【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. 추후 업데이트

 

 

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. AlphaGeometry (DDAR mode)

 

==========================
 * 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
==========================

 

○ 풀이. AlphaGeometry (AlphaGeometry mode)

 

==========================
 * From theorem premises:
A B C D E F G H I : Points
AB = AC [00]
∠CBA = ∠ACB [01]
BD ⟂ AB [02]
DE = DB [03]
DF = DB [04]
E,F,A are collinear [05]
DG = DB [06]
C,B,G are collinear [07]
DH = DB [08]
H,F,C are collinear [09]
DI = DB [10]
E,C,I are collinear [11]

 * Auxiliary Constructions:
: Points


 * Proof steps:
001. DF = DB [04] & DI = DB [10] & DH = DB [08] & DG = DB [06] & DE = DB [03] ⇒  H,F,I,G are concyclic [12]
002. DF = DB [04] & DI = DB [10] & DH = DB [08] & DG = DB [06] & DE = DB [03] ⇒  H,B,I,G are concyclic [13]
003. DF = DB [04] & DE = DB [03] ⇒  D is the circumcenter of \Delta BFE [14]
004. D is the circumcenter of \Delta BFE [14] & BD ⟂ AB [02] ⇒  ∠ABF = ∠BEF [15]
005. ∠ABF = ∠BEF [15] & E,F,A are collinear [05] ⇒  ∠ABF = ∠BEA [16]
006. E,F,A are collinear [05] & ∠ABF = ∠BEF [15] ⇒  ∠AFB = ∠EBA [17]
007. ∠ABF = ∠BEA [16] & ∠AFB = ∠EBA [17] (Similar Triangles)⇒  AB:AF = AE:AB [18]
008. AB:AF = AE:AB [18] & AB = AC [00] ⇒  CA:FA = EA:CA [19]
009. E,F,A are collinear [05] ⇒  ∠EAC = ∠FAC [20]
010. CA:FA = EA:CA [19] & ∠EAC = ∠FAC [20] (Similar Triangles)⇒  ∠CEA = ∠ACF [21]
011. DF = DB [04] & DI = DB [10] & DH = DB [08] & DG = DB [06] & DE = DB [03] ⇒  F,E,I,G are concyclic [22]
012. F,E,I,G are concyclic [22] ⇒  ∠FEI = ∠FGI [23]
013. H,F,C are collinear [09] & ∠CEA = ∠ACF [21] & ∠FEI = ∠FGI [23] & E,F,A are collinear [05] & E,C,I are collinear [11] ⇒  ∠IGF = ∠(CA-HF) [24]
014. DG = DB [06] & DI = DB [10] ⇒  D is the circumcenter of \Delta BGI [25]
015. D is the circumcenter of \Delta BGI [25] & BD ⟂ AB [02] ⇒  ∠ABG = ∠BIG [26]
016. ∠CBA = ∠ACB [01] & ∠ABG = ∠BIG [26] & C,B,G are collinear [07] ⇒  ∠GIB = ∠ACB [27]
017. H,B,I,G are concyclic [13] ⇒  ∠HIB = ∠HGB [28]
018. ∠HIB = ∠HGB [28] & C,B,G are collinear [07] ⇒  ∠HIB = ∠(GH-BC) [29]
019. ∠GIB = ∠ACB [27] & ∠HIB = ∠(GH-BC) [29] ⇒  ∠GIH = ∠(AC-GH) [30]
020. ∠IGF = ∠(CA-HF) [24] & ∠GIH = ∠(AC-GH) [30] ⇒  ∠(FG-HI) = ∠FHG [31]
021. H,F,C are collinear [09] & ∠(FG-HI) = ∠FHG [31] ⇒  ∠HFG = ∠GHI [32]
022. H,F,I,G are concyclic [12] & ∠HFG = ∠GHI [32] ⇒  HG = GI
==========================

 

 

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. 추후 업데이트

 

 

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. AlphaGeometry (DDAR mode)

 

==========================
 * 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
==========================

 

○ 풀이. AlphaGeometry (AlphaGeometry mode)

 

==========================
 * From theorem premises:
A B C G D E F H I J K : Points
∠ACG = ∠GCB [00]
∠BAG = ∠GAC [01]
B,D,C are collinear [02]
DG ⟂ BC [03]
A,C,E are collinear [04]
EG ⟂ AC [05]
B,F,A are collinear [06]
FG ⟂ AB [07]
D,H,E are collinear [08]
B,G,H are collinear [09]
D,F,I are collinear [10]
G,I,C are collinear [11]
J,A,H are collinear [12]
B,J,C are collinear [13]
K,A,I are collinear [14]
B,C,K are collinear [15]

