Let $ABC$ be an acute triangle with circumcenter $O$. Let $D$ be the foot of the altitude from $A$ to $BC$ and let $M$ be the midpoint of $OD$. The points $O_b$ and $O_c$ are the circumcenters of triangles $AOC$ and $AOB$, respectively. If $AO=AD$, prove that points $A$, $O_b$, $M$ and $O_c$ are concyclic. Marin Hristov and Bozhidar Dimitrov, Bulgaria
Problem
Source: JBMO 2023 Problem 4
Tags: geometry, circumcircle, Circumcenter
VicKmath7
26.06.2023 10:27
Let $AM \cap BC=P, OO_c \cap BC=X, OO_b \cap BC=Y$. We have that $\angle AO_bY=180^{o}-2\gamma=\angle AXY$, so $AO_bXY$ is cyclic and similarly $AO_cXY$ is cyclic. Hence, we are left to prove that $AMXY$ is cyclic. We have that $AM$ is the perpendicular bisector of $OD$ and $\angle AOP=90^{o}$, so $PO^2=PM \cdot PA$. We are left to see that $PO^2=PX \cdot PY$, which will mean that $PM \cdot PA=PX \cdot PY$, i.e. that $AMXY$ is cyclic. Indeed, $\angle POX=\angle OAC=\angle XYO$, so $PO$ touches $(OXY)$, as wanted.
Orestis_Lignos
26.06.2023 10:41
Let $X,Y$ be the reflections of $A$ across $O_c,O_b$ respectively, and $S$ be the midpoint of arc $BC$ in circle $(ABC)$. Note that, $AD=AO=OS$ and $AD \parallel OS,$ hence quadrilateral $ADSO$ is a parallelogram, therefore points $A,M,S$ are collinear and so $M$ is the midpoint of $AS$. We have the following Claim. Claim: Quadrilateral $AXSY$ is cyclic. Proof: Note that $\angle AOX=\angle AOY=90^\circ$, and so $O \in XY$. Moreover, if $XY$ intersects $BC$ at point $T$, then $\angle AOT=\angle ADT=90^\circ$, and since $AD=AO$ we obtain that $AT$ bisects $\angle DAO$. Since $AD,AO$ are isogonal, this means $T \in AS$. To finish, note that $\angle BXY=\angle BXO=180^\circ-\angle BAO=90^\circ+\angle C=\angle C+\angle ACY=\angle BCY,$ hence $BXCY$ is cyclic. Thus, $TA \cdot TS=TB \cdot TC=TX \cdot TY,$ as desired $\blacksquare$ Back to the problem, since $AXSY$ is cyclic and $O_c,M,O_b$ are the midpoints of $AX,AS,AY$ respectively, we obtain that $AO_cMO_b$ is cyclic, too, as desired.
newinolympiadmath
26.06.2023 13:02
can you please mention motivations behind constructing such points?
motannoir
26.06.2023 19:17
We will do inversion at A The problem restates as follows Let triangle $ABC$,$D$ antipode, $O$ the symmetric of $A$ wrt $BC$,we have that $AO=AD$ and let $M$ be the antipode of A in $(AOD)$. Let $O_B$ and $O_C$ the symmetric of A wrt $BO$ and $CO$. Prove that $O_B,O_C,M$ are collinear Solution: by a homothety we relabel $O_B,O_C$ as the foot from A to $BO,CO$ and we want to prove that $O_B,O_C,S$ are collinear where S is the centre of $(AMD)$. But since $AO=AD$ we have that (AS is the bisector of $\angle BAC$ Now by midlines and parralelism we can prove that S is on BC Now we finish with menelaus on $\triangle BCO$ with $O_C-S-O_B$ . After sine law is enough to prove that $2sinBsinC=1$ but this is easy to prove bc of the condition $AO=AD$ which translates as the $dist(A,BC)=R$ and we do law of sines again and we finish.
Assassino9931
27.06.2023 17:55
This marvelous geometry problem was proposed by Marin Hristov (Marinchoo) and Bozhidar Dimitrov (Strudan_Borisov). With time consuming problems 1-3 many people unfortunately did not attempt this problem thorougly, but overall it was very appreciated! (Still, we did see five complete solutions and one very nice trig bash attempt, as well as a student with 5 points for which I do not know the details.) Here is a rephrased version of the authors' solution which has ended up as the official solution.
