In triangle ABC, the incenter is I and the circumcenter is O. Let AI intersects (ABC) second time at P . The line passes through I and perpendicular to AI intersects BC at X. The feet of the perpendicular from X to IO is Y. Prove that A,P,X,Y cyclic.
Problem
Source: 2024 Turkey TST P1
Tags: geometry, incenter, circumcircle
19.03.2024 02:51
XP,XA meets ⊙ABC again at E,D,respectively.Easy to know ∠ADI=∠PEI=π2. Then if we let K be a point on the ray IO such that IK=2IO,we will have ∠XYK=∠XAK=∠XPK=π2.◻
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19.03.2024 05:11
We proceed by complex bashing. Identify the incircle of △ABC with the unit circle, and let it touch sides ¯BC,¯CA,¯AB at D,E,F respectively. Rename I to J. Then |d|=|e|=|f|=1a=2efe+fb=2dfd+fc=2ded+ej=0o=2def(d+e+f)(d+e)(d+f)(e+f)p=2def(d+e)(d+f)Now the coordinate of X satisfies ¯x=−xef=2d−xd2⟹x=2defef−d2and then we find the coordinate of Y as y=12(x+o¯x¯o)=defde+df+efWe have the vectors a−p=2ef(d−e)(d−f)(d+e)(d+f)(e+f)y−x=def(d2+2de+2df+ef)(d2−ef)(de+df+ef)a−y=ef(de+df+2ef)(e+f)(de+df+ef)p−x=2d2ef(2d+e+f)(d+e)(d+f)(d2−ef)Then (a−p)(y−x)(a−y)(p−x)=(d−e)(d−f)(d2+2de+2df+ef)d(2d+e+f)(de+df+2ef)which is real. ◼
19.03.2024 10:23
Point Circles! Let Z be on AP s.t. ZI2=ZA⋅ZP, now clearly Z,X lie on the radical axis of of point circle I and (ABC) and therefore OI⊥XZ, so now let OI∩XZ=Y′ then we have Y′ on OI and also ∡XY′I=90∘ therefore Y≡Y′ and we’re done by power of point on Z
19.03.2024 10:33
egxa wrote: In triangle ABC, the incenter is I and the circumcenter is O. Let AI intersects (ABC) second time at P . The line passes through I and perpendicular to AI intersects BC at X. The feet of the perpendicular from X to IO is Y. Prove that A,P,X,Y cyclic.
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19.03.2024 13:45
Let K=I⊥AX and M=I⊥BX. So IMYXK is cyclic,now we can prove two things: 1.AKCB is cyclic. Proof:∠BIM=∠PIC⟹∠IBC=∠CIX so IX2=XC⋅XB,at the same time IX2=XK⋅XA⟹AKCB is cyclic. 2.P,M,K are collinear. Proof:It's well known that PA∩BC,A,M,K are concyclic(if P,M,K are collinear),rest is easy angle chasing. If we prove that ∠KYO=∠OYP we're done,but OP∥IM⟹OPYK cyclic,OP=OK.
19.03.2024 14:28
Any solution with isogonal lines ?
19.03.2024 16:36
math_comb01 wrote: Point Circles! Let Z be on AP s.t. ZI2=ZA⋅ZP, now clearly Z,X lie on the radical axis of of point circle I and (ABC) and therefore OI⊥XZ, so now let OI∩XZ=Y′ then we have Y′ on OI and also ∡XY′I=90∘ therefore Y≡Y′ and we’re done by power of point on Z Nice use of point circles... I couldn't prove that Z was on XY. Could you explain why X is on the radical axis of circle I and (ABC)?
