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.

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 $\angle CAY=\angle BAR=\angle BAX$

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 $Y$?

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 $X$.

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 $.

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 $\angle BAX = \angle CAY.$

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

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

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.

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 $\angle BAX=\angle CAY$

inversion with center a and radius AB.AC/2

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

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.

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 $BAC$.

Dear Yetti and Mathlinkers, your idea of the circle (AIJ) is marvellous... it intersect (O) in a remarkable point... Sincerely Jean-Louis

Dear Mathlinkers, you can see http://jl.ayme.pagesperso-orange.fr/Docs/Serbia%202013%20Problem%203.pdf Sincerely Jean-Louis

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.

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.

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 $C$.

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.

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Le%20produit%20AB.AC.pdf p. 21... Sincerely Jean-Louis

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?

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.

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.

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.

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.

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.

Is there any solution without using inversion？

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.

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

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.

guys you are posting same inversion solution

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

$ $

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.

storage

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 $\angle A$.