IMO 기하 문제 풀이 (2000년 ~ 2004년)
추천글 : 【기하학】 IMO 기하 문제 풀이 종합
IMO 2000, Problem 1. Two circles G1 and G2 intersect at two points M and N. Let AB be the line tangent to these circles at A and B, respectively, so that M lies closer to AB than N. Let CD be the line parallel to AB and passing through the point M, with C on G1 and D on G2. Lines AC and BD meet at E; lines AN and CD meet at P; lines BN and CD meet at Q. Show that EP = EQ.
○ 풀이. AlphaGeometry (DDAR mode)
==========================
* From theorem premises:
A B G1 G2 M N C D E P Q : Points
AG_1 ⟂ AB [00]
BA ⟂ G_2B [01]
G_1M = G_1A [02]
G_2M = G_2B [03]
G_1N = G_1A [04]
G_2N = G_2B [05]
∠MAN = ∠MAN [06]
G_1C = G_1A [07]
CM ∥ AB [08]
∠CAN = ∠CAN [09]
G_2D = G_2B [10]
DM ∥ AB [11]
∠DBN = ∠DBN [12]
E,D,B are collinear [13]
E,A,C are collinear [14]
P,C,D are collinear [15]
P,A,N are collinear [16]
C,D,Q are collinear [17]
N,Q,B are collinear [18]
DQ:ND = DQ:ND [19]
* Auxiliary Constructions:
: Points
* Proof steps:
001. G_1N = G_1A [04] & G_1C = G_1A [07] & G_1M = G_1A [02] ⇒ A,C,N,M are concyclic [20]
002. A,C,N,M are concyclic [20] ⇒ ∠CAN = ∠CMN [21]
003. G_2M = G_2B [03] & G_2D = G_2B [10] & G_2N = G_2B [05] ⇒ N,D,B,M are concyclic [22]
004. N,D,B,M are concyclic [22] ⇒ ∠NBD = ∠NMD [23]
005. N,D,B,M are concyclic [22] ⇒ ∠DNB = ∠DMB [24]
006. E,D,B are collinear [13] & E,A,C are collinear [14] & ∠CAN = ∠CMN [21] & CM ∥ AB [08] & ∠NBD = ∠NMD [23] & DM ∥ AB [11] ⇒ ∠EBN = ∠EAN [25]
007. ∠EBN = ∠EAN [25] ⇒ E,A,N,B are concyclic [26]
008. E,A,N,B are concyclic [26] ⇒ ∠EAB = ∠ENB [27]
009. E,A,N,B are concyclic [26] ⇒ ∠ENA = ∠EBA [28]
010. G_2N = G_2B [05] & G_2D = G_2B [10] ⇒ G_2 is the circumcenter of \Delta BND [29]
011. G_2 is the circumcenter of \Delta BND [29] & G_2B ⟂ BA [01] ⇒ ∠DBA = ∠DNB [30]
012. DM ∥ AB [11] & CM ∥ AB [08] ⇒ MC ∥ MD [31]
013. MC ∥ MD [31] ⇒ C,D,M are collinear [32]
014. C,D,Q are collinear [17] & ∠DNB = ∠DBA [30] & DM ∥ AB [11] & C,D,M are collinear [32] ⇒ ∠DNB = ∠BDQ [33]
015. N,Q,B are collinear [18] & ∠DBN = ∠DBN [12] ⇒ ∠DBN = ∠DBQ [34]
016. ∠DNB = ∠BDQ [33] & ∠DBN = ∠DBQ [34] (Similar Triangles)⇒ BD:BN = BQ:BD [35]
017. G_1C = G_1A [07] & G_1M = G_1A [02] ⇒ G_1 is the circumcenter of \Delta ACM [36]
018. G_1 is the circumcenter of \Delta ACM [36] & AG_1 ⟂ AB [00] ⇒ ∠BAC = ∠AMC [37]
019. E,A,C are collinear [14] & ∠BAC = ∠AMC [37] & CM ∥ AB [08] ⇒ ∠BAE = ∠MAB [38]
020. G_2D = G_2B [10] & G_2M = G_2B [03] ⇒ G_2 is the circumcenter of \Delta BDM [39]
021. G_2 is the circumcenter of \Delta BDM [39] & G_2B ⟂ BA [01] ⇒ ∠ABD = ∠BMD [40]
022. E,D,B are collinear [13] & ∠ABD = ∠BMD [40] & DM ∥ AB [11] ⇒ ∠MBA = ∠ABE [41]
023. ∠BAE = ∠MAB [38] & ∠MBA = ∠ABE [41] (Similar Triangles)⇒ BM = BE [42]
024. ∠BAE = ∠MAB [38] & ∠MBA = ∠ABE [41] (Similar Triangles)⇒ AM = AE [43]
025. G_2N = G_2B [05] & G_2M = G_2B [03] ⇒ G_2 is the circumcenter of \Delta BNM [44]
026. G_2 is the circumcenter of \Delta BNM [44] & G_2B ⟂ BA [01] ⇒ ∠ABM = ∠BNM [45]
027. ∠ABM = ∠BNM [45] & ∠DNB = ∠DMB [24] & DM ∥ AB [11] ⇒ ∠DNB = ∠BNM [46]
028. N,D,B,M are concyclic [22] & ∠DNB = ∠BNM [46] ⇒ DB = BM [47]
029. BD:BN = BQ:BD [35] & BM = BE [42] & DB = BM [47] ⇒ EB:NB = QB:EB [48]
030. N,Q,B are collinear [18] & E,D,B are collinear [13] & ∠NBD = ∠NBD [12] ⇒ ∠QBE = ∠NBE [49]
031. EB:NB = QB:EB [48] & ∠QBE = ∠NBE [49] (Similar Triangles)⇒ ∠BQE = ∠NEB [50]
