【국제수학올림피아드】 IMO 기하 문제 풀이 (2010년 ~ 2014년)

IMO 기하 문제 풀이 (2010년 ~ 2014년)

 

추천글 : 【기하학】 IMO 기하 문제 풀이 종합 

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IMO 2010, Problem 2. Let I be the incenter of a triangle ABC and let Γ be its circumcircle. Let line AI intersect Γ again at D. Let E be a point on arc BDC and F a point on side BC such that

 

∠BAF = ∠CAE < 1/2 ∠BAC.

 

Finally, let G be the midpoint of IF. Prove that DG and EI intersect on Γ.

 

○ 풀이. Evan Chen의 풀이 

○ 풀이. AlphaGeometry를 사용할 때, 새로운 점을 추가하는 AlphaGeometry 모드로 풀립니다. (∵ 더 어려운 문제) (ref) 

 

==========================
 * From theorem premises:
A B C D E F G H I J : Points
DA = DB [00]
DB = DC [01]
∠ACE = ∠ECB [02]
∠BAE = ∠EAC [03]
DF = DA [04]
F,E,A are collinear [05]
C,B,G are collinear [06]
DH = DA [07]
∠CAH = ∠GAB [08]
I,G,E are collinear [09]
IE = IG [10]
I,F,J are collinear [11]
J,H,E are collinear [12]

 * Auxiliary Constructions:
K L M : Points
B,K,E are collinear [13]
KE = KB [14]
DL = DF [15]
F,L,D are collinear [16]
M,E,A are collinear [17]
ME = MA [18]

 * Proof steps:
001. DA = DB [00] & DF = DA [04] & DB = DC [01] ⇒  C,B,F,A are concyclic [19]
002. DA = DB [00] & DH = DA [07] & DF = DA [04] & C,B,F,A are concyclic [19] ⇒  C,B,H,A are concyclic [20]
003. M,E,A are collinear [17] & ME = MA [18] ⇒  M is midpoint of EA [21]
004. I,G,E are collinear [09] & IE = IG [10] ⇒  I is midpoint of EG [22]
005. M is midpoint of EA [21] & I is midpoint of EG [22] ⇒  MI ∥ AG [23]
006. B,K,E are collinear [13] & KE = KB [14] ⇒  K is midpoint of EB [24]
007. M is midpoint of EA [21] & K is midpoint of EB [24] ⇒  MK ∥ AB [25]
008. ∠CAH = ∠GAB [08] & AG ∥ IM [23] & AB ∥ KM [25] ⇒  ∠CAH = ∠IMK [26]
009. C,B,H,A are concyclic [20] ⇒  ∠CBA = ∠CHA [27]
010. I is midpoint of EG [22] & K is midpoint of EB [24] ⇒  IK ∥ GB [28]
011. ∠CBA = ∠CHA [27] & IK ∥ GB [28] & C,B,G are collinear [06] & AB ∥ KM [25] ⇒  ∠IKM = ∠CHA [29]
012. ∠CAH = ∠IMK [26] & ∠IKM = ∠CHA [29] (Similar Triangles)⇒  IM:KM = CA:HA [30]
013. F,E,A are collinear [05] & ∠BAE = ∠EAC [03] ⇒  ∠BAF = ∠FAC [31]
014. C,B,F,A are concyclic [19] & ∠BAF = ∠FAC [31] ⇒  BF = FC [32]
015. C,B,F,A are concyclic [19] ⇒  ∠BCF = ∠BAF [33]
016. ∠BCF = ∠BAF [33] & F,E,A are collinear [05] ⇒  ∠BCF = ∠BAE [34]
017. ∠BAE = ∠EAC [03] & ∠ACE = ∠ECB [02] & ∠BCF = ∠BAE [34] (Angle chase)⇒  ∠CEA = ∠FCE [35]
018. F,E,A are collinear [05] & ∠FCE = ∠CEA [35] ⇒  ∠FCE = ∠CEF [36]
019. ∠FCE = ∠CEF [36] ⇒  FC = FE [37]
020. BF = FC [32] & FC = FE [37] ⇒  F is the circumcenter of \Delta CBE [38]
