I needed 10 minutes to sort the trash out of the configuration. After that everything was trivial. If T is the point of intersection of the circumcircles of triangles BMR and CNR, then we must show that T lies on BC, i. e. that we have < BTR + < CTR = $180^{\circ}$. Since BMRT is a cyclic quadrilateral, we have < BTR = $180^{\circ}$ - < BMR, so < BTR = < AMR. Similarly < CTR = < ANR. So we have to show that < AMR + < ANR = $180^{\circ}$, or, in other words, we must prove that the quadrilateral AMRN is cyclic. Well, the points M and N lie on the circle with diameter BC, and the center of this circle is obviously the midpoint O of BC. Thus, MO = NO, so that the triangle MON is isosceles, and hence the line OR, being the angle bisector of the angle MON, coincides with the perpendicular bisector of the segment MN. Hence, the point R lies on the perpendicular bisector of the segment MN. In other words, the point R is the point of intersection of the perpendicular bisector of the segment MN with the angle bisector of the angle MAN. But a well-known fact states that the perpendicular bisector of a side of a triangle meets the angle bisector of the opposite angle at a point on the circumcircle of the triangle. Hence, the point R lies on the circumcircle of triangle MAN, and therefore, the quadrilateral AMRN is cyclic. Darij

One can also prove the cyclicity with the following argument (for those unfamiliar with that well-known fact): We know that $MR = NR$ and$ \angle MAR = \angle NAR$. By sine law, ${{MR}\over{\sin MAR}}={{AR}\over{\sin AMR}}$ and ${{NR}\over{\sin NAR}}={{AR}\over{\sin ANR}}~.$ We thus obtain $\sin AMR = \sin ANR~.$ However $\angle AMR \neq \angle ANR$, therefore $\angle AMR + \angle ANR = 180^o$ and the quadrilateral $AMRN$ is cyclic.

For those unfamiliar with said well-known fact, it's trivial to derive: In $\Delta AMN$, with circumcircle $\omega$, let the internal bisector of $A$ meet $\omega$ at $R'$. Since $R'$ is the midpoint of arc$MN$ (not containing $A$), the perpendicular bisector of segment $MN$ also intersects $\omega$ at $R'$. However, because the intersector of the internal bisector of $A$ meets the perpendicular bisector of $MN$ at a unique point, $R' = R$, implying that $ARMN$ is cyclic. Hence, you didn't need to know this off the top of your head because it's simple to derive.

let's say it all can be done after some easy calculation and by applying inversion respect to a circumference having center in A and such that M'=B and N'=C... not difficult, just some more elegant (apart fro ugly trigonometry)... i'm wondering how many can have done it like me

further more we can get that $R$ is the incenter of $\triangle MON$ and common point of the two circumcircles $P$ lying on BC is on the bisector of $\angle BAC$

Valentin Vornicu wrote: 1. Let $ABC$ be an acute-angled triangle with $AB\neq AC$. The circle with diameter $BC$ intersects the sides $AB$ and $AC$ at $M$ and $N$ respectively. Denote by $O$ the midpoint of the side $BC$. The bisectors of the angles $\angle BAC$ and $\angle MON$ intersect at $R$. Prove that the circumcircles of the triangles $BMR$ and $CNR$ have a common point lying on the side $BC$. Just a reaction from a victim of this problem. One half of all participants who had solutions to this problem got 6 of 7 points, the reasons being i. a. - a lack of the proof that the common point lies on the segment BC, not just on the line BC; - missing proofs for implicitely used statements about which point lies in which halfplane; - a definition of the second common point of the two circumcircles without considering the (only theoretically) possible case when they touch. Of course, if the coordinators have nothing better to do, they are allowed to bitch around about this, but I believe it is under the niveau of a reasonable IMO... just my two pennies. I have got 6 points on the problem, I was really under stress and randalated, but I still have 27 points and get silver... darij

darij grinberg wrote: a victim of this problem LOOOL Are you mad ???? 6 is good ! and 27 is good too !! I wish I had 27 (All my team too )