 * Auxiliary Constructions:
: Points


 * Proof steps:
001. B,D,C are collinear [02] & B,A,F are collinear [06] & FG ⟂ AB [07] & DG ⟂ BC [03] ⇒  ∠GDB = ∠GFB [16]
002. ∠GDB = ∠GFB [16] ⇒  B,D,G,F are concyclic [17]
003. B,D,G,F are concyclic [17] ⇒  ∠BDF = ∠BGF [18]
004. B,D,G,F are concyclic [17] ⇒  ∠BGD = ∠BFD [19]
005. B,D,C are collinear [02] & A,C,E are collinear [04] & EG ⟂ AC [05] & DG ⟂ BC [03] ⇒  ∠GDC = ∠CEG [20]
006. B,D,C are collinear [02] & A,C,E are collinear [04] & ∠GCB = ∠ACG [00] ⇒  ∠GCD = ∠ECG [21]
007. ∠GDC = ∠CEG [20] & ∠GCD = ∠ECG [21] (Similar Triangles)⇒  GD = GE [22]
008. ∠GDC = ∠CEG [20] & ∠GCD = ∠ECG [21] (Similar Triangles)⇒  CD = CE [23]
009. ∠GDC = ∠CEG [20] & ∠GCD = ∠ECG [21] (Similar Triangles)⇒  ∠DGC = ∠CGE [24]
010. A,C,E are collinear [04] & A,B,F are collinear [06] & FG ⟂ AB [07] & EG ⟂ AC [05] ⇒  ∠GEA = ∠AFG [25]
011. A,C,E are collinear [04] & B,F,A are collinear [06] & ∠GAC = ∠BAG [01] ⇒  ∠GAE = ∠FAG [26]
012. ∠GEA = ∠AFG [25] & ∠GAE = ∠FAG [26] (Similar Triangles)⇒  GE = GF [27]
013. GD = GE [22] & GE = GF [27] ⇒  G is the circumcenter of \Delta DEF [28]
014. GD = GE [22] & GE = GF [27] ⇒  GD = GF [29]
015. B,D,C are collinear [02] & DG ⟂ BC [03] ⇒  GD ⟂ DC [30]
016. G is the circumcenter of \Delta DEF [28] & GD ⟂ DC [30] ⇒  ∠CDE = ∠DFE [31]
017. D,H,E are collinear [08] & B,G,H are collinear [09] & ∠BDF = ∠BGF [18] & B,D,C are collinear [02] & ∠CDE = ∠DFE [31] ⇒  ∠FEH = ∠FGH [32]
018. ∠FEH = ∠FGH [32] ⇒  G,F,H,E are concyclic [33]
019. A,B,F are collinear [06] & A,C,E are collinear [04] & FG ⟂ AB [07] & EG ⟂ AC [05] ⇒  ∠AFG = ∠AEG [34]
020. ∠AFG = ∠AEG [34] ⇒  F,G,A,E are concyclic [35]
021. F,G,A,E are concyclic [35] & G,F,H,E are concyclic [33] ⇒  G,F,H,A are concyclic [36]
022. B,D,C are collinear [02] & A,C,E are collinear [04] & EG ⟂ AC [05] & DG ⟂ BC [03] ⇒  ∠GDC = ∠GEC [37]
023. ∠GDC = ∠GEC [37] ⇒  D,G,C,E are concyclic [38]
024. D,G,C,E are concyclic [38] ⇒  ∠DCG = ∠DEG [39]
025. D,F,I are collinear [10] & G,I,C are collinear [11] & ∠DCG = ∠DEG [39] & B,D,C are collinear [02] & ∠CDE = ∠DFE [31] ⇒  ∠IFE = ∠IGE [40]
026. ∠IFE = ∠IGE [40] ⇒  F,G,I,E are concyclic [41]
027. F,G,A,E are concyclic [35] & F,G,I,E are concyclic [41] ⇒  F,G,I,A are concyclic [42]
028. G,F,H,E are concyclic [33] & G,F,H,A are concyclic [36] & F,G,I,E are concyclic [41] & F,G,I,A are concyclic [42] & F,G,A,E are concyclic [35] ⇒  F,I,G,A,E,H are concyclic [43]
029. G,F,H,E are concyclic [33] & G,F,H,A are concyclic [36] & F,G,I,E are concyclic [41] & F,G,I,A are concyclic [42] & F,G,A,E are concyclic [35] ⇒  F,I,A,H are concyclic [44]
030. F,I,A,H are concyclic [44] ⇒  ∠FIA = ∠FHA [45]
031. A,I,K are collinear [14] & F,D,I are collinear [10] & ∠FIA = ∠FHA [45] ⇒  ∠AHF = ∠KID [46]
032. GD = GE [22] & CD = CE [23] ⇒  DE ⟂ CG [47]
033. D,H,E are collinear [08] & G,I,C are collinear [11] & DG ⟂ BC [03] & DE ⟂ CG [47] ⇒  ∠(DH-GI) = ∠(CB-DG) [48]
034. B,A,F are collinear [06] & AB ⟂ FG [07] ⇒  BF ⟂ GF [49]
