【국제수학올림피아드】 IMO 기하 문제 풀이 (2005년 ~ 2009년)

IMO 기하 문제 풀이 (2005년 ~ 2009년)

 

추천글 : 【기하학】 IMO 기하 문제 풀이 종합 

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IMO 2005, Problem 1. Six points are chosen on the sides of an equilateral triangle ABC : A₁, A₂ on BC, B₁, B₂ on CA and C₁, C₂ on AB, such that they are the vertices of a convex hexagon A₁A₂B₁B₂C₁C2 with equal side lengths. Prove that the lines A₁B₂, B₁C₂ and C₁A₂ are concurrent.

 

○ 풀이. 추후 업데이트

 

 

IMO 2005, Problem 5. Let ABCD be a fixed convex quadrilateral with BC = DA and BC ∦ DA. Let two variable points E and F lie on the sides BC and DA, respectively, and satisfy BE = DF. The lines AC and BD meet at P, the lines BD and EF meet at Q, the lines EF and AC meet at R. Prove that the circumcircles of the triangles PQR, as E and F vary, have a common point other than P.

 

○ 풀이. AlphaGeometry (DDAR mode)

 

==========================
 * From theorem premises:
A B C D E F P Q R O1 O2 M : Points
DA = BC [00]
C,E,B are collinear [01]
FD = EB [02]
F,D,A are collinear [03]
P,C,A are collinear [04]
D,P,B are collinear [05]
F,Q,E are collinear [06]
D,Q,B are collinear [07]
R,C,A are collinear [08]
R,F,E are collinear [09]
O_1A = O_1P [10]
O_1P = O_1D [11]
O_2B = O_2P [12]
O_2P = O_2C [13]
O_2M = O_2P [14]
O_1M = O_1P [15]

 * Auxiliary Constructions:
: Points


 * Proof steps:
001. O_2B = O_2P [12] & O_2M = O_2P [14] & O_2P = O_2C [13] ⇒  C,M,P,B are concyclic [16]
002. C,M,P,B are concyclic [16] ⇒  ∠CMB = ∠CPB [17]
003. C,M,P,B are concyclic [16] ⇒  ∠CMP = ∠CBP [18]
004. C,M,P,B are concyclic [16] ⇒  ∠CPM = ∠CBM [19]
005. O_1A = O_1P [10] & O_1M = O_1P [15] & O_1P = O_1D [11] ⇒  P,M,D,A are concyclic [20]
006. P,M,D,A are concyclic [20] ⇒  ∠PDM = ∠PAM [21]
007. P,M,D,A are concyclic [20] ⇒  ∠PMA = ∠PDA [22]
008. P,M,D,A are concyclic [20] ⇒  ∠PMD = ∠PAD [23]
009. ∠CMB = ∠CPB [17] & P,C,A are collinear [04] & D,P,B are collinear [05] & ∠PDM = ∠PAM [21] ⇒  ∠DMA = ∠BMC [24]
010. ∠CMP = ∠CBP [18] & D,P,B are collinear [05] & ∠PMA = ∠PDA [22] ⇒  ∠DAM = ∠BCM [25]
011. ∠DMA = ∠BMC [24] & ∠DAM = ∠BCM [25] (Similar Triangles)⇒  DA:BC = DM:BM [26]
012. DA:BC = DM:BM [26] & DA = BC [00] ⇒  MB = MD [27]
013. F,A,D are collinear [03] & C,P,A are collinear [04] & ∠PMD = ∠PAD [23] ⇒  ∠(FD-PC) = ∠DMP [28]
014. C,P,A are collinear [04] & C,E,B are collinear [01] & ∠CPM = ∠CBM [19] ⇒  ∠PCE = ∠PMB [29]
015. ∠(FD-PC) = ∠DMP [28] & ∠PCE = ∠PMB [29] ⇒  ∠(FD-CE) = ∠DMB [30]
016. F,D,A are collinear [03] & C,E,B are collinear [01] & ∠(FD-CE) = ∠DMB [30] ⇒  ∠FDM = ∠EBM [31]
017. FD = EB [02] & DM = BM [27] & ∠FDM = ∠EBM [31] (SAS)⇒  ME = MF [32]
018. FD = EB [02] & DM = BM [27] & ∠FDM = ∠EBM [31] (SAS)⇒  ∠FME = ∠DMB [33]
019. MB = MD [27] & ME = MF [32] ⇒  MB:MD = ME:MF [34]
020. MB:MD = ME:MF [34] & ∠FME = ∠DMB [33] (Similar Triangles)⇒  ∠BDM = ∠EFM [35]
021. F,Q,E are collinear [06] & D,Q,B are collinear [07] & ∠EFM = ∠BDM [35] ⇒  ∠QFM = ∠QDM [36]
022. ∠QFM = ∠QDM [36] ⇒  F,D,M,Q are concyclic [37]
023. F,D,M,Q are concyclic [37] ⇒  ∠FDM = ∠FQM [38]
024. R,C,A are collinear [08] & P,C,A are collinear [04] & R,F,E are collinear [09] & F,Q,E are collinear [06] & ∠FDM = ∠FQM [38] & F,D,A are collinear [03] & ∠PMD = ∠PAD [23] ⇒  ∠RPM = ∠RQM [39]
025. ∠RPM = ∠RQM [39] ⇒  R,Q,M,P are concyclic
==========================

