Intersection of circumcircles of MNP and BOC

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Djile

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Djile

#1 h Apr 8, 2013, 3:13 PM • 4 Y

Y by Adventure10, Mango247, Rounak_iitr, and 1 other user

Let $M$ , $N$ and $P$ be midpoints of sides $BC, AC$ and $AB$ , respectively, and let $O$ be circumcenter of acute-angled triangle $ABC$ . Circumcircles of triangles $BOC$ and $MNP$ intersect at two different points $X$ and $Y$ inside of triangle $ABC$ . Prove that $\angle BAX=\angle CAY.$

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yetti

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yetti

#2 h Apr 9, 2013, 7:28 AM • 2 Y

Y by Adventure10, Mango247

Let $[BC] = a, [CA] = b, [AB] = c$ and $\measuredangle CAB = \widehat A$ . Label the 3 defined circles as $\odot(ABC) \equiv (O)$ , $\odot(MNP) \equiv (K)$ and $\odot(OBC) \equiv (Q)$ with radii $[OA]=R$ , $[KM]=\frac{R}{2}$ and $[QO]=\frac{R}{2 \cos \widehat A}$ , resp.
Let $(K)$ cut $AB$ again at $F$ . $(K)$ is 9-point circle of $\triangle ABC$ $\implies$ $F$ is foot of C-altitude of this triangle. Power of $A$ to $(K)$ is then $p(A, (K)) = \overline{AP} \cdot \overline{AF} = \tfrac{1}{2}bc \cos \widehat A$ .
Let $(Q)$ cut $AB$ again at $W$ . $\measuredangle AWC = \pi - \measuredangle COB = \pi - 2 \widehat A$ $\implies$ $\triangle AWC$ is W-isosceles and $WN \perp CA$ . Power of $A$ to $(Q)$ is then $p(A, (Q)) = \overline{AB} \cdot \overline{AW} = \frac{bc}{2 \cos \widehat A}$ .
Combined, $\frac{p(A, (K))}{p(A, (Q))} = \cos^2 \widehat A = \frac{[KM]^2}{[QO]^2}$ $\implies$ $\frac{[AK]^2}{[AQ]^2} = \frac{p(A, (K)) + [KM]^2}{p(A, (Q)) + [QO]^2} = \frac{[KM]^2}{[QO]^2}$ $\implies$ $\frac{[AK]}{[AQ]} = \frac{[KM]}{[QO]}$ . Let internal and external bisectors of $\widehat A$ cut center line $KQ$ at $I, J$ , resp.
It is well known that $AK, AQ$ are isogonals WRT $\widehat A$ (see link (1) below) $\implies$ $AI, AJ$ bisect $\measuredangle QAK$ $\implies$ $-\frac{\overline{IK}}{\overline{IQ}} = \frac{\overline{JK}}{\overline{JQ}} = \frac{[AK]}{[AQ]} = \frac{[KM]}{[QO]}$ $\implies$ $I, J$ are internal and external similarity centers of circles $(K), (Q)$ .
Powers of $I$ to $(K), (Q)$ are in ratio $\frac{p(I, (K))}{p(I, (Q))} = \frac{[IK] ^2 - [KM^2]}{[IQ]^2 - [QO]^2} = \frac{[KM^2]}{[QO]^2} = \frac{p(A, (K))}{p(A, (Q))}$ and likewise, $\frac{p(J, (K))}{p(J, (Q))} = \frac{p(A, (K))}{p(A, (Q))}$ $\implies$ circle $(Z) \equiv \odot(AIJ)$ is coaxal with $(K), (Q)$ (see link (2) below) $\implies$
$X, Y \in (Z)$ . Then $\measuredangle IAX = \measuredangle IJX = \measuredangle IJY = \measuredangle IAY$ $\implies$ $\measuredangle XAB = \measuredangle IAB - \measuredangle IAX = \measuredangle IAC - \measuredangle IAY = \measuredangle YAC$ . $\blacksquare$

(1) - http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=289760&p=1567386#p1567386, hidden text.
(2) - http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=121062&p=687577#p687577.

