Prove that the circumcentre of AB_1C_1 lies on the line AO_1

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Prove that the circumcentre of AB_1C_1 lies on the line AO_1

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Source: Tuymaada 2009, Senior League, Second Day, Problem 3

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orl

#1 h Jul 19, 2009, 11:36 PM • 3 Y

Y by narutomath96, Adventure10, Mango247

A triangle $ABC$ is given. Let $B_1$ be the reflection of $B$ across the line $AC$ , $C_1$ the reflection of $C$ across the line $AB$ , and $O_1$ the reflection of the circumcentre of $ABC$ across the line $BC$ . Prove that the circumcentre of $AB_1C_1$ lies on the line $AO_1$ .

Proposed by A. Akopyan

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yetti

#2 h Jul 20, 2009, 6:20 AM • 1 Y

Y by Adventure10

orl wrote:

A triangle $ABC$ is given. Let $B_1$ be the reflection of $B$ across the line $AC$ , $C_1$ the reflection of $C$ across the line $AB$ , and $O_1$ the reflection of the circumcentre of $ABC$ across the line $BC$ . Prove that the circumcentre of $AB_1C_1$ lies on the line $AO_1$ .

Let $(O)$ be circumcircle of $\triangle ABC$ . Let parallel to $BC$ through $A$ cut $(O)$ again at $Z.$
By Ptolemy for isosceles trapezoid $AZBC,$ $AZ \cdot BC = AB^2 - AC^2.$

Let $D \equiv BC_1 \cap CB_1.$ $\triangle ABC \cong \triangle ABC_1 \cong \triangle AB_1C$ are congruent $\Longrightarrow$
$\angle BDC = 2\pi - (3 \angle A + \angle B + \angle C) = \pi - 2 \angle A = \pi - \angle COB$ $\Longrightarrow$
$OBDC$ is cyclic, let $(P)$ be its circumcircle.

Let $CD$ cut $(O)$ again at $E$ $\Longrightarrow$
$\angle BDE = \pi - 2 \angle A,$ $\angle DEB = \angle A$ $\Longrightarrow$ $\triangle BDE$ is isosceles with $BD = ED$ and
$\angle CBE = \pi - (\angle A + 2 \angle B) = \angle C - \angle B.$ Consequently, $AZ = CE = CD - ED = CD - DB.$
Let $K, L$ be midpoints of $BD, CD,$ respectively. Then

$AB_1^2 - AC_1^2 + KB_1^2 - KD^2 + LD^2 - LC_1^2 = AB^2 - AC^2 + BC_1 \cdot DB - CB_1 \cdot DC =$
$= AZ \cdot BC - BC \cdot (CD - DB) = AZ \cdot BC - BC \cdot CE = 0$

By Carnot theorem for $\triangle DB_1C_1,$ perpendiculars to $B_1C_1, C_1D, DB_1$ from $A, L, K$ are concurrent at $P.$

Let $N$ be 9-point circle center of $\triangle ABC$ ; $N$ is midpoint of $AO_1.$
Kosnita point $Ko \equiv N^*$ is isogonal conjugate of $N$ (well known, see hidden stuff below) and $Ko \in AP.$
Triangles $\triangle ABC, \triangle AB_1C_1$ have common internal bisector of $\angle A.$
Since $AKoP$ is A-altitude of $\triangle AB_1C_1$ passing through its orthocenter $\Longrightarrow$
its isogonal $ANO_1$ WRT $\angle C_1AB_1$ passes through its circumcenter. $\square$

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#3 h Jul 21, 2009, 6:41 PM • 2 Y

Y by Adventure10, Mango247

Let us to present an elementary solution of this nice problem.