021. F is the circumcenter of \Delta CBE [38] & K is midpoint of EB [24] ⇒  ∠ECB = ∠EFK [39]
022. M,E,A are collinear [17] & F,E,A are collinear [05] & ∠ACE = ∠ECB [02] & ∠ECB = ∠EFK [39] ⇒  ∠MFK = ∠ACE [40]
023. M,E,A are collinear [17] & F,E,A are collinear [05] & ∠BAE = ∠EAC [03] & AB ∥ KM [25] ⇒  ∠KMF = ∠EAC [41]
024. ∠MFK = ∠ACE [40] & ∠KMF = ∠EAC [41] (Similar Triangles)⇒  KF:KM = CE:EA [42]
025. ∠MFK = ∠ACE [40] & ∠KMF = ∠EAC [41] (Similar Triangles)⇒  KF:FM = CE:CA [43]
026. IM:KM = CA:HA [30] & KF:KM = CE:EA [42] & KF:FM = CE:CA [43] (Ratio chase)⇒  EA:HA = IM:FM [44]
027. ∠BAE = ∠EAC [03] & ∠HAC = ∠BAG [08] (Angle chase)⇒  ∠GAE = ∠EAH [45]
028. M,E,A are collinear [17] & F,E,A are collinear [05] & ∠EAH = ∠GAE [45] & AG ∥ IM [23] ⇒  ∠EAH = ∠IMF [46]
029. EA:HA = IM:FM [44] & ∠EAH = ∠IMF [46] (Similar Triangles)⇒  ∠EHA = ∠IFM [47]
030. DA = DB [00] & DH = DA [07] & DF = DA [04] & C,B,H,A are concyclic [20] & C,B,F,A are concyclic [19] ⇒  C,H,F,A are concyclic [48]
031. C,H,F,A are concyclic [48] ⇒  ∠CHA = ∠CFA [49]
032. J,H,E are collinear [12] & J,F,I are collinear [11] & ∠EHA = ∠IFM [47] & M,E,A are collinear [17] & F,E,A are collinear [05] & ∠CHA = ∠CFA [49] ⇒  ∠HCF = ∠HJF [50]
033. ∠HCF = ∠HJF [50] ⇒  C,F,H,J are concyclic [51]
034. DA = DB [00] & DF = DA [04] & DL = DF [15] & DH = DA [07] & C,B,H,A are concyclic [20] & C,B,F,A are concyclic [19] & C,F,H,J are concyclic [51] & C,H,F,A are concyclic [48] ⇒  J,H,F,L are concyclic [52]
035. J,H,F,L are concyclic [52] ⇒  ∠JFH = ∠JLH [53]
036. DA = DB [00] & DF = DA [04] & DL = DF [15] ⇒  D is the circumcenter of \Delta FBL [54]
037. D is the circumcenter of \Delta FBL [54] & F,L,D are collinear [16] ⇒  BF ⟂ BL [55]
038. DH = DA [07] & DF = DA [04] & DL = DF [15] ⇒  D is the circumcenter of \Delta FHL [56]
039. D is the circumcenter of \Delta FHL [56] & F,L,D are collinear [16] ⇒  FH ⟂ HL [57]
040. BF ⟂ BL [55] & FH ⟂ HL [57] ⇒  ∠FBL = ∠LHF [58]
041. C,F,H,J are concyclic [51] & C,H,F,A are concyclic [48] & C,B,H,A are concyclic [20] & C,B,F,A are concyclic [19] & DA = DB [00] & DF = DA [04] & DL = DF [15] & DH = DA [07] ⇒  B,J,F,L are concyclic [59]
042. B,J,F,L are concyclic [59] ⇒  ∠BFJ = ∠BLJ [60]
043. J,F,I are collinear [11] & ∠JFH = ∠JLH [53] & ∠FBL = ∠LHF [58] & ∠BFJ = ∠BLJ [60] ⇒  FJ ⟂ JL [61]
044. F,L,D are collinear [16] & DL = DF [15] ⇒  D is midpoint of FL [62]
045. FJ ⟂ JL [61] & D is midpoint of FL [62] ⇒  FD = JD [63]
046. FD = JD [63] & DF = DA [04] ⇒  DA = DJ
==========================