 

○ 풀이. AlphaGeometry (AlphaGeometry mode)

 

==========================
 * From theorem premises:
A B C D E F G H I J K L : Points
DA = BC [00]
C,B,E are collinear [01]
FD = EB [02]
D,A,F are collinear [03]
D,G,B are collinear [04]
C,A,G are collinear [05]
D,B,H are collinear [06]
F,E,H are collinear [07]
I,F,E are collinear [08]
C,A,I are collinear [09]
JG = JD [10]
JA = JG [11]
KG = KC [12]
KB = KG [13]
JL = JG [14]
KL = KG [15]

 * Auxiliary Constructions:
: Points


 * Proof steps:
001. JG = JD [10] & JL = JG [14] & JA = JG [11] ⇒  D,A,L,G are concyclic [16]
002. D,A,L,G are concyclic [16] ⇒  ∠DLA = ∠DGA [17]
003. D,A,L,G are concyclic [16] ⇒  ∠DAL = ∠DGL [18]
004. D,A,L,G are concyclic [16] ⇒  ∠DAG = ∠DLG [19]
005. KG = KC [12] & KL = KG [15] & KB = KG [13] ⇒  C,L,B,G are concyclic [20]
006. C,L,B,G are concyclic [20] ⇒  ∠CLB = ∠CGB [21]
007. C,L,B,G are concyclic [20] ⇒  ∠CLG = ∠CBG [22]
008. C,L,B,G are concyclic [20] ⇒  ∠CBL = ∠CGL [23]
009. ∠DLA = ∠DGA [17] & D,G,B are collinear [04] & C,A,G are collinear [05] & ∠CLB = ∠CGB [21] ⇒  ∠BLC = ∠DLA [24]
010. ∠DAL = ∠DGL [18] & D,G,B are collinear [04] & ∠CLG = ∠CBG [22] ⇒  ∠LCB = ∠LAD [25]
011. ∠BLC = ∠DLA [24] & ∠LCB = ∠LAD [25] (Similar Triangles)⇒  CB:AD = CL:AL [26]
012. ∠BLC = ∠DLA [24] & ∠LCB = ∠LAD [25] (Similar Triangles)⇒  BC:DA = BL:DL [27]
013. BC:DA = BL:DL [27] & DA = BC [00] ⇒  BL = DL [28]
014. ∠DAG = ∠DLG [19] & C,A,G are collinear [05] ⇒  ∠DAC = ∠DLG [29]
015. ∠CBL = ∠CGL [23] & C,A,G are collinear [05] ⇒  ∠BCA = ∠BLG [30]
016. ∠DAC = ∠DLG [29] & ∠BCA = ∠BLG [30] ⇒  ∠ADL = ∠CBL [31]
017. C,B,E are collinear [01] & D,A,F are collinear [03] & ∠CBL = ∠ADL [31] ⇒  ∠EBL = ∠FDL [32]
018. FD = EB [02] & BL = DL [28] & ∠EBL = ∠FDL [32] (SAS)⇒  LE = LF [33]