We denote as $M,\ N,$ the midpoints of the side-segments $AC,\ AB$ respectively and let be the points $D\equiv AC_{1}\cap OM$ and $E\equiv AB_{1}\cap ON.$

It is easy to show that $DE\parallel B_{1}C_{1}$ $,(1)$ because of $\frac {AD}{AE} = \frac {AC}{AB} = \frac {AC_{1}}{AB_{1}}$ from the similar right triangles $\bigtriangleup MAD,\ \bigtriangleup NAE$ and the similar isosceles triangles $\bigtriangleup ABB_{1},\ \bigtriangleup ACC_{1}.$

So, from $(1)$ we conclude that $DE\parallel KL$ $,(2)$ where $K,\ L,$ are the midpoints of the segments $AC_{1},\ AB_{1},$ respectively.

$\bullet$ Let $BB',\ CC'$ be, the altitudes of the given triangle $\bigtriangleup ABC$ $($ where $B'\equiv AC\cap BB_{1}$ and $C'\equiv AB\cap CC_{1}$ $)$

and we denote the points $P\equiv CC'\cap OM$ and $Q\equiv BB'\cap ON.$

Because of $\angle CPM = 90^{o} - \angle PCM = 90^{o} - \angle OCB = 90^{o} - \angle O_{1}CB = \angle CO_{1}O,$

we conclude that the quadrilateral $OPCO_{1}$ is cyclic and then we have that $\angle OPO_{1} = \angle OCO_{1}$ $,(3)$

From $(3)$ $\Longrightarrow$ $\angle OPO_{1} = 2\angle OCB = 2\angle PCA = \angle CAD$ $\Longrightarrow$ $\angle MAD = \angle MPK'$ $,(4)$ where $K'\equiv AD\cap O_{1}P.$

From $(4)$ and because of $PM\perp AC,$ we conclude that $PK'\equiv O_{1}K'\perp AD$ and by the same way we can prove that $QL'\equiv O_{1}L'\perp AE,$ where $L'\equiv AE\cap O_{1}Q.$

$\bullet$ Because of the similar right triangles $\bigtriangleup K'PD,\ \bigtriangleup L'QE$ $($ from $\angle K'PD = \angle MAD = \angle NAE = \angle L'QE$ $),$

we have that $\frac {K'D}{L'E} = \frac {PD}{QE}$ $,(5)$

Because of the similar triangles $\bigtriangleup APD,\ \bigtriangleup AQE,$ $($ from $\angle PDA = \angle QEA$ and

$\angle PAD = \angle PAC + \angle CAD = \angle PCA + \angle CAD = \angle QBA + \angle BAE$ $\Longrightarrow$ $\angle PAD = \angle QAB + \angle BAE = \angle QAE$ $),$

we have that $\frac {AD}{AE} = \frac {PD}{QE}$ $,(6)$

From $(5),\ (6)$ $\Longrightarrow$ $\frac {K'D}{L'E} = \frac {AD}{AE}$ $\Longrightarrow$ $K'L'\parallel DE$ $,(7)$

From $(2),\ (7)$ $\Longrightarrow$ $KL\parallel K'L'$ $\Longrightarrow$ $\frac {AK}{AK'} = \frac {AL}{AL'}$ $(8)$

$\bullet$ We consider now, the points $K,\ A,\ K',$ as the orthogonal projections of $O',\ A,\ O_{1}$ respectively, on the line segment $AC_{1},$

where $O'$ is the circumcenter of the triangle $\bigtriangleup AB_{1}C_{1}.$

Similarly, we consider the points $L,\ A,\ L',$ as the orthogonal projections of $O',\ A,\ O_{1}$ respectively, on the line segment $AB_{1}.$

Hence, because of $(8),$ we conclude that the points $O',\ A,\ O_{1},$ are collinear and the proof is completed.

$($ As well known, if the orthogonal ( or parallel ) projections of three given points, on two lines intersect each other, form segments with the same ratio then, these points are collinear with the same ratio $).$

$\bullet$ This proof is dedicated to Jean-Louis Ayme.

Kostas Vittas.