19.03.2024 16:58
Lemma: In triangle ABC, D,E,F are altitudes from A,B,C respectively and H is the orthocenter. M is the midpoint of BC and O is the circumcenter of (ABC). OM intersects (DEFM) at M,S. If K is the perpendicular from O to AD, then SMKH is an isosceles trapezoid. Proof: Let N9 be the circumcenter of the nine point circle. N9 is the midpoint of OH. N9 lies on the perpendicular bisector of SM since S,M∈(N9). Also HK⊥KO so N9 is the circumcenter of (HKO). Hence N9 also lies on the perpendicular bisector of HK. And HK⊥BC⊥OM gives that HK∥SM so SMKH is an isosceles trapezoid. Let D,E,F be the perpendiculars from I to BC,CA,AB respectively. Let S be the Sharky-Devil Point where X,S,A are collinear. Let's invert from I with radius ID. It's well known that the perpendicular from D to EF, I and S are collinear. (DISYX)→DX∗Y∗S∗ which is the perpendicular from D to EF. (ABCP)→(A∗B∗C∗P∗) so P∗ lies on the nine point circle of DEF. IO is the euler line of DEF and Y∗ lies on the perpendicular from D to EF thus Y∗ is the orthocenter of DEF. We should prove that X∗Y∗A∗P∗ is cyclic. It's true from lemma.
26.03.2024 15:58
SBYT wrote: we will have ∠XYK=∠XAK=∠XPK=π2.◻ Can you please explain how?
26.03.2024 16:23
XI is a tangent line on ⊙BIC.XI2=XB⋅XC=XD⋅XA=XE⋅XP.So ∠XEI=∠XDI=π2. Then by ADIK and IEPK are both right angled trapezoid.So ∠XAK=∠XPK=π2.
26.03.2024 16:56
Thanks a lot
26.03.2024 17:40
Solved with GrantStar: Let R be the second intersection of ¯XA with (ABC). By radical axis on (ARI), (BIC) and (ABC) we have that ¯XI is the radical axis of (ARI) and (BIC), hence ∠IRA=90∘. Similarly, let T be the second intersection of ¯XP with (ABC). By radical axis on (ITP), (IBC) and (ABC) it follows that IX is the radical axis of(IBC) and (ITP), hence ∠ITP=90∘. Let I′ be the reflection of I over O. Since the perpendicular bisector of ¯AR passes through O, the perpendicular to ¯AR through A passes through I′. Similarly, the perpendicular to ¯TP passes through I′. So, APXY is cyclic with diameter XI′.
02.04.2024 18:45
Let S and D be the foot of the perpendiculars from I to lines AX and BC resperctively. Claim1: S is the A−SharkyDevil of triangle ABC. Since AX is perpendicular to AP and P is the center of circle (BIC) XI2=XB.XC Similarly XI is tangent to the circle with diameter AI XI2=XS.XA Thus XB.XC=XS.XA so S lies on (ABC) hence S is the A−SharkyDevil of triangle ABC. Claim2: It is known that points S,D,P are collinear(properties of the SharkyDevil point ) Claim3: S,I,D,Y,X are concyclic ∠ISX + ∠IDX=90+90=180 thus S,I,D,X are concyclic ∠XDI=90=∠XYI thus X,Y,D,I are concyclic Claim4: S,O,P,Y are concyclic Since ID is parallel to OP ∠YOP=∠YID=∠YSD=∠YSP Claim5: A,X,Y,P are concyclic Since S,O,P,Y are concyclic and OS=OP ∠SXI=∠SYI=∠SYO=∠OYP=∠IYP and ∠SXI=90−∠XAP and ∠XYP=∠IYP+90 hence ∠XYP=180−∠XAP Q.E.D
27.04.2024 17:06
Nice geo : Let AI and XY meet at Z and let w be the circumcircle of ABC. Claim: XI2=pow(X,w) Consider the circle centered at P passing through B, I, and C. Notice XI⊥AP so XI is tangent to this circle. Then XI2=XB⋅XC=pow(X,w). Claim: Line XY is the radical axes of I and w Since pow(X,I)=XI2=pow(X,w) and XY⊥IO the result follows. Claim: The points A, P, X, and Y lie on a circle It is sufficient to show that ZP⋅ZA=ZX⋅ZY. Since Z lies on the radical axes of I and w we get that ZI2=ZP⋅ZA. But triangle ZIY and ZXI are similar so ZI2=ZX⋅ZY. The result follows.
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28.04.2024 00:06
utterly stupid solution: Let XY,AP concur at Z. Then △XZI∼△IZY, giving ZX⋅ZY=ZI2, so we want to prove ZI2=ZA⋅ZP, equivalent to showing pow(Z,(O))−pow(Z,(I))=r2−R2. By linearity of power of a point, and noting that XY is parallel to the radical axis of the incenter and circumcenter, it suffices to show that X also satisfies this power relation. However, it is trivial to note that XI2=XB⋅XC since XI is tangent to (BIC) by angle chasing, this finishes.
