Adventure10

Dear all my friends.

For reading easiness, I post here the figures of the configurations of all as I wrote about the shobber's problem.

Kostas Vittas.

Attachments:

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Amir.S

#15 h Jan 2, 2007, 6:42 PM • 2 Y

Y by Adventure10, Mango247

To Dear vittasko :Thank you very much for writing the rest of my solution.

I was very busy I couldn't type long posts.

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#16 h Nov 5, 2007, 8:45 AM • 2 Y

Y by Adventure10, Mango247

shobber wrote:

Let $K$ and $M$ be points on the side $AB$ of a triangle $\triangle{ABC}$ , and let $L$ and $N$ be points on the side $AC$ . The point $K$ is between $M$ and $B$ , and the point $L$ is between $N$ and $C$ . If $\frac {BK}{KM} = \frac {CL}{LN}$ , then prove that the orthocentres of the triangles $\triangle{ABC}$ , $\triangle{AKL}$ and $\triangle{AMN}$ lie on one line.

Proof. Let $ABCD$ be a tetrahedron and the points $\|\begin{array}{c} \{M,K\}\subset (AB)\\\ \{N,L\}\subset (AC)\\\ \{U,V\}\subset (AD)\end{array}$

so the the planes $(MUN)$ , $(KVL)$ , $(BDC)$ are parallely.

The orthocenters of the triangles $MUN$ , $KVL$ , $BDC$ are collinearly.

For $D\rightarrow A$ obtain $U\rightarrow A$ and $V\rightarrow A$ , i.e. in this limit case

the orthocenters of the triangles $MAN$ , $KAL$ , $BAC$ are collinearly.

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#17 h Nov 5, 2007, 4:11 PM • 2 Y

Y by Adventure10, Mango247

Virgil Nicula wrote:

shobber wrote:

Let $K$ and $M$ be points on the side $AB$ of a triangle $\triangle{ABC}$ , and let $L$ and $N$ be points on the side $AC$ . The point $K$ is between $M$ and $B$ , and the point $L$ is between $N$ and $C$ . If $\frac {BK}{KM} = \frac {CL}{LN}$ , then prove that the orthocentres of the triangles $\triangle{ABC}$ , $\triangle{AKL}$ and $\triangle{AMN}$ lie on one line.

Proof. Let $ABCD$ be a tetrahedron and the points $\|\begin{array}{c} \{M,K\}\subset (AB) \\ \ \{N,L\}\subset (AC) \\ \ \{U,V\}\subset (AD)\end{array}$

so the the planes $(MUN)$ , $(KVL)$ , $(BDC)$ are parallely.

The orthocenters of the triangles $MUN$ , $KVL$ , $BDC$ are collinearly.

For $D\rightarrow A$ obtain $U\rightarrow A$ and $V\rightarrow A$ , i.e. in this limit case

the orthocenters of the triangles $MAN$ , $KAL$ , $BAC$ are collinearly.

I think it is true $($ to be the planes $[MUN],$ $[KVL],$ $[BDC]$ parallel each other $),$ only when $MN\parallel KL\parallel BC$ and then, the line connecting the orthocenters of the triangles $\bigtriangleup MUN,$ $\bigtriangleup KVL,$ $\bigtriangleup BDC,$ passes through the point $A.$

But in this particular case, we don’t need the clever artifice of tetrahedron $ABCD.$

Also, the problem doesn’t state that the pairs of points $K,$ $M$ and $L,$ $N,$ must be in the same place, with respect to the line segment $BC.$

Kostas Vittas.

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#18 h Jul 10, 2008, 3:48 PM • 2 Y

Y by Adventure10, Mango247

Lemma: Let $\mathcal{A}=A_1A_2...A_n$ and $\mathcal{B}=B_1B_2...B_n$ be directly similar polygons. Construct $A_iB_iC_i$ that are directly similar. Then $\mathcal{A}$ , $\mathcal{B}$ , and $\mathcal{C}$ all have the same spiral center (and they are directly similar).

Problem: Let $ABC$ be a triangle, let $B_1$ , $C_1$ on $AB$ and $AC$ . Construct $B_2$ and $C_2$ on segments $AB_1$ and $AC_1$ such that $\frac{BB_1}{B_1B_2}=\frac{CC_1}{C_1C_2}$ . Prove that the orthocenters of $ABC$ , $AB_1C_1$ , and $AB_2C_2$ are collinear.

Proof: Let $S$ be the point that such that $BB_2$ maps to $CC_2$ under a direct similarity. Then $BB_1B_2$ and $CC_1C_2$ are directly similar degenerate triangles constructed atop these similar polygons, so $BB_1B_2$ ; $CC_1C_2$ are directly similar wrt $S$ .