032. E,D,B are collinear [13] & E,C,A are collinear [14] & ∠EAB = ∠ENB [27] & ∠BQE = ∠NEB [50] & N,Q,B are collinear [18] & DM ∥ AB [11] & C,D,M are collinear [32] ⇒ ∠QED = ∠DCE [51]
033. C,D,Q are collinear [17] & E,D,B are collinear [13] & DM ∥ AB [11] & C,D,M are collinear [32] ⇒ ∠QDE = ∠CDE [52]
034. ∠QED = ∠DCE [51] & ∠QDE = ∠CDE [52] (Similar Triangles)⇒ EQ:DQ = EC:ED [53]
035. P,C,D are collinear [15] & E,C,A are collinear [14] & E,D,B are collinear [13] & ∠EAB = ∠ENB [27] & ∠BQE = ∠NEB [50] & N,Q,B are collinear [18] & DM ∥ AB [11] & C,D,M are collinear [32] ⇒ ∠PCE = ∠QED [54]
036. G_1M = G_1A [02] & G_1N = G_1A [04] ⇒ G_1 is the circumcenter of \Delta AMN [55]
037. G_1 is the circumcenter of \Delta AMN [55] & AG_1 ⟂ AB [00] ⇒ ∠MAB = ∠MNA [56]
038. C,D,M are collinear [32] & P,C,D are collinear [15] & ∠MNA = ∠MAB [56] & DM ∥ AB [11] ⇒ ∠MNA = ∠AMP [57]
039. P,A,N are collinear [16] & ∠MAN = ∠MAN [06] ⇒ ∠MAN = ∠MAP [58]
040. ∠MNA = ∠AMP [57] & ∠MAN = ∠MAP [58] (Similar Triangles)⇒ AM:AN = AP:AM [59]
041. G_1C = G_1A [07] & G_1N = G_1A [04] ⇒ G_1 is the circumcenter of \Delta ACN [60]
042. G_1 is the circumcenter of \Delta ACN [60] & AG_1 ⟂ AB [00] ⇒ ∠BAC = ∠ANC [61]
043. C,D,M are collinear [32] & ∠ANC = ∠BAC [61] & DM ∥ AB [11] ⇒ ∠ANC = ∠MCA [62]
044. A,C,N,M are concyclic [20] & ∠ANC = ∠MCA [62] ⇒ AC = MA [63]
045. AM:AN = AP:AM [59] & AC = MA [63] & AM = AE [43] ⇒ PA:EA = EA:AN [64]
046. E,A,C are collinear [14] & P,A,N are collinear [16] & ∠CAN = ∠CAN [09] ⇒ ∠EAN = ∠EAP [65]
047. PA:EA = EA:AN [64] & ∠EAN = ∠EAP [65] (Similar Triangles)⇒ ∠AEN = ∠EPA [66]
048. E,C,A are collinear [14] & C,D,Q are collinear [17] & E,D,B are collinear [13] & ∠ENA = ∠EBA [28] & ∠AEN = ∠EPA [66] & P,A,N are collinear [16] & DM ∥ AB [11] & C,D,M are collinear [32] ⇒ ∠PEC = ∠QDE [67]
049. ∠PCE = ∠QED [54] & ∠PEC = ∠QDE [67] (Similar Triangles)⇒ EP:DQ = EC:ED [68]
050. EQ:DQ = EC:ED [53] & EP:DQ = EC:ED [68] ⇒ EP:DQ = EQ:DQ [69]
051. EP:DQ = EQ:DQ [69] & DQ:ND = DQ:ND [19] ⇒ EQ = EP
==========================
○ 풀이. AlphaGeometry (AlphaGeometry mode)
==========================
* From theorem premises:
A B C D E F G H I J K : Points
AC ⟂ AB [00]
BA ⟂ DB [01]
DE = DB [02]
CE = CA [03]
DF = DB [04]
CF = CA [05]
GE ∥ AB [06]
CG = CA [07]
∠FAG = ∠FAG [08]
HE ∥ AB [09]
DH = DB [10]
∠HBF = ∠HBF [11]
H,B,I are collinear [12]
A,I,G are collinear [13]
H,G,J are collinear [14]
A,F,J are collinear [15]
H,G,K are collinear [16]
B,F,K are collinear [17]
HK:EK = HK:EK [18]
* Auxiliary Constructions:
: Points
* Proof steps:
001. EH ∥ AB [09] & EG ∥ AB [06] ⇒ EG ∥ EH [19]
002. EG ∥ EH [19] ⇒ H,E,G are collinear [20]
003. DF = DB [04] & DH = DB [10] & DE = DB [02] ⇒ H,B,F,E are concyclic [21]
004. H,B,F,E are concyclic [21] ⇒ ∠HBF = ∠HEF [22]
005. CE = CA [03] & CG = CA [07] & CF = CA [05] ⇒ A,G,E,F are concyclic [23]
006. A,G,E,F are concyclic [23] ⇒ ∠GAF = ∠GEF [24]
007. A,I,G are collinear [13] & I,B,H are collinear [12] & ∠HBF = ∠HEF [22] & EH ∥ AB [09] & ∠GAF = ∠GEF [24] & EG ∥ AB [06] ⇒ ∠IAF = ∠IBF [25]
008. ∠IAF = ∠IBF [25] ⇒ A,B,I,F are concyclic [26]
009. A,B,I,F are concyclic [26] ⇒ ∠ABF = ∠AIF [27]
010. A,B,I,F are concyclic [26] ⇒ ∠ABI = ∠AFI [28]
011. DH = DB [10] & DF = DB [04] ⇒ D is the circumcenter of \Delta BHF [29]
012. D is the circumcenter of \Delta BHF [29] & DB ⟂ BA [01] ⇒ ∠HBA = ∠HFB [30]
013. H,E,G are collinear [20] & H,G,K are collinear [16] & ∠HFB = ∠HBA [30] & AB ∥ EG [06] ⇒ ∠HFB = ∠BHK [31]
014. B,F,K are collinear [17] & ∠HBF = ∠HBF [11] ⇒ ∠HBF = ∠HBK [32]