Igor wrote: LOOOL Are you mad ???? 6 is good ! Well, but don't forget I did never think to get a point rejected in a simple geometry problem! dg

ltsin wrote: further more we can get that $R$ is the incenter of $\triangle MON$ and common point of the two circumcircles $P$ lying on BC is on the bisector of $\angle BAC$ these results are more interesting than the problem but still easy ... (and proving the first part gives another proof of the problem that automatically proves the intersection lies on the *segment* BC). Let the perpendiculars from C and B to AB and AC be C' and B' (i'm too lazy to think of M and N ) first let A' be the point where angle bisector of <BAC intersects BC. <C'RA'=180-<C'RA=180-<C'B'A (because MNAR is concyclic) = 180-<ABC (because B'C'BC is concyclic) so <C'RA' + <C'BA' = 180 proving the concylicity of RC'A'B. To prove R is the incentre of B'C'O it suffices to show <B'C'O=<C'B'O=2*<RC'B'=<B'AC', i.e. <C'OB'=180 - 2*<BAC (since OB'=OC'). Let A'' be the foot of the altitude from A to BC --> since B'C'OA'' is concylic (nine point circle), <C'OB'=<C'A''B'=2*<C'A''A (because H is the incentre of A'B'C') =2*<C'CA (since C'A''AC is concylic)=2*(90 - <BAC) = 180 - 2*<BAC as required.

darij grinberg wrote: Igor wrote: LOOOL Are you mad ???? 6 is good ! Well, but don't forget I did never think to get a point rejected in a simple geometry problem! dg Well, you had to... How many people got 0 on this problem, eh ? Antony

a small notice, now that everyone is cooled off this problem was proposed by two Romanians, can you guess who? :D

Valentin Vornicu wrote: a small notice, now that everyone is cooled off this problem was proposed by two Romanians, can you guess who? :D You and Branzei? darij

darij grinberg wrote: Valentin Vornicu wrote: a small notice, now that everyone is cooled off this problem was proposed by two Romanians, can you guess who? :D You and Branzei? darij close enough. it's actually Dinu Serbanescu (Team Leader of the JBMO Team) and me :P

Consider the inversion $f$ with center $A$ and power $AM \times AB$. Then $f(M)=B$, $f(N)=C$. Lets show that $f(R)=X$, with $X$ on the side $BC$. Let $\Gamma$ the circumcircle of $ABC$, $\Gamma_1$ the circumcircle of $AMN$ and $T=AR\cap \Gamma$. As in the previous solutions, $MR=NR$ and $AMRN$ is cyclic. Also: $\triangle ANM \sim \triangle ABC$ and $R$, $T$ are homologous (because $R$ is on $\Gamma_1$ and on the bisector of $\angle MAN$). Then \[\frac{AR}{AT}= \frac{AM}{AC}.\quad(1)\] From similar triangles, it immediately follows that \[\frac{AX}{AC}=\frac{AB}{AT} .\qquad (2)\] ($\triangle AXB \sim \triangle ACT$). From (1) and (2), \[ AX\times AR=AB\times AM\] and then $f(R)=X$.

ltsin wrote: further more we can get that $R$ is the incenter of $\triangle MON$ and common point of the two circumcircles $P$ lying on BC is on the bisector of $\angle BAC$ In fact you can prove this fact without knowing AMRN is cyclic. Given this it's easy to prove that it is cyclic, and solve the problem this way. <OMB = <B as OB = OM, and <AMN = <C since M is the foot of the perpendicular from C. So <NMO = <A. Let I be the incentre of MON; then <IMN = <INM = (1/2)<A. So <IMA + <INA = <A + <B + <C = 180, thus AMIN is cyclic, so <IAN = <IMN = <INM = <IAM, so I lies on the bisector of <A. We also know that I lies on the bisector of <MON. As AB <> AC the bisectors of <A and <MON do not coincide, so they have a single intersection and R must coincide with I. It follows that AMRN is cyclic.

because MO=NO, MN and OR, they are vertical, so MR=RN. because AM is not =AN , also AR=AR, so <AMR+<ANR=180', so AMRM is cyclic by construction, so <AMR=<RNC, assume a point P on BC, and MRPB is cyclic by construction, so <AMR=<BPR, because <AMR=<RNC, so <BPR=<RNC, so PRNC is cyclic by construction. OK OK? I'm sorry, I'm not good at English, I am a Chinese student, English is not my languege..

zzq-19871011 wrote: OK OK? I'm sorry, I'm not good at English, I am a Chinese student, English is not my languege.. Your English is great, don't worry about it.