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t=289760.pdf (9kb)

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jayme

#4 h Jul 22, 2009, 3:55 AM • 2 Y

Y by Adventure10, Mango247

Dear Kostas,
thank you for your dedication. Your attached figure is very interesting and has much more to say.
Sincerely
Jean-Louis

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#5 h Jul 26, 2009, 7:30 PM • 1 Y

Y by Adventure10

First we will prove the lemma:

Let $ABC$ be a triangle and $H$ be its orthocenter. $BH,CH$ meet $AC,AB$ at $B_1,C_1$ . $B_2,C_2$ are reflection points of $A$ through $B_1,C_1$ . $P$ be the reflection point of H through $B_2C_2$ .Let the reflections of $BH,CH$ through $AB,AC$ meet at $Q$ .Then $A,P,Q$ are collinear.
We see that $PB_2 \| AC_3,PC_2 \| PB_3,B_2C_2 \| B_3C_3$ .Then $AP,B_2C_3,B_3C_2$ are concurrent.Now we need to prove that $AQ,B_2C_3,B_3C_2$ are also concurrent.
Let $AB \cap CQ = B_4,AC \cap BQ = C_4$ .We have $\angle{B_4BC_4} = \angle {C_4CB_4}$ then $B,C,B_4,C_4$ lie on a circle.Apply Reim's theorem to the tranversals $BC_1B_4,CB_1C_4$ we conclude that $B_4C_4 \| B_1C_1$ .Let $B_1C_1 \cap B_2C_2 = R$ in the infinity line,then $B_4,C_4,R$ are collinear.Apply Desargue's theorem to the triangles $AB_2C_2$ and $QC_3B_3$ we have $AQ,B_2C_3,B_3C_2$ are concurrent.
Return to our problem,apply the previous lemma to triangle $AB'C'$ , we have completed the proof.

Figure of the lemma:
Image not found

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#6 h Jul 29, 2009, 3:56 PM • 3 Y

Y by chronondecay, Adventure10, Mango247

Let $H$ be the orthocenter of $ABC$ . Let $AB_1$ and $CH$ meet at $P$ , and $AC_1$ and $BH$ meet at $Q$ . Let $K_1$ and $K_2$ be the circumcircles of $BHC$ and $AB_1C_1$ , respectively.

We have $\angle HBC = 90^{o} - \angle BCA$ and $\angle BCH = 90^{o} - \angle ABC$ , so $180^{o} - \angle CHB = \angle CAB$ .
Now as $O$ is the circumcentre of $ABC$ , $\angle COB = 2\angle CAB$ . Therefore $\angle BO_1C = 2(180^{o} - \angle CHB)$ . Using that $O_1C = O_1B$ , we get that $O_1$ must be the center of $K_1$ .

Consider the quadrilateral $B_1AHC$ . We have $\angle HCA = 90 - \angle CAB = \angle ABH = \angle HB_1A$ . Therefore $B_1AHC$ is cyclic.
Similarly, $C_1AHB$ is cyclic. Let $K_3$ and $K_4$ be their circumcircles, respectively.

Now, the radical axis of $K_2$ and $K_3$ is $aB_1$ , and the radical axis of $K_1$ and $K_3$ is $CH$ . Then $P$ is the radical center of $C_1$ , $C_2$ and $C_3$ , and it must be in the radical axis of $C_1$ and $C_2$ .
Similarly we can get that $Q$ lies on the radical axis of those circumferences. Therefore, $PQ$ is perpendicular to $O_1O_2$ .

Now, $\angle QC_1P = \angle HCA = 90 - \angle CAB = \angle ABH = \angle QB_1P$ . Then $PQB_1C_1$ is cyclic, and $AQP$ is similar to $AC_1B_1$ .
As $\angle C_1AO_2 = 90^{o} - \angle AB_1C_1 = 90^{o} - \angle PQA$ , we get that $O_2A$ is perpendicular to $PQ$ .

But then both $O_2A$ and $O_2O_1$ are perpendicular to $PQ$ , and $A$ , $O_1$ and $O_2$ are collinear.