 

 

IMO 2010, Problem 4. Let P be a point interior to triangle ABC (with CA ≠ CB). The lines AP, BP and CP meet again its circumcircle Γ at K, L, M, respectively. The tangent line at C to Γ meets the line AB at S. Show that from SC = SP follows MK = ML.

 

○ 풀이. Evan Chen의 풀이 

○ 풀이. AlphaGeometry를 사용할 때, 새로운 점을 추가하지 않는 DD+AR 모드로 풀립니다. (ref)

 

==========================
 * From theorem premises:
S C P O A B M L K : Points
SC = SP [00]
∠PCS = ∠SPC [01]
CO ⟂ CS [02]
OA = OC [03]
OB = OC [04]
S,B,A are collinear [05]
OM = OC [06]
P,M,C are collinear [07]
OL = OC [08]
P,L,B are collinear [09]
OK = OC [10]
P,K,A are collinear [11]

 * Auxiliary Constructions:
: Points


 * Proof steps:
001. OA = OC [03] & OM = OC [06] & OB = OC [04] & OL = OC [08] ⇒  M,L,B,A are concyclic [12]
002. OB = OC [04] & OK = OC [10] & OM = OC [06] & OA = OC [03] & OL = OC [08] & M,L,B,A are concyclic [12] ⇒  L,K,A,M are concyclic [13]
003. OB = OC [04] & OK = OC [10] & OM = OC [06] & OA = OC [03] & OL = OC [08] & M,L,B,A are concyclic [12] ⇒  L,K,C,M are concyclic [14]
004. OB = OC [04] & OA = OC [03] ⇒  O is the circumcenter of \Delta CBA [15]
005. O is the circumcenter of \Delta CBA [15] & CO ⟂ CS [02] ⇒  ∠SCA = ∠CBA [16]
006. A,S,B are collinear [05] & ∠CBA = ∠SCA [16] ⇒  ∠CBS = ∠SCA [17]
007. S,A,B are collinear [05] ⇒  ∠ASC = ∠BSC [18]
008. S,A,B are collinear [05] ⇒  ∠PSB = ∠PSA [19]
009. ∠CBS = ∠SCA [17] & ∠ASC = ∠BSC [18] (Similar Triangles)⇒  SA:SC = SC:SB [20]
010. SA:SC = SC:SB [20] & SC = SP [00] ⇒  SA:PS = PS:SB [21]
011. SA:PS = PS:SB [21] & ∠PSB = ∠PSA [19] (Similar Triangles)⇒  ∠SPB = ∠PAS [22]
012. L,A,M,B are concyclic [12] ⇒  ∠AML = ∠ABL [23]
013. P,K,A are collinear [11] & ∠SPB = ∠PAS [22] & P,L,B are collinear [09] & S,B,A are collinear [05] & ∠AML = ∠ABL [23] ⇒  ∠AML = ∠KPS [24]
014. OM = OC [06] & OL = OC [08] ⇒  O is the circumcenter of \Delta CML [25]
015. O is the circumcenter of \Delta CML [25] & CO ⟂ CS [02] ⇒  ∠SCM = ∠CLM [26]
016. P,M,C are collinear [07] & ∠PCS = ∠SPC [01] & ∠SCM = ∠CLM [26] ⇒  ∠MLC = ∠SPM [27]
017. L,K,C,M are concyclic [14] ⇒  ∠CLK = ∠CMK [28]
018. P,M,C are collinear [07] & ∠CLK = ∠CMK [28] ⇒  ∠CLK = ∠PMK [29]
019. ∠MLC = ∠SPM [27] & ∠CLK = ∠PMK [29] ⇒  ∠MLK = ∠(PS-KM) [30]
020. ∠AML = ∠KPS [24] & ∠MLK = ∠(PS-KM) [30] ⇒  ∠(MA-LK) = ∠PKM [31]
021. ∠(MA-LK) = ∠PKM [31] & P,K,A are collinear [11] ⇒  ∠MAK = ∠LKM [32]
022. L,K,A,M are concyclic [13] & ∠MAK = ∠LKM [32] ⇒  LM = MK
==========================

 

 

IMO 2011, Problem 6. Let ABC be an acute triangle with circumcircle Γ. Let ℓ be a tangent line to Γ, and let ℓ_a, ℓ_b, ℓ_c be the lines obtained by reflecting ℓ in the lines BC, CA, and AB, respectively. Show that the circumcircle of the triangle determined by the lines ℓ_a, ℓ_b, and ℓ_c is tangent to the circle Γ.

 

○ 풀이. Evan Chen의 풀이 

○ 풀이. AlphaGeometry로 풀리지 않았습니다. (ref)

 

 

IMO 2012, Problem 1. Given triangle ABC the point J is the centre of the excircle opposite the vertex A. This excircle is tangent to the side BC at M, and to the lines AB and AC at K and L, respectively. The lines LM and BJ meet at F, and the lines KM and CJ meet at G. Let S be the point of intersection of the lines AF and BC, and let T be the point of intersection of the lines AG and BC. Prove that M is the midpoint of ST.

 

○ 풀이. Evan Chen의 풀이 

○ 풀이. AlphaGeometry를 사용할 때, 새로운 점을 추가하지 않는 DD+AR 모드로 풀립니다. (ref)

 

==========================
 * From theorem premises:
A B C J M L K F G S T : Points
∠ACJ = ∠JCB [00]
∠JAB = ∠CAJ [01]
B,C,M are collinear [02]
JM ⟂ BC [03]
L,C,A are collinear [04]
JL ⟂ AC [05]
B,K,A are collinear [06]
JK ⟂ AB [07]
J,B,F are collinear [08]
L,F,M are collinear [09]
G,K,M are collinear [10]
J,G,C are collinear [11]
F,S,A are collinear [12]
B,C,S are collinear [13]
T,G,A are collinear [14]
T,B,C are collinear [15]