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leader

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leader

#3 h Apr 9, 2013, 10:09 PM • 7 Y

Y by ZacPower123, karitoshi, Said, Adventure10, and 3 other users

i wish i figured this out during the contest
let $D,E$ be feet of altitudes from $B,C$ to $AC,AB$ then consider inversion of pole $A$ with power $AN*AD$ it pictures the Euler's circle of $ADE$ into circle $OBC$ and circle $MNP$ to itself.
so if $R$ is the point where circle $MNP$ and Euler's circle of $ADE$ intersect than $R$ is pictured into $X$ or $Y$ .
WLOG let $R$ be pictured into $X$
circle $MNP$ is to $\triangle ADE$ what circle $OBC$ is to $\triangle ABC$ so using the composition of a homotethy with center $A$ and coefficient $AD/AB$ and the symmetry wrt the internal bisector of $\angle BAC$ we picture $ADE$ into $ABC$ and $R$ into $Y$ .
so $\angle CAY=\angle BAR=\angle BAX$

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yunxiu

571 posts

yunxiu

#4 h Apr 10, 2013, 3:27 AM • 1 Y

Y by Adventure10

leader wrote:

so using the composition of a homotethy with center $A$ and coefficient $AD/AB$ and the symmetry wrt the internal bisector of $\angle BAC$ we picture $ADE$ into $ABC$ and $R$ into $Y$ .
so $\angle CAY=\angle BAR=\angle BAX$

Of course this can "picture $ADE$ into $ABC$ ",but why $R$ into $Y$ ?

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leader

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leader

#5 h Apr 10, 2013, 8:53 AM • 2 Y

Y by Adventure10, Mango247

yunxiu wrote:

leader wrote:

so using the composition of a homotethy with center $A$ and coefficient $AD/AB$ and the symmetry wrt the internal bisector of $\angle BAC$ we picture $ADE$ into $ABC$ and $R$ into $Y$ .
so $\angle CAY=\angle BAR=\angle BAX$

Of course this can "picture $ADE$ into $ABC$ ",but why $R$ into $Y$ ?

well $R$ is the interscetion point of Euler's circle of $ADE$ (which pictures into circle $MNP$ )and the circle that passes through the circumcenter of $ADE$ and $D,E$ (which pictures into circle $OBC$ ) so $R$ pictures into $X$ or $Y$

let $NO$ intersect $AB$ at $S$ than $\angle CSN=\angle ASN=90-\angle BAC$ so $S$ is on circle $OBC$ now $AD<AB$ and $AN<AS$ so $A$ does not have equal power to circles $MNP$ and $OBC$ so $X,Y,A$ are not collinear. also if $Q$ is the other intersection of Euler's circle of $ADE$ and Euler's than $Q$ lies on $AX$ or $AY$ .
if $Q$ is on $AX$ than $A-Q-R$ which implies $A-X-Y$ a contradiction.
if $Q$ is on $AY$ than if $Q$ pictures into $Y$ (and $R$ to $X$ ) than both $AX,AY$ must be bisectors of $\angle BAC$ but then $A-X-Y$ a contradiction.
so we finally get that $R$ pictures into $Y$ and $Q$ to $X$ .

This post has been edited 1 time. Last edited by leader, Apr 11, 2013, 10:15 AM

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simplependulum

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simplependulum

#6 h Apr 11, 2013, 3:44 AM • 2 Y

Y by Adventure10, Mango247

Everything becomes trivial if we notice the inversion about the circle centered at $A$ with radius $\sqrt{AN\cdot AB}$ together with the reflection w.r.t. the angle bisector of $\angle BAC$ .

