This problem was discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=84104 , http://www.mathlinks.ro/Forum/viewtopic.php?t=47101 and http://www.mathlinks.ro/Forum/viewtopic.php?t=549 (where the problem is given in a slightly modified version). However, let me give a different solution: Let U and V be the circumcenters of triangles ABC and KBN, respectively. We will use directed angles modulo 180°. Since the point V is the circumcenter of triangle KBN, i. e. the center of a circle passing through the points K, B and N, the central angle theorem for directed angles modulo 180° yields < KBV = 90° - < BNK. On the other hand, since the points C, A, K and N lie on one circle, < CNK = < CAK. Thus, < (AB; BV) = < KBV = 90° - < BNK = 90° - < CNK = 90° - < CAK = 90° - < (CA; AB), so that < (AB; BV) + < (CA; AB) = 90°, what simplifies to < (CA; BV) = 90°. Thus, $BV\perp CA$. On the other hand, since the center of a circle lies on the perpendicular bisector of each chord of the circle, the points U and O both lie on the perpendicular bisector of the segment CA (in fact, the point U is the center of the circumcircle of triangle ABC, and the segment CA is a chord in this circumcircle; the point O is the center of the circle through the points C, A, K and N, which also has CA as chord; so both points U and O lie on the perpendicular bisector of the segment CA). Hence, the line UO is the perpendicular bisector of the segment CA (the line UO really exists, i. e. the points U and O don't coincide - else, we would have K = N = B, contradicting the problem condition). Thus, $UO\perp CA$. Combining this with $BV\perp CA$, we obtain BV || UO. Now, if we consider not the triangle ABC with the points K and N on its sidelines AB and BC, but the triangle KBN with the points A and C on its sidelines KB and BN instead, then we similarly obtain BU || VO. From BV || UO and BU || VO, it follows that the quadrilateral BUOV is a parallelogram. Hence, its diagonals BO and UV bisect each other, i. e. the midpoint R of the segment BO is also the midpoint of the segment UV. The points B and M are the two common points of the circumcircles of triangles ABC and KBN; the centers of these two circumcircles are U and V. Thus, since the two common points of two circles are symmetric to each other with respect to the center line, it follows that the points B and M are symmetric to each other with respect to the line UV. Since the point R lies on the line UV, we thus get RB = RM. Hence, the point M lies on the circle with center R and radius RB. But since R is the midpoint of the segment BO, the circle with center R and radius RB is the circle with diameter BO; thus, the point M lies on the circle with diameter BO, so that < OMB = 90°, and the problem is solved. Darij

Here is my solution: Points B and M are the end-points of a common chord of two circles ABC and KBN. Centers of these two circles are on the perpendicular bisector p of BM. Circle OBM must also have it's center X on p. So X is the midpoint of OB. XM =XB = XO makes angle BMO = 90 Thank you. M.T.

armpist wrote: Here is my solution: Points B and M are the end-points of a common chord of two circles ABC and KBN. Centers of these two circles are on the perpendicular bisector p of BM. Circle OBM must also have it's center X on p. So X is the midpoint of OB. XM =XB = XO makes angle BMO = 90 Nice solution. Alas, 90% of the solution is missing... Darij

But you managed to navigate over the gaps and voids, Darij Grinberg! Consider yourself lucky, at least 10% is there. As you remember, J. Steiner most of the times made statements without any proof whatsoever. Also listen to this: I hope that posterity will judge me kindly, not only as to the things which I have explained, but also to those which I have intentionally omitted so as to leave to others the pleasure of discovery. La Geometrie. It was another old-timer Rene Descartes talking. Can you appreciate this? I think I can, and hope you do too. Thank you. M.T.