 * Auxiliary Constructions:
: Points


 * Proof steps:
001. B,K,A are collinear [06] & B,C,M are collinear [02] & JM ⟂ BC [03] & JK ⟂ AB [07] ⇒  ∠BKJ = ∠BMJ [16]
002. ∠BKJ = ∠BMJ [16] ⇒  J,B,K,M are concyclic [17]
003. J,B,K,M are concyclic [17] ⇒  ∠JBM = ∠JKM [18]
004. J,B,K,M are concyclic [17] ⇒  ∠JBK = ∠JMK [19]
005. B,C,M are collinear [02] & L,C,A are collinear [04] & JL ⟂ AC [05] & JM ⟂ BC [03] ⇒  ∠JMC = ∠CLJ [20]
006. J,G,C are collinear [11] & B,C,M are collinear [02] & L,C,A are collinear [04] & ∠JCB = ∠ACJ [00] ⇒  ∠JCM = ∠LCJ [21]
007. ∠JMC = ∠CLJ [20] & ∠JCM = ∠LCJ [21] (Similar Triangles)⇒  JM = JL [22]
008. ∠JMC = ∠CLJ [20] & ∠JCM = ∠LCJ [21] (Similar Triangles)⇒  CM = CL [23]
009. ∠JMC = ∠CLJ [20] & ∠JCM = ∠LCJ [21] (Similar Triangles)⇒  ∠MJC = ∠CJL [24]
010. B,K,A are collinear [06] & L,C,A are collinear [04] & JL ⟂ AC [05] & JK ⟂ AB [07] ⇒  ∠JKA = ∠ALJ [25]
011. B,K,A are collinear [06] & L,C,A are collinear [04] & ∠JAB = ∠CAJ [01] ⇒  ∠JAK = ∠LAJ [26]
012. ∠JKA = ∠ALJ [25] & ∠JAK = ∠LAJ [26] (Similar Triangles)⇒  JK = JL [27]
013. JM = JL [22] & JK = JL [27] ⇒  J is the circumcenter of \Delta LMK [28]
014. JM = JL [22] & JK = JL [27] ⇒  JM = JK [29]
015. B,C,M are collinear [02] & JM ⟂ BC [03] ⇒  JM ⟂ MB [30]
016. J is the circumcenter of \Delta LMK [28] & JM ⟂ MB [30] ⇒  ∠BML = ∠MKL [31]
017. J,B,F are collinear [08] & ∠JBM = ∠JKM [18] & B,C,M are collinear [02] & G,K,M are collinear [10] & ∠BML = ∠MKL [31] & L,F,M are collinear [09] ⇒  ∠KLF = ∠KJF [32]
018. ∠KLF = ∠KJF [32] ⇒  J,L,K,F are concyclic [33]
019. L,C,A are collinear [04] & B,K,A are collinear [06] & JL ⟂ AC [05] & JK ⟂ AB [07] ⇒  ∠ALJ = ∠AKJ [34]
020. ∠ALJ = ∠AKJ [34] ⇒  J,L,K,A are concyclic [35]
021. L,C,A are collinear [04] & B,C,M are collinear [02] & JL ⟂ AC [05] & JM ⟂ BC [03] ⇒  ∠CLJ = ∠CMJ [36]
022. ∠CLJ = ∠CMJ [36] ⇒  J,L,C,M are concyclic [37]
023. J,L,C,M are concyclic [37] ⇒  ∠JLM = ∠JCM [38]
024. J,L,C,M are concyclic [37] ⇒  ∠JCL = ∠JML [39]
025. J,G,C are collinear [11] & ∠JLM = ∠JCM [38] & L,F,M are collinear [09] & B,C,M are collinear [02] & ∠BML = ∠MKL [31] & G,K,M are collinear [10] ⇒  ∠GKL = ∠GJL [40]
026. ∠GKL = ∠GJL [40] ⇒  J,G,L,K are concyclic [41]
027. J,L,K,F are concyclic [33] & J,L,K,A are concyclic [35] & J,L,K,G are concyclic [41] ⇒  K,A,J,L,G,F are concyclic [42]
028. J,L,K,F are concyclic [33] & J,L,K,A are concyclic [35] & J,L,K,G are concyclic [41] ⇒  K,A,G,F are concyclic [43]
029. K,A,G,F are concyclic [43] ⇒  ∠KGA = ∠KFA [44]
030. G,K,M are collinear [10] & T,G,A are collinear [14] & ∠KGA = ∠KFA [44] ⇒  ∠MGT = ∠KFA [45]
031. JM = JL [22] & CM = CL [23] ⇒  LM ⟂ JC [46]
032. J,G,C are collinear [11] & LM ⟂ JC [46] & L,F,M are collinear [09] ⇒  JG ⟂ LF [47]
033. B,K,A are collinear [06] & JK ⟂ AB [07] ⇒  JK ⟂ BK [48]
034. JG ⟂ LF [47] & JK ⟂ BK [48] ⇒  ∠GJK = ∠(LF-BK) [49]
035. J,G,C are collinear [11] & B,K,A are collinear [06] & ∠GJK = ∠(LF-BK) [49] ⇒  ∠(LF-JG) = ∠BKJ [50]
036. J is the circumcenter of \Delta LMK [28] & JK ⟂ BK [48] ⇒  ∠BKL = ∠KML [51]