015. ∠HFB = ∠BHK [31] & ∠HBF = ∠HBK [32] (Similar Triangles)⇒ BH:BF = BK:BH [33]
016. CG = CA [07] & CE = CA [03] ⇒ C is the circumcenter of \Delta AGE [34]
017. C is the circumcenter of \Delta AGE [34] & AC ⟂ AB [00] ⇒ ∠BAG = ∠AEG [35]
018. A,I,G are collinear [13] & ∠BAG = ∠AEG [35] & EG ∥ AB [06] ⇒ ∠EAB = ∠BAI [36]
019. DH = DB [10] & DE = DB [02] ⇒ D is the circumcenter of \Delta BHE [37]
020. D is the circumcenter of \Delta BHE [37] & DB ⟂ BA [01] ⇒ ∠ABH = ∠BEH [38]
021. I,B,H are collinear [12] & ∠ABH = ∠BEH [38] & EH ∥ AB [09] ⇒ ∠EBA = ∠ABI [39]
022. ∠EAB = ∠BAI [36] & ∠EBA = ∠ABI [39] (Similar Triangles)⇒ BE = BI [40]
023. ∠EAB = ∠BAI [36] & ∠EBA = ∠ABI [39] (Similar Triangles)⇒ AE = AI [41]
024. H,E,G are collinear [20] & ∠HFB = ∠HBA [30] & AB ∥ EG [06] ⇒ ∠HFB = ∠BHE [42]
025. H,B,F,E are concyclic [21] & ∠HFB = ∠BHE [42] ⇒ HB = BE [43]
026. BH:BF = BK:BH [33] & BE = BI [40] & HB = BE [43] ⇒ BI:BF = BK:BI [44]
027. I,B,H are collinear [12] & B,F,K are collinear [17] & ∠HBF = ∠HBF [11] ⇒ ∠IBF = ∠IBK [45]
028. BI:BF = BK:BI [44] & ∠IBF = ∠IBK [45] (Similar Triangles)⇒ ∠BIF = ∠IKB [46]
029. I,H,B are collinear [12] & H,E,G are collinear [20] & I,G,A are collinear [13] & ∠ABF = ∠AIF [27] & ∠BIF = ∠IKB [46] & B,F,K are collinear [17] & AB ∥ EG [06] ⇒ ∠KIH = ∠HGI [47]
030. H,E,G are collinear [20] & H,G,K are collinear [16] & H,B,I are collinear [12] & AB ∥ EG [06] ⇒ ∠KHI = ∠GHI [48]
031. ∠KIH = ∠HGI [47] & ∠KHI = ∠GHI [48] (Similar Triangles)⇒ IK:HK = IG:HI [49]
032. H,E,G are collinear [20] & H,G,J are collinear [14] & I,G,A are collinear [13] & I,H,B are collinear [12] & ∠ABF = ∠AIF [27] & ∠BIF = ∠IKB [46] & B,F,K are collinear [17] & AB ∥ EG [06] ⇒ ∠JGI = ∠KIH [50]
033. CG = CA [07] & CF = CA [05] ⇒ C is the circumcenter of \Delta AGF [51]
034. C is the circumcenter of \Delta AGF [51] & AC ⟂ AB [00] ⇒ ∠BAG = ∠AFG [52]
035. H,E,G are collinear [20] & H,G,J are collinear [14] & ∠BAG = ∠AFG [52] & AB ∥ EG [06] ⇒ ∠JGA = ∠AFG [53]
036. A,F,J are collinear [15] & ∠FAG = ∠FAG [08] ⇒ ∠JAG = ∠FAG [54]
037. ∠JGA = ∠AFG [53] & ∠JAG = ∠FAG [54] (Similar Triangles)⇒ AJ:AG = AG:AF [55]
038. CE = CA [03] & CF = CA [05] ⇒ C is the circumcenter of \Delta AEF [56]
039. C is the circumcenter of \Delta AEF [56] & AC ⟂ AB [00] ⇒ ∠EAB = ∠EFA [57]
040. ∠EAB = ∠EFA [57] & AB ∥ EG [06] ⇒ ∠AEG = ∠EFA [58]
041. A,G,E,F are concyclic [23] & ∠AEG = ∠EFA [58] ⇒ AG = EA [59]
042. AG:AF = AJ:AG [55] & AG = EA [59] & AE = AI [41] ⇒ AI:AF = AJ:AI [60]
043. A,I,G are collinear [13] & A,F,J are collinear [15] & ∠GAF = ∠GAF [08] ⇒ ∠IAF = ∠IAJ [61]
044. AI:AF = AJ:AI [60] & ∠IAF = ∠IAJ [61] (Similar Triangles)⇒ ∠AIF = ∠IJA [62]
045. I,G,A are collinear [13] & H,E,G are collinear [20] & H,G,K are collinear [16] & H,B,I are collinear [12] & ∠ABI = ∠AFI [28] & ∠AIF = ∠IJA [62] & A,F,J are collinear [15] & AB ∥ EG [06] ⇒ ∠JIG = ∠KHI [63]
046. ∠JGI = ∠KIH [50] & ∠JIG = ∠KHI [63] (Similar Triangles)⇒ IJ:HK = IG:HI [64]
047. IK:HK = IG:HI [49] & IJ:HK = IG:HI [64] ⇒ IJ:HK = IK:HK [65]
048. IJ:HK = IK:HK [65] & HK:EK = HK:EK [18] ⇒ IK = IJ
==========================
IMO 2000, Problem 6. Let AH1, BH2, and CH3 be the altitudes of an acute triangle ABC. The incircle ω of triangle ABC touches the sides BC, CA and AB at T1, T2 and T3, respectively. Consider the reflections of the lines H1H2, H2H3, and H3H1 with respect to the lines T1T2, T2T3, and T3T1. Prove that these images form a triangle whose vertices lie on ω.