035. G is the circumcenter of \Delta DEF [28] & BF ⟂ GF [49] ⇒  ∠BFD = ∠FED [50]
036. B,G,H are collinear [09] & ∠BDF = ∠BGF [18] & B,D,C are collinear [02] & ∠CDE = ∠DFE [31] & ∠BFD = ∠FED [50] & B,F,A are collinear [06] & ∠BGD = ∠BFD [19] ⇒  ∠DGH = ∠HGF [51]
037. GD = GF [29] & ∠DGH = ∠HGF [51] (SAS)⇒  ∠GDH = ∠HFG [52]
038. GD = GF [29] & ∠DGH = ∠HGF [51] (SAS)⇒  HD = HF [53]
039. D,H,E are collinear [08] & G,I,C are collinear [11] & DE ⟂ CG [47] ⇒  DH ⟂ GI [54]
040. DH ⟂ GI [54] & BF ⟂ GF [49] ⇒  ∠(DH-GF) = ∠(GI-BF) [55]
041. G,I,C are collinear [11] & B,A,F are collinear [06] & ∠GDH = ∠HFG [52] & D,H,E are collinear [08] & ∠(DH-GF) = ∠(GI-BF) [55] ⇒  ∠(GI-BF) = ∠(DG-FH) [56]
042. ∠(DH-GI) = ∠(CB-DG) [48] & ∠(GI-BF) = ∠(DG-FH) [56] ⇒  ∠(DH-BF) = ∠(CB-FH) [57]
043. G,I,C are collinear [11] & ∠DGC = ∠CGE [24] ⇒  ∠DGI = ∠IGE [58]
044. GD = GE [22] & ∠DGI = ∠IGE [58] (SAS)⇒  ID = IE [59]
045. GD = GE [22] & ID = IE [59] ⇒  GI ⟂ DE [60]
046. F,G,I,E are concyclic [41] & F,G,I,A are concyclic [42] ⇒  I,G,A,E are concyclic [61]
047. I,G,A,E are concyclic [61] ⇒  ∠IGE = ∠IAE [62]
048. F,I,G,A are concyclic [42] ⇒  ∠FGI = ∠FAI [63]
049. A,B,F are collinear [06] & A,I,K are collinear [14] & B,C,K are collinear [15] & B,D,C are collinear [02] & ∠(DH-BF) = ∠(CB-FH) [57] & D,H,E are collinear [08] & GI ⟂ DE [60] & ∠IGE = ∠IAE [62] & G,I,C are collinear [11] & A,C,E are collinear [04] & FG ⟂ AB [07] & EG ⟂ AC [05] & ∠FGI = ∠FAI [63] ⇒  ∠AFH = ∠IKD [64]
050. ∠AHF = ∠KID [46] & ∠AFH = ∠IKD [64] (Similar Triangles)⇒  AH:DI = AF:DK [65]
051. F,G,A,E are concyclic [35] ⇒  ∠FGA = ∠FEA [66]
052. B,A,F are collinear [06] & ∠BAG = ∠GAC [01] & ∠FGA = ∠FEA [66] & A,C,E are collinear [04] ⇒  ∠GFE = ∠(BF-GA) [67]
053. GD = GF [29] ⇒  ∠GDF = ∠DFG [68]
054. B,A,F are collinear [06] & ∠BGD = ∠BFD [19] & ∠GDF = ∠DFG [68] ⇒  ∠DFG = ∠GBF [69]
055. ∠GFE = ∠(BF-GA) [67] & ∠DFG = ∠GBF [69] ⇒  ∠BGA = ∠DFE [70]
056. D,F,I are collinear [10] & B,G,H are collinear [09] & ∠DFE = ∠BGA [70] ⇒  ∠IFE = ∠HGA [71]
057. F,I,G,A,E,H are concyclic [43] & ∠IFE = ∠HGA [71] ⇒  IE = HA [72]
058. GD = GF [29] & HD = HF [53] ⇒  GH ⟂ DF [73]
059. G,F,H,E are concyclic [33] & G,F,H,A are concyclic [36] ⇒  G,A,E,H are concyclic [74]
060. G,A,E,H are concyclic [74] ⇒  ∠GEA = ∠GHA [75]
061. F,G,A,H are concyclic [36] ⇒  ∠FGH = ∠FAH [76]
062. J,A,H are collinear [12] & GH ⟂ DF [73] & ∠GEA = ∠GHA [75] & A,C,E are collinear [04] & B,G,H are collinear [09] & FG ⟂ AB [07] & EG ⟂ AC [05] & ∠FGH = ∠FAH [76] & B,F,A are collinear [06] ⇒  JA ∥ DF [77]
063. B,D,C are collinear [02] & B,J,C are collinear [13] ⇒  B,J,D are collinear [78]
064. JA ∥ DF [77] & B,J,D are collinear [78] & B,F,A are collinear [06] ⇒  DB:FB = DJ:FA [79]
065. B,D,C are collinear [02] & ∠CDE = ∠DFE [31] & ∠BFD = ∠FED [50] & B,F,A are collinear [06] & ∠BGD = ∠BFD [19] ⇒  ∠BGD = ∠FDB [80]
066. B,D,G,F are concyclic [17] & ∠BGD = ∠FDB [80] ⇒  BD = FB [81]
067. AH:DI = AF:DK [65] & ID = IE [59] & IE = HA [72] & DB:FB = DJ:FA [79] & BD = FB [81] ⇒  DJ = DK
==========================