Special Cases

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Collinearity.pdf (30kb)

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#7 h Aug 19, 2009, 8:34 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Solution. First, let's introduce some new points:
$\bigoplus \ \ \ \boxed{\begin{array}{l} M - \mathrm{ midpoint \ of }\ [AC], \ N - \mathrm{midpoint \ of}\ [AB], \ P - \mathrm{midpoint \ of} \ [BC] \\ \\ M_1 - \mathrm{midpoint \ of}\ [AC_1], \ N_1 - \mathrm{midpoint \ of}\ [AB_1] \\ \\ O_2 = \mathrm{sim}_{AC}\ O, \ O_3 = \mathrm{sim}_{AB}\ O, \ O' = O_3 M_1 \ \cap\ O_2 N_1 \\ \\ S = \mathrm{pr}_{O'M_1} \ O_1, \ T = \mathrm{pr}_{O'N_1} \ O_1\end{array}}$

Now, it's obvious to observe that $O'$ is the circumcenter of $\triangle{AB_1 C_1}.$ We have:

$O_1\in AO' \ \Longleftrightarrow\ \frac {\mathrm{d}(A;O'O_3)}{\mathrm{d}(A;O'O_2)} = \frac {\mathrm{d}(O_1;O'O_3)}{\mathrm{d}(O_1;O'O_2)} \ \Longleftrightarrow\ \frac {AM_1}{AN_1} = \frac {OS}{OT} \ \Longleftrightarrow$

$\Longleftrightarrow \ \frac {AC}{AB} = \frac {O_1O_3}{O_1O_2}\cdot\frac {\cos{\widehat{SO_1O_3}}}{\cos{\widehat{TO_1O_2}}}$

$\bullet$ Note that: $\triangle{ABC}\ \equiv\ \triangle{O_1 O_2 O_3}\ \Longrightarrow\ \frac {O_1O_3}{O_1O_2} = \frac {AC}{AB}\dots\dots (*)$

$\bullet$ We also have:

$\angle{SO_1O_3} = \angle{(O_3O_1;SO_1)} = \angle{(AC; AC_1)} = \angle{(AB_1; AB)} =$

$= \angle{(O_1T; O_1O_2)} = \angle{TO_1O_2}\ \Longrightarrow\ \cos{\widehat{SO_1O_3}} = \cos{\widehat{TO_1O_2}}\dots\dots (**)$

According to $(*)$ and $(**)$ we get the result.

I denote:

$\boxed{\begin{array}{l} \mathrm{sim_{YZ}\ X} - \mathrm{the \ reflection \ of} \ X \ \mathrm{across \ line} \ YZ \\ \\ \mathrm{d(X,YZ)} - \mathrm{the\ distance\ between}\ X \ \mathrm{and} \ YZ \\ \\ \mathrm{pr_{YZ}\ X} - \mathrm{the \ projection\ of\ point}\ X\ \mathrm{onto\ line}\ YZ \end{array}}$

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#8 h Sep 3, 2009, 8:12 AM • 2 Y

Y by narutomath96, Adventure10

Quote:

A triangle $ABC$ is given. Let $B_1$ be the reflection of $B$ across the line $AC$ , $C_1$ the reflection of $C$ across the line $AB$ , and $O_1$ the reflection of the circumcentre of $ABC$ across the line $BC$ . Prove that the circumcentre of $AB_1C_1$ lies on the line $AO_1$ .

Proposed by A. Akopyan

We note that if we let $ABC$ be a triangle with $H$ is its orthocenter. Denote $H_a,H_b,H_c$ respectively by the projections of $H$ onto $BC,CA,AB$ . Let $P_b,P_c$ respectively be the intersections of $H_aH_c$ with $HB$ ; $H_aH_b$ with $HC$ . Let $K_b,K_c$ respectively be the intersections of $(HAB)\cap ([AC])$ ; $(HAC)\cap ([AB])$ . Denote $I_a,O_a,$ $\mathcal {E}$ respectively by the circumcenters of $(AK_bK_c)$ , $(HBC)$ and the Euler circle wrt $\triangle ABC$ . Then $\mathcal {E}$ $,O_a,I_a\in d_a$ . (Where $([PQ])$ here means the circle with diameter $PQ$ )

