IMO 2024 P4
Let ABC
be a triangle with AB < AC < BC
. Let the incentre and incircle of triangle
ABC
be I
and \omega
, respectively. Let X
be the point on line BC
different from C
such that the line
through X
parallel to AC
is tangent to \omega
. Similarly, let Y
be the point on line BC
different from
B
such that the line through Y
parallel to AB
is tangent to \omega
. Let AI
intersect the circumcircle of
triangle ABC
again at P \ne A
. Let K
and L
be the midpoints of AC
and AB
, respectively.
Prove that \angle KIL + \angle Y P X = 180 \degree
.
AlphaGeometry Declaration Statement
General comment:
-
We will consider the orientation of the problem as in the diagram. Other orientations will be considered similarly.
-
All angle calculations are directed angles modulo
\pi
.
Problem formalization:
-
We add the center for each circle when it is necessary for the construction.
-
We add points of tangency when it is necessary for the construction.
-
We manually construct an approximate diagram.
Solution:
-
Just like in AlphaGeometry 1.0, all auxiliary points in the solutions below are automatically generated by a language model.
-
All angle chasing done in the proof is Gaussian elimination.
d(AB) - d(CD)
is the same as the directed angle fromAB
toCD
modulo\pi
. -
We will annotate similar triangles and congruent triangles pairs manually (shown in red).
AlphaGeometry Solution (reversed proof)
Formalization:
triangle a b c; a b < a c; a c < b c; incenter i a b c; coll x b c; x != c; foot d i b c; perp x t1 t1 i; cong i t1 i d; para x t1 a c; coll y b c; y != b; perp y t2 t2 i; cong i t2 i d; para y t2 a b; circle o a b c; line_circle_intersection p a i o a; p != a; midpoint k a c; midpoint l a b; ? angeq i l i k p x p y 1 -1 1 -1 180
Auxiliary points: Construct e = satisfying: e i b are collinear, |l e| = |l b| Reverse proof: 93. d(i l) - d(i k) + d(p x) - d(p y) + 180 = 0 (by calculation) ╰-- 92. ∠b p y = ∠i k a (by similar triangles sas) ↳(similar triangles BPY and AKI, proved using 89, 90 and 91) | ╰-- 91. b→p→y and i→k→a are of the same clock direction. (by diagram check) | ╰-- 90. ∠p b y = ∠i a k (by calculation) | | ╰-- 60. ∠a p o = ∠o a p (by isosceles triangle) | | | ╰-- 4. |o p| = |o a| (by assumption) | | ╰-- 59. ∠c b o = ∠o c b (by isosceles triangle) | | | ╰-- 58. |o b| = |o c| (by calculation) | | | ╰-- 7. |o a| = |o c| (by assumption) | | | ╰-- 3. |o a| = |o b| (by assumption) | | ╰-- 57. p a i are collinear (by assumption) | | ╰-- 18. b x d y c are collinear (by merge lines) | | | ╰-- 17. y b c are collinear (by assumption) | | | ╰-- 16. b x d c are collinear (by merge lines) | | | ╰-- 15. x b c are collinear (by assumption) | | | ╰-- 14. b d c are collinear (by assumption) | | ╰-- 13. ∠a c o = ∠o a c (by isosceles triangle) | | | ╰-- 12. |o c| = |o a| (by calculation) | | | ╰-- 7. |o a| = |o c| (by assumption) | | ╰-- 6. ∠o b p = ∠b p o (by isosceles triangle) | | | ╰-- 5. |o b| = |o p| (by calculation) | | | ╰-- 4. |o p| = |o a| (by assumption) | | | ╰-- 3. |o a| = |o b| (by assumption) | | ╰-- 1. a k c are collinear (by assumption) | ╰-- 89. |a k| / |a i| = |b p| / |b y| (by calculation) | ╰-- 88. |a c| / |a b| = |a k| / |a l| (by similar triangles aa) ↳(similar triangles AKL and ACB, proved using 74 and 87) | | ╰-- 87. ∠c b a = ∠k l a (by calculation) | | | ╰-- 86. ∠l k o = ∠l a o (by inscribed angle) | | | | ╰-- 85. l k o a are concyclic (by inscribed angle) | | | | ╰-- 84. ∠o l a = ∠o k a (by calculation) | | | | ╰-- 83. ∠b l o = ∠o l a (by similar triangles sss) ↳(similar triangles ALO and BLO, proved using 80, 81 and 82) | | | | | ╰-- 82. b→l→o