05.07.2024 20:58
Let Q be the intersection of lines XY and AP. Since ∠XYI=∠XIA=90∘, it follows that QY⋅QX=QI2. Since line XY is the radical axis of (ABC) and I, it then follows that QX⋅QY=QA⋅QP, implying APXY cyclic.
13.08.2024 15:35
I will use #2's diagram. Let midpoint of major arc BAC is Na. Radical axis on (DYXE),(ABC),(APX) we should show that XY,DE,AP concurrent. Miquel of complate quadrilateral ADEP. We should show that O,I,AE∩DP collinear and this is Pascal on PNaEAA′D where A′ is a antipode.
29.08.2024 15:49
I will provide two different solutions from similarity and linearity of PoP + Coaxality Lemma. Let DEF be the orthic triangle of ABC. Let A′ be the reflection of A across IO and I′ be the reflection of I across BC. It is known that P,A′,I′ are collinear and AIO∼AI′P. Q is the antipode of P in (ABC). G is the midpoint of BC, S=QI∩(ABC). A′1 is the reflection of A across QP. M is the midpoint of OX. Solution 1 Claim 1: IYXD∼QA′PA′1 Proof. ∠YIX=∠A′AP=∠A′QP⟹PA′Q∼XYI, ∠QPA′1=∠QPA=∠DIP=∠IXD⟹QA′1P∼IDX. Combining gives the desired result.◻ Claim 2: AA′Q∼IA′O Proof. ∠IOA′=∠APA′=∠AQA′ and ∠AA′Q=∠IAO=∠IA′O.◻ Claim 3: S∈PX Proof. There are various ways to do this. SBI∼SIC⟹SBSC=IB2IC2=XBXC which gives the desired result due to ex-angle bisector theorem.◻ Main Claim: AIY∼I′PG Proof. ∠AIY=180−∠AIO=180−∠II′P=∠GPI′. Note that IDIY1=QA′1QA′=QAQA′2=OIR PI′PG=AIIY⟺PI′IO⋅IOPG=AI⋅OIID⋅R⟺PIAO⋅IOPG=AI⋅OIID⋅R⟺AIIF=PCPGwhich is true because AIF∼CFG.◻ Now, ∠AYX=90+∠AYI=90+∠I′GP=90+∠DIG=∠AD′C=180−∠APS as desired. Solution 2 Because IYXD and OYXG are cyclic, it is sufficient to prove that AI2PI2=AM2−MO2PG⋅PO. AM2−MO2=12(AO2+AX2−XO2)=12(AO2+AI2+IX2−XO2)=12(AO2+AI2+IX2−GO2−GX2)= 12(AO2+AI2+PG2−GO2−PG⋅PQ)=12(AO2+AI2−GO2−GB2)=12AI2. On the other hand, AI2PI2⋅PG⋅PO=AI2PI2⋅12PI2=12AI2 as desired.
05.09.2024 21:31
Let XY meet AP at K. We are going to show KP.KA=KY.KX to finish the problem. Let R be the radius of (ABC). By power of a point: KP.KA=KO2−R2. Also we know that XI is tangent to (BIC) thus: XI2=XB.XC. Also XB.XC=XO2−R2. Substract these two equations: KP.KA−XI2=KO2−XO2. Now, we know that OI⊥XK. Hence: KO2−XO2=KI2−XI2. So we conclude: KP.KA−XI2=KI2−XI2⟹KP.KA=KI2. By euclid, KI2=KY.KX hence: KP.KA=KY.KX and we are done. ◼
05.11.2024 20:09
Similar to some solutions... From angle chasing we got XI2=XB⋅XC. That means X lies on radical axis of circles (I) and (ABC). OI perpendicular to XY means XY is radical axis line of circles (I) and (ABC). Let XY∩AP at K. It is obvious that KA⋅KP=KI2.From euclid theorem for right angles KI2=KY⋅KX.Which means APXY is cyclic.