037. J,B,F are collinear [08] & ∠JBM = ∠JKM [18] & B,C,M are collinear [02] & G,K,M are collinear [10] & ∠BML = ∠MKL [31] & L,F,M are collinear [09] & ∠BKL = ∠KML [51] & B,K,A are collinear [06] & ∠JBK = ∠JMK [19] ⇒  ∠MJF = ∠FJK [52]
038. JM = JK [29] & ∠MJF = ∠FJK [52] (SAS)⇒  ∠JMF = ∠FKJ [53]
039. J,G,C are collinear [11] & ∠JMF = ∠FKJ [53] & L,F,M are collinear [09] & ∠JCL = ∠JML [39] & L,C,A are collinear [04] & ∠ACJ = ∠JCB [00] ⇒  ∠(JG-BC) = ∠JKF [54]
040. ∠(LF-JG) = ∠BKJ [50] & ∠(JG-BC) = ∠JKF [54] ⇒  ∠(LF-BC) = ∠BKF [55]
041. J,G,L,K are concyclic [41] & J,L,K,A are concyclic [35] ⇒  J,G,K,A are concyclic [56]
042. J,G,K,A are concyclic [56] & J,L,K,A are concyclic [35] ⇒  J,G,L,A are concyclic [57]
043. A,J,L,G are concyclic [57] ⇒  ∠ALJ = ∠AGJ [58]
044. J,G,K,A are concyclic [56] ⇒  ∠KAG = ∠KJG [59]
045. T,B,C are collinear [15] & B,C,M are collinear [02] & T,G,A are collinear [14] & B,K,A are collinear [06] & ∠(LF-BC) = ∠BKF [55] & JC ⟂ LM [46] & ∠ALJ = ∠AGJ [58] & L,C,A are collinear [04] & J,G,C are collinear [11] & JL ⟂ AC [05] & JK ⟂ AB [07] & ∠KAG = ∠KJG [59] & L,F,M are collinear [09] ⇒  ∠MTG = ∠FKA [60]
046. ∠MGT = ∠KFA [45] & ∠MTG = ∠FKA [60] (Similar Triangles)⇒  MG:AF = MT:AK [61]
047. J,L,K,A are concyclic [35] ⇒  ∠JKL = ∠JAL [62]
048. B,K,A are collinear [06] & ∠JKL = ∠JAL [62] & L,C,A are collinear [04] & ∠BAJ = ∠JAC [01] ⇒  ∠(BK-JA) = ∠JKL [63]
049. JM = JK [29] ⇒  ∠JMK = ∠MKJ [64]
050. B,K,A are collinear [06] & J,B,F are collinear [08] & ∠JMK = ∠MKJ [64] & G,K,M are collinear [10] & ∠JBK = ∠JMK [19] ⇒  ∠KBJ = ∠JKG [65]
051. ∠(BK-JA) = ∠JKL [63] & ∠KBJ = ∠JKG [65] ⇒  ∠LKG = ∠AJB [66]
052. J,B,F are collinear [08] & ∠LKG = ∠AJB [66] ⇒  ∠AJF = ∠LKG [67]
053. K,A,J,L,G,F are concyclic [42] & ∠AJF = ∠LKG [67] ⇒  AF = LG [68]
054. J,G,C are collinear [11] & ∠MJC = ∠CJL [24] ⇒  ∠LJG = ∠GJM [69]
055. JM = JL [22] & ∠LJG = ∠GJM [69] (SAS)⇒  GL = GM [70]
056. J,B,F are collinear [08] & ∠JBM = ∠JKM [18] & B,C,M are collinear [02] & G,K,M are collinear [10] & ∠BML = ∠MKL [31] & L,F,M are collinear [09] & ∠BKL = ∠KML [51] & B,K,A are collinear [06] & ∠JBK = ∠JMK [19] ⇒  ∠BJK = ∠MJB [71]
057. J,B,K,M are concyclic [17] & ∠BJK = ∠MJB [71] ⇒  BK = MB [72]
058. JM = JK [29] & BK = MB [72] ⇒  JB ⟂ MK [73]
059. J,L,K,F are concyclic [33] & J,L,K,A are concyclic [35] ⇒  J,K,F,A are concyclic [74]
060. J,L,K,F are concyclic [33] & J,L,K,A are concyclic [35] ⇒  A,J,L,F are concyclic [75]
061. A,J,L,F are concyclic [75] ⇒  ∠ALJ = ∠AFJ [76]
062. J,K,F,A are concyclic [74] ⇒  ∠KAF = ∠KJF [77]
063. F,S,A are collinear [12] & G,K,M are collinear [10] & JB ⟂ MK [73] & ∠ALJ = ∠AFJ [76] & L,C,A are collinear [04] & J,B,F are collinear [08] & JL ⟂ AC [05] & JK ⟂ AB [07] & ∠KAF = ∠KJF [77] & B,K,A are collinear [06] ⇒  SA ∥ MK [78]
064. B,C,S are collinear [13] & B,C,M are collinear [02] ⇒  B,S,M are collinear [79]
065. SA ∥ MK [78] & B,S,M are collinear [79] & B,K,A are collinear [06] ⇒  MB:KB = MS:KA [80]
066. MG:AF = MT:AK [61] & AF = LG [68] & GL = GM [70] & MB:KB = MS:KA [80] & BK = MB [72] ⇒  MS = MT
==========================