Note that $AB = AC$ cannot hold since $AO = AD$ would imply that $O$ is the midpoint of $BC$, which is not possible for an acute triangle. So we may assume without loss of generality that $AB < AC$. Let $M_b$ and $M_c$ be the midpoints of $AC$ and $AB$, respectively. Since $\angle AM_bO = \angle AM_cO = 90^{\circ} = \angle AMO$ (the latter since $AO = AD$), the pentagon $AM_bOMM_c$ is cyclic.
Next, notice that $AM$ is the perpendicular bisector of $OD$, $O_bO_c$ is the perpendicular bisector of $AO$ and $M_bM_c$ is the perpendicular bisector of $AD$. Hence these three lines are concurrent - denote their common point by $T$.
The quadrilateral $AO_bMO_C$ is cyclic if and only if $AT \cdot TM = O_bT \cdot O_cT$. From the cyclic $AM_bMM_c$ we have $AT \cdot TM = M_bT \cdot M_cT$. Hence it now suffices to argue $M_bT \cdot M_cT = O_bT \cdot O_cT$ -- or equivalently, that $M_b$, $M_c$, $O_b$ and $O_c$ are concyclic.
We assume that $\angle AOB < 90^{\circ}$ and $\angle AOC > 90^{\circ}$ so that $O_c$ is in the interior of triangle $AOB$ and $O_b$ in external to the triangles $AOC$ (the other cases are analogous and if $\angle AOB = 90^{\circ}$ or $\angle AOC = 90^{\circ}$, then $M_b\equiv O_b$ or $M_c \equiv O_c$ and we are automatically done). We have
\[ \angle M_cM_bO_b = 90^{\circ} + \angle AM_bM_c = 90^{\circ} + \angle ACB\]as well as (since $O_cO_b$ is a perpendicular bisector of $AO$ and hence bisects $\angle AO_CO$)
\[ \angle M_cO_cO_b = 180^{\circ} - \angle OO_cO_b = 90^{\circ} + \frac{\angle AO_cM_c}{2} = 90^{\circ} + \frac{\angle AO_cB}{4} = 90^{\circ} + \frac{\angle AOB}{2} = 90^{\circ} + \angle ACB \]and therefore $O_bM_bO_cM_c$ is cyclic, as desired.
Having in mind the difficulty of the problem, the coordinators and leaders decided to award 2 points for showing $AM_bOMM_c$ is cyclic and 4 points for showing $M_bO_cM_cO_b$ is cyclic, respectively 3 points for identifying $T$ as a concurrency point of the three lines and 1 point for the PoP finish. So one can actually obtain $6$ points by just angle chasing which is rewarding for the people who have attempted the problem with a good strategy! But as I said above, the main reason the scores on this problem are quite low is that problems 1-3 were quite time consuming, so this is a lesson to be learned for leaders when voting for the competition problems.
The optimal strategy is to consider $T = AM \cap O_bO_c$, having in mind it is the circumcenter of $AOD$, and show $AT \cdot TM = O_bT \cdot O_cT$. Have in mind that when $AB < AC$ we have $\angle OAM = \angle MAD = \frac{\angle ABC - \angle ACB}{2}$.