○ 풀이. 추후 업데이트
IMO 2002, Problem 2. Let BC be a diameter of circle ω with center O. Let A be a point of circle ω such that 0° < ∠AOB < 120°. Let D be the midpoint of arc AB not containing C. Line ℓ passes through O and is parallel to line AD. Line ℓ intersects line AC at J. The perpendicular bisector of segment OA intersects circle ω at E and F. Prove that J is the incenter of triangle CEF.
○ 풀이. AlphaGeometry (DDAR mode)
==========================
* From theorem premises:
B C O A D E F J : Points
OB = OC [00]
C,O,B are collinear [01]
OA = OB [02]
OD = OB [03]
DA = DB [04]
∠EOA = ∠OAE [05]
∠AEO = ∠AEO [06]
OE = OB [07]
EO = EA [08]
∠FOA = ∠OAF [09]
OF = OB [10]
FO = FA [11]
C,A,J are collinear [12]
JO ∥ AD [13]
* Auxiliary Constructions:
: Points
* Proof steps:
001. ∠EOA = ∠OAE [05] & ∠AEO = ∠AEO [06] (Similar Triangles)⇒ OA:OE = AO:AE [14]
002. OA:OE = AO:AE [14] & OE = OB [07] & OA = OB [02] ⇒ AO = AE [15]
003. ∠FOA = ∠OAF [09] (Similar Triangles)⇒ OF:OA = AF:AO [16]
004. OF = OB [10] & OA = OB [02] ⇒ OA = OF [17]
005. OA = OB [02] & OD = OB [03] ⇒ OD = OA [18]
006. OD = OA [18] (SSS)⇒ ∠ODA = ∠DAO [19]
007. OA = OB [02] & DA = DB [04] ⇒ AB ⟂ OD [20]
008. OA = OB [02] & OB = OC [00] ⇒ O is the circumcenter of \Delta CAB [21]
009. O is the circumcenter of \Delta CAB [21] & C,O,B are collinear [01] ⇒ AC ⟂ AB [22]
010. C,A,J are collinear [12] & ∠ODA = ∠DAO [19] & AB ⟂ OD [20] & AC ⟂ AB [22] & AD ∥ JO [13] ⇒ ∠AOJ = ∠OJA [23]
011. ∠AOJ = ∠OJA [23] ⇒ AO = AJ [24]
012. AO = AE [15] & OF:OA = AF:AO [16] & OA = OF [17] & AO = AJ [24] ⇒ O,E,F,J are concyclic [25]
013. O,E,F,J are concyclic [25] ⇒ ∠FOJ = ∠FEJ [26]
014. FO = FA [11] & OF = OB [10] & OE = OB [07] ⇒ OE = FA [27]
015. AE = AO [15] & OA = OF [17] & OE = FA [27] (SSS)⇒ AE ∥ FO [28]
016. OD = OB [03] & OF = OB [10] & OB = OC [00] & OA = OB [02] & OE = OB [07] ⇒ E,A,F,D are concyclic [29]
017. E,A,F,D are concyclic [29] ⇒ ∠AEF = ∠ADF [30]
018. FO = FA [11] & OF = OB [10] & OE = OB [07] & EO = EA [08] ⇒ AE = AF [31]
019. AE = AF [31] ⇒ ∠AEF = ∠EFA [32]
020. OD = OB [03] & OF = OB [10] & OB = OC [00] & OA = OB [02] & OE = OB [07] ⇒ E,A,F,C are concyclic [33]
021. E,A,F,C are concyclic [33] ⇒ ∠EFA = ∠ECA [34]
022. ∠AEF = ∠ADF [30] & ∠AEF = ∠EFA [32] & ∠EFA = ∠ECA [34] ⇒ ∠ECA = ∠ADF [35]
023. OF = OB [10] & OD = OB [03] ⇒ OD = OF [36]
024. OD = OF [36] ⇒ ∠ODF = ∠DFO [37]
025. ∠ODF = ∠DFO [37] & AB ⟂ OD [20] & AC ⟂ AB [22] ⇒ ∠(CA-DF) = ∠DFO [38]
026. ∠ECA = ∠ADF [35] & ∠(CA-DF) = ∠DFO [38] ⇒ ∠(CE-DF) = ∠(AD-FO) [39]
027. C,A,J are collinear [12] & AD ∥ JO [13] & AB ⟂ OD [20] & AC ⟂ AB [22] ⇒ ∠DOJ = ∠JAD [40]
028. JO ∥ AD [13] ⇒ ∠DJO = ∠JDA [41]
029. ∠DOJ = ∠JAD [40] & ∠DJO = ∠JDA [41] (Similar Triangles)⇒ DO = JA [42]
030. OF = OB [10] & OE = OB [07] & EO = EA [08] ⇒ EA = OF [43]
031. C,A,J are collinear [12] & AB ⟂ OD [20] & AC ⟂ AB [22] & AE ∥ FO [28] ⇒ ∠JAE = ∠DOF [44]
032. DO = JA [42] & EA = OF [43] & ∠JAE = ∠DOF [44] (SAS)⇒ ∠AJE = ∠ODF [45]
033. ∠FOJ = ∠FEJ [26] & AE ∥ FO [28] & JO ∥ AD [13] & ∠(CE-DF) = ∠(AD-FO) [39] & ∠AJE = ∠ODF [45] & C,A,J are collinear [12] & AB ⟂ OD [20] & AC ⟂ AB [22] ⇒ ∠CEJ = ∠JEF
==========================