019. FD = EB [02] & BL = DL [28] & ∠EBL = ∠FDL [32] (SAS)⇒  ∠ELF = ∠BLD [34]
020. CB:AD = CL:AL [26] & DA = BC [00] & LE = LF [33] ⇒  LC:LA = LE:LF [35]
021. D,G,B are collinear [04] & ∠DLA = ∠DGA [17] & C,A,G are collinear [05] ⇒  ∠LDG = ∠LAC [36]
022. D,G,B are collinear [04] & ∠CLB = ∠CGB [21] & C,A,G are collinear [05] ⇒  ∠(DG-LB) = ∠ACL [37]
023. ∠LDG = ∠LAC [36] & ∠(DG-LB) = ∠ACL [37] ⇒  ∠BLD = ∠CLA [38]
024. ∠ELF = ∠BLD [34] & ∠BLD = ∠CLA [38] ⇒  ∠CLA = ∠ELF [39]
025. LC:LA = LE:LF [35] & ∠CLA = ∠ELF [39] (Similar Triangles)⇒  ∠LAC = ∠LFE [40]
026. I,A,C are collinear [09] & I,F,E are collinear [08] & ∠LAC = ∠LFE [40] ⇒  ∠LAI = ∠LFI [41]
027. ∠LAI = ∠LFI [41] ⇒  I,A,L,F are concyclic [42]
028. I,A,L,F are concyclic [42] ⇒  ∠ILA = ∠IFA [43]
029. D,B,H are collinear [06] & D,G,B are collinear [04] & I,F,E are collinear [08] & F,E,H are collinear [07] & ∠ILA = ∠IFA [43] & D,A,F are collinear [03] & ∠DAL = ∠DGL [18] ⇒  ∠HGL = ∠HIL [44]
030. ∠HGL = ∠HIL [44] ⇒  I,G,L,H are concyclic
==========================

 

 

IMO 2007, Problem 4. In triangle ABC the bisector of ∠BCA meets the circumcircle again at R, the perpendicular bisector of BC at P, and the perpendicular bisector of AC at Q. The midpoint of BC is K and the midpoint of AC is L. Prove that the triangles RPK and RQL have the same area.

 

○ 풀이. AlphaGeometry (DDAR mode)

 

==========================
 * From theorem premises:
A B C O R L K P Q L1 K1 : Points
OB = OC [00]
OA = OB [01]
OR = OA [02]
∠BAR = ∠RBA [03]
RA = RB [04]
A,L,C are collinear [05]
LC = LA [06]
B,K,C are collinear [07]
KC = KB [08]
P,K,O are collinear [09]
P,C,R are collinear [10]
L,Q,O are collinear [11]
Q,C,R are collinear [12]
L_1,C,R are collinear [13]
L_1L ⟂ CR [14]
K_1,C,R are collinear [15]
CR ⟂ K_1K [16]