 

 

IMO 2012, Problem 5. Let ABC be a triangle with ∠BCA = 90°, and let D be the foot of the altitude from C. Let X be a point in the interior of the segment CD. Let K be the point on the segment AX such that BK = BC. Similarly, let L be the point on the segment BX such that AL = AC. Let M = AL ∩ BK. Prove that MK = ML.

 

○ 풀이. Evan Chen의 풀이 

○ 풀이. AlphaGeometry를 사용할 때, 새로운 점을 추가하는 AlphaGeometry 모드로 풀립니다. (∵ 더 어려운 문제) (ref)

 

 

IMO 2013, Problem 3. Let the excircle of triangle ABC opposite the vertex A be tangent to the side BC at the point A₁. Define the points B₁ on CA and C₁ on AB analogously, using the excircles opposite B and C, respectively. Suppose that the circumcenter of triangle A₁B₁C₁ lies on the circumcircle of triangle ABC. Prove that triangle ABC is right-angled.

 

○ 풀이. Evan Chen의 풀이 

○ 풀이. AlphaGeometry 풀이 대상은 아니었습니다.

 

 

IMO 2013, Problem 4. Let ABC be an acute triangle with orthocenter H, and let W be a point on the side BC, between B and C. The points M and N are the feet of the altitudes drawn from B and C, respectively. Suppose ω₁ is the circumcircle of triangle BWN and X is a point such that WX is a diameter of ω1. Similarly, ω₂ is the circumcircle of triangle CWM and Y is a point such that WY is a diameter of ω₂. Show that the points X, Y , and H are collinear.

 

○ 풀이. Evan Chen의 풀이 

○ 풀이. AlphaGeometry를 사용할 때, 새로운 점을 추가하지 않는 DD+AR 모드로 풀립니다. (ref)

 

==========================
 * From theorem premises:
A B C H M N W O1 O2 X Y : Points
BH ⟂ AC [00]
AH ⟂ BC [01]
B,H,M are collinear [02]
C,A,M are collinear [03]
N,C,H are collinear [04]
B,A,N are collinear [05]
B,C,W are collinear [06]
O_1N = O_1W [07]
O_1B = O_1N [08]
O_2M = O_2W [09]
O_2C = O_2M [10]
O_1X = O_1W [11]
W,X,O_1 are collinear [12]
O_2Y = O_2W [13]
O_2,Y,W are collinear [14]