○ 풀이. AlphaGeometry (AlphaGeometry mode)
==========================
* From theorem premises:
A B C D E F G H : Points
CA = CB [00]
C,B,A are collinear [01]
CD = CA [02]
CE = CA [03]
ED = EA [04]
∠FCD = ∠CDF [05]
∠DFC = ∠DFC [06]
CF = CA [07]
FC = FD [08]
∠GCD = ∠CDG [09]
∠DGC = ∠DGC [10]
CG = CA [11]
GC = GD [12]
H,B,D are collinear [13]
HC ∥ DE [14]
* Auxiliary Constructions:
: Points
* Proof steps:
001. ∠GCD = ∠CDG [09] & ∠DGC = ∠DGC [10] (Similar Triangles)⇒ CD:CG = DC:DG [15]
002. CG = CA [11] & CD = CA [02] ⇒ CD = CG [16]
003. CD:CG = DC:DG [15] & CD = CG [16] ⇒ DC = DG [17]
004. ∠FCD = ∠CDF [05] & ∠DFC = ∠DFC [06] (Similar Triangles)⇒ CD:CF = DC:DF [18]
005. CF = CA [07] & CD = CA [02] ⇒ CD = CF [19]
006. CD:CF = DC:DF [18] & CD = CF [19] ⇒ DC = DF [20]
007. CE = CA [03] & CD = CA [02] ⇒ CD = CE [21]
008. CD = CE [21] (SSS)⇒ ∠CDE = ∠DEC [22]
009. CD = CA [02] & ED = EA [04] ⇒ DA ⟂ CE [23]
010. CD = CA [02] & ED = EA [04] (SSS)⇒ ∠CDE = ∠EAC [24]
011. CD = CA [02] & CA = CB [00] ⇒ C is the circumcenter of \Delta ADB [25]
012. C is the circumcenter of \Delta ADB [25] & C,B,A are collinear [01] ⇒ AD ⟂ BD [26]
013. H,D,B are collinear [13] & ∠CDE = ∠DEC [22] & DA ⟂ CE [23] & AD ⟂ BD [26] & DE ∥ CH [14] ⇒ ∠DCH = ∠CHD [27]
014. ∠DCH = ∠CHD [27] ⇒ DC = DH [28]
015. DC = DG [17] & DC = DF [20] & DC = DH [28] ⇒ C,H,F,G are concyclic [29]
016. C,H,F,G are concyclic [29] ⇒ ∠HCG = ∠HFG [30]
017. CF = CA [07] & CG = CA [11] & GC = GD [12] ⇒ DG = FC [31]
018. DC = DF [20] & DG = FC [31] & CD = CG [16] (SSS)⇒ CG ∥ DF [32]
019. CA = CB [00] & CF = CA [07] & CE = CA [03] ⇒ F,E,A,B are concyclic [33]
020. CD = CA [02] & CG = CA [11] & CA = CB [00] & CE = CA [03] & F,E,A,B are concyclic [33] ⇒ E,F,D,G are concyclic [34]
021. E,F,D,G are concyclic [34] ⇒ ∠GEF = ∠GDF [35]
022. DC = DF [20] & FC = FD [08] & CD = CF [19] (SSS)⇒ ∠CDF = ∠DFC [36]
023. CG = CA [11] & CF = CA [07] & FC = FD [08] ⇒ DF = GC [37]
024. DC = DG [17] & DF = GC [37] & CD = CF [19] (SSS)⇒ CF ∥ DG [38]
025. ∠GEF = ∠GDF [35] & ∠CDF = ∠DFC [36] & CF ∥ DG [38] ⇒ ∠FDC = ∠GEF [39]
026. E,A,F,B are concyclic [33] ⇒ ∠EAB = ∠EFB [40]
027. ∠EAB = ∠EFB [40] & ∠CDE = ∠EAC [24] & C,B,A are collinear [01] & DE ∥ CH [14] ⇒ ∠DCH = ∠EFB [41]
028. ∠FDC = ∠GEF [39] & ∠DCH = ∠EFB [41] ⇒ ∠(DF-CH) = ∠(EG-BF) [42]
029. H,D,B are collinear [13] & DE ∥ CH [14] & DA ⟂ CE [23] & AD ⟂ BD [26] ⇒ ∠HCE = ∠EDH [43]
030. H,D,B are collinear [13] & DA ⟂ CE [23] & AD ⟂ BD [26] ⇒ ∠HEC = ∠EHD [44]
031. ∠HCE = ∠EDH [43] & ∠HEC = ∠EHD [44] (Similar Triangles)⇒ EC = HD [45]
032. H,D,B are collinear [13] & DA ⟂ CE [23] & AD ⟂ BD [26] & DF ∥ CG [32] ⇒ ∠HDF = ∠ECG [46]
033. EC = HD [45] & DF = GC [37] & ∠HDF = ∠ECG [46] (SAS)⇒ ∠DHF = ∠CEG [47]
034. ∠HCG = ∠HFG [30] & CH ∥ DE [14] & CG ∥ DF [32] & ∠(DF-CH) = ∠(EG-BF) [42] & ∠DHF = ∠CEG [47] & H,B,D are collinear [13] & DA ⟂ CE [23] & AD ⟂ BD [26] ⇒ ∠BFH = ∠HFG
==========================
IMO 2003, Problem 4. Let ABCD be a cyclic quadrilateral. Let P, Q and R be the feet of perpendiculars from D to lines BC, CA and AB, respectively. Show that PQ = QR if and only if the bisectors of angles ABC and ADC meet on segment AC.