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mathuz

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mathuz

#7 h Apr 16, 2013, 3:16 PM • 2 Y

Y by Adventure10, Mango247

Djile wrote:

Let $M$ , $N$ and $P$ be midpoints of sides $BC, AC$ and $AB$ , respectively, and let $O$ be circumcenter of acute-angled triangle $ABC$ . Circumcircles of triangles $BOC$ and $MNP$ intersect at two different points $X$ and $Y$ inside of triangle $ABC$ . Prove that $\angle BAX=\angle CAY.$

It's very nice geometry problem! HERE MY SOLUTION:
Let $\omega$ is circumcircle $\bigtriangleup BOC$ , $E_1$ is Euler's circle of $ABC$ and $E_2$ is Euler's line $AEF$ .
(here, points $E,F$ intersection $\omega$ and $AB,AC$ second times.)
Point $O$ and $O_1$ circumcenters triangle $ABC$ and $AEF$ , $\omega_1$ is circumcircle of triangle $EO_1F.$
[similar problem here AllRussian2000 http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2889224&sid=980f5b7e1eb7ee5a960469a7dbb0b3f9#p2889224]

Hence, that triangle's $ABC$ and $AEF$ similar.
Let $E_1 \cap \omega=X,Y$ and $E_2 \cap \omega_1 =Y_1,X_1.$
From the triangle's $ABC$ and $AEF$ similar, that $\angle BAX= \angle FAX_1, \angle EAY_1 = \angle FAY.$ (*)

this is my prove, END.

Denote, $l(X,Y)$ is radical axis circles $X$ and $Y$ and $EPNF$ - cyclic and circumcircle is $\phi.$
We have five circles - $\omega , \omega_1, \phi, E_1$ and $E_2:$
(1) from $\phi, E_1, \omega$ , that $PN, EY, XY$ concurrent;
(2) from $\omega, E_1, E_2$ , that $PN, XY, l(\omega,E_2)$ concurrent;
(3) from $\omega, \omega_1, E_2$ , that $l(\omega,E_2), X_1Y_1, EF$ concurrent;
(4) from $\omega_1, E_2,E_1$ , that $l(\omega_1,E_1), X_1Y_1, PN$ concurrent;
Hence, (1),(2),(3),(4) all intersect at point $K.$ Such that $KY.KX=KN.KP=KX_1.KY_1$ $\Longrightarrow$ $XYX_1Y_1$ is cyclic.

From (*), we have triangle's $AXY$ and $AX_1Y_1$ similar and $\angle AXY=\angle AX_1Y_1, \angle AYX=\angle AY_1X_1$ and $AX.AY_1 = AY.AX_1.$

Hence, that $A, X, Y_1$ and $A, Y, X_1$ collinear and $\angle BAX = \angle CAY.$

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mathuz

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mathuz

#8 h Apr 17, 2013, 1:15 AM • 2 Y

Y by Adventure10 and 1 other user

sorry, to be small mistake $E_2$ is Euler's circle, not Euler's line.

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duanby

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duanby

#9 h Apr 22, 2013, 10:39 AM • 2 Y

Y by Adventure10, Mango247

My solution:
Let E,FBe the foot of B,C
Let S,T be the Points of X,Y with inversion A and power $\ AE*AC$
then $\angle{ESF}=\angle A,\angle{BSC}=180-\angle{A}$
so $\angle{ESF}+\angle{BSC}=180$ so the isogonal conjugate of S
wrt BCEF exists and by angle chasing we know that is Y!
so Y is the isogonal conjugate of S wrt ABC
So done

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HAPPYNEWYEAR1729

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HAPPYNEWYEAR1729

#10 h Apr 26, 2013, 6:11 AM • 1 Y

Y by Adventure10

As $NA$ and $AO_a$ are isogonal lines. So it remains to prove that $\angle NAX= \angle O_aAY$ . which can be easily done by angle chasing.