Let $H$ be the homothety mapping $M$ to $C$ and $N$ to $B$ Let $OR$ intersect $MN$ at $P$. Since $P$ lies on the angle bisector of $\angle MON$, $\angle MOP = \angle NOP$. Also since $OM=ON$, $\Delta OMP \cong \Delta ONP$ and thus $MP=PN$, meaning that $P$ is the midpoint of $MN$. So $H$ also maps $P$ to $Q$. Let $AR$ intersect $MN$ at $Q$ and $BC$ at $D$. By the Angle Bisecctor Theorem, $\frac{MQ}{QN} = \frac{AM}{AN}$. By the Power of a Point Theorem, $AM \cdot AB = AN \cdot AC$, so $\frac{AM}{AN} = \frac{AC}{AB}$. Again by the Inscribed Angle Theorem, $\frac{AC}{AB} = \frac{DC}{BD}$. So $\frac{MQ}{QN} = \frac{DC}{BD}$. Finally, $H$ maps $Q$ to $D$. Since $PO$ and $QD$ both pass through $R$, $R$ must be the center of $H$. Thus $B,R,N$ and $C,R,M$ are collinear. Then $\angle BMR = \angle CNR = \pi / 2$. Let $R'$ be the foot of the perpendicular from $R$ to $BC$. $\angle RR'B = \pi / 2 = \angle RR'C$. Thus $BMRR'$ and $CNRR'$ are cyclic. So the circumcircles of $BMR$ and $CNR$ pass through $R'$, which lies on $BC$, as desired. QED.

Well triangle MOR and NOR are congruent, so MR=RN. Now let $\omega$ be the circumcircle of triangle AMN. Let the angle bisector of angle MAN intersect $\omega$ at $R'$. arc MR' = R'N hence MR'=R'N. Consider the locus of R', it lies on angle bisector of MAN and the perpendicular bisector of MN, there's only 1 intersection, hence R' and R must be the same point => AMNR is a cyclic quadrilateral. Let the circumcircle of BMR intersect BC at S. MRSB is another cyclic quadrilateral. angle AMR = RSB by supplementary angles. Angle AMR and ANR are supplementary by cyclic quad AMRN => RSB and ANR are supplementary => angle RSC and RNC are supplementary => RNCS is a cylic quadrilateral. Hence the circumcircles of BRM and RNC intersect at S on BC as desired.

The idea is to keep reducing and thinking backwards! The desired point should be the intersection of the $A$-angle bisector with $BC$. (Think of radical axis) Hence, we should define $R'$ as the intersection of the $A$-angle bisector with $BC$. Let $(BMR')$ and $(CNR')$ intersect at $D'$. It now suffices to show $OD'$ bisects $\angle MON$. At this point the angles $\angle MOD'$ and $\angle D'ON$ are very difficult to find. Instead we may strive for wishful thinking. Since $OD'$ supposedly bisects $\angle MON$, what if $D'$ was the incenter of $MON$? This automatically gives us $\angle MOD' = \angle D'ON$. Thankfully this turns out to true and is very easy to prove, just recall that $BMNC$ is cyclic and the rest is simple angle chasing (note that $AMD'N$ is also cyclic because of Miquel's theorem.) $\blacksquare$

Claim: $AMNR$ is cyclic. Proof. Since $OM=ON$, $\Delta OMN$ is isoceles. Thus, $OR$ lies on the perpendicular bisector of $MN$, leading to $RM = RN$. However, $R$ also lies on the angle bisector of $\angle MAN$, so the result follows by the Incenter-Excenter Lemma in $\Delta AMN$. $\blacksquare$ Let $D = (BMR) \cap BC$. It becomes clear that $R$ is the Miquel point of $\Delta ABC$, so $(CND)$ passes through $R$, giving the desired result.

O is the center of the circle with diameter $BC$. $\therefore MO=NO \Rightarrow MN\perp OR$. Point R is the intersection of the angular bisector of $\angle BAC$ and the perpendicular bisector of $MN$. Hence $AMRN$ is cyclic. Now, let circumcircle of $\triangle AMR$ intersect $BC$ at $I$. $\angle ANR =\angle AMR = \angle NIC$ (This is called miquel point) . Hence, $NRIC$ is collinear.