 

 

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. 추후 업데이트 

 

 

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. 추후 업데이트

 

 

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. AlphaGeometry (DDAR mode)

 

==========================
 * 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
==========================

 

○ 풀이. AlphaGeometry (AlphaGeometry mode)

 

==========================
 * From theorem premises:
A B C D E F G H I J K : Points
BD ⟂ AC [00]
AD ⟂ BC [01]
C,E,A are collinear [02]
D,E,B are collinear [03]
D,C,F are collinear [04]
F,A,B are collinear [05]
G,C,B are collinear [06]
HF = HG [07]
HB = HF [08]
IE = IG [09]
IC = IE [10]
HJ = HG [11]
G,H,J are collinear [12]
IK = IG [13]
I,G,K are collinear [14]

 * Auxiliary Constructions:
: Points


 * Proof steps:
001. HF = HG [07] & HJ = HG [11] ⇒  H is the circumcenter of \Delta GFJ [15]
002. H is the circumcenter of \Delta GFJ [15] & G,H,J are collinear [12] ⇒  FG ⟂ FJ [16]
003. IE = IG [09] & IK = IG [13] ⇒  I is the circumcenter of \Delta GEK [17]
004. I is the circumcenter of \Delta GEK [17] & I,G,K are collinear [14] ⇒  EG ⟂ EK [18]
005. C,E,A are collinear [02] & D,E,B are collinear [03] & BD ⟂ AC [00] & EG ⟂ EK [18] ⇒  ∠GEK = ∠AED [19]
006. D,E,B are collinear [03] & C,E,A are collinear [02] & BD ⟂ AC [00] ⇒  DE ⟂ CE [20]
007. G,C,B are collinear [06] & AD ⟂ BC [01] ⇒  DA ⟂ GC [21]
008. DE ⟂ CE [20] & DA ⟂ GC [21] ⇒  ∠EDA = ∠ECG [22]
009. DE ⟂ CE [20] & DA ⟂ GC [21] ⇒  ∠(DE-GC) = ∠(CE-DA) [23]
010. IE = IG [09] & IK = IG [13] & IC = IE [10] ⇒  G,C,E,K are concyclic [24]
011. G,C,E,K are concyclic [24] ⇒  ∠GCE = ∠GKE [25]
012. I,G,K are collinear [14] & D,E,B are collinear [03] & ∠EDA = ∠ECG [22] & C,E,A are collinear [02] & G,C,B are collinear [06] & ∠GCE = ∠GKE [25] ⇒  ∠GKE = ∠ADE [26]
013. ∠GEK = ∠AED [19] & ∠GKE = ∠ADE [26] (Similar Triangles)⇒  GE:EA = KE:DE [27]
014. DE ⟂ CE [20] & EG ⟂ EK [18] ⇒  ∠DEK = ∠CEG [28]
015. DE ⟂ CE [20] & EG ⟂ EK [18] ⇒  ∠DEG = ∠CEK [29]
016. C,E,A are collinear [02] & D,E,B are collinear [03] & ∠DEK = ∠CEG [28] ⇒  ∠GEA = ∠KED [30]
017. GE:EA = KE:DE [27] & ∠GEA = ∠KED [30] (Similar Triangles)⇒  ∠GAE = ∠KDE [31]
018. GE:EA = KE:DE [27] & ∠GEA = ∠KED [30] (Similar Triangles)⇒  ∠GEK = ∠(AG-DK) [32]