○ 풀이. AlphaGeometry (DDAR mode)
==========================
* From theorem premises:
A B C O B1 D1 X D P Q R : Points
OB = OC [00]
OA = OB [01]
OB_1 = OA [02]
B_1C = B_1A [03]
OD_1 = OA [04]
D_1C = D_1A [05]
B,X,B_1 are collinear [06]
A,X,C are collinear [07]
OD = OA [08]
D_1,X,D are collinear [09]
P,B,C are collinear [10]
DP ⟂ BC [11]
QC:QD = QC:QD [12]
A,Q,C are collinear [13]
DQ ⟂ AC [14]
A,B,R are collinear [15]
DR ⟂ AB [16]
* Auxiliary Constructions:
: Points
* Proof steps:
001. OD_1 = OA [04] & OD = OA [08] & OB_1 = OA [02] ⇒ A,D_1,D,B_1 are concyclic [17]
002. A,D_1,D,B_1 are concyclic [17] & OB = OC [00] & OA = OB [01] & OD_1 = OA [04] & OB_1 = OA [02] ⇒ A,B,C,D are concyclic [18]
003. A,B,C,D are concyclic [18] ⇒ ∠BAD = ∠BCD [19]
004. A,B,C,D are concyclic [18] ⇒ ∠ABD = ∠ACD [20]
005. A,B,C,D are concyclic [18] ⇒ ∠CAD = ∠CBD [21]
006. A,B,R are collinear [15] & A,Q,C are collinear [13] & DQ ⟂ AC [14] & DR ⟂ AB [16] ⇒ ∠ARD = ∠AQD [22]
007. ∠ARD = ∠AQD [22] ⇒ A,Q,D,R are concyclic [23]
008. A,Q,D,R are concyclic [23] ⇒ ∠ADQ = ∠ARQ [24]
009. OA = OB [01] & OB = OC [00] ⇒ OC = OA [25]
010. OC = OA [25] & D_1C = D_1A [05] ⇒ CA ⟂ OD_1 [26]
011. OC = OA [25] & D_1C = D_1A [05] (SSS)⇒ ∠CD_1O = ∠OD_1A [27]
012. OC = OA [25] & B_1C = B_1A [03] ⇒ CA ⟂ OB_1 [28]
013. OC = OA [25] & B_1C = B_1A [03] (SSS)⇒ ∠CB_1O = ∠OB_1A [29]
014. ∠BAD = ∠BCD [19] & ∠ADQ = ∠ARQ [24] & A,B,R are collinear [15] & CA ⟂ OD_1 [26] & AC ⟂ B_1O [28] & DQ ⟂ AC [14] ⇒ ∠RQD = ∠BCD [30]
015. CA ⟂ OD_1 [26] & AC ⟂ B_1O [28] & A,Q,C are collinear [13] & DQ ⟂ AC [14] & DR ⟂ AB [16] ⇒ ∠(D_1O-AQ) = ∠(DR-AB) [31]
016. A,Q,C are collinear [13] & ∠ACD = ∠ABD [20] ⇒ ∠(AQ-DC) = ∠ABD [32]
017. ∠(D_1O-AQ) = ∠(DR-AB) [31] & ∠(AQ-DC) = ∠ABD [32] ⇒ ∠(D_1O-DC) = ∠RDB [33]
018. ∠(D_1O-DC) = ∠RDB [33] & CA ⟂ OD_1 [26] & AC ⟂ B_1O [28] & DQ ⟂ AC [14] ⇒ ∠QDR = ∠CDB [34]
019. ∠RQD = ∠BCD [30] & ∠QDR = ∠CDB [34] (Similar Triangles)⇒ QD:QR = DC:BC [35]
020. OB = OC [00] & OA = OB [01] & OD_1 = OA [04] & OB_1 = OA [02] ⇒ D_1,B_1,B,C are concyclic [36]
021. OB = OC [00] & OA = OB [01] & OD_1 = OA [04] & OB_1 = OA [02] ⇒ D_1,B_1,A,B are concyclic [37]
022. D_1,B_1,B,C are concyclic [36] ⇒ ∠D_1B_1B = ∠D_1CB [38]
023. D_1,B_1,A,B are concyclic [37] ⇒ ∠D_1B_1B = ∠D_1AB [39]
024. B,X,B_1 are collinear [06] & ∠D_1B_1B = ∠D_1CB [38] & ∠CD_1O = ∠OD_1A [27] & CA ⟂ OD_1 [26] & AC ⟂ B_1O [28] & ∠D_1B_1B = ∠D_1AB [39] ⇒ ∠ABX = ∠XBC [40]
025. ∠ABX = ∠XBC [40] & A,X,C are collinear [07] ⇒ AX:AB = XC:BC [41]
026. A,D_1,D,B_1 are concyclic [17] & OB = OC [00] & OA = OB [01] & OD_1 = OA [04] & OB_1 = OA [02] ⇒ D_1,B_1,C,D are concyclic [42]
027. D_1,B_1,C,D are concyclic [42] ⇒ ∠D_1B_1C = ∠D_1DC [43]
028. D_1,B_1,A,D are concyclic [17] ⇒ ∠D_1B_1A = ∠D_1DA [44]
029. D_1,X,D are collinear [09] & ∠D_1B_1C = ∠D_1DC [43] & ∠CB_1O = ∠OB_1A [29] & CA ⟂ OD_1 [26] & AC ⟂ B_1O [28] & ∠D_1B_1A = ∠D_1DA [44] ⇒ ∠ADX = ∠XDC [45]
030. ∠ADX = ∠XDC [45] & A,X,C are collinear [07] ⇒ AX:AD = XC:DC [46]
031. AX:AB = XC:BC [41] & AX:AD = XC:DC [46] ⇒ DC:BC = AD:AB [47]
032. P,B,C are collinear [10] & A,Q,C are collinear [13] & DQ ⟂ AC [14] & DP ⟂ BC [11] ⇒ ∠CPD = ∠CQD [48]
033. ∠CPD = ∠CQD [48] ⇒ P,Q,D,C are concyclic [49]
034. P,Q,D,C are concyclic [49] ⇒ ∠PQC = ∠PDC [50]
035. ∠PQC = ∠PDC [50] & A,Q,C are collinear [13] & ∠ABD = ∠ACD [20] ⇒ ∠ABD = ∠QPD [51]
036. CA ⟂ OD_1 [26] & AC ⟂ B_1O [28] & A,Q,C are collinear [13] & P,B,C are collinear [10] & DQ ⟂ AC [14] & DP ⟂ BC [11] ⇒ ∠(D_1O-AQ) = ∠DPB [52]
037. A,Q,C are collinear [13] & P,B,C are collinear [10] & ∠CAD = ∠CBD [21] ⇒ ∠QAD = ∠PBD [53]
038. ∠(D_1O-AQ) = ∠DPB [52] & ∠QAD = ∠PBD [53] ⇒ ∠(D_1O-AD) = ∠PDB [54]