Let $K$ be the intersection of lines $AR$ and $BC$. We show that $BMRK$ and $CNRK$ are cyclic, leading to the desired result. We will show that $BMRK$ is cyclic, and a symmetrical argument can be used to show $CNRK$ is cyclic. Claim: $AMRN$ is cyclic. Let $R'$ be the phantom point corresponding to the second intersection of $(AMN)$ and line $AK$. Since $\angle R'AM = \angle R'AN$, we know that $R'$ must be the arc midpoint of arc $MN$, meaning that it is the intersection of the perpendicular bisector of $MN$ and line $AK$. However, since triangle $OMN$ is isosceles, $R$ is also the intersection of the perpendicular bisector of $MN$ and line $AK$. Thus, $R' = R$ and $AMRN$ is cyclic. Simple angle chasing gives us that $\angle RMN = \frac12 \angle A$ and $\angle OMN = \angle A$, and thus $\angle OMR = \frac12 \angle A = \angle BAK$. Putting things together, we have (using directed angles) \[ \angle BMR = \angle BMO + \angle OMR = \angle OBM + \angle BAK = \angle KBA + \angle BAK = \angle BKA = \angle BKR. \]Thus, we are done.

The given conditions gives us that $R$ s the midpoint of $\widehat{MN}$ in ${AMN}$. Now letting $(BMR) \cap (CNR) = X \ne R$, just some angle chasing gives us $B-X-C$. $\square$

Note that $R$ lies on the perpendicular bisector of $MN$, and the angle bisector of $MAN$, so it is just the arc midpoint on $(MAN)$, so $(AMNR)$ is cyclic, the finish is then obvious from miquel.

It is easy to see that the arc midpoint of $MN$ wrt $(AMN)$ lies on the bisectors of $\angle MON$ and $\angle MAN$ so $R$ is said arc midpoint. Then since $R \in (AMN)$, $R$ is the Miquel point in $\triangle ABC$ and $(BMR)$ and $(CMR)$ must concur on $BC$.

Radical axis of three circles are concurrent on point A

storage

this thread is so old some users have spaces in their names. 20 year thread

By the three-tangents lemma $R$ is the midpoint of arc $\widehat{MN}$ of $(AMN)$. Then $\angle BMR = ANR$ so $\angle BMR + \angle RNC = 180^\circ$, so there exists a point on $BC$ that lies on both $(BMR)$ and $(CNR)$.

By Miquels, we can see that the condition we need to prove is equivalent to $AMRN$ being cyclic. But this is just saying that the intersection of the angle bisector of $\angle MAN$ in $\triangle MAN$ and the perpendicular bisector of $MN$ lies on $(AMN)$ which is obvious.

Take $Q$ to be the point at which one of the circle hits $BC$, then we need to prove that $QRNC$ is cyclic. This reduces to proving $AMNR$ is cyclic. $MRO$ and $NRO$ are congruent, so $MR=RN$. $AR$ is the angle bisector and as it hits the circumcircle at the $MN$ arc midpoint. $R$ has to lie on $AMN$

Observe by the Tri-Tangent Lemma that $OM$ and $ON$ are tangent to $(AMN).$ The intersection of the angle bisector of $\angle MON$ and $(AMN)$ must be the midpoint of small arc $MN$ which is $R$ by Fact 5, so $AMRN$ concyclic. Let $P$ be the intersection of $BC$ and $(BMR).$ Then \[\angle RPC=180-\angle RPB=\angle BMR=180-\angle AMR=\angle ANR=180-\angle RNC\]so $RPCN$ cyclic and we are done.

Clearly $MR=RN$ by congruent triangles $\triangle MOR \cong \triangle NOR.$ Therefore by the Law of Sines $$1 = \frac{MR}{RN} = \frac{\sin \angle MAR \cdot \frac{AR}{\sin \angle RMA}}{\sin \angle NAR \cdot \frac{AR}{\sin \angle RNA}} \implies \sin \angle RNA = \sin \angle RMA.$$Hence either $\angle RNA = \angle RMA$ or $RNA=180^\circ - \angle RMA.$ If the former holds, we have that $\triangle MAR \cong \triangle NAR$ so $AM=AN$ and thus $AB=AC,$ a contradiction. Therefore the latter holds and $MRNA$ is a cyclic quadrilateral, and the rest follows by Miquel's Theorem. QED