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leader

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leader

#11 h May 3, 2013, 5:37 PM • 1 Y

Y by Adventure10

here's another way(no transformations).
since the Euler's circle and the bisector of $\angle BAC$ on the side of $NP$ with circle $BOC$ intersect outside $ABC$ $X,Y$ are not on the bisector of $\angle BAC$ . Consider point $K$ such that $\angle BAK=\angle CAX$ and $AK*AX=AN*AB=AP*AC$ it's easy to prove $AKN\sim ABX$ , $AKP\sim ACX$ , $AKB\sim ANX$ and $AKC\sim APX$ so ow $\angle NKP=\angle XBA+\angle XCA=\angle BOC-\angle BAC=\angle BAC$ so $K$ is on the Euler's circle.
but $\angle KBA+\angle KCA=\angle PXA+\angle NXA=\angle BAC$ so $\angle BKC=2*\angle BAC$ and $K$ is also on circle $BOC$ since $K$ is different from $X$ $K=Y$ and $\angle BAX=\angle CAY$

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nima-amini

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nima-amini

#12 h Dec 31, 2013, 12:35 PM • 2 Y

Y by Adventure10, Mango247

inversion with center a
and radius AB.AC/2

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nkalosidhs

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nkalosidhs

#13 h Dec 27, 2014, 2:08 PM • 2 Y

Y by Adventure10, Mango247

Can we prove spiral similarity of the triangles AXB and ANY?

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XmL

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XmL

#14 h May 24, 2016, 5:11 AM • 2 Y

Y by Adventure10, Mango247

See http://www.artofproblemsolving.com/community/c2671h1246951 for a generalization as well as a direct solution for this nice problem. Preview: $A$ lies on the circle of similitude of the two circles.

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anantmudgal09

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anantmudgal09

#15 h Jun 5, 2016, 9:39 PM • 4 Y

Y by Drunken_Master, Anar24, k12byda5h, Adventure10

An inversion about $A$ with radius $\sqrt{\frac{1}{2}\cdot AB.AC}$ followed by reflection in bisector maps these two circles two one another. Since the intersection points aren't fixed points under this operation, $X,Y$ are mapped to each other giving $AX,AY$ isogonal in angle $BAC$ .

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jayme

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jayme

#16 h Feb 11, 2017, 9:11 AM • 3 Y

Y by WizardMath, Adventure10, Mango247

Dear Yetti and Mathlinkers,

your idea of the circle (AIJ) is marvellous... it intersect (O) in a remarkable point...

Sincerely
Jean-Louis

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jayme

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jayme

#17 h Feb 18, 2017, 10:50 AM • 1 Y

Y by Adventure10

Dear Mathlinkers,
you can see

http://jl.ayme.pagesperso-orange.fr/Docs/Serbia%202013%20Problem%203.pdf

Sincerely
Jean-Louis

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Mosquitall

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Mosquitall

#18 h Apr 29, 2017, 3:41 PM • 2 Y

Y by Adventure10, Mango247

One need to prove that $\angle (AB, (MNP)) = \angle (AC, (BCO))$ . And similarly $\angle (AC, (MNP)) = \angle (AB, (BCO))$ . So now it's particular case of inversion + similarity principle.

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omarius

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omarius

#19 h Jan 12, 2018, 11:39 AM • 2 Y

Y by Adventure10, Mango247

anantmudgal09 wrote:

An inversion about $A$ with radius $\sqrt{\frac{1}{2}\cdot AB.AC}$ followed by reflection in bisector maps these two circles two one another. Since the intersection points aren't fixed points under this operation, $X,Y$ are mapped to each other giving $AX,AY$ isogonal in angle $BAC$ .

Sorry but I can't see exactly why it swaps the circles ? Obviously it takes B to N and C to P but we need another point , no ? Thank you.

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WizardMath

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WizardMath

#20 h Feb 3, 2018, 7:58 PM • 1 Y

Y by Adventure10

jayme wrote:

Dear Yetti and Mathlinkers,

your idea of the circle (AIJ) is marvellous... it intersect (O) in a remarkable point...

Sincerely
Jean-Louis

I just saw this today. I remember having proved it quite some time back. Nice observation jayme! The point is the extremely famous Miquel point of the quadrilateral formed by the feet of the $B,C$ altitudes and $B$ and $C$ .

