Let $AO \cap (ABC) = Y$. Since $\angle AKE = \angle ACB$, we have $KBLC$ is cyclic. So by PoP, we have $AE\cdot EY = BE \cdot EC = KE \cdot EL$, implying that $ALYK$ is cyclic. Now note that $\angle AXK = \angle ALK = \angle ABC$, giving $BDXK$ is cyclic. Similarly $\angle AYK = \angle ALK = \angle ABC$, giving $BEYK$ is cyclic. Thus by PoP, we have $AE\cdot AY = AB \cdot AK = AD \cdot AX$, implying that $DEYX$ is cyclic, so the circumcircles of $ABC , DEX$ and $AKL$ have a common point, which is $Y$.

Let $A'$ be the antipode of $A$ in $(ABC)$. The main claim is that all three circles pass through $A'$. Claim 1. $A'$ lies on $\omega$. Proof. Notice that $\angle A'EL=\angle LCA'=90^{\circ}$ and thus $A'ELC$ is cyclic. This gives that $\angle LA'A=\angle ACB$. Additionally, we have $\angle ALK=180^{\circ}-\angle OAL-\angle AEL=180^{\circ}-(90^{\circ}-\angle ABC)-90^{\circ}=\angle ABC$. Thus $$\angle AKL=180^{\circ}-\angle ALK-\angle KAL=180^{\circ}-\angle ABC-\angle BAC=\angle ACB.$$ Therefore $\angle AKL=\angle LA'A$ which gives that $AKA'L$ is cyclic, as needed. Claim 2. $A'$ lies on $(DEX)$. Proof. In cyclic quadrilateral $AXA'L$ we have $$\angle LA'X=180^{\circ}-\angle XAL=180^{\circ}-\left(90^{\circ}-\angle ACB\right)=90^{\circ}+\angle ACB.$$ But recall that $\angle LA'A=\angle ACB$ and therefore $\angle EA'X=\angle LA'X-\angle LA'A=90^{\circ}=\angle XDE$, which proves that $DEA'X$ is cyclic, as needed. As $A'$ lies on $(ABC)$ by definition, our proof is done. @below(s) As a Romanian TST taker (I didn't make the team), I am not aware of something similar to Problem 4 appearing in our tests from recent years.

Let $A'$ be the antipode of $A$ in $(ABC)$, $ALK \sim ABC$. Transformation: reflect everything wrt angle bisector of $A$ and perform a homothety centered at $A$ which takes the reflection of $ABC$ to $ALK$ . $BC$ is the simson line of $A'$ wrt $AKL \implies A'$ is on $(ALK)$. Note that $AD$ and $AE$ are isogonal conjugates. Thus $D \longrightarrow E$, $A' \longrightarrow X$ $\implies ADE \sim AA'X \implies A' \in (DXE)$.

Firstly, $\angle BKL\equiv \angle AKE=90^\circ-\angle KAE\equiv 90^\circ-\angle BAO=90^\circ-(90^\circ-\gamma)=\gamma=\angle BCA\equiv\angle BCL$, so $BKCL$ is cyclic. Now, let the second intersection of $(ABC)$ and $(AKL)$ be $Z$. Then, the pairwise radical axes of $(ABZC)$, $(AKXZL)$ and $(BKCL)$, which are the lines $AZ$, $BC$ and $KL$, are concurrent at their radical center. Since $BC\cap KL=E$, we get that $E\in AZ$, i.e. $A, O, E, Z$ are collinear. Now, $\angle ZXL=\angle ZAL\equiv\angle OAC=90^\circ-\beta=\angle BAD\equiv\angle KAX=\angle KLX$. Therefore, $XZ\parallel KL$, so $XZ\perp AEZ$. Since $\angle XDE+\angle XZE=180^\circ$, we get that $Z\in (DEX)$.