Indeed, consider the inversion through pole $A$ , power $k = \overline {AH}\cdot \overline {AH_a}$ $= \overline {AH_c}\cdot \overline {AB}$ $= \overline {AH_b}\cdot \overline {AC}$ . We have $\mathcal {I}(A,k):$ $H_c\mapsto B,$ $H_b\mapsto B,$ $H\mapsto H_a$ . Therefore by $\mathcal {I}(A,k)$ , $H_cH_a\mapsto (AHB)$ , $H_aH_b\mapsto (AHC)$ , $HB\mapsto ([AC])$ and $HC\mapsto ([AB])$ . As the result, $P_b\mapsto K_b$ and $P_c\mapsto K_c$ $\Longrightarrow$ $P_bP_c\mapsto (AK_bK_c)$ , which leads to the fact that $I_a\in d_a$ $(*)$ . In the other hand, we have $\overline {P_bK_b}\cdot \overline{P_bA} =$ $\overline {P_bH_c}\cdot \overline {P_bH_a} =$ $\overline {P_bH}\cdot \overline {P_bB}$ $\Longrightarrow$ $\mathcal {P}_{P_b/(AK_bK_c)} =$ $\mathcal {P}_{P_b/(H_aH_bH_c)} =$ $\mathcal {P}_{P_b/(HBC)}$ . With the same argument for $P_c$ . Then it is followed that $P_bP_c$ is the radical axes of $(H_aH_bH_c)$ , $(HBC)$ and $(AK_bK_c)$ $\Longrightarrow$ $\mathcal {E},$ $O_a$ and $I_a$ are collinear and $\overline {\mathcal {E}O_aI_a}\perp P_bP_c$ $(**)$ . From $(*)$ and $(**)$ , we obtain result. $\square$

Back to our problem. Let $K_b,K_c$ respectively be the intersections of $(HAB)\cap ([AC])$ ; $(HAC)\cap ([AB])$ and $H_a,H_b,H_c$ respectively be the projections from $H$ onto $BC,CA,AB$ .In order to prove the circumcenter $I_a$ of $\triangle AB_1C_1$ $\in AO_1$ and also by the notice mentioned above, which is $A,O_a,O_1$ are collinear- where $O_a$ is the circumcenter of $\triangle AK_bK_c$ ; we will prove that $A,I_a,O_a$ are collinear. Indeed:

Consider the inversion through pole $A$ , power $k = \overline {AH}\cdot \overline {AH_a}$ $= \overline {AH_c}\cdot \overline {AB}$ $= \overline {AH_b}\cdot \overline {AC}$ . We have $\mathcal {I}(A,k):$ $(AK_bK_c)\mapsto K_bK_c$ $\equiv P_bP_c$ $(*)$ - where $P_b,P_c$ respectively are the intersections of $H_aH_c,HB$ and $H_aH_b,HC$ . In the other hand, since $C_1$ is the reflection of $C$ wrt $AB$ $\Longrightarrow$ $(C_1A,C_1B)$ $\equiv (CB,CA)$ $\equiv (HA,HB)$ $\pmod \pi$ , which implies that $C_1\in (HAB)$ , also note that $C_1\in CH_c$ . Now, we have known that $\mathcal {I}(A,k):$ $(AHB)\mapsto H_cH_a$ and $CH_c\mapsto ([AB])$ . Therefore if $C_1\mapsto Q_c$ then $Q_c\equiv H_aH_c\cap ([AB])$ . With the same argument, we also conclude $Q_b\equiv H_aH_b\cap ([AC])$ . $\mathcal {I}(A,k):$ $(AC_1B_1)\mapsto Q_cQ_b$ $(**)$ .