019. ∠GAE = ∠KDE [31] & C,E,A are collinear [02] & D,E,B are collinear [03] & ∠DEG = ∠CEK [29] & ∠GEK = ∠(AG-DK) [32] ⇒  GA ⟂ DK [33]
020. FG ⟂ FJ [16] & GA ⟂ DK [33] ⇒  ∠JFG = ∠(DK-AG) [34]
021. C,E,A are collinear [02] & D,E,B are collinear [03] & BD ⟂ AC [00] ⇒  ∠CEB = ∠DEA [35]
022. C,E,A are collinear [02] & D,E,B are collinear [03] & BD ⟂ AC [00] ⇒  ∠CED = ∠BEA [36]
023. D,E,B are collinear [03] & C,E,A are collinear [02] & ∠(DE-GC) = ∠(CE-DA) [23] & G,C,B are collinear [06] ⇒  ∠CBE = ∠DAE [37]
024. ∠CEB = ∠DEA [35] & ∠CBE = ∠DAE [37] (Similar Triangles)⇒  CE:DE = EB:EA [38]
025. CE:DE = EB:EA [38] & ∠CED = ∠BEA [36] (Similar Triangles)⇒  ∠CDE = ∠BAE [39]
026. CE:DE = EB:EA [38] & ∠CED = ∠BEA [36] (Similar Triangles)⇒  ∠ECD = ∠EBA [40]
027. FG ⟂ FJ [16] & DE ⟂ CE [20] ⇒  ∠(FJ-DE) = ∠(GF-CE) [41]
028. D,C,F are collinear [04] & F,A,B are collinear [05] & ∠CDE = ∠BAE [39] & D,E,B are collinear [03] & C,E,A are collinear [02] & ∠(FJ-DE) = ∠(GF-CE) [41] ⇒  ∠JFG = ∠DFA [42]
029. HF = HG [07] & HJ = HG [11] & HB = HF [08] ⇒  G,F,B,J are concyclic [43]
030. HF = HG [07] & HJ = HG [11] & HB = HF [08] ⇒  H is the circumcenter of \Delta GBJ [44]
031. G,F,B,J are concyclic [43] ⇒  ∠GFB = ∠GJB [45]
032. H is the circumcenter of \Delta GBJ [44] & G,H,J are collinear [12] ⇒  GB ⟂ BJ [46]
033. G,H,J are collinear [12] & F,A,B are collinear [05] & ∠GFB = ∠GJB [45] & GB ⟂ BJ [46] & AD ⟂ BC [01] & G,C,B are collinear [06] ⇒  ∠JGF = ∠DAF [47]
034. ∠JFG = ∠DFA [42] & ∠JGF = ∠DAF [47] (Similar Triangles)⇒  FJ:DF = GF:FA [48]
035. F,A,B are collinear [05] & ∠CDE = ∠BAE [39] & D,E,B are collinear [03] & C,E,A are collinear [02] & ∠ECD = ∠EBA [40] ⇒  DC ⟂ FA [49]
036. FG ⟂ FJ [16] & DC ⟂ FA [49] ⇒  ∠(FJ-DC) = ∠GFA [50]
037. D,C,F are collinear [04] & F,A,B are collinear [05] & ∠(FJ-DC) = ∠GFA [50] ⇒  ∠JFD = ∠GFA [51]
038. FJ:DF = GF:FA [48] & ∠JFD = ∠GFA [51] (Similar Triangles)⇒  ∠JFG = ∠(DJ-AG) [52]
039. ∠JFG = ∠(DK-AG) [34] & ∠JFG = ∠(DJ-AG) [52] ⇒  ∠(DJ-GA) = ∠(DK-GA) [53]
040. ∠(DJ-GA) = ∠(DK-GA) [53] ⇒  DJ ∥ DK [54]
041. DJ ∥ DK [54] ⇒  D,K,J 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. 추후 업데이트

 

 

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. 추후 업데이트

 

입력: 2024.05.05 20:32