This post has been edited 2 times. Last edited by WizardMath, Feb 3, 2018, 8:03 PM

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MarkBcc168

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MarkBcc168

#21 h May 28, 2018, 9:52 AM • 7 Y

Y by datcsp03, guptaamitu1, gvole, Adventure10, Mango247, Onetho, GeoKing

First projective sol

Replace labels $M, N, P$ by $M_a, M_b, M_c$ since I like this label. Let $D, E, F$ be the feet of altitude from $A, B, C$ of $\Delta ABC$ . Let $K$ be the Dumpty point of $\Delta ABC$ which is a midpoint of $A$ -symmedian chord. Clearly $K\in\odot(BOC)$ .

Radical axis on $\odot(M_aM_bM_c), \odot(BCEF), \odot(BOC)$ gives $BC, EF, XY$ are concurrent at $P$ and radical axis on $\odot(AM_bM_c), \odot(BOC), \odot(M_aM_bM_c)$ gives $M_bM_c, XY, OK$ are concurrent at $Q$ .

Claim : $\angle BAP=\angle CAQ$ .

Proof : Let $AK$ cut $BC$ and $M_bM_c$ at $L', L$ . Let $AQ$ cut $BC$ at $Q'$ . It suffices to show that $(B, C; Q', L') = (C, B; P, M_a)$ . We do the following sequence of projections.
$\begin{align*} (B, C; Q', L') &= (M_b, M_c; Q, L)\\ &= K(M_b, M_c; O, A)\\ &= A(B, C; O, A)\\ &= A(C, B; D, {\infty}_{BC})\\ &=-\frac{CD}{BD}\\ &=(C, B; P, M_a) \end{align*}$ as desired.

Finally by Desargues Involution Theorem on line $\overline{XYPQ}$ and quadrilateral $M_bM_cEF$ , we get involution swapping $(AP, AQ), (AX, AY), (AB, AC).$ By above claim, this involution must be isogonality w.r.t. $\angle BAC$ so we are done.

This post has been edited 1 time. Last edited by MarkBcc168, May 28, 2018, 9:53 AM
Reason: link

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jayme

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jayme

#22 h Mar 27, 2019, 9:13 AM • 1 Y

Y by Adventure10

Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/Le%20produit%20AB.AC.pdf p. 21...

Sincerely
Jean-Louis

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datcsp03

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datcsp03

#23 h Apr 17, 2020, 4:35 AM

Y by

MarkBcc168 wrote:

First projective sol

Replace labels $M, N, P$ by $M_a, M_b, M_c$ since I like this label. Let $D, E, F$ be the feet of altitude from $A, B, C$ of $\Delta ABC$ . Let $K$ be the Dumpty point of $\Delta ABC$ which is a midpoint of $A$ -symmedian chord. Clearly $K\in\odot(BOC)$ .

Radical axis on $\odot(M_aM_bM_c), \odot(BCEF), \odot(BOC)$ gives $BC, EF, XY$ are concurrent at $P$ and radical axis on $\odot(AM_bM_c), \odot(BOC), \odot(M_aM_bM_c)$ gives $M_bM_c, XY, OK$ are concurrent at $Q$ .

Claim : $\angle BAP=\angle CAQ$ .

Proof : Let $AK$ cut $BC$ and $M_bM_c$ at $L', L$ . Let $AQ$ cut $BC$ at $Q'$ . It suffices to show that $(B, C; Q', L') = (C, B; P, M_a)$ . We do the following sequence of projections.
$\begin{align*} (B, C; Q', L') &= (M_b, M_c; Q, L)\\ &= K(M_b, M_c; O, A)\\ &= A(B, C; O, A)\\ &= A(C, B; D, {\infty}_{BC})\\ &=-\frac{CD}{BD}\\ &=(C, B; P, M_a) \end{align*}$ as desired.

Finally by Desargues Involution Theorem on line $\overline{XYPQ}$ and quadrilateral $M_bM_cEF$ , we get involution swapping $(AP, AQ), (AX, AY), (AB, AC).$ By above claim, this involution must be isogonality w.r.t. $\angle BAC$ so we are done.

I'm new to aops. I'm wondering where to find the article like yours "desargues involution theorem" on aops?