there exists nice solutions here. hmmmm... i dont see any solution with inversion. i want to write it . problem:a circle with center$O$ passes through the vertices $B,C$ of triangle $ABC$and intersects the segments $AB,AC$ again at distinct points $F,S$respectively.the circumcircles $ABC$ and $AFS$,intersects at exactly two distinct points $M,A$.prove that: $\angle AMO=90$ ok... lets invert the shape at a center $A$ and with the radius of power of $A$ to circle $(O)$. i have been said that i dont like to invertall the shape,because it may make it too busy. then i just invert the important part of our shape. before inversion: first iwant to say that $AM,SF,BC$ are concurrent.suppose $AM, SF$ intersects at $M\"$. $M\"$ is on the radical axis of circles $ABC,AFS$. it is also on radical axis of circles $(O),AFS$. it means that power of $M\"$ to circles $ABC,(O)$ is eaqual .and $M\",B,C$ should be colinear. after inversion: $S,F$ changes to $C,B$ respectively.and invert of M,should be colinear with them. then $M\"$ should be the invert of $M$. suppose $K$ is a point inwhich $AK$ is tangent to circle$(O)$. the polar of $A$ to circle $(O)$ both goes through $K$ and $M\"$.then the polar is $KM\"$. it is known that the polarof point $A$ to circle $(O)$ is prependicular to $AO$. name the intersection $AO,KM\"$ , $O'$. triangles $AKO'$ and $AKO$, are simmilar to each other.then we have: $AO'.AO=AK^2$. and it means that $O'$ is the inversive form of $O$. ok... now after inversion we see that: $\angle AO'M\"=90$ and it shows before inversion we should have: $\angle AMO=90$ and every thing is ok.

Sorry to repost on this. I think I've a bit different solution. Let $Q = CK \cap AN$ and $R = KN \cap AC$. Then the line through $Q$ and $R$ is the polar of $B$ wrt circle with center $O$. Since $\angle BKC = \angle ANB = 90^{\circ}$, then $Q$ belongs to the circumcircle of $BKN$, and $BQ$ is a diameter. Let $S = QR \cap BO$, since $QR \bot BO$ then $S$ also belongs to the circumcircle. Hence $BNSQKM$ is a hexagon inscribed in that circumcircle. By the Radical Axis Theorem $MB$ intersect $AC$ in $R$, and finally by Pascal's Theorem $M, Q, O$ are collinear, and we are done.

Denote $L\in AC\cap BM$ and $S\in AN\cap CK$. Prove easily that $L\in KN$ and the point $M$ is the Miquel's point of the complete quadrilateral $ACNKBL$ (the intersection between at least the circumcircles of the triangles $ABC$ and $BKN$). Because this quadrilateral is inscribed in the circle $C(O)$ it is well-known that the center $O$ is the orthocenter of the triangle $BLS$ (the quadrupoint $OBLS$ is orthocentrically). In conclusion, $OM\perp LB$, i.e. $OM\perp MB$.

An equivalent enunciation. Let $\triangle ABC$ and the points $M\in BC\ ,\ P\in AM$. The circumcircles $C(O_{b})$, $C(O_{c})$ of the triangles $MPB$, $MPC$ cut again the lines $AB$, $AC$ in the points $X$, $Y$ respectively. Prove that $\{\begin{array}{ccc}P\in BY & \Longleftrightarrow & O_{c}X\perp AB\\\ P\in CX & \Longleftrightarrow & O_{b}Y\perp AC\end{array}$. What happen if $P\in CX\cap BY$ ?

Let $ AC\cap NK=M'$, $ AN\cap KC=X$. $ OX\cap BM'=M^*$. $ BO\cap XM'=H$. Call the circumcircle of $ \triangle AKN$: $ \Gamma$. Let $ \omega$ be the circle center $ B$ that is orthogonal to $ \Gamma$. Theorem: $ XM'$ is the polar of $ B$ wrt $ \Gamma$. $ XB$ is the polar of $ M'$ wrt $ \Gamma$. Lemma 1: $ OXM^*\perp BM'$ and $ M'XH\perp BO$ Proof: Since $ X$ is on the polar of $ B$ and $ M'$, then $ B$,$ M'$ are on the polar of $ X$ so $ BM'$ is the polar of $ X$. Now both facts follow since the line through a point and the center of the circle is perpendicular to that point's polar. Lemma 2: Let $ XM'\cap \Gamma=Y,Z$. $ \omega$ is the circle center $ B$ with radius $ BY$. Proof: Since $ M'YXZ$ is the polar of $ B$, so $ BY$ and $ BZ$ are tangents to $ \Gamma$. But then the circle center $ B$, radius $ BY=BZ$ is orthogonal to $ \Gamma$. Lemma 3: $BM'*BM^*=r_\omega^2$ Proof: $ \angle OHM'=\angle HM^*M'=90^\circ$, so $ OHM^*M'$ is cyclic. Bypower of a point, $ BM'*BM^*=BH*BO$. Since $ \angle BYO=\angle YHB=90^\circ$ and a simple triangle similarity, $ r_\omega^2=BY^2=BH*BO$. The result follows. Lemma 4: $ r_\omega^2=BM*BM'$. Proof: Consider an inversion about $ \omega$. $ \Gamma$ maps to itself. The circumcircle of $ \triangle BNKM$ maps to line since $ B$ is the center of inversion. Since $ N$ maps to $ C$ and $ K$ maps to $ A$, the inversive image is $ AC$. Likewise, the circumcircle of $ \triangle BCAM$ maps to $ NK$. Since $ M$ is the intersection of these two circumcircles, the intersection of these two lines is the inversive image of $ M$. The result follows. By lemmas 3 and 4, $ BM=BM^*$. But $ M$ and $ M^*$ are both on segment $ BM'$, so $ M\equiv M^*$, and the perpendicularity follows.