039. ∠(D_1O-AD) = ∠PDB [54] & CA ⟂ OD_1 [26] & AC ⟂ B_1O [28] & DQ ⟂ AC [14] ⇒ ∠QDP = ∠ADB [55]
040. ∠ABD = ∠QPD [51] & ∠QDP = ∠ADB [55] (Similar Triangles)⇒ QD:PQ = AD:AB [56]
041. QD:QR = DC:BC [35] & DC:BC = AD:AB [47] & QD:PQ = AD:AB [56] ⇒ QD:PQ = QD:QR [57]
042. QC:QD = QC:QD [12] & QD:PQ = QD:QR [57] ⇒ PQ = QR
==========================
○ 풀이. AlphaGeometry (AlphaGeometry mode)
==========================
* From theorem premises:
A B C D E F G H I J K : Points
DB = DC [00]
DA = DB [01]
DE = DA [02]
∠ECA = ∠CAE [03]
DF = DA [04]
∠FCA = ∠CAF [05]
E,G,B are collinear [06]
C,A,G are collinear [07]
DH = DA [08]
H,F,G are collinear [09]
I,C,B are collinear [10]
HI ⟂ BC [11]
CH:JH = CH:JH [12]
C,J,A are collinear [13]
HJ ⟂ AC [14]
K,A,B are collinear [15]
HK ⟂ AB [16]
* Auxiliary Constructions:
: Points
* Proof steps:
001. DB = DC [00] & DA = DB [01] & DH = DA [08] ⇒ H,C,A,B are concyclic [17]
002. H,C,A,B are concyclic [17] ⇒ ∠CBH = ∠CAH [18]
003. H,C,A,B are concyclic [17] ⇒ ∠CAB = ∠CHB [19]
004. H,C,A,B are concyclic [17] ⇒ ∠ABH = ∠ACH [20]
005. H,C,A,B are concyclic [17] ⇒ ∠ACB = ∠AHB [21]
006. C,J,A are collinear [13] & K,A,B are collinear [15] & HK ⟂ AB [16] & HJ ⟂ AC [14] ⇒ ∠AJH = ∠AKH [22]
007. ∠AJH = ∠AKH [22] ⇒ K,J,A,H are concyclic [23]
008. K,J,A,H are concyclic [23] ⇒ ∠KJA = ∠KHA [24]
009. ∠CBH = ∠CAH [18] & ∠KJA = ∠KHA [24] & C,J,A are collinear [13] ⇒ ∠JKH = ∠CBH [25]
010. C,J,A are collinear [13] & AC ⟂ HJ [14] ⇒ CJ ⟂ JH [26]
011. K,A,B are collinear [15] & HK ⟂ AB [16] ⇒ KH ⟂ KA [27]
012. CJ ⟂ JH [26] & KH ⟂ KA [27] ⇒ ∠(CJ-KA) = ∠JHK [28]
013. ∠(CJ-KA) = ∠JHK [28] & C,J,A are collinear [13] & K,A,B are collinear [15] & ∠CAB = ∠CHB [19] ⇒ ∠CHB = ∠JHK [29]
014. ∠JKH = ∠CBH [25] & ∠CHB = ∠JHK [29] (Similar Triangles)⇒ JH:KJ = CH:CB [30]
015. DB = DC [00] & DA = DB [01] & DE = DA [02] ⇒ C,E,A,B are concyclic [31]
016. C,E,A,B are concyclic [31] ⇒ ∠EBA = ∠ECA [32]
017. C,E,A,B are concyclic [31] ⇒ ∠CBE = ∠CAE [33]
018. E,G,B are collinear [06] & ∠EBA = ∠ECA [32] & ∠ECA = ∠CAE [03] & ∠CBE = ∠CAE [33] ⇒ ∠CBG = ∠GBA [34]
019. ∠CBG = ∠GBA [34] & C,A,G are collinear [07] ⇒ CG:CB = AG:AB [35]
020. C,B,A,H are concyclic [17] & DB = DC [00] & DA = DB [01] & DF = DA [04] ⇒ C,A,H,F are concyclic [36]
021. C,A,H,F are concyclic [36] ⇒ ∠CAH = ∠CFH [37]
022. C,A,H,F are concyclic [36] ⇒ ∠CAF = ∠CHF [38]
023. F,G,H are collinear [09] & ∠CAH = ∠CFH [37] & ∠FCA = ∠CAF [05] & ∠CAF = ∠CHF [38] ⇒ ∠CHG = ∠GHA [39]
024. ∠CHG = ∠GHA [39] & C,A,G are collinear [07] ⇒ CG:CH = AG:AH [40]
025. CG:CB = AG:AB [35] & CG:CH = AG:AH [40] ⇒ AH:AB = CH:CB [41]
026. I,C,B are collinear [10] & C,J,A are collinear [13] & HJ ⟂ AC [14] & HI ⟂ BC [11] ⇒ ∠HIC = ∠HJC [42]
027. ∠HIC = ∠HJC [42] ⇒ I,C,J,H are concyclic [43]
028. I,C,J,H are concyclic [43] ⇒ ∠IJC = ∠IHC [44]
029. ∠ABH = ∠ACH [20] & ∠IJC = ∠IHC [44] & C,J,A are collinear [13] ⇒ ∠JIH = ∠ABH [45]
030. I,C,B are collinear [10] & BC ⟂ HI [11] ⇒ IC ⟂ IH [46]
031. IC ⟂ IH [46] & CJ ⟂ JH [26] ⇒ ∠ICJ = ∠IHJ [47]
032. ∠ICJ = ∠IHJ [47] & I,C,B are collinear [10] & C,J,A are collinear [13] & ∠ACB = ∠AHB [21] ⇒ ∠AHB = ∠JHI [48]
033. ∠JIH = ∠ABH [45] & ∠AHB = ∠JHI [48] (Similar Triangles)⇒ JH:IJ = AH:AB [49]
034. JH:KJ = CH:CB [30] & AH:AB = CH:CB [41] & JH:IJ = AH:AB [49] ⇒ JH:IJ = JH:KJ [50]
035. CH:JH = CH:JH [12] & JH:IJ = JH:KJ [50] ⇒ IJ = KJ
==========================
IMO 2004, Problem 1. Let ABC be an acute-angled triangle with AB ≠ AC. The circle with diameter BC intersects the sides AB and AC at M and N respectively. Denote by O the midpoint of the side BC. The bisectors of the angles ∠BAC and ∠MON intersect at R. Prove that the circumcircles of the triangles BMR and CNR have a common point lying on the side BC.