egxa
02.07.2023 19:57
Assassino9931 wrote: as well as a student with 5 points for which I do not know the details.) I have 5 points. I didn't see only concurrency on the circumcentre. I did the all of the PoP so rest of the question is only see 3 PoP equality
Assassino9931
27.12.2023 20:04
Here goes a nice trig solution, I cannot remember if it the same as the approach of Nina Susic (SRB1), but definitely the starting idea is the same. It's a shame I did not figure this out in the free 20 min I had before discussing marking schemes with coordinators and other leaders! Let $O_bO_c \cap AM = T$, then by Power of a Point it suffices to show $AT \cdot TM = O_bT \cdot O_cT$. Note that $AM$ is the perpendicular bisector of $OD$ (by the problem condition $AO = AD$) and $O_bO_c$ is the perpendicular bisector of $AO$, so $T$ is the circumcenter of $AOD$ - this makes calculating the abovementioned segments very promising! One may want to express them in terms of elements of $ABC$, but in the process it becomes clear that expressing them via $AT$ and elements of $ABC$ makes things cleaner and we shall do this here. We remark that $AM$ is the angle bisector of $\angle OAD$ (due to $AO = AD$), so $\angle OAM = \angle DAM = \frac{\beta - \gamma}{2}$, where as usual $\angle ABC = \beta$, $\angle ACB = \gamma$ and without loss of generality $AB < AC$. From the right triangle $TOM$ and $TO = AT$ we obtain $TM = AT\cos(\beta - \gamma)$. Note that $AO_cOO_b$ is a kite, so $\angle AO_cO_b = \angle O_bO_cO = 90^{\circ}-\angle AOO_c = 90^{\circ} - \gamma$; analogously $\angle AO_bO_c = 90^{\circ} - \beta$. Hence the Sine Law in triangles $ATO_c$ and $ATO_b$ yields $$ O_bT \cdot O_cT = \frac{AT \cdot \sin\left(\frac{3\beta-\gamma}{2}\right)}{\sin(90^{\circ} - \beta)} \cdot \frac{AT \cdot \sin\left(\frac{3\gamma-\beta}{2}\right)}{\sin(90^{\circ} - \gamma)} = \frac{AT^2 \cdot \sin\left(\frac{3\gamma-\beta}{2}\right)\sin\left(\frac{3\beta-\gamma}{2}\right)}{\cos\beta\cos\gamma} $$hence the desired equality is equivalent to $\cos(\beta-\gamma)\cos\beta\cos\gamma = \sin\left(\frac{3\gamma-\beta}{2}\right)\sin\left(\frac{3\beta-\gamma}{2}\right)$. On the other hand, the Sine Law in $ABC$ yields $AD = AB\sin\beta = 2AO\sin\beta\sin\gamma$, so $AD = AO$ is equivalent to $\sin\beta\sin\gamma = \frac{1}{2}$. We have $\sin\left(\frac{3\gamma-\beta}{2}\right)\sin\left(\frac{3\beta-\gamma}{2}\right) = \frac{1}{2}(\cos(2\beta-2\gamma) - \cos(\beta+\gamma)) = \frac{1}{2}(2\cos^2(\beta - \gamma) - 1 - \cos\beta\cos\gamma + \sin\beta\sin\gamma) = \frac{1}{2}(2\cos^2(\beta - \gamma) - 2\sin\beta\sin\gamma - \cos\beta\cos\gamma + \sin\beta\sin\gamma) = \cos^2(\beta - \gamma) - \frac{1}{2}\cos(\beta-\gamma)$. Hence by ignoring a factor of $\cos(\beta-\gamma)$, it now suffices to show $\cos\beta\cos\gamma = \cos(\beta - \gamma) - \frac{1}{2}$. However, the right-hand side is $\cos\beta\cos\gamma + \sin\beta\sin\gamma - \frac{1}{2} = \cos\beta\cos\gamma$ and we are done!