and o→l→a are of the same clock direction. (by diagram check) | | | | | ╰-- 81. |o l| / |o b| = |o l| / |o a| (by calculation) | | | | | | ╰-- 3. |o a| = |o b| (by assumption) | | | | | ╰-- 80. |b l| / |b o| = |a l| / |a o| (by calculation) | | | | | ╰-- 52. |a l| = |b l| (by assumption) | | | | | ╰-- 3. |o a| = |o b| (by assumption) | | | | ╰-- 79. ∠c k o = ∠o k a (by similar triangles sss) ↳(similar triangles AKO and CKO, proved using 76, 77 and 78) | | | | | ╰-- 78. c→k→o and o→k→a are of the same clock direction. (by diagram check) | | | | | ╰-- 77. |o k| / |o c| = |o k| / |o a| (by calculation) | | | | | | ╰-- 7. |o a| = |o c| (by assumption) | | | | | ╰-- 76. |c k| / |c o| = |a k| / |a o| (by calculation) | | | | | ╰-- 75. |a k| = |c k| (by assumption) | | | | | ╰-- 7. |o a| = |o c| (by assumption) | | | | ╰-- 2. a l b are collinear (by assumption) | | | | ╰-- 1. a k c are collinear (by assumption) | | | ╰-- 79. ∠c k o = ∠o k a (by similar triangles sss) see above ↳(similar triangles AKO and CKO, proved using 76, 77 and 78) | | | ╰-- 59. ∠c b o = ∠o c b (by isosceles triangle) see above | | | ╰-- 13. ∠a c o = ∠o a c (by isosceles triangle) see above | | | ╰-- 11. ∠a b o = ∠o a b (by isosceles triangle) | | | | ╰-- 10. |o b| = |o a| (by calculation) | | | | ╰-- 3. |o a| = |o b| (by assumption) | | | ╰-- 1. a k c are collinear (by assumption) | | ╰-- 74. ∠c a b = ∠k a l (by calculation) | | ╰-- 2. a l b are collinear (by assumption) | | ╰-- 1. a k c are collinear (by assumption) | ╰-- 73. |c y| / |y i| = |c i| / |i a| (by similar triangles aa) ↳(similar triangles CYI and CIA, proved using 70 and 71) | | ╰-- 71. ∠y i c = ∠i a c (by calculation) | | | ╰-- 44. ∠b c i = ∠i c a (by assumption) | | | ╰-- 32. ∠b a i = ∠i a c (by assumption) | | | ╰-- 28. ∠d y i = ∠i y t2 (by similar triangles Ssa) ↳(similar right triangles DYI and T2YI, proved using 24, 26 and 27) | | | | ╰-- 27. d→i→y and i→t2→y are of the same clock direction. (by diagram check) | | | | ╰-- 26. ∠y t2 i = ∠i d y (by calculation) | | | | | ╰-- 25. y t2 ⊥ t2 i (by assumption) | | | | | ╰-- 21. i d ⊥ b c (by assumption) | | | | | ╰-- 18. b x d y c are collinear (by merge lines) see above | | | | ╰-- 24. |i y| / |i d| = |i y| / |i t2| (by calculation) | | | | ╰-- 23. |i t2| = |i d| (by assumption) | | | ╰-- 22. y t2 || a b (by assumption) | | | ╰-- 18. b x d y c are collinear (by merge lines) see above | | ╰-- 70. ∠y c i = ∠i c a (by calculation) | | ╰-- 44. ∠b c i = ∠i c a (by assumption) | | ╰-- 18. b x d y c are collinear (by merge lines) see above | ╰-- 72. |c y| / |c i| = |c i| / |c a| (by similar triangles aa) ↳(similar triangles CYI and CIA, proved using 70 and 71) | | ╰-- 71. ∠y i c = ∠i a c (by calculation) see above | | ╰-- 70. ∠y c i = ∠i c a (by calculation) see above | ╰-- 69. |p b| = |p c| (by isosceles triangle) | | ╰-- 68. ∠c b p = ∠p c b (by calculation) | | ╰-- 60. ∠a p o = ∠o a p (by isosceles triangle) see above | | ╰-- 59. ∠c b o = ∠o c b (by isosceles triangle) see above | | ╰-- 57. p a i are collinear (by assumption) | | ╰-- 32. ∠b a i = ∠i a c (by assumption) | | ╰-- 13. ∠a c o = ∠o a c (by isosceles triangle) see above | | ╰-- 11. ∠a b o = ∠o a b (by isosceles triangle) see above | | ╰-- 9. ∠o c p = ∠c p o (by isosceles triangle) | | | ╰-- 8. |o c| = |o p| (by calculation) | | | ╰-- 7. |o a| = |o c| (by assumption) | | | ╰-- 4. |o p| = |o a| (by assumption) | | ╰-- 6. ∠o b p = ∠b p o (by isosceles triangle) see above | ╰-- 63. |l a| / |a e| = |p c| / |c i| (by similar triangles aa) ↳(similar triangles ALE and IPC, proved using 61 and 62) | | ╰-- 62. ∠a e l = ∠p i c (by