A bit easy for problem 3 perhaps? Standard angle chasing and power of a point give the solution quickly. Also the common point is not very hard to guess

steppewolf wrote: A bit easy for problem 3 perhaps? Standard angle chasing and power of a point give the solution quickly. Also the common point is not very hard to guess Yes, I did as you say in exam. With PoP it is easy to find that $AE$ is radical axis of $\omega$ and $(ABC)$

steppewolf wrote: A bit easy for problem 3 perhaps? Standard angle chasing and power of a point give the solution quickly. Also the common point is not very hard to guess This is my problem. When I've discovered this property I wasn't sure should I sent this to the JBMO because I've thought it's maybe too easy for P1. Few testsolvers convinced me that it is ok for somewhat easier P1. I have no idea how this finished as P3.

MathsLion wrote: steppewolf wrote: A bit easy for problem 3 perhaps? Standard angle chasing and power of a point give the solution quickly. Also the common point is not very hard to guess This is my problem. When I've discovered this property I wasn't sure should I sent this to the JBMO because I've thought it's maybe too easy got P1. Few testsolvers convinced me that it is ok for somewhat easier P1. I have no idea how this finished as P3. Congratulations! The problem is quite nice for P1 or even an easier P2, such problems usually look like this. I found the problem placement unusual, not the problem itself

Basically same soln. Let $X'$ the antipode of $A$ in $(ABC)$. We have that $<ABC=<KLA$, so $KBLC$ is cyclic. By PoP, $EA.EX'=EB.EC=EK.EL$, so $X'$ is on $AKL$. It is left only to show that $X'$ is on $(DEX)$, so we need only $<AX'X=90$, so $<AKX=90$ is sufficient. So we need $BDXK$ to be cyclic. But $<AXK=<ALK=<ABD$, so done. Comment: due to the similarity of $AKL$ and $ABC$, the center of $(AKL)$ is on $AD$.

MathsLion wrote: steppewolf wrote: A bit easy for problem 3 perhaps? Standard angle chasing and power of a point give the solution quickly. Also the common point is not very hard to guess This is my problem. When I've discovered this property I wasn't sure should I sent this to the JBMO because I've thought it's maybe too easy got P1. Few testsolvers convinced me that it is ok for somewhat easier P1. I have no idea how this finished as P3. I think this year number of problems should be different. For instance, $P1$ was harder than both $P2$ and $P3$ (I think).

@above In my opinion, problem 1 is deemed to be hard because - a lot of students taking the exam weren't familiar with the integer part and - part b statement looked a bit scary for some of them who didn't even know what to prove. Still, as problem 2 or 3 it would have been too easy. On the other hand it's hard for a 1, since 1's are aimed to be accessible by all students. Here, part a) is solvable by the majority of the students, in contrast with part b). Problem 2 was perfect for its placement, I think. The only drawback is that IMO 2011 #1 has a striking similarity with it, and as steppewolf points out, students who've seen that problem beforehand would have an advantage over others. Problem 3 wasn't that easy, in my opinion. It is tricky to locate the concurrence point, and involves a bit of work to prove that it actually is the desired point. Oh well, maybe it would have suited better as problem 2, although then some would argue that it is too hard for a 2. Overall, I think this years JBMO was pretty solid and ''unconventional''. The only issue seems to be the unoriginality of some problems. Apart from problem 2, I was told that problem 4 was similar/same with a Romania TST problem. I can't confirm whether this is true or not, though.

Orestis_Lignos wrote: @above In my opinion, problem 1 is deemed to be hard because - a lot of students taking the exam weren't familiar with the integer part and - part b statement looked a bit scary for some of them who didn't even know what to prove. Still, as problem 2 or 3 it would have been too easy. On the other hand it's hard for a 1, since 1's are aimed to be accessible by all students. Here, part a) is solvable by the majority of the students, in contrast with part b). Problem 2 was perfect for its placement, I think. The only drawback is that IMO 2011 #1 has a striking similarity with it, and as steppewolf points out, students who've seen that problem beforehand would have an advantage over others. Problem 3 wasn't that easy, in my opinion. It is tricky to locate the concurrence point, and involves a bit of work to prove that it actually is the desired point. Oh well, maybe it would have suited better as problem 2, although then some would argue that it is too hard for a 2. Overall, I think this years JBMO was pretty solid and ''unconventional''. The only issue seems to be the unoriginality of some problems. Apart from problem 2, I was told that problem 4 was similar/same with a Romania TST problem. I can't confirm whether this is true or not, though. It would be REALLY unfair towards all of us, the other competitors, if $P4$ has been given (or a similar problem has been given) to a Romania TST, even the possibility of that happening is just wrong.