Wow...you can pretty much angle chase this to death... Let $ AC$, $ NK$, $ BM$ intersect at $ R$, the radical center of the circles. $ \angle NMR=\angle RKB=180-\angle AKN=180-\angle RCN$, so $ NCRM$ is cyclic. Note if $ \angle COK=2z$, then $ \angle CAK=\angle CMR=\angle CNR=\angle KNB=\angle KMB=z$, so $ \angle CMK=180-\angle KMB-\angle CMR=180-\angle COK$, so $ COKM$ is cyclic. Let $ \angle KMO=\angle OCK=y$, then in triangle $ OCK$, $ 2z+y+y=180$, so $ y+z=90=\angle OMB$.

orl wrote: A circle with center $ O$ passes through the vertices $ A$ and $ C$ of the triangle $ ABC$ and intersects the segments $ AB$ and $ BC$ again at distinct points $ K$ and $ N$ respectively. Let $ M$ be the point of intersection of the circumcircles of triangles $ ABC$ and $ KBN$ (apart from $ B$). Prove that $ \angle OMB = 90^{\circ}$. Here's a slightly different solution without use of the radial center:

Sorry for reviving this topic again, but here is another neat solution to this problem:

armpist wrote: Here is my solution: Points B and M are the end-points of a common chord of two circles ABC and KBN. Centers of these two circles are on the perpendicular bisector p of BM. Circle OBM must also have it's center X on p. So X is the midpoint of OB. XM =XB = XO makes angle BMO = 90 Thank you. M.T. I check this problem for many times But the following answer is weird... I think this answer has somthing wrong... Or it isn't all right... Could someone help me please??

Ashegh proved the problem with inversion of center $A$. But there is a simpler solution by using inversion with center $B$. Now consider an inversion with center $B$ and an arbitrary radius of inversion. This would transform $A'. C', M'$ to be collinear and $K, N', M'$ also collinear in such a way that $A'C'N'K'$ forms a homothetic circle to $ACNK$. Now, to find the inverse of $O$, consider the polar of $B$ with respect to $ACNK$, that is $XY$ where $BX, BY$ are tangents from $B$ to the circle $ACNK$. Let after the inversion, the inverse images of $X,Y$ be $X', Y'$. Now since $B, O, X, Y$ are concyclic, it implies $O', X', Y'$ are collinear . However, as $O$ was equidistant from $X,Y$ ; it implies that $O'$ is the midpoint of $X'Y'$. Now, $B$ is the pole of $KM$. THis implies polar of $B$ passes through $M'$. But $X'Y'$ is the polar of $B$ with respect to $A'C'N'K'$. Therefore, angles $OBM = BO'M' = BO'X' =BO'Y' = 90$ degrees. (As tangents from an exterior point are equal).

The argument of antiparallel lines also works here and let me to say more details about the solution by armpist's way. So, we restate as $O'$ the circumcenter of $ACNK$ and let $O,\ O''$ be, the circumcenters of $\triangle ABC,\ \triangle KBN,$ respectively. Because of the line segments $KN,\ AC,$ are antiparallel each other, with respect to $\angle B$ of the triangles $\triangle ABC,\ \triangle KBN$ respectively, we have that $BO\perp KN$ $,(1)$ and $BO''\perp AC$ $,(2)$ But, we have also $OO'\perp AC$ $,(3)$ and $O'O''\perp KN$ $,(4)$ From $(1),\ (2),\ (3),\ (4),$ we conclude that $BOO'O''$ is a parallelogram and let be the point $X\equiv BO'\cap OO''.$ Because of now, the line segment $OO''$ is the midperpendicular of $BM,$ we have $XM = XB = XO'$ $,(5)$ From $(5)$ we conclude that $\triangle MBO'$ is a right triangle with $\angle BMO' = 90^{o}$ and the proof is completed. Kostas Vittas.