In the other hand, we have $(H_bA,H_bQ_c)\equiv (H_aA,H_aQ_c)$ $\pmod \pi$ but $(H_aA,H_aH_c)\equiv (CA,CH_c)$ $\equiv (H_bA,CH_c)$ $\pmod \pi$ $\Longrightarrow$ $CH_c\| Q_cH_b$ . With the same argument, we also have $Q_bH_c\| P_bH_b$ . Therefore, by the Pappus theorem apply for $\begin{vmatrix}H_c & Q_c & P_b \\ H_b & P_c & Q_b\end{vmatrix}$ , we have $H_cP_c\cap Q_cH_b\equiv \infty$ , $H_cQ_b\cap P_bH_b\equiv \infty$ , thus $Q_cQ_b\cap P_bP_c\equiv \infty$ , which implies that $Q_cQ_b\|P_cP_c$ . From $(*)$ and $(**)$ , we conclude that $(AB_1C_1)$ tangents to $(AK_bK_c)$ at $A$ , which leads to the fact that $A,I_a,O_a$ are collinear. Our proof is completed then. $\square$

Attachments:

tuymaada yakutMO09.pdf (31kb)

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#9 h Sep 13, 2009, 7:35 AM • 3 Y

Y by narutomath96, Adventure10, Mango247

It’s easy to see that $AO_1$ passes through the center $N$ of nine-point circle of triangle $ABC.$
We want to have the center of $(AB_1C_1)$ lies on $AN$ therefore the A-altitude of triangle $AB_1C_1$ must pass through the isogonal conjugate point $K$ of $N$ wrt triangle $ABC$ ( Kosnita point).
We know that $AK$ passes through the circumcenter $J$ of triangle $BOC$ .
So $AJ\perp B_1C_1 \Leftrightarrow AB^2_1 - AC^2_1 = JB^2_1 - JC^2_1$
$\Leftrightarrow 4R^2(\sin^2 C - \sin^2 B) = (B_1C^2 + CJ^2 - 2B_1C.JC.\cos \angle B_1CJ - BC_1^2 - BJ^2 + 2BC_1.BJ.\cos \angle C_1BJ)$
$\Leftrightarrow 4R^2(\sin^2 C - \sin^2 B) = 2BC.R_{BOC}(\cos \angle C_1BJ - \cos \angle B_1CJ)$
$\Leftrightarrow 4R^2(\sin^2 C - \sin^2 B) = 2R.\sin A.\frac {R}{\cosh}.(\cos(2A + 2B - 90^o) - \cos (2A + 2C - 90^o))$
$\Leftrightarrow 2(\sin C - \sin B)(\sin C + \sin B) = \tan A(\sin2B - \sin2C)$ (right!)
We are done.

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XmL

#10 h Apr 16, 2013, 2:38 AM • 3 Y

Y by narutomath96, Adventure10, Mango247

Sorry for digging up old posts but here's my solution to this problem. See attachment..

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leader

#11 h Apr 16, 2013, 1:39 PM • 6 Y

Y by Mahi, MariusBocanu, Vietnamisalwaysinmyheart, kun1417, Adventure10, Mango247

if $M$ is the midpoint of $AB$ and $D=BB_{1}\cap AC$ then clearly $AMO\sim ADC$ where $O$ is the circumcenter of $ABC$ . so $BO*BD=BM*BC$ so $BB_{1}*BO=BA*BC$ but $\angle O_{1}BA=\angle B_{1}BC_{1}$ so $BB_{1}C_{1}\sim BO_{1}A$ and $\angle BAO_{1}=\angle BB_{1}C_{1}$ if $L$ is the circumcenter of $AB_{1}C_{1}$ then
$\angle C_{1}AO_{1}=\angle C_{1}AB+\angle BAO_{1}=\angle BAC+\angle BB_{1}C_{1}=90-\angle BB_{1}A+\angle BB_{1}C_{1}=90-\angle C_{1}B_{1}A=\angle C_{1}AL$ so $L$ is on $AO_{1}$

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#12 h May 24, 2014, 1:02 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

Well, I used the same similar triangle pair of leader, but my finishing is a bit different.
Proof

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#13 h Oct 18, 2014, 8:54 AM • 5 Y

Y by Isogonics, DanDumitrescu, Siddharth03, lofiboy, Adventure10

This problem is the combination of two properties of nine point center

$(1)$

Let $O'$ be the reflection of the circumcenter $O$ of $\triangle ABC$ in $BC$ .
Then the nine point center of $\triangle ABC$ lie on $AO'$ .