Orestis_Lignos wrote: The only issue seems to be the unoriginality of some problems. Apart from problem 2, I was told that problem 4 was similar/same with a Romania TST problem. I can't confirm whether this is true or not, though. Really? Which year's TST? It is very unfair, if it's true that it is given in TST (or similar one).

As I said, I don't know! I was just said this is true, but I can't confirm it! Please, please don't take my word for granted by no means, unless it is indeed proven it is true.

I have a slightly different approach. I named F the concurrence point of the 2 big circles, and then proved that it lies on AE and on the DEX circle.

hakN wrote: Let $AO \cap (ABC) = Y$. Since $\angle AKE = \angle ACB$, we have $KBLC$ is cyclic. So by PoP, we have $AE\cdot EY = BE \cdot EC = KE \cdot EL$, implying that $ALYK$ is cyclic. After this observe that $D$ and $E$ are isogonal wrt $\angle A$ in $\triangle AKL$. Since $AE$ is perpendicularto $KL$ $\implies$ circumcenter of $\omega$ lies on $AX$ $\implies$ $\angle AYX=90°$ also $\angle EDX=90°$ $\implies$ $DEYX$ is cyclic.

celilcelil wrote: hakN wrote: Let $AO \cap (ABC) = Y$. Since $\angle AKE = \angle ACB$, we have $KBLC$ is cyclic. So by PoP, we have $AE\cdot EY = BE \cdot EC = KE \cdot EL$, implying that $ALYK$ is cyclic. After this observe that $D$ and $E$ are isogonal wrt $\angle A$ in $\triangle AKL$. Since $AE$ is perpendicularto $KL$ $\implies$ circumcenter of $\omega$ lies on $AX$ $\implies$ $\angle AYX=90°$ also $\angle EDX=90°$ $\implies$ $DEYX$ is cyclic. Wrote exactly the same. And this completes the proof.

Let $AO\cap \odot (ABC) = Y$. Claim. $AKYL$ is cyclic Proof. Since $AY$ is a diameter of $\odot (ABC)$ we have $\angle ACY = \angle ABY = \frac{\pi}{2}$. Note that $\angle KBY = \angle KEY = \angle YEL = \angle LCY = \frac{\pi}{2}$. Hence, $KBEY$ and $ECLY$ are cyclical. So, $\angle EYL = \angle ECL = \angle AYB = \angle AKE$. $\square$ It remains to show that $XDEY$ is cyclic. Now, notice that $ELCY$ is cyclic. Thus $\angle AXK = \angle ALK = \angle AYC = \angle ABC \Rightarrow BKXD$ $-$ cyclic. Finally, $\angle BKX = \pi - \angle BDK = \frac{\pi}{2} \Rightarrow \angle XYA = \pi - \angle AKX = \frac{\pi}{2} \Rightarrow \angle XDE + \angle XYE = \frac{\pi}{2} + \frac{\pi}{2} = \pi \Rightarrow XDEY$ is cyclic. $\blacksquare$

Perhaps surprisingly, if I am not mistaken, it turns out that one can give a very quick solution which does not show where exactly is the common point located (i.e. we do not rely on it being the antipode of $A$) and uses only angle chasing. Here is (shortened as much as possible version of) contestant BUL2's solution. Let $Y$ be the second common point of the circumcircles of $ABC$ and $AKXL$ - it suffices to show that $DEYX$ is cyclic. These two circumcircles give $\angle YKE = \angle YKL = \angle YAL = \angle YAC = \angle YBC = \angle YBE$ and so $KYEB$ is cyclic. Therefore $\angle DEY = \angle BEY = 180^{\circ} - \angle AKY = 180^{\circ} - \angle AXY = 180^{\circ} - \angle DXY$ and we are done.