Dear Mathlinkers, 1. let 0, 1, 2 be the circumcircles of ABC, AKNC, KBN, AKD 2. According to the pivot theorem, 0, 1 and 2 concur at the Miquel point of the delta determined by the triangle DCN and the line AKB i.e. at M. 3. Let U, V the second points of intersection of MA, MN with 1. 4. With my favourite theorem, AV, NU and BD are parallel. 5. the quadrilateral AVNU being a isosceles trapeze, OM perpendicular to BD and passes trough E. Sincerely Jean-Louis

Let $L$ be the intersection of $AN$ with $KC$. Let $J$ be the intersection of $KN$ and $AC$. It is well known that $O$ is the orthocenter of triangle $BJL$, so $OL \perp BJ$. We know prove that $M$ lies on $BJ$. Let $M'$ be the intersecion of $BJ$ with the circumcircle of $ABC$. By power of point $J$ WRT the circle passing through point $A,C,K$ and $N$ we have: $JA*JC = JK*JN$ By power of point $J$ WRT to the circumcircle of $ABC$ we have: $JA* JC = JM' * JB$ and therefore $JK*JN= JM'*JB$ So $M'$ lies on circumcircle of $BNK$ $=> M=M'$ We now have $OL \perp BM$. We will show that $O-L-M$ to conclude. Because of power of point $L$ WRT to the circle passing through $KNCA$ we have $LA * LN = LK* LC$ (1) Since $\angle {BMA} = 180 - C$ and $\angle{NMB}=\angle{NKB}=\angle{BCA}=C$ we have that $\angle{NMA}= 180-2C$. Since $O$ is the circumcenter of $ANC$, $\angle{ANO}=2\angle{ACN}= 2C$ and therefore cuadrilateral $MNOA$ is cyclic. Call $C_1$ the circle passing through it. Since $A=\angle{BAC}= \angle{BNK}$ we have $\angle{BMK}=180-A$. Since $A=\angle{BAC}= \angle{CMB}$ we have $\angle{CKM}= 180-2A$ Since $O$ is the circumcenter of $AKC$, $\angle{KOC}=2\angle{KAC}= 2A$ and therefore cuadrilateral $MKOC$ is cyclic. Call $C_2$ the circle passing through it. It is clear that $BM$ is the radical axis of circles $C_1$ and $C_2$. By (1), $L$ has the same power WRT to $C_1$ and $C_2$. Hence, $L$ lies on $BM$ and $B-L-M$ which concludes the proof.

My solution: Since $M$ is the second intersection of circles $ACB$ and $BKN$, $M$ is the Miquel's point of cyclic quadrilateral $ACNK$ . Then it is well known that $OM\bot MB$, hence $\angle OMB=90^\circ$. Would this be sufficient at the IMO?

Agr_94_Math wrote: Now, $B$ is the pole of $KM$. THis implies polar of $B$ passes through $M'$. But $X'Y'$ is the polar of $B$ with respect to $A'C'N'K'$. Therefore, angles $OBM = BO'M' = BO'X' =BO'Y' = 90$ degrees. (As tangents from an exterior point are equal). can you please expain how $B$ is the pole of $KM$?

Let $R=\overline{KC} \cap \overline{AN}$ and $P=\overline{KN} \cap \overline{AC}$. Note that $M$ is the Miquel Point of $KNCA$. It is well known that under inversion under $(KNCA)$, $R$ inverts to $M$. By Brocard's, it follows that $OM \perp BP$, so $\angle OMB=90$.

$M$ is the Miquel Point of $ABCD$. Then apply the Master Miquel Theorem.