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DrYouKnowWho

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DrYouKnowWho

#24 h Apr 5, 2021, 10:43 AM

Y by

omarius wrote:

anantmudgal09 wrote:

An inversion about $A$ with radius $\sqrt{\frac{1}{2}\cdot AB.AC}$ followed by reflection in bisector maps these two circles two one another. Since the intersection points aren't fixed points under this operation, $X,Y$ are mapped to each other giving $AX,AY$ isogonal in angle $BAC$ .

Sorry but I can't see exactly why it swaps the circles ? Obviously it takes B to N and C to P but we need another point , no ? Thank you.

To answer your question, we present a more clear version of anant mudgal's solution.
solution

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JAnatolGT_00

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JAnatolGT_00

#25 h Jul 26, 2021, 10:49 PM

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Perform inversion with center $A$ and radius $\sqrt{\frac{1}{2}|AB|\cdot |AC|}$ followed by symmetry wrt bisector of angle $BAC$ . Then $B,N$ are swapped, $C,M$ are swapped. Also note that $O$ is swapped with projection of $A$ onto $BC,$ so $(BOC)$ is swapped with nine-point circle. Hence $X,Y$ are swapped implying desired assertion.

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bluedragon17

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bluedragon17

#26 h Apr 9, 2022, 7:48 PM

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We invert about $A$ with radius $\sqrt{\frac{AB\cdot AC}{2}}$ ,
Note that it follows easily by similar triangles that $O\longleftrightarrow K$
where $K$ is the projection of $A$ over $BC$ .

$\Rightarrow (BOC) \longleftrightarrow (MNP)$
The result now follows since $X$ and $Y$ are the intersections of these two circles.

This post has been edited 1 time. Last edited by bluedragon17, Apr 9, 2022, 7:48 PM

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hydo2332

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hydo2332

#27 h Apr 16, 2022, 7:28 PM

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Let $D$ be the foot of the perpendicular from $A$ to $BC$ and let $D_a$ be the Dumpty point of vertice $A$ . We invert wrt $A$ and radius $\sqrt{AB \cdot AC}$ , reflect through the bisector of angle $\angle{A}$ , and take an homothety through $A$ with ratio $\frac 12$ . Notice that $O \rightarrow D, M \rightarrow D_a, N \rightarrow C, P \rightarrow B$ . Therefore the Nine point circle swaps with $\circ{BOC}$ so $X$ swaps with $Y$ and we're done.

This post has been edited 1 time. Last edited by hydo2332, Apr 16, 2022, 7:30 PM

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Alcumusgrinder07

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Alcumusgrinder07

#28 h Sep 11, 2022, 3:53 PM

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We will invert around $A$ with radius $\sqrt{\frac{AB\cdot AC}{2}}$ , and we see that the points $B,N$ $C,M$ are swapped with eachother. We also have that the center is swapped with the projection of $A$ onto $BC$ therefore we have that the circumcenter of $BOC$ is swapped with the 9 point circle so $X$ is swapped with $Y$ so we are done.

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WALKERSkeren

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WALKERSkeren

#29 h Nov 16, 2022, 2:43 PM • 1 Y

Y by Mango247

Is there any solution without using inversion？

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HamstPan38825

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HamstPan38825

#30 h Aug 13, 2023, 7:40 PM

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We perform $\sqrt{\frac{bc}2}$ inversion, which swaps $N, B$ , and $P, C$ . Then notice that $M$ goes to the intersection of the $A$ -symmedian and $(APN)$ , which lies on $(OBC)$ by symmedian properties. Hence $(MPN)$ and $(OBC)$ are swapped, so $\overline{AX}$ and $\overline{AY}$ are isogonal.