Outline. Show that $BKLC$ is cyclic by trivial angle chase. Let $P=(ABC)\cap (AKL)$. By radical axis, $AE$ passes through $P$. Also, obviously the centre of $(AKL)$ lies on $AX$. Thus, $AX$ is the diameter of $(AKL)$. Now we have $\measuredangle EPX=\measuredangle APX=90^{\circ}=\measuredangle EDX$. Thus, $EDXP$ is cyclic.

Both L567 (with whom I solved the problem) and I found this much easier than P1. Let $A'$ be the $A$ antipode in $\triangle ABC$. Notice that $BKLC$ is cyclic and therefore $$EK \cdot EL = BE \cdot EC = AE \cdot EA'$$ meaning that $A'$ lies on the circumcircle of $\triangle AKL$. Moreover, $AD$ contains the center of the circumcircle of $\triangle ABC$ as $AD$ and $AE$ are isogonal in $\angle AKL$ meaning that $$\angle EA'X = \angle AA'X = 90^{\circ} = \angle XDE$$ We ultimately get that $A'$ is the desired concurrency point. $\blacksquare$

Let $AO \cap \odot(ABC)=F$. We claim that $AKFL$ is cyclic. To prove it, notice that $BE \cdot EC= AE \cdot EF$. However, $BE = \frac{AE \cos{C}}{\sin{B}}= AL\cos{C}$. Similarly, $EC= AK \cos{B}$. Therefore, $AL \cos{B} \cdot AK \cos{C}= EK \cdot EL = EA \cdot EF$. As a result, $AKFL$ Is cyclic $\blacksquare$. To finish this off, we prove that $DXFE$ is cyclic. However, we know that $\angle{FLA}=90-B = \angle{FKL} \implies \angle{KFE}=B$. We also know that $\angle{KXF}=\angle{KXA}=90-B \implies \angle{XFE}=90$, as desired $\blacksquare$. Remarks: We can avoid using trig, instead using similar triangles.

Let $F$ be the reflection of $A$ over $O$. We prove that $F$ is the desired point of concurrency. Claim: $AKLF$ is cyclic. Proof: Notice that $LEFC$ is cyclic because $\angle LEF = \angle LCF = \angle ACF = 90^\circ.$ $$\angle KLF = \angle ELF = \angle ECF = \angle BAF = \angle KAF.$$ Claim: $AX$ is the diameter of $(AKL)$. Proof: $\angle KAX = \angle DAB$ and $\angle AXK = \angle ALK = \angle ALE = \angle ABD \implies \angle AKX = \angle ADB = 90^\circ.$ Claim: $DEXF$ is cyclic. Proof: $$\angle EFX = \angle AFX = \angle AKX = \angle ADB = \angle EDX = 90^\circ.$$ Thus $(ABC) , (DEX ) , (AKL)$ intersect at a common point $F$.

Lukaluce wrote: Let $ABC$ be an acute scalene triangle with circumcenter $O$. Let $D$ be the foot of the altitude from $A$ to the side $BC$. The lines $BC$ and $AO$ intersect at $E$. Let $s$ be the line through $E$ perpendicular to $AO$. The line $s$ intersects $AB$ and $AC$ at $K$ and $L$, respectively. Denote by $\omega$ the circumcircle of triangle $AKL$. Line $AD$ intersects $\omega$ again at $X$. Prove that $\omega$ and the circumcircles of triangles $ABC$ and $DEX$ have a common point. Let $Y$ be the diametrically opposite point of $A$ with respect to the circumcircle of $\triangle ABC$. We claim that $Y$ is the desired concurrence point. Lemma 1: $Y$ lies on the circumcircle of $\triangle ALK$. Proof: We see that $\angle AXK = \angle ALK = 90^\circ - \angle EAL = 90^\circ - \angle KAW = \angle ABD$, implying that the points $B, D, X, K$ are concyclic. Therefore $\angle AKX = 90^\circ$. Similarly, we see that the points $C, E, L, Y$ are concyclic. Therefore, $\angle KLY = \angle ELY = \angle ECY = \angle BAY = \angle KAY$ which means that the points $A, K, L, Y$ and $X$ are concyclic. This also means that $\angle AYX = 180^\circ - \angle AKX = 90^\circ$. Therefore, $\angle XDE + \angle AYX = 180^\circ$ implies that the points $D, E, X, Y$ are concyclic, implying that the point $Y$ lies on the circumcircles of $\triangle ABC$, $\triangle DEX$ and $\triangle AKL$ as desired.