GrantStar
20.01.2024 08:15
The key step is to $\sqrt{bc}$ invert. Let $A'$ be the reflection of $A$ over $BC$ or the inverse of $O$ and $D'$ be the antipode of $A$. $(AOC)$ is sent to $A'C$ so $O_b$ is sent to the reflection of $A$ over $A'C$. Similarly, $O_C$ is sent to the reflection of $A$ over $A'B$ $M$ is sent to the antipode of $A$ on the circumcircle of isosceles triangle $AA'D'$. Thus taking a homothety at $A$ with scale factor $\frac 12$, it suffices to prove the foot from $A$ to $A'B$, the foot from $A$ to $A'C$, and the circumcenter of $AA'D'$ are collinear. Call these points $B',C', E$ respectively. Note that $E$ is where the bisector of $\angle BAC$ hits $BC$. Now, be Menelaus' in $\triangle A'BC$ it suffies to show \[1=\frac{A'B'}{B'B} \cdot \frac{BE}{EC} \cdot \frac{CC'}{C'A'}.\]But by angle bisector theorem $\frac{BE}{EC}=\frac{AB}{AC}$ and by power of a point at $A'$ using $BB'DA$ and $CC'DA$ cyclic we find \[A'B\cdot A'B' = A'D\cdot A'A=A'C'\cdot A'C \iff \frac{A'B'}{A'C'}=\frac{A'C}{A'B}=\frac{AC}{AB}\]so it suffices to show $CC'=BB'$. By power of a point again, it suffices to show $A'C'=A'B=AB$ since that would imply $A;C=A'B'$ and subtracting lengths finish. Tho show this, note $ADC'C$ cycle gives $\angle A'AC'=\angle DCA'=\angle BCA$ so \[A'C'=AA'\cdot \sin(\angle A'AC')=2AD\cdot \sin(\angle BCA) =2AO\cdot \sin(\angle BCA)=AB\]which finishes. Remark: This solution may seem complicated, but every step of it is not that difficult and pretty motivated. Nice problem!
bin_sherlo
31.03.2024 00:17
Lemma: In isosceles triangle $ABC$ with $BA=BC$, let $H$ be the orthocenter. $AH\cap BC=D$. Let $\omega$ be the circle centered at $C$ with radius $CM$ where $M$ is the midpoint of $AC$. The tangent at $D$ to $(M)$ intersects $\omega$ at $K,L$. The reflections of $C$ to $ML,MK$ are $X,Y$. So $X,H,Y$ are collinear. Proof: We have $MX=XL=LC=CK=KY=YM=MC\implies KCMY, MCLX, LKYX$ are parallelograms. $MYX\sim CKL$. We can carry $MXY$ onto $CKL$ by removing the vertices parallel to $AC$ with length $MC$. Let $HT=MC$ and $HT\parallel AC$. We get that $HMXY\sim TCKL$. $MHCE$ is cyclic since it's a rectangle. Also $\angle HDC=90=\angle CMH$ thus $TCHMD$ are cyclic which gives us that $\angle MDT=90$ so $TD$ is tangent to $(M)$. Hence $KDTL$ are collinear which gives that $X,Y,H$ are collinear. By inverting from $O$ with radius $OA$, we have $\angle AO_B^*O=\angle OAO_B^*=O_BOA\implies AO=AO_B^*$ Similarily $AO=AO_C^*$ Also we have $AO=AD$. Thus $(O_B^*O_C^*OD)$ on the circle centered at $A$ with radius $AO$. Circle with diameter $AO$ goes to the line tangent to $(O)$ at $A$ thus $M^*=DO\cap AA$. By inverting from $A$ with radius $AO$, we want to prove that $O_B^*,O_C^*,H$ are collinear where $H=AM^*\cap (ADO)$. $OA\cap (A)=O'$. This holds because of the lemma as desired.$\blacksquare$
In the inverted configuration,
$A \rightarrow M,O\rightarrow C,M^*\rightarrow B,H\rightarrow H,\rightarrow O_B^*\rightarrow Y, O_C^*\rightarrow X, D\rightarrow D, B\rightarrow K,C\rightarrow L$
tugra_ozbey_eratli
29.04.2024 23:09