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kamatadu

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kamatadu

#31 h Dec 30, 2023, 6:14 PM

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Perform a $\sqrt{\dfrac{bc}{2}}$ -inversion. This gives us that,
$\odot(BOC)\leftrightarrow \odot(MNP).$
This simply means that $X\leftrightarrow Y$ which clearly finishes. :yoda:

Remark: @2above: "Is there any solution without using inversion?". Me: Proceeds to post another solution using inversion. :rotfl:

This post has been edited 2 times. Last edited by kamatadu, Dec 30, 2023, 6:17 PM

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eibc

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eibc

#32 h Apr 19, 2024, 2:28 PM

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Let $\Psi$ be the transformation consisting of an inversion about the circle centered at $A$ with radius $\sqrt{\tfrac{AB \cdot AC}{2}}$ , followed by a reflection over the bisector of $\angle A$ . The two pairs $(B, N)$ and $(C, P)$ swap under $\Psi$ . Additionally, an inversion about the circle centered at $A$ with radius $\sqrt{AB \cdot AC}$ followed by a reflection over the bisector of $\angle A$ swaps $M$ and the second intersection of the $A$ -symmedian with $(ABC)$ , so if $Q$ is the $A$ -dumpty point, then $\Psi$ swaps $M$ and $Q$ . It's well-known that $Q$ lies on $(BOC)$ , so under $\Psi$ , $(MNP)$ and $(BOC)$ swap, and hence $AX$ and $AY$ are isogonal in $\angle A$ , which finishes.

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egxa

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egxa

#33 h Apr 20, 2024, 8:53 AM • 1 Y

Y by bin_sherlo

guys you are posting same inversion solution

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cj13609517288

1930 posts

cj13609517288

#34 h Sep 10, 2024, 6:09 PM

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buh

$(BOC)$ and $(MNP)$ map to each other under a $\sqrt{bc/2}$ inversion, so their two intersections swap. $\blacksquare$

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Mathandski

774 posts

Mathandski

#35 h Sep 20, 2024, 5:25 AM

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Subjective Rating (MOHs) $\text{[math]}$

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Eka01

204 posts

Eka01

#36 h Oct 12, 2024, 4:42 AM

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It is well known that the nine point circle and $(BOC)$ swap under $\sqrt{\frac{bc}{2}}$ inversion, so their intersections swap as well, implying $AX$ and $AY$ are isogonal.

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onyqz

195 posts

onyqz

#37 h Dec 20, 2024, 7:26 PM

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storage
solution

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cursed_tangent1434

661 posts

cursed_tangent1434

#38 h Jan 2, 2025, 3:22 AM

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Really easy. Note that under $\frac{\sqrt{AB \cdot AC}}{2}$ inversion centered at $A$ followed by a reflection across the $\angle A-$ bisector, circles $\omega_9$ and $(BOC)$ map to each other. Thus, their intersections map to each other, which implies that $AX$ and $AY$ are isogonal with respect to $\angle A$ .

This post has been edited 1 time. Last edited by cursed_tangent1434, Jan 2, 2025, 3:22 AM
Reason: wrong point names oops

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Ilikeminecraft

685 posts

Ilikeminecraft

#39 h Feb 19, 2025, 7:33 PM

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Take $\sqrt{AP\cdot AN}$ inversion centered at $A$ and then reflect across the angle bisector of $A.$ Then:
$O$ maps to the foot from $A$ to $PN,$ $B$ maps to midpoint of $AP,$ $C$ maps to midpoint of $AN,$ so $(BOC)$ maps to 9 point circle of $APN.$ Furthermore, the foot from $A$ to $BC$ maps to the circumcenter of $(APN).$ Hence, by taking a 2 times homothety centered at $A,$ we get that the nine point circle and $(BOC)$ map to each other. However, since we are also reflecting across angle bisector, we get $AX$ and $AY$ are isogonal with respect to $A.$ This finishes.

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bjump

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bjump

#40 h Apr 30, 2025, 12:38 AM

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Take a $\tfrac{\sqrt{AB \cdot AC}}{2}$ inversion at $A$ followed by a reflection over the angle bisector of $\angle BAC$ . $C \leftrightarrow P$ , $B\leftrightarrow N$ , $M$ goes to the $A$ -Symmedians intersection with $(ANP)$ . $O$ goes to the foot from $A$ to $\triangle ABC$ . So the diagram inverts back to itself under this transformation because $(B'O'C')$ is the nine point circle of $\triangle ABC$ . Finishing.

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