 * Auxiliary Constructions:
: Points


 * Proof steps:
001. OB = OC [00] & KC = KB [08] ⇒  BC ⟂ KO [17]
002. OA = OB [01] & OB = OC [00] ⇒  OC = OA [18]
003. OC = OA [18] & LC = LA [06] ⇒  AC ⟂ LO [19]
004. L,Q,O are collinear [11] & L,A,C are collinear [05] & B,K,C are collinear [07] & P,K,O are collinear [09] & BC ⟂ KO [17] & AC ⟂ LO [19] ⇒  ∠QLC = ∠CKP [20]
005. OA = OB [01] & OR = OA [02] & OB = OC [00] ⇒  A,B,C,R are concyclic [21]
006. OA = OB [01] & OR = OA [02] & OB = OC [00] ⇒  O is the circumcenter of \Delta RCA [22]
007. OA = OB [01] & OR = OA [02] & OB = OC [00] ⇒  O is the circumcenter of \Delta RCB [23]
008. OA = OB [01] & OR = OA [02] & OB = OC [00] ⇒  OR = OC [24]
009. A,B,C,R are concyclic [21] ⇒  ∠ABC = ∠ARC [25]
010. A,B,C,R are concyclic [21] ⇒  ∠ABR = ∠ACR [26]
011. K_1,C,R are collinear [15] & L_1,C,R are collinear [13] & P,C,R are collinear [10] & Q,C,R are collinear [12] & L,A,C are collinear [05] & B,K,C are collinear [07] & ∠ABC = ∠ARC [25] & ∠BAR = ∠RBA [03] & ∠ABR = ∠ACR [26] ⇒  ∠QCL = ∠KCP [27]
012. ∠QLC = ∠CKP [20] & ∠QCL = ∠KCP [27] (Similar Triangles)⇒  PK:LQ = PC:QC [28]
013. K_1,C,R are collinear [15] & L_1,C,R are collinear [13] & P,C,R are collinear [10] & Q,C,R are collinear [12] & L_1L ⟂ CR [14] & CR ⟂ K_1K [16] ⇒  ∠KK_1P = ∠QL_1L [29]
014. A,L,C are collinear [05] & LC = LA [06] ⇒  L is midpoint of CA [30]
015. O is the circumcenter of \Delta RCA [22] & L is midpoint of CA [30] ⇒  ∠ARC = ∠AOL [31]
016. OA = OB [01] & RA = RB [04] (SSS)⇒  ∠RAO = ∠OBR [32]
017. B,K,C are collinear [07] & KC = KB [08] ⇒  K is midpoint of CB [33]
018. O is the circumcenter of \Delta RCB [23] & K is midpoint of CB [33] ⇒  ∠BRC = ∠BOK [34]
019. P,K,O are collinear [09] & K_1,C,R are collinear [15] & L_1,C,R are collinear [13] & P,C,R are collinear [10] & Q,C,R are collinear [12] & L,Q,O are collinear [11] & ∠ARC = ∠AOL [31] & ∠RAO = ∠OBR [32] & ∠BRC = ∠BOK [34] ⇒  ∠KPK_1 = ∠L_1QL [35]
020. ∠KK_1P = ∠QL_1L [29] & ∠KPK_1 = ∠L_1QL [35] (Similar Triangles)⇒  K_1K:LL_1 = PK:LQ [36]
021. OR = OC [24] ⇒  ∠ORC = ∠RCO [37]
022. K_1,C,R are collinear [15] & L_1,C,R are collinear [13] & P,C,R are collinear [10] & Q,C,R are collinear [12] & ∠ORC = ∠RCO [37] ⇒  ∠ORQ = ∠PCO [38]
023. L,Q,O are collinear [11] & K_1,C,R are collinear [15] & L_1,C,R are collinear [13] & P,C,R are collinear [10] & Q,C,R are collinear [12] & P,K,O are collinear [09] & ∠ARC = ∠AOL [31] & ∠RAO = ∠OBR [32] & ∠BRC = ∠BOK [34] ⇒  ∠OQR = ∠CPO [39]
024. ∠ORQ = ∠PCO [38] & ∠OQR = ∠CPO [39] (Similar Triangles)⇒  RO:CO = RQ:CP [40]
025. K_1,C,R are collinear [15] & L_1,C,R are collinear [13] & P,C,R are collinear [10] & Q,C,R are collinear [12] & ∠ORC = ∠RCO [37] ⇒  ∠ORP = ∠QCO [41]
026. P,K,O are collinear [09] & K_1,C,R are collinear [15] & L_1,C,R are collinear [13] & P,C,R are collinear [10] & Q,C,R are collinear [12] & L,Q,O are collinear [11] & ∠ARC = ∠AOL [31] & ∠RAO = ∠OBR [32] & ∠BRC = ∠BOK [34] ⇒  ∠OPR = ∠CQO [42]
027. ∠ORP = ∠QCO [41] & ∠OPR = ∠CQO [42] (Similar Triangles)⇒  RO:CO = RP:CQ [43]
028. PK:LQ = PC:QC [28] & K_1K:LL_1 = PK:LQ [36] & RO:CO = RQ:CP [40] & OR = OC [24] & RO:CO = RP:CQ [43] ⇒  KK_1:LL_1 = RQ:RP
==========================