$(2)$

Let $A', B', C'$ be the reflection of $A, B, C$ in $BC, CA, AB$ .
Then $\triangle A'B'C'$ is homothety with the pedal triangle of the nine point center of $\triangle ABC$ .

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#14 h Oct 19, 2014, 12:16 PM • 3 Y

Y by umaru, Adventure10, and 1 other user

Just consider point $F$ such that $<FO1B=<BAC$ and $<O1BF=<CBA$ .Now,we have spiral similarity with center B that pictures $BFO1$ in $BCA$ so we have that $BO1A$ is similar to $BFC$ and we have that $FO1C$ is similar to $AB1C1$ ,so from here it is pure angle chasing,so we are finished.

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#15 h Jul 2, 2016, 5:41 PM • 3 Y

Y by baladin, Adventure10, Mango247

An extension.

Let $ABC$ be a triangle inscribed incircle $(O)$ . $P$ is an any point. $PC,PB$ cut $(O)$ again at $M,N$ . $E,F$ are reflections of $B,C$ through $AM,AN$ . $K,L$ are circumcenters of triangles $AEF$ and $PBC$ . $G$ is inversion image of $L$ through $(O)$ . Prove that $\angle KAG=180^\circ-|\angle NAB-\angle MAC|$ .

See the figure in the case $P$ outside $(O)$ . When $PB,PC$ are tangent to $(O)$ we get Tuymaada problem.

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#16 h Jul 3, 2016, 12:56 AM • 5 Y

Y by buratinogigle, baladin, brokendiamond, enhanced, Adventure10

buratinogigle wrote:

An extension.

Let $ABC$ be a triangle inscribed incircle $(O)$ . $P$ is an any point. $PC,PB$ cut $(O)$ again at $M,N$ . $E,F$ are reflections of $B,C$ through $AM,AN$ . $K,L$ are circumcenters of triangles $AEF$ and $PBC$ . $G$ is inversion image of $L$ through $(O)$ . Prove that $\angle KAG=180^\circ-|\angle NAB-\angle MAC|$ .

By simple angle chasing $\Longrightarrow$ $\measuredangle EMP$ $=$ $\measuredangle FNP,$ so combining $\tfrac{EM}{PM}$ $=$ $\tfrac{BM}{PM}$ $=$ $\tfrac{CN}{PN}$ $=$ $\tfrac{FN}{PN}$ and $\triangle BPL$ $\stackrel{+}{\sim}$ $\triangle BMO$ $\stackrel{+}{\sim}$ $\triangle BEA$ we get $\triangle PME$ $\stackrel{+}{\sim}$ $\triangle LOA$ $\stackrel{+}{\sim}$ $\triangle PNF,$ hence $\measuredangle (MN, EF)$ $=$ $\measuredangle (OL,AL)$ $\Longrightarrow$ $\measuredangle (AL, \perp EF)$ $=$ $\measuredangle (BC, MN)$ $(\star)$ $.$ On the other hand, it's easy to see that the angle between the bisector of $\angle BAC$ and $\angle EAF$ is equal to $\measuredangle (BC, MN),$ so from $(\star)$ and notice $AG,$ $AL$ are isogonal conjugate WRT $\angle A$ (well-known) we conclude that $\measuredangle (AG, AK)$ $=$ $\measuredangle (BC, MN).$

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#17 h Jul 4, 2016, 7:59 AM • 2 Y

Y by Adventure10, Mango247

Another proof. Let $O_a, O_b, O_c, J$ be the circumcenters of triangles $BHC, CHA, AHB$ , $AB_1C_1$ . By simple angle chasing, $\angle O_bOO_c=\angle OO_bJ=\angle OO_cJ=180^\circ-\angle BAC$ . From lemma in my post at here, we get $J$ lies on Euler line of triangle $OO_bO_c$ or $J,A,O_a$ are collinear.