calculation) | | | ╰-- 57. p a i are collinear (by assumption) | | | ╰-- 54. ∠l a e = ∠a e l (by isosceles triangle) | | | | ╰-- 53. |l e| = |l a| (by calculation) | | | | ╰-- 52. |a l| = |b l| (by assumption) | | | | ╰-- 49. |l e| = |l b| (by assumption) | | | ╰-- 51. ∠l e b = ∠e b l (by isosceles triangle) | | | | ╰-- 50. |l b| = |l e| (by calculation) | | | | ╰-- 49. |l e| = |l b| (by assumption) | | | ╰-- 47. e i b are collinear (by assumption) | | | ╰-- 44. ∠b c i = ∠i c a (by assumption) | | | ╰-- 32. ∠b a i = ∠i a c (by assumption) | | | ╰-- 20. ∠a b i = ∠i b c (by assumption) | | | ╰-- 2. a l b are collinear (by assumption) | | ╰-- 61. ∠a l e = ∠i p c (by calculation) | | ╰-- 60. ∠a p o = ∠o a p (by isosceles triangle) see above | | ╰-- 59. ∠c b o = ∠o c b (by isosceles triangle) see above | | ╰-- 57. p a i are collinear (by assumption) | | ╰-- 51. ∠l e b = ∠e b l (by isosceles triangle) see above | | ╰-- 47. e i b are collinear (by assumption) | | ╰-- 20. ∠a b i = ∠i b c (by assumption) | | ╰-- 11. ∠a b o = ∠o a b (by isosceles triangle) see above | | ╰-- 9. ∠o c p = ∠c p o (by isosceles triangle) see above | | ╰-- 2. a l b are collinear (by assumption) | ╰-- 56. |b a| / |a e| = |b y| / |y i| (by similar triangles aa) ↳(similar triangles ABE and YBI, proved using 48 and 55) | ╰-- 55. ∠a e b = ∠b i y (by calculation) | | ╰-- 54. ∠l a e = ∠a e l (by isosceles triangle) see above | | ╰-- 51. ∠l e b = ∠e b l (by isosceles triangle) see above | | ╰-- 28. ∠d y i = ∠i y t2 (by similar triangles Ssa) see above ↳(similar right triangles DYI and T2YI, proved using 70 and 71) | | ╰-- 22. y t2 || a b (by assumption) | | ╰-- 20. ∠a b i = ∠i b c (by assumption) | | ╰-- 18. b x d y c are collinear (by merge lines) see above | | ╰-- 2. a l b are collinear (by assumption) | ╰-- 48. ∠a b e = ∠i b y (by calculation) | ╰-- 47. e i b are collinear (by assumption) | ╰-- 20. ∠a b i = ∠i b c (by assumption) | ╰-- 18. b x d y c are collinear (by merge lines) see above ╰-- 67. ∠c p x = ∠i l a (by similar triangles sas) ↳(similar triangles ALI and CPX, proved using 64, 65 and 66) | ╰-- 66. c→p→x and i→l→a are of the same clock direction. (by diagram check) | ╰-- 65. ∠p c x = ∠i a l (by calculation) | | ╰-- 60. ∠a p o = ∠o a p (by isosceles triangle) see above | | ╰-- 59. ∠c b o = ∠o c b (by isosceles triangle) see above | | ╰-- 57. p a i are collinear (by assumption) | | ╰-- 15. x b c are collinear (by assumption) | | ╰-- 11. ∠a b o = ∠o a b (by isosceles triangle) see above | | ╰-- 9. ∠o c p = ∠c p o (by isosceles triangle) see above | | ╰-- 2. a l b are collinear (by assumption) | ╰-- 64. |a l| / |a i| = |c p| / |c x| (by calculation) | ╰-- 63. |l a| / |a e| = |p c| / |c i| (by similar triangles aa) see above ↳(similar triangles ALE and IPC, proved using 61 and 62) | ╰-- 56. |b a| / |a e| = |b y| / |y i| (by similar triangles aa) see above ↳(similar triangles ABE and YBI, proved using 48 and 55) | ╰-- 46. |c x| / |x i| = |c i| / |i d| (by similar triangles aa) ↳(similar triangles CXI and IXD, proved using 43 and 45) | | ╰-- 45. ∠x i c = ∠c d i (by calculation) | | | ╰-- 44. ∠b c i = ∠i c a (by assumption) | | | ╰-- 39. ∠t1 x i = ∠i x d (by similar triangles Ssa) ↳(similar right triangles T1XI and DXI, proved using 35, 37 and 38) | | | | ╰-- 38. i→t1→x and d→i→x are of the same clock direction. (by diagram check) | | | | ╰-- 37. ∠x d i = ∠i t1 x (by calculation) | | | | | ╰-- 36. x t1 ⊥ t1 i (by assumption) | | | | | ╰-- 21. i d ⊥ b c (by assumption) | | | | | ╰-- 16. b x d c are collinear (by merge lines) see above | | | | ╰-- 35. |i x| / |i t1| = |i x| / |i d| (by