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 ω the circumcircle of triangle AKL. Line AD intersects ω again at X. Prove that ω and the circumcircles of triangles ABC and DEX have a common point.
Problem
Source: JBMO 2021
Tags: Junior, Balkan, geometry, circumcircle, concurrence
01.07.2021 16:43
Let AO∩(ABC)=Y. Since ∠AKE=∠ACB, we have KBLC is cyclic. So by PoP, we have AE⋅EY=BE⋅EC=KE⋅EL, implying that ALYK is cyclic. Now note that ∠AXK=∠ALK=∠ABC, giving BDXK is cyclic. Similarly ∠AYK=∠ALK=∠ABC, giving BEYK is cyclic. Thus by PoP, we have AE⋅AY=AB⋅AK=AD⋅AX, implying that DEYX is cyclic, so the circumcircles of ABC,DEX and AKL have a common point, which is Y.
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01.07.2021 16:58
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 ω. Proof. Notice that ∠A′EL=∠LCA′=90∘ and thus A′ELC is cyclic. This gives that ∠LA′A=∠ACB. Additionally, we have ∠ALK=180∘−∠OAL−∠AEL=180∘−(90∘−∠ABC)−90∘=∠ABC. Thus ∠AKL=180∘−∠ALK−∠KAL=180∘−∠ABC−∠BAC=∠ACB. Therefore ∠AKL=∠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 ∠LA′X=180∘−∠XAL=180∘−(90∘−∠ACB)=90∘+∠ACB. But recall that ∠LA′A=∠ACB and therefore ∠EA′X=∠LA′X−∠LA′A=90∘=∠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.
01.07.2021 17:10
Let A′ be the antipode of A in (ABC), ALK∼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⟹A′ is on (ALK). Note that AD and AE are isogonal conjugates. Thus D⟶E, A′⟶X ⟹ADE∼AA′X⟹A′∈(DXE).
01.07.2021 17:52
Firstly, ∠BKL≡∠AKE=90∘−∠KAE≡90∘−∠BAO=90∘−(90∘−γ)=γ=∠BCA≡∠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∩KL=E, we get that E∈AZ, i.e. A,O,E,Z are collinear. Now, ∠ZXL=∠ZAL≡∠OAC=90∘−β=∠BAD≡∠KAX=∠KLX. Therefore, XZ∥KL, so XZ⊥AEZ. Since ∠XDE+∠XZE=180∘, we get that Z∈(DEX).
01.07.2021 18:06
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
01.07.2021 18:43
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 ω and (ABC)
01.07.2021 18:44
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.
01.07.2021 19:28
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
01.07.2021 19:35
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.
01.07.2021 20:24
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).
01.07.2021 21:02
@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.
01.07.2021 21:22
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.
01.07.2021 21:34
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).
01.07.2021 23:47
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.
02.07.2021 00:23
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.
02.07.2021 08:43
hakN wrote: Let AO∩(ABC)=Y. Since ∠AKE=∠ACB, we have KBLC is cyclic. So by PoP, we have AE⋅EY=BE⋅EC=KE⋅EL, implying that ALYK is cyclic. After this observe that D and E are isogonal wrt ∠A in △AKL. Since AE is perpendicularto KL ⟹ circumcenter of ω lies on AX ⟹ \angle AYX=90° also \angle EDX=90° \implies DEYX is cyclic.
02.07.2021 09:00
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.
02.07.2021 10:01
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
03.07.2021 19:19
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.
03.07.2021 19:58
03.07.2021 23:23
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.
04.07.2021 12:48
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 XDEWe ultimately get that A' is the desired concurrency point. \blacksquare
05.07.2021 07:37
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.
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05.07.2021 08:53
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.
06.07.2021 09:34
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.
09.07.2021 23:08
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
26.07.2021 18:27
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!
24.09.2021 18:05
Quote: Let ABC be an acute scalene triangle with circumcenter O. How this is possible ? acute scalene ?
24.09.2021 20:37
Scalene=Different side length
16.01.2022 17:42
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.
25.03.2023 17:46
10.05.2023 06:38
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.
01.06.2023 21:25
\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\}
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02.06.2023 08:08
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.