I couldn't solve this question in the exam but it was easy with complex bash $Solution:$ Let $(ABC)$ be the unit circle and $BC$ parallel to imaginary axis, so that $$|a|=|b|=|c|=1,bc=1,a-d=1$$$$d=\frac{a+b+c-\frac{bc}{a}}{2}=\frac{a+b+c-\frac{1}{a}}{2}=a-1\implies a(b+c)=(a-1)^2$$$$o_b=\frac{ac}{a+c},o_c=\frac{ab}{a+b},m=\frac{d}{2}=\frac{a-1}{2}$$so we want to show that $$\frac{o_b-a}{o_c-a}\cdot\frac{o_c-m}{o_b-m}=\overline{\frac{o_b-a}{o_c-a}\cdot\frac{o_c-m}{o_b-m}}$$$$\iff$$$$\frac{\frac{-a^2}{a+c}}{\frac{-a^2}{a+b}}\cdot\frac{\frac{ab+a+b-a^2}{2(a+b)}}{\frac{ac+a+c-a^2}{2(a+c)}}=\overline{\frac{\frac{-a^2}{a+c}}{\frac{-a^2}{a+c}}\cdot\frac{\frac{ab+a+b-a^2}{2(a+b)}}{\frac{ac+a+c-a^2}{2(a+c)}}}$$$$\iff$$$$\frac{ab+a+b-a^2}{ac+a+c-a^2}=\overline{\frac{ab+a+b-a^2}{ac+a+c-a^2}}$$$$\iff$$$$\frac{ab+a+b-a^2}{ac+a+c-a^2}=\frac{ac+abc+a^2c-bc}{ab+abc+a^2b-bc}$$$$\iff$$$$(ab+a+b-a^2)(ab+abc+a^2b-bc)=(ac+a+c-a^2)(ac+abc+a^2c-bc)$$$$\iff$$$$[(a-a^2)+(ab+b)][(abc-bc)+(a^2b+ab)]=[(a-a^2)+(ac+c)][(abc-bc)+(a^2c+ac)]$$$$\iff$$$$(ba^2-b^2c)(a^2-1)-b^2a(a+1)^2=(ca^2-c^2b)(a^2-1)-c^2a(a+1)^2$$$$\iff$$$$(a^2-1)(ba^2-b^2c+c^2b-ca^2)+a(a+1)^2(c^2-b^2)=0$$$$\iff$$$$(c-b)[(a^2-1)(bc-a^2)+a(b+c)(a+1)^2]=0$$We know that $b\neq c$, $bc=1$ and $a(b+c)=(a-1)^2$ so, $$(c-b)[(a^2-1)(bc-a^2)+a(b+c)(a+1)^2]=0$$$$\iff$$$$[(a^2-1)(1-a^2)+(a-1)^2(a+1)^2]=(a^2-1)^2-(a^2-1)^2=0$$Which is true. We are done $\blacksquare$
KST2003
30.05.2024 10:08
Let $N$, $Q$, $P$ be the midpoints of $AB$, $AD$ and $AC$ respectively. Then $QM \parallel AO$, so \[ \angle PQM = 90^\circ - \angle OAD = 90^\circ - (\angle A - 2(90^\circ - \angle B)) = (90^\circ - \angle B) + \angle C = \angle PAO_B, \]and \[ \frac{AO_B}{QM} = \frac{AO_B}{AO/2} = \frac{1}{\cos \angle C} = \frac{AP}{AQ}, \]so it follows that $\triangle PAO_B \sim \triangle PQM$. By spiral similarity, we then have $\triangle PQA \sim \triangle PO_BM$, so $\angle PMO_B = 90^\circ$. Similarly, $\angle NMO_C = 90^\circ$. We also have $\angle AMO = 90^\circ$, so $M$ lies on $(ANOP)$. Therefore, \[ \measuredangle O_BMO_C = \measuredangle O_BMP + \measuredangle PMN + \measuredangle NMO_C = \measuredangle PMN = \measuredangle CAB, \]and \[ \measuredangle O_BAO_C = \measuredangle O_BAO + \measuredangle OAO_C = \measuredangle ACB + \measuredangle CBA = \measuredangle CAB, \]so we are done.
bin_sherlo
11.06.2024 23:16
Let $(AOB)\cap BC=K, (AOC)\cap BC=L,AM\cap BC=N$ $\textbf{Claim:} \ KO\perp AC, \ LO\perp AB$. \[\angle OKC=90-\angle C \ \text{and} \ \angle BLO=90-\angle B\]$\textbf{Claim:} \ A,O_B,O_C,K,L$ are cyclic. \[\angle O_BKO_C=\angle O_BKO=\angle A=\angle OLO_C=\angle O_BLO_C\implies O_B,O_C,K,L \ \text{are cyclic.}\]\[\angle O_CAO_B=\angle O_BOO_C=180-\angle A=180-\angle O_BKO_C\implies A\in (O_BO_CKL)\]$\textbf{Claim:} \ A,M,K,L$ are cyclic. \[\angle NLO=\angle KLO=90-\angle B=180-(90+\angle B)=180-\angle NOA-\angle AOO_C=\angle KON\]\[\implies NM.NA=NO^2=NK.NL\]These claims give that $A,O_B,O_C,K,L,M$ are cyclic as desired.$\blacksquare$
Attachments:
Orestis_Lignos
28.06.2024 09:50
This was G6 at last year's shortlist.