calculation) | | | | ╰-- 34. |i t1| = |i d| (by assumption) | | | ╰-- 33. x t1 || a c (by assumption) | | | ╰-- 21. i d ⊥ b c (by assumption) | | | ╰-- 16. b x d c are collinear (by merge lines) see above | | ╰-- 43. ∠x c i = ∠d c i (by calculation) | | ╰-- 16. b x d c are collinear (by merge lines) see above | | ╰-- 15. x b c are collinear (by assumption) | ╰-- 42. |b x| / |x i| = |b i| / |i a| (by similar triangles aa) ↳(similar triangles BXI and BIA, proved using 31 and 40) | | ╰-- 40. ∠x i b = ∠i a b (by calculation) | | | ╰-- 39. ∠t1 x i = ∠i x d (by similar triangles Ssa) see above ↳(similar right triangles T1XI and DXI, proved using 35, 37 and 38) | | | ╰-- 33. x t1 || a c (by assumption) | | | ╰-- 32. ∠b a i = ∠i a c (by assumption) | | | ╰-- 20. ∠a b i = ∠i b c (by assumption) | | | ╰-- 16. b x d c are collinear (by merge lines) see above | | ╰-- 31. ∠x b i = ∠i b a (by calculation) | | ╰-- 20. ∠a b i = ∠i b c (by assumption) | | ╰-- 15. x b c are collinear (by assumption) | ╰-- 41. |b x| / |b i| = |b i| / |b a| (by similar triangles aa) ↳(similar right triangles BXI and BIA, proved using 31 and 40) | | ╰-- 40. ∠x i b = ∠i a b (by calculation) see above | | ╰-- 31. ∠x b i = ∠i b a (by calculation) see above | ╰-- 30. |b y| / |y i| = |b i| / |i d| (by similar triangles aa) ↳(similar triangles BIY and BDI, proved using 19 and 29) | ╰-- 29. ∠y i b = ∠b d i (by calculation) | | ╰-- 28. ∠d y i = ∠i y t2 (by similar triangles Ssa) see above ↳(similar right triangles DYI and T2YI, proved using 24, 26 and 27) | | ╰-- 22. y t2 || a b (by assumption) | | ╰-- 21. i d ⊥ b c (by assumption) | | ╰-- 20. ∠a b i = ∠i b c (by assumption) | | ╰-- 18. b x d y c are collinear (by merge lines) see above | | ╰-- 16. b x d c are collinear (by merge lines) see above | ╰-- 19. ∠y b i = ∠d b i (by calculation) | ╰-- 18. b x d y c are collinear (by merge lines) see above | ╰-- 16. b x d c are collinear (by merge lines) see above ╰-- 13. ∠a c o = ∠o a c (by isosceles triangle) see above ╰-- 11. ∠a b o = ∠o a b (by isosceles triangle) see above ╰-- 9. ∠o c p = ∠c p o (by isosceles triangle) see above ╰-- 6. ∠o b p = ∠b p o (by isosceles triangle) see above ╰-- 2. a l b are collinear (by assumption) ╰-- 1. a k c are collinear (by assumption)download Download the full proof
Intuitive Explanation of the Proof (Human-written)
Summary of the key steps
-
Triangle
AKI
is similar to triangleBPY
(s-a-s)-
∠AIK = ∠BYP
-
∠AKI = ∠BPY
-
-
Triangle
ALI
is similar to triangleCPX
(s-a-s)-
∠ALI = ∠CXP
-
∠AIL = ∠CPX
-
-
Triangle
BXI
is similar to triangleBIA
(a-a)-
BX/XI = BI/IA
-
-
Triangle
CYI
is similar to triangleCIA
(a-a)-
CY/YI = CI/IA
-
Intuitions behind the auxiliary construction
The problem asks the relationship between ∠KIL
and ∠XPY
. The former is an angle formed by a midpoint and the incenter, which usually do not go well together and cannot be computed by the angles of the main triangle ABC
. Thus, this necessitates the need for auxiliary point construction. To this end, the model constructs E
, which is a point on the line BI
such that ∠AEB = 90\degree
. In addition, this also creates pairs of similar triangles ABE
and YBI
, ALE
and IPC
. These pairs of similar triangles would create new pairs of equal angles and equal side length ratio. Thus, the point E
gives purposes to the midpoint L
of AB
.
To complete the proof, we need to prove ∠AIK = ∠BYP
and ∠AIL = ∠CPX
. To this end, we need to prove that triangle AKI
is similar to triangle BPY
and triangle ALI
is similar to triangle CPX
, which is done by side length ratio chasing, which is obtained from the pairs of similar triangles above.