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#18 h Dec 24, 2016, 1:14 PM • 2 Y

Y by Adventure10, Mango247

Let $B_2,C_2,C_3,B_3$ be projections of $B,C,C_1,B_1,$ onto lines $AC_1,AB_1.$ Also let $U \equiv B_2B \cap C_2C.$

Easy angle chasing shows us that $\angle O_1BC=\angle O_1BC=\angle ACC_2=\angle ABB_2=\frac{\pi}{2}-\alpha$ while $\angle B_2AB=\angle CAC_2=\alpha.$ Hence by Jacobi theorem on hexagon $AB_2BO_1CC_2$ we deduce that $B_2C,BC_2,AO_1$ concur at $V.$

By Dual of Desargues Involution theorem it follows that $AB_2 \mapsto AC_2, AB \mapsto AC, AU \mapsto AV$ is involution which then coincides with reflection over A-angle bisector. Hence $\angle BAV=\angle CAU$ so easy angle chasing gives us $AV \perp B_2C_2.$ Since $V$ lies on $AO_1$ it follows that $AO_1 \perp B_2C_2.$

Note that $\frac{AB_2}{AC_2}=\frac{AB}{AC}=\frac{AB_1}{AC_1} \implies C_1B_2C_2B_1$ is cyclic. But $C_1C_3B_3B_1$ is also cyclic, hence by Reim's theorem it follows that $B_2C_2 \parallel B_3C3 \implies AO_1 \perp B_3C_3 \implies$ center of $\odot(AB_1C_1)$ lies on $AO_1$ as desired.

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#19 h Sep 29, 2020, 8:12 PM • 1 Y

Y by Mango247

Complex numbers rule!!!
Let $(ABC)$ be the unit circle.Denote with $x$ the affixe of point $X$ .We have:
$b_1=a+c-a\overline{b}c,c_1=a+b-ab\overline{c},o_1=b+c$ By making a translation which takes $A$ to the origin we have to show that the circumcenter $o_2$ of $0(b_1-a)(c_1-a)$ lies on the line $0(o_1-a)$ .
It is known that the circumcenter of $0xy$ is $o=\frac{xy(\overline{x}-\overline{y})}{\overline{x}y-x\overline{y}}$ ,so
$o_2=\frac{(c-a\overline{b}c)(b-ab\overline{c})(\overline{c-b+a(b\overline{c}-\overline{b}c)})}{(\overline{c-a\overline{b}c})(b-ab\overline{c})-(c-a\overline{b}c)\overline{b-ab\overline{c}}}=\frac{cb(1-\frac{a}{b})(1-\frac{a}{c})\frac{b-c}{bc}(1-\frac{\overline{a(b+c)}}{\overline{bc}})}{\frac{b}{c}(1-\frac{b}{a})(1-\frac{a}{c})-\frac{c}{b}(1-\frac{c}{a})(1-\frac{a}{b})}=\frac{\frac{b-c}{bc}(-1+\frac{\frac{b+c}{bca}}{\frac{1}{bc}})}{\frac{b}{ac^2}-\frac{c}{ab^2}}=\frac{bc(b-c)(b+c-a)}{b^3-c^3}=(b+c-a)\cdot \frac{bc}{b^2+bc+c^2}$ .
Now,just note that $\frac{o_2}{o_1-a}=\frac{bc}{b^2+bc+c^2}=\frac{\frac{1}{bc}}{\frac{1}{b^2}+\frac{1}{bc}\frac{1}{c^2}}=\overline{\frac{bc}{b^2+bc+c^2}}\Rightarrow \frac{o_2}{o_1-a}\in\mathbb{R}\Rightarrow o_2 \text{ is on } 0(o_1-a)$ $\square$

This post has been edited 1 time. Last edited by Doru2718, Apr 3, 2021, 6:56 PM
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