○ 풀이. 추후 업데이트
IMO 2004, Problem 5. In a convex quadrilateral ABCD, the diagonal BD bisects neither the angle ABC nor the angle CDA. The point P lies inside ABCD and satisfies
∠PBC = ∠DBA and ∠PDC = ∠BDA.
Prove that ABCD is a cyclic quadrilateral if and only if AP = CP.
○ 풀이. AlphaGeometry (DDAR mode)
==========================
* From theorem premises:
A B C O D P : Points
OA = OB [00]
OB = OC [01]
OD = OA [02]
∠PBC = ∠ABD [03]
∠PDC = ∠ADB [04]
* Auxiliary Constructions:
: Points
* Proof steps:
001. OA = OB [00] & OB = OC [01] ⇒ OC = OA [05]
002. OB = OC [01] & OA = OB [00] & OD = OA [02] ⇒ B,C,D,A are concyclic [06]
003. OB = OC [01] & OA = OB [00] & OD = OA [02] ⇒ OC = OD [07]
004. B,C,D,A are concyclic [06] ⇒ ∠BDC = ∠BAC [08]
005. B,C,D,A are concyclic [06] ⇒ ∠BCD = ∠BAD [09]
006. OC = OA [05] ⇒ ∠OCA = ∠CAO [10]
007. OB = OC [01] ⇒ ∠OCB = ∠CBO [11]
008. OC = OD [07] ⇒ ∠OCD = ∠CDO [12]
009. OA = OB [00] ⇒ ∠OAB = ∠ABO [13]
010. ∠PBC = ∠ABD [03] & ∠PDC = ∠ADB [04] & ∠BCD = ∠BAD [09] & ∠OCB = ∠CBO [11] & ∠OCD = ∠CDO [12] (Angle chase)⇒ ∠BPD = ∠BOD [14]
011. ∠BPD = ∠BOD [14] ⇒ B,O,P,D are concyclic [15]
012. B,O,P,D are concyclic [15] ⇒ ∠BPO = ∠BDO [16]
013. ∠PBC = ∠ABD [03] & ∠BDC = ∠BAC [08] & ∠OCA = ∠CAO [10] & ∠OCB = ∠CBO [11] & ∠OCD = ∠CDO [12] & ∠OAB = ∠ABO [13] & ∠BPO = ∠BDO [16] (Angle chase)⇒ ∠POA = ∠COP [17]
014. OC = OA [05] & ∠POA = ∠COP [17] (SAS)⇒ PC = PA
==========================
○ 풀이. AlphaGeometry (AlphaGeometry mode)
==========================
* From theorem premises:
A B C D E F : Points
DA = DB [00]
DB = DC [01]
DE = DA [02]
∠FEC = ∠AEB [03]
∠FBC = ∠ABE [04]
* Auxiliary Constructions:
: Points
* Proof steps:
001. DA = DB [00] & DB = DC [01] ⇒ DC = DA [05]
002. DA = DB [00] & DE = DA [02] & DB = DC [01] ⇒ B,A,C,E are concyclic [06]
003. B,A,C,E are concyclic [06] ⇒ ∠BCA = ∠BEA [07]
004. B,A,C,E are concyclic [06] ⇒ ∠BAC = ∠BEC [08]
005. DB = DC [01] ⇒ ∠DBC = ∠BCD [09]
006. DA = DB [00] ⇒ ∠DBA = ∠BAD [10]
007. DC = DA [05] ⇒ ∠DCA = ∠CAD [11]
008. DE = DA [02] & DA = DB [00] ⇒ DB = DE [12]
009. DB = DE [12] ⇒ ∠DBE = ∠BED [13]
010. ∠FBC = ∠ABE [04] & ∠FEC = ∠AEB [03] & ∠BCA = ∠BEA [07] & ∠BAC = ∠BEC [08] & ∠DBC = ∠BCD [09] & ∠DBA = ∠BAD [10] & ∠DBE = ∠BED [13] & ∠DCA = ∠CAD [11] (Angle chase)⇒ ∠BFE = ∠BDE [14]
011. ∠BFE = ∠BDE [14] ⇒ E,B,D,F are concyclic [15]
012. E,B,D,F are concyclic [15] ⇒ ∠EBD = ∠EFD [16]
013. ∠FEC = ∠AEB [03] & ∠BCA = ∠BEA [07] & ∠BAC = ∠BEC [08] & ∠DBC = ∠BCD [09] & ∠DBA = ∠BAD [10] & ∠DCA = ∠CAD [11] & ∠EBD = ∠EFD [16] (Angle chase)⇒ ∠CDF = ∠FDA [17]
014. DC = DA [05] & ∠CDF = ∠FDA [17] (SAS)⇒ FC = FA
==========================
입력: 2024.05.05 18:42
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