Let $KN \cap AC=L$,$AN \cap KC=R$ Note that: $M$ is the Miquel - point of $ACNKBL$ But $AKNC$ is cyclic $\implies M \in LB$.(This is since the radical axes of $\odot ABC, \odot AKNC, \odot BKN \implies BM, KN, AC$ are concurrents in $L$). By chasing angles: $\angle BKM = \angle BNM = \angle MLC$ Let $\angle BMK=\angle BNK=\angle KAC=\angle NML=\alpha$ $\implies \angle KOR=\angle ROC=\alpha$ Lemma: Let $ABCD$ be a cyclic quadrilateral, inscribed in a circle $\omega$ centered at $O$.Let $AB \cap CD=P$ and $AD \cap BC=Q$. Let the intersection of the diagonals $AC$ and $BD$ be $R$. Let $M$ be the Miquel point of $ABCDPQ$. Then: $OM \perp PQ$ Proof: By chasing angles: $M \in (BOD)$. In a similar way, we can prove that $M \in (AOC)$. The line $AC$ is sent to the circle $(OAC)$ and the line $BD$ is sent to the circle $(OBD)$. The lines $AC$ and $BD$ intersect at $R \implies $ the image of $R=R^*$ is $R^* \in \odot AOC \cap \odot DOB=M$. $\implies O,R,M$ are collinears By polars: $r \equiv PQ \implies OR \perp PQ \implies OM \perp PQ$ In the problem: $OM \perp LB \implies \angle OMB=90$ $\blacksquare$.

It's clear that $M$ is the Miquel Point of $ACNK$. But $ACNK$ is also cyclic, so Master Miquel finishes. $\blacksquare$

HamstPan38825 wrote: $M$ is the Miquel Point of $ABCD$. Then apply the Master Miquel Theorem. ike.chen wrote: It's clear that $M$ is the Miquel Point of $ACNK$. But $ACNK$ is also cyclic, so Master Miquel finishes. $\blacksquare$ what is master miquel theorem?

math12345678 wrote: HamstPan38825 wrote: $M$ is the Miquel Point of $ABCD$. Then apply the Master Miquel Theorem. ike.chen wrote: It's clear that $M$ is the Miquel Point of $ACNK$. But $ACNK$ is also cyclic, so Master Miquel finishes. $\blacksquare$ what is master miquel theorem? This handout covers some the most essential properties of Complete Quadrilaterals.

Notice that $M$ is the Miquel point of cyclic quadrilateral $ACNK$ and $\overline{MB},\overline{KN},$ and $\overline{AC}$ concur by Radical Axis. We are done as $\overline{OM}\perp\overline{BL}$ where $L=\overline{KN}\cap\overline{AC}.$ $\square$

Let the intersection of $MK$ and $BC$ be $Q$ and let the intersection of $MO$ and $BC$ be $P$. Can you help me by proving that $(C,Q;P,B)=-1$?

We, first of all, join $MC$ and denote the intersection of $MC$ with circle having center $O$ as $L$ since points $A,K,L,C$ are concyclic we have $\angle{MLK}=A$ and also since points $M,B,C,A$ are concylic we have $\angle{BMC}=A$ now this gives $MB \parallel K$. so we just want to show that $OM \perp KL$ notice $OL=OK$ and if $\angle{BCM}=\angle{LKN}$ (angle subtended by chord $LN$) and hence $\angle{MBC}=180^{\circ}-(\angle{BCM}+A)$ and since $K,M,B,N$ are concyclic points we have $\angle{MKN}=\angle{BCM}+A$ or $\angle{LKN}+A$ which gives $\angle{MKL}=A$ which gives $ML=MK$ which gives quadrilateral $MKOL$ is a kite and hence $KL \perp OM$ and hence $OM \perp MB$ as desired $\blacksquare$

OTIS Version of IMO 1985/5 wrote: Let $ABCD$ be a cyclic quadrilateral with center $O$. Lines $AD$ and $BC$ meet at $P$. The circumcircles of $PAB$ and $PDC$ meet at $M$. Prove that $\angle OMP=90^{\circ}$. Note that $M$ is the Miquel point of $ABCD$. Then, by Master Miquel, $\angle OMP=90^{\circ}$ as required.

Define $AC \cap KN=D$ and $AN \cap CK=E$. By Master Miquel, $K$ lies on $OE \cap BD$, and by Brocard we have $OE \perp BD$, giving us our result. $\blacksquare$

Light-weight angle chasing solution

As $M$ is the miquel point of $ACKN$, the result is true by Master Miquel Theorem.

Note that $M$ is the miquel point of the cyclic quadrilateral $ACNK$ so the result is pretty well known.

$M$ is the miquel point of degenerate quad $ACBD.$ Then, by Master Miquel Theorem, $M$ and $Q$ are inverses, so $OMQ$ is collinear. By radical axis, $M, P, R$ are collinear. By Brocard's, $MPR\perp OQ.$