Here is a really fast solution I found on the car ride back from the beach: The key claim is Claim: $X$ is the $A$-antipode wrt $(AKL)$. Proof. Let $H$ be the orthocenter of $\triangle ABC$. It is well-known that $O$ and $H$ are isogonal conjugates wrt $\triangle ABC$, so $AD$ and $AE$ are isogonal lines. Moreover, since $K$ and $L$ lie on sides $AB$ and $AC$, respectively, lines $AD$ and $AE$ are still isogonal wrt $\triangle AKL$. But note that $AE$ is the $A$-altitude of $\triangle AKL$. Thus, the orthocenter of $\triangle AKL$ lies on $AE$, implying that the circumcenter of $\triangle AKL$ lies on $AD$, further implying that $AX$ is a diameter of $(AKL)$, implying the result. $\square$ Now, let $Y$ be the $A$-antipode wrt $(ABC)$. It suffices to show that quadrilaterals $LAXY$ and $XDEY$ are cyclic. For $LAXY$, just note that $\measuredangle XLA=\measuredangle XYA=90^\circ$ by Thales. For $XDEY$, just note that $\measuredangle XDE=90^\circ=\measuredangle XYA=\measuredangle XYE$ also by Thales. Done! $\blacksquare$

Too easy for JBMO P3: Let $AF$ be a diameter of $ABC$. Due to Spain MO 2016 P3, we could prove that $AKFL$ is cyclic and its circumference lies on line $AD$ or $A, X$ are diametrically opposite (exactly what that problem requested). Hence $\widehat{ABC}=90$ and $XDEF$ is also cyclic. Q.E.D!

Quote: Let $ABC$ be an acute scalene triangle with circumcenter $O$. How this is possible ? acute scalene ?

Scalene=Different side length

Let AE meet circumcircle of ABC at P. AP is diameter so ∠LCP = 90 and we have ∠LEP = 90 so LCPE is cyclic. ∠CPL = ∠CEL = ∠BEK and ∠ABC = ∠APC so ∠AKL = ∠APL so AKPL is cyclic. we will prove P is the common point. for that we will prove DEPX is cyclic. ∠PXD = ∠PKA = ∠PLC = ∠PEC so DEPX is cyclic. we're Done.

Claim: $BKCL$ is cyclic. Note that $\angle BKE=90-\angle KAE$. Similarly, $\angle LCE=90-\angle LAD$. Since $AD$ and $AE$ are isogonal, $\angle BKE=\angle LCE$, as desired. Thus, $LE\cdot KE=CE\cdot BE$. Hence, $E$ lies on the radical axis of $(AKL)$ and $(ABC)$, so the second intersection of $(AKL)$ and $(ABC)$, which we call $P$, lies on line $AE$. Thus, it suffices to show that $DEPX$ is cyclic, which is the same as $\angle XPE=90$, which is the same as showing that $AX$ is a diameter of $(AKL)$. However, from cyclic $ALXK$, $\angle AXK=\angle ALK$. However, $$\angle ALK=90-\angle LAE=90-\angle BAD,$$ so $\angle AXK=90-\angle BAD,$ and hence $\angle AKX=90$. Thus, $AX$ is a diameter of $(AKL)$, as desired.