 

○ 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]
∠BAE = ∠EBA [03]
EA = EB [04]
C,F,A are collinear [05]
FC = FA [06]
G,C,B are collinear [07]
GC = GB [08]
G,H,D are collinear [09]
E,C,H are collinear [10]
I,F,D are collinear [11]
E,C,I are collinear [12]
J,C,E are collinear [13]
FJ ⟂ CE [14]
E,C,K are collinear [15]
CE ⟂ KG [16]

 * Auxiliary Constructions:
: Points


 * Proof steps:
001. DB = DC [00] & DA = DB [01] ⇒  DA = DC [17]
002. DA = DC [17] & FC = FA [06] ⇒  AC ⟂ DF [18]
003. DB = DC [00] & GC = GB [08] ⇒  BC ⟂ DG [19]
004. G,C,B are collinear [07] & G,H,D are collinear [09] & I,F,D are collinear [11] & C,F,A are collinear [05] & AC ⟂ DF [18] & BC ⟂ DG [19] ⇒  ∠CGH = ∠IFC [20]
005. DB = DC [00] & DA = DB [01] & DE = DA [02] ⇒  E,C,A,B are concyclic [21]
006. DB = DC [00] & DA = DB [01] & DE = DA [02] ⇒  D is the circumcenter of \Delta EAC [22]
007. DB = DC [00] & DA = DB [01] & DE = DA [02] ⇒  D is the circumcenter of \Delta ECB [23]
008. DB = DC [00] & DA = DB [01] & DE = DA [02] ⇒  DC = DE [24]
009. E,C,A,B are concyclic [21] ⇒  ∠BCE = ∠BAE [25]
010. E,C,A,B are concyclic [21] ⇒  ∠ECA = ∠EBA [26]
011. E,C,I are collinear [12] & C,F,A are collinear [05] & G,C,B are collinear [07] & E,C,H are collinear [10] & ∠BCE = ∠BAE [25] & ∠BAE = ∠EBA [03] & ∠ECA = ∠EBA [26] ⇒  ∠ICF = ∠GCH [27]
012. ∠CGH = ∠IFC [20] & ∠ICF = ∠GCH [27] (Similar Triangles)⇒  IF:HG = IC:HC [28]
013. C,F,A are collinear [05] & FC = FA [06] ⇒  F is midpoint of CA [29]
014. I,F,D are collinear [11] & DF ⟂ AC [18] ⇒  IF ⟂ CA [30]
015. F is midpoint of CA [29] & IF ⟂ CA [30] ⇒  IC = IA [31]
016. G,C,B are collinear [07] & GC = GB [08] ⇒  G is midpoint of CB [32]
017. G,H,D are collinear [09] & DG ⟂ BC [19] ⇒  HG ⟂ CB [33]
018. G is midpoint of CB [32] & HG ⟂ CB [33] ⇒  HC = HB [34]
019. E,C,K are collinear [15] & E,C,H are collinear [10] & J,C,E are collinear [13] & E,C,I are collinear [12] & FJ ⟂ CE [14] & CE ⟂ KG [16] ⇒  ∠GKH = ∠IJF [35]
020. D is the circumcenter of \Delta EAC [22] & F is midpoint of CA [29] ⇒  ∠EAD = ∠(CE-DF) [36]
021. DA = DB [01] & EA = EB [04] (SSS)⇒  ∠EAD = ∠DBE [37]
022. D is the circumcenter of \Delta ECB [23] & G is midpoint of CB [32] ⇒  ∠DBE = ∠(DG-CE) [38]