 * Auxiliary Constructions:
: Points


 * Proof steps:
001. O_1N = O_1W [07] & O_1X = O_1W [11] ⇒  O_1 is the circumcenter of \Delta XNW [15]
002. O_1 is the circumcenter of \Delta XNW [15] & W,X,O_1 are collinear [12] ⇒  NX ⟂ NW [16]
003. NX ⟂ NW [16] & BH ⟂ AC [00] ⇒  ∠XNW = ∠(AC-BH) [17]
004. B,H,M are collinear [02] & C,A,M are collinear [03] & BH ⟂ AC [00] ⇒  ∠BMC = ∠AMH [18]
005. B,H,M are collinear [02] & C,A,M are collinear [03] & BH ⟂ AC [00] ⇒  ∠BMA = ∠CMH [19]
006. AH ⟂ BC [01] & BH ⟂ AC [00] ⇒  ∠HAC = ∠CBH [20]
007. B,H,M are collinear [02] & C,A,M are collinear [03] & ∠CBH = ∠HAC [20] ⇒  ∠CBM = ∠HAM [21]
008. ∠BMC = ∠AMH [18] & ∠CBM = ∠HAM [21] (Similar Triangles)⇒  CM:HM = BM:AM [22]
009. CM:HM = BM:AM [22] & ∠BMA = ∠CMH [19] (Similar Triangles)⇒  ∠MBA = ∠MCH [23]
010. CM:HM = BM:AM [22] & ∠BMA = ∠CMH [19] (Similar Triangles)⇒  ∠BAM = ∠CHM [24]
011. N,C,H are collinear [04] & N,B,A are collinear [05] & ∠XNW = ∠(AC-BH) [17] & ∠MBA = ∠MCH [23] & B,H,M are collinear [02] & C,A,M are collinear [03] ⇒  ∠HNA = ∠XNW [25]
012. O_1B = O_1N [08] & O_1N = O_1W [07] & O_1X = O_1W [11] ⇒  B,X,W,N are concyclic [26]
013. O_1B = O_1N [08] & O_1N = O_1W [07] & O_1X = O_1W [11] ⇒  O_1 is the circumcenter of \Delta XBW [27]
014. B,X,W,N are concyclic [26] ⇒  ∠BXW = ∠BNW [28]
015. B,X,W,N are concyclic [26] ⇒  ∠BXN = ∠BWN [29]
016. O_1 is the circumcenter of \Delta XBW [27] & W,X,O_1 are collinear [12] ⇒  BW ⟂ XB [30]
017. N,B,A are collinear [05] & W,X,O_1 are collinear [12] & ∠BXW = ∠BNW [28] & BW ⟂ XB [30] & AH ⟂ BC [01] & B,C,W are collinear [06] ⇒  ∠HAN = ∠XWN [31]
018. ∠HNA = ∠XNW [25] & ∠HAN = ∠XWN [31] (Similar Triangles)⇒  NH:NX = NA:NW [32]
019. N,C,H are collinear [04] & ∠BAM = ∠CHM [24] & C,A,M are collinear [03] & B,H,M are collinear [02] & ∠MBA = ∠MCH [23] ⇒  BA ⟂ NC [33]
020. NX ⟂ NW [16] & BA ⟂ NC [33] ⇒  ∠XNC = ∠(NW-BA) [34]
021. N,C,H are collinear [04] & N,B,A are collinear [05] & ∠XNC = ∠(NW-BA) [34] ⇒  ∠HNX = ∠ANW [35]
022. NH:NX = NA:NW [32] & ∠HNX = ∠ANW [35] (Similar Triangles)⇒  ∠(HX-AW) = ∠XNW [36]
023. O_2M = O_2W [09] & O_2Y = O_2W [13] ⇒  O_2 is the circumcenter of \Delta YMW [37]
024. O_2 is the circumcenter of \Delta YMW [37] & O_2,Y,W are collinear [14] ⇒  MY ⟂ MW [38]
025. BH ⟂ AC [00] & MY ⟂ MW [38] ⇒  ∠(MW-AC) = ∠(MY-BH) [39]
026. BH ⟂ AC [00] & MY ⟂ MW [38] ⇒  ∠YMW = ∠(AC-BH) [40]
027. B,H,M are collinear [02] & C,A,M are collinear [03] & BH ⟂ AC [00] & MY ⟂ MW [38] ⇒  ∠YMW = ∠HMA [41]
028. O_2C = O_2M [10] & O_2M = O_2W [09] & O_2Y = O_2W [13] ⇒  W,Y,C,M are concyclic [42]
029. O_2C = O_2M [10] & O_2M = O_2W [09] & O_2Y = O_2W [13] ⇒  O_2 is the circumcenter of \Delta YCW [43]
030. W,Y,C,M are concyclic [42] ⇒  ∠WYC = ∠WMC [44]
031. O_2 is the circumcenter of \Delta YCW [43] & O_2,Y,W are collinear [14] ⇒  CW ⟂ YC [45]
032. W,O_2,Y are collinear [14] & C,A,M are collinear [03] & ∠WYC = ∠WMC [44] & CW ⟂ YC [45] & ∠BXN = ∠BWN [29] & B,C,W are collinear [06] & NX ⟂ NW [16] & BW ⟂ XB [30] & AH ⟂ BC [01] ⇒  ∠YWM = ∠HAM [46]
033. ∠YMW = ∠HMA [41] & ∠YWM = ∠HAM [46] (Similar Triangles)⇒  YM:HM = WM:AM [47]
034. C,A,M are collinear [03] & B,H,M are collinear [02] & ∠(MW-AC) = ∠(MY-BH) [39] ⇒  ∠WMA = ∠YMH [48]
035. YM:HM = WM:AM [47] & ∠WMA = ∠YMH [48] (Similar Triangles)⇒  ∠WAM = ∠YHM [49]
036. YM:HM = WM:AM [47] & ∠WMA = ∠YMH [48] (Similar Triangles)⇒  ∠YMW = ∠(HY-AW) [50]
037. ∠WAM = ∠YHM [49] & C,A,M are collinear [03] & B,H,M are collinear [02] & ∠YMW = ∠(AC-BH) [40] & ∠YMW = ∠(HY-AW) [50] ⇒  YH ⟂ WA [51]
038. YH ⟂ WA [51] & NX ⟂ NW [16] ⇒  ∠(HY-AW) = ∠XNW [52]
039. ∠(HX-AW) = ∠XNW [36] & ∠(HY-AW) = ∠XNW [52] ⇒  ∠(YH-WA) = ∠(XH-WA) [53]
040. ∠(YH-WA) = ∠(XH-WA) [53] ⇒  YH ∥ XH [54]
041. HY ∥ HX [54] ⇒  Y,X,H are collinear
==========================

 

 

IMO 2014, Problem 3. Convex quadrilateral ABCD has ∠ABC = ∠CDA = 90°. Point H is the foot of the perpendicular from A to BD. Points S and T lie on sides AB and AD, respectively, such that H lies inside triangle SCT and

 

∠CHS − ∠CSB = 90°, ∠THC − ∠DTC = 90°.

 

Prove that line BD is tangent to the circumcircle of triangle TSH.

 

○ 풀이. Evan Chen의 풀이 

○ 풀이. AlphaGeometry 풀이 대상은 아니었습니다.