$\angle A=\alpha$ , $\angle B$=$\beta$ , $\angle C$= $\gamma$ Let $AO \cap (ABC) = F$ this means that $F\in (ABC)$ and that $AF$ is the diameter of $(ABC)$ $=>$ $\angle ACF=\angle ABF=90$ $\angle ACF=\angle LCF=90$ , $\angle LEF=90$ $\angle LCF+\angle LEF= 90+90=180$ $<=>$ $LCEF-cyclic$ $\angle ACB=\angle ACE=\angle LCE=\angle LFE=\angle LFA=\angle AFL=\gamma$ $=>$ $\angle AFL=\gamma$ $...(1)$ By $ABCF-cyclic$ we get: $\angle ACB=\angle AFB=\gamma$ From the $\triangle ABF$ we have: $\angle ABF + \angle AFB + \angle BAF = 180$ 90+\gamma$+ \angle BAF = 180$ $\angle BAF=90-\gamma$ $\angle BAF=\angle KAF=\angle KAE=90-\gamma$ $=>$ $\angle KAE=90-\gamma$ From the $\triangle AKE$ $\angle AEK +\angle KAE +\angle AKE=180$ $90+90-\gamma+ \angle AKE=180$ $180-\gamma+ \angle AKE=180$ $\angle AKE=\gamma$ $\angle AKE=\angle AKL=\gamma$ $=>$ $\angle AKL=\gamma$ $...(2)$ Combining $(1)$ with $(2)$ we get: $\angle AFL=\gamma=\angle AKL$ $\angle AFL=\angle AKL$ $<=>$ From the $\triangle ACD$ $\angle ADC + \angle ACD + \angle CAD= 180$ $90+\angle ACB + \angle CAD= 180$ $90+ \gamma + \angle CAD= 180$ $\angle CAD=90-\gamma$ $\angle CAD=\angle CAX= \angle LAX=90-\gamma$ $=>$ $\angle LAX=90-\gamma$ $AKFL-cyclic$ $=>$ $F\in$ $w$ $ALFX-is$ $cyclic$ $\because$ $A,K,F,L\in$ $w$ By $ALFX$ we get: $\angle LAX + \angle LFX=180$ $90-\gamma+ \angle LFX=180$ $\angle LFX=\gamma+90$ $\angle LFX=\angle LFE + \angle EFX$ $\angle LFX=\angle LCE + \angle EFX$ $\gamma+90=\gamma + \angle EFX$ $\angle EFX=90$ $\angle EFX+\angle EDX=90+90=180$ $=>$ $\angle EFX+\angle EDX=180$ $<=>$ $DEFX-cyclic$ $=>$ $F\in$ $(DEX)$ Since $F\in (ABC)$ , $F\in$ $w$ and $F\in$ $(DEX)$ we get that $w$ $\cap$ $(ABC)$ $\cap$ $(DEX)$=$\{F\}$

I claim that this desired point is the antipode of $A$ on $(ABC)$. Denote this point as $A'$ Claim 1: I claim that $(AKL)$ passes through $A'$. Proof: Consider the circumcenter $O$ of $(AKL)$. I show that the perpendicular bisector of $AA'$ passes through $O$. We can do this by showing the perpendicular $OM$ to $AA'$ satisfies $AM = R$. Notice that the ratio of similarity between $ABC$, $AKL$ is $\frac{AE}{AD}$. Letting $O'$ be the center of $ABC$, the perpendicular from $O'M'$ to $AD$ gives $AM' = \frac{AM'}{AO'} AO' = \frac{AD}{AE} AO'$ since $AED, AO'M'$ are similar. Then $AM = AO' = R$, so we are done. In addition, notice $M = O'$, $OO'$ is the perpendicular bisector of $AA'$. To finish, notice $\frac{XA}{OA} = \frac{A'A}{O'A}$ gives $XA'$ parallel to $OO'$, both are perpendicular to $AA'$, so since quadrilateral $DEXA'$ has two opposing right angles, we are done.