wizixez
22.12.2024 22:45
What an easy problem for p4
Professor33
05.01.2025 11:38
Assassino9931 wrote: This marvelous geometry problem was proposed by Marin Hristov (Marinchoo) and Bozhidar Dimitrov (Strudan_Borisov). With time consuming problems 1-3 many people unfortunately did not attempt this problem thorougly, but overall it was very appreciated! (Still, we did see five complete solutions and one very nice trig bash attempt, as well as a student with 5 points for which I do not know the details.) Here is a rephrased version of the authors' solution which has ended up as the official solution.
Note that $AB = AC$ cannot hold since $AO = AD$ would imply that $O$ is the midpoint of $BC$, which is not possible for an acute triangle. So we may assume without loss of generality that $AB < AC$. Let $M_b$ and $M_c$ be the midpoints of $AC$ and $AB$, respectively. Since $\angle AM_bO = \angle AM_cO = 90^{\circ} = \angle AMO$ (the latter since $AO = AD$), the pentagon $AM_bOMM_c$ is cyclic.
Next, notice that $AM$ is the perpendicular bisector of $OD$, $O_bO_c$ is the perpendicular bisector of $AO$ and $M_bM_c$ is the perpendicular bisector of $AD$. Hence these three lines are concurrent - denote their common point by $T$.
The quadrilateral $AO_bMO_C$ is cyclic if and only if $AT \cdot TM = O_bT \cdot O_cT$. From the cyclic $AM_bMM_c$ we have $AT \cdot TM = M_bT \cdot M_cT$. Hence it now suffices to argue $M_bT \cdot M_cT = O_bT \cdot O_cT$ -- or equivalently, that $M_b$, $M_c$, $O_b$ and $O_c$ are concyclic.
We assume that $\angle AOB < 90^{\circ}$ and $\angle AOC > 90^{\circ}$ so that $O_c$ is in the interior of triangle $AOB$ and $O_b$ in external to the triangles $AOC$ (the other cases are analogous and if $\angle AOB = 90^{\circ}$ or $\angle AOC = 90^{\circ}$, then $M_b\equiv O_b$ or $M_c \equiv O_c$ and we are automatically done). We have
\[ \angle M_cM_bO_b = 90^{\circ} + \angle AM_bM_c = 90^{\circ} + \angle ACB\]as well as (since $O_cO_b$ is a perpendicular bisector of $AO$ and hence bisects $\angle AO_CO$)
\[ \angle M_cO_cO_b = 180^{\circ} - \angle OO_cO_b = 90^{\circ} + \frac{\angle AO_cM_c}{2} = 90^{\circ} + \frac{\angle AO_cB}{4} = 90^{\circ} + \frac{\angle AOB}{2} = 90^{\circ} + \angle ACB \]and therefore $O_bM_bO_cM_c$ is cyclic, as desired.
Having in mind the difficulty of the problem, the coordinators and leaders decided to award 2 points for showing $AM_bOMM_c$ is cyclic and 4 points for showing $M_bO_cM_cO_b$ is cyclic, respectively 3 points for identifying $T$ as a concurrency point of the three lines and 1 point for the PoP finish. So one can actually obtain $6$ points by just angle chasing which is rewarding for the people who have attempted the problem with a good strategy! But as I said above, the main reason the scores on this problem are quite low is that problems 1-3 were quite time consuming, so this is a lesson to be learned for leaders when voting for the competition problems.
The optimal strategy is to consider $T = AM \cap O_bO_c$, having in mind it is the circumcenter of $AOD$, and show $AT \cdot TM = O_bT \cdot O_cT$. Have in mind that when $AB < AC$ we have $\angle OAM = \angle MAD = \frac{\angle ABC - \angle ACB}{2}$.
loved ur eulidean solution