Let $O$ , $O_1$ , $O_2$ be the circumcenters of $ABC$ , $AB_1C_1$ , $AB_2C_2$ . Clearly, $OBC$ , $OB_1C_1$ , and $OB_2C_2$ are directly similar since $OB=OC$ , $O_1B_1=O_1C_1$ , $O_2B_2=O_2C_2$ and $\angle BOC=\angle B_1O_1C_1=\angle B_2O_2C_2=2\angle BAC$ . Therefore, $OO_1O_2$ is directly similar to $BB_1B_2$ , so they are are both degenerate.

Let $P$ , $P_1$ , and $P_2$ be the midpoints of $BC$ , $B_1C_1$ , and $B_2C_2$ . Then $BPC$ , $B_1P_1C_1$ , and $B_2P_2C_2$ are directly similar degenerate triangles, so $BB_1B_2$ and $PP_1P_2$ are directly similar, so they are both degenerate.

Let $G$ , $G_1$ , and $G_2$ be the centroids of $ABC$ , $A_1B_1C_1$ . A homothety center $A$ with ratio $\frac{2}{3}$ maps $PP_1P_2$ to $GG_1G_2$ so $GG_1G_2$ is directly similar to $PP_1P_2$ .

Let $T$ be the spiral center that sends $GG_1G_2$ to $OO_1O_2$ (note $T\ne S$ necessarily). Now by the euler line, $HOG$ , $H_1O_1G_1$ , and $H_2O_2G_2$ are directly similar triangles, so $OO_1O_2$ and $HH_1H_2$ are directly similar, and the result follows.

Comment: This is a great example of the power of spiral similarity! With the proper knowledge of that topic, this problem is straightforward. Surely, this problem is subject to generalization. That is, we can take any triangle centers of $ABC$ , $AB_1C_1$ , $AB_2C_2$ that are directly similar to $BB_1B_2$ and follow the same logic.

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#19 h Jan 26, 2009, 7:22 AM • 1 Y

Y by Adventure10

Here is my solution
Set the ration $\frac {MK}{KB} = \frac {NL}{LC} = k$
Call orthorcenters of triangles $\Delta AMN, \Delta AKL,\Delta ABC$ respect are $H_{1}, H_{2},H_{3}$ .
And $O_{1}, O_{2},O_{3}$ are circumcenters of triangles $\Delta ABC, \Delta AMN, \Delta AKL$ respect.
By CANADA MO 2003, easy to prove that $O_{1}, O_{2},O_{3}$ are collinear and $\frac {O_{1}O_{2}}{O_{2}O_{3}} = k$
$\Rightarrow \vec{O_{1}O_{2}} = k\vec{O_{2}}{O_{3}}$
So, we use indentions flowing:
$\vec{O_{1}H_{1}} = \vec{O_{1}A} + \vec{O_{1}B} + \vec{O_{1}C}$
$\vec{O_{2}H_{2}} = \vec{O_{2}A} + \vec{O_{2}M} + \vec{O_{2}N}$
$\vec{O_{3}H_{3}} = \vec{O_{3}A} + \vec{O_{3}K} + \vec{O_{3}L}$
Hence, we get
$ \vec{H_{1}H_{2}} = k\vex{H_{2}}{H_{3}}$
I think the Lemma of Kostas Vittas can prove by using vector(very short)

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PROF65

#20 h Feb 26, 2019, 8:05 PM • 1 Y

Y by Adventure10

an other approach :
let $S$ the Miquel points of $ABC$ wrt $M-N$ since $\frac{BK}{KM}=\frac{CL}{LN}\iff \frac{\mathcal{P}_{(ABC)}(K)}{\mathcal{P}_{(AMN)}(K}= \frac{\mathcal{P}_{(ABC)}(L)}{\mathcal{P}_{(AMN)}(L)}$ then $SAKL$ is cyclic but the Steiner line of $S$ is the same wrt $AMN,ABC,AKL$ hence their orthocenters are collinear .

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#21 h Jul 31, 2023, 3:33 PM

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Relabel the points such that the condition reads $\frac{BB_1}{BB_2} = \frac{CC_1}{CC_2}$ . The idea is that the Miquel point $M = (AB_2C_2) \cap (ABC)$ is also concyclic with $AC_1B$ . So by spiral similarity, it suffices to show that the orthocenters are corresponding parts of the similarity. This immediately follows by checking the $A$ -antipodes and midpoints are both corresponding parts too.

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#22 h Mar 25, 2024, 1:08 AM

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After getting that 3 circumcenters are collinear, then just reflect this line about angle bisector of $\angle {A}$ , we get desired conclusion!

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