023. G,H,D are collinear [09] & E,C,K are collinear [15] & E,C,H are collinear [10] & J,C,E are collinear [13] & E,C,I are collinear [12] & I,F,D are collinear [11] & ∠EAD = ∠(CE-DF) [36] & ∠EAD = ∠DBE [37] & ∠DBE = ∠(DG-CE) [38] ⇒  ∠GHK = ∠JIF [39]
024. ∠GKH = ∠IJF [35] & ∠GHK = ∠JIF [39] (Similar Triangles)⇒  GK:JF = GH:IF [40]
025. I,F,D are collinear [11] & E,I,C are collinear [12] & E,C,H are collinear [10] & G,H,D are collinear [09] & ∠EAD = ∠(CE-DF) [36] & ∠EAD = ∠DBE [37] & ∠DBE = ∠(DG-CE) [38] ⇒  ∠DIE = ∠CHD [41]
026. DC = DE [24] ⇒  ∠ECD = ∠DEC [42]
027. E,I,C are collinear [12] & E,C,H are collinear [10] & ∠DEC = ∠ECD [42] ⇒  ∠DEI = ∠HCD [43]
028. ∠DIE = ∠CHD [41] & ∠DEI = ∠HCD [43] (Similar Triangles)⇒  ED:CD = EI:CH [44]
029. I,F,D are collinear [11] & E,I,C are collinear [12] & E,C,H are collinear [10] & G,H,D are collinear [09] & ∠EAD = ∠(CE-DF) [36] & ∠EAD = ∠DBE [37] & ∠DBE = ∠(DG-CE) [38] ⇒  ∠DIC = ∠EHD [45]
030. E,C,I are collinear [12] & E,C,H are collinear [10] & ∠DCE = ∠CED [42] ⇒  ∠DCI = ∠HED [46]
031. ∠DIC = ∠EHD [45] & ∠DCI = ∠HED [46] (Similar Triangles)⇒  CD:ED = CI:EH [47]
032. IF:HG = IC:HC [28] & IC = IA [31] & HC = HB [34] & GK:JF = GH:IF [40] & ED:CD = EI:CH [44] & DC = DE [24] & CD:ED = CI:EH [47] ⇒  GK:FJ = EI:EH
==========================

 

 

IMO 2008, Problem 1. Let H be the orthocenter of an acute-angled triangle ABC. The circle Γ_A centered at the midpoint of BC and passing through H intersects the sideline BC at points A₁ and A₂. Similarly, define the points B₁, B₂, C₁, and C₂. Prove that six points A₁, A₂, B₁, B₂, C₁, C₂ are concyclic.

 

○ 풀이. 추후 업데이트

 

 

IMO 2008, Problem 6. Let ABCD be a convex quadrilateral with BA ≠ BC. Denote the incircles of triangles ABC and ADC by ω₁ and ω₂ respectively. Suppose that there exists a circle ω tangent to ray BA beyond A and to the ray BC beyond C, which is also tangent to the lines AD and CD. Prove that the common external tangents to ω₁ and ω₂ intersect on ω.

 

○ 풀이. 추후 업데이트 

 

 

IMO 2009, Problem 2. Let ABC be a triangle with circumcenter O. The points P and Q are interior points of the sides CA and AB respectively. Let K, L, M be the midpoints of BP, CQ, PQ, respectively, and let Γ be the circumcircle of △KLM. Suppose that PQ is tangent to Γ. Prove that OP = OQ.

 

○ 풀이. 추후 업데이트

 

입력: 2024.05.05 19:16