 

 

IMO 2014, Problem 4. Let P and Q be on segment BC of an acute triangle ABC such that ∠PAB = ∠BCA and ∠CAQ = ∠ABC. Let M and N be points on AP and AQ, respectively, such that P is the midpoint of AM and Q is the midpoint of AN. Prove that BM and CN meet on the circumcircle of △ABC.

 

○ 풀이. Evan Chen의 풀이 

○ 풀이. AlphaGeometry를 사용할 때, 새로운 점을 추가하는 AlphaGeometry 모드로 풀립니다. (∵ 더 어려운 문제) (ref)

 

==========================
 * From theorem premises:
A B C D E F G H I : Points
D,C,B are collinear [00]
∠DAB = ∠BCA [01]
∠(AD-BC) = ∠BAC [02]
C,B,E are collinear [03]
∠EAC = ∠CBA [04]
∠(BC-AE) = ∠BAC [05]
F,D,A are collinear [06]
DA = DF [07]
EA = EG [08]
G,E,A are collinear [09]
F,H,B are collinear [10]
H,G,C are collinear [11]
IA = IB [12]
IB = IC [13]

 * Auxiliary Constructions:
J : Points
IJ = IA [14]
I,J,A are collinear [15]

 * Proof steps:
001. IA = IB [12] & IJ = IA [14] & IB = IC [13] ⇒  J,C,B,A are concyclic [16]
002. IA = IB [12] & IJ = IA [14] & IB = IC [13] ⇒  I is the circumcenter of \Delta JCA [17]
003. J,C,B,A are concyclic [16] ⇒  ∠JCA = ∠JBA [18]
004. D,C,B are collinear [00] & F,D,A are collinear [06] & ∠BCA = ∠DAB [01] ⇒  ∠DCA = ∠(FD-BA) [19]
005. ∠JCA = ∠JBA [18] & ∠DCA = ∠(FD-BA) [19] ⇒  ∠(JB-FD) = ∠JCD [20]
006. C,B,E are collinear [03] & ∠BCA = ∠DAB [01] ⇒  ∠ECA = ∠DAB [21]
007. D,C,B are collinear [00] & ∠EAC = ∠CBA [04] ⇒  ∠EAC = ∠DBA [22]
008. ∠ECA = ∠DAB [21] & ∠EAC = ∠DBA [22] (Similar Triangles)⇒  EC:EA = DA:DB [23]
009. E,C,B are collinear [03] & D,C,B are collinear [00] & ∠(BC-AE) = ∠BAC [05] & ∠(AD-BC) = ∠BAC [02] ⇒  ∠ADE = ∠DEA [24]
010. ∠ADE = ∠DEA [24] ⇒  AD = AE [25]
011. EC:EA = DA:DB [23] & DA = DF [07] & AD = AE [25] & EA = EG [08] ⇒  EC:EG = DF:DB [26]
012. E,C,B are collinear [03] & G,E,A are collinear [09] & F,D,A are collinear [06] & D,C,B are collinear [00] & ∠(BC-AE) = ∠BAC [05] & ∠(AD-BC) = ∠BAC [02] ⇒  ∠CEG = ∠FDB [27]
013. EC:EG = DF:DB [26] & ∠CEG = ∠FDB [27] (Similar Triangles)⇒  ∠ECG = ∠DFB [28]
014. F,H,B are collinear [10] & H,G,C are collinear [11] & ∠(JB-FD) = ∠JCD [20] & F,D,A are collinear [06] & D,C,B are collinear [00] & ∠ECG = ∠DFB [28] & C,B,E are collinear [03] ⇒  ∠BHC = ∠BJC [29]
015. ∠BHC = ∠BJC [29] ⇒  H,C,B,J are concyclic [30]
016. H,C,B,J are concyclic [30] & J,C,B,A are concyclic [16] ⇒  J,H,A,B are concyclic [31]
017. H,C,B,J are concyclic [30] & J,C,B,A are concyclic [16] ⇒  J,C,H,A are concyclic [32]
018. J,H,A,B are concyclic [31] ⇒  ∠AHJ = ∠ABJ [33]
019. IA = IB [12] & IJ = IA [14] ⇒  I is the circumcenter of \Delta JBA [34]
020. I is the circumcenter of \Delta JBA [34] & I,J,A are collinear [15] ⇒  BJ ⟂ AB [35]
021. I is the circumcenter of \Delta JCA [17] & I,J,A are collinear [15] ⇒  CJ ⟂ AC [36]
022. BJ ⟂ AB [35] & CJ ⟂ AC [36] ⇒  ∠ABJ = ∠JCA [37]
023. J,C,H,A are concyclic [32] ⇒  ∠JCA = ∠JHA [38]
024. ∠AHJ = ∠ABJ [33] & ∠ABJ = ∠JCA [37] & ∠JCA = ∠JHA [38] ⇒  HJ ⟂ HA [39]
025. I,J,A are collinear [15] & IJ = IA [14] ⇒  I is midpoint of AJ [40]
026. HJ ⟂ HA [39] & I is midpoint of AJ [40] ⇒  AI = HI
==========================

 

입력: 2024.05.05 20:32