IMO 2019 Q2 #
In triangle ABC, point A₁ lies on side BC and point B₁ lies on side AC. Let P and
Q be points on segments AA₁ and BB₁, respectively, such that PQ is parallel to AB.
Let P₁ be a point on line PB₁, such that B₁ lies strictly between P and P₁, and
∠PP₁C = ∠BAC. Similarly, let Q₁ be a point on line QA₁, such that A₁ lies strictly
between Q and Q₁, and ∠CQ₁Q = ∠CBA.
Prove that points P, Q, P₁, and Q₁ are concyclic.
We follow Solution 1 from the
official solutions.
Letting the rays AA₁ and BB₁ intersect the circumcircle of ABC at A₂ and B₂
respectively, we show with an angle chase that P, Q, A₂, B₂ are concyclic and let ω be
the circle through those points. We then show that C, Q₁, A₂, A₁ are concyclic, and
then that Q₁ lies on ω, and similarly that P₁ lies on ω, so the required four points are
concyclic.
Note that most of the formal proof is actually proving nondegeneracy conditions needed for that
angle chase / concyclicity argument, where an informal solution doesn't discuss those conditions
at all. Also note that (as described in Geometry.Euclidean.Angle.Oriented.Basic) the oriented
angles used are modulo 2 * π, so parts of the angle chase that are only valid for angles modulo
π (as used in the informal solution) are represented as equalities of twice angles, which we write
as (2 : ℤ) • ∡ _ _ _ = (2 : ℤ) • ∡ _ _ _.
A configuration satisfying the conditions of the problem. We define this structure to avoid passing many hypotheses around as we build up information about the configuration; the final result for a statement of the problem not using this structure is then deduced from one in terms of this structure.
- A : Pt
- B : Pt
- C : Pt
- A₁ : Pt
- B₁ : Pt
- P : Pt
- Q : Pt
- P₁ : Pt
- Q₁ : Pt
- angle_PP₁C_eq_angle_BAC : EuclideanGeometry.angle self.P self.P₁ self.C = EuclideanGeometry.angle self.B self.A self.C
- angle_CQ₁Q_eq_angle_CBA : EuclideanGeometry.angle self.C self.Q₁ self.Q = EuclideanGeometry.angle self.C self.B self.A
Instances For
The configuration has symmetry, allowing results proved for one point to be applied for another (where the informal solution says "similarly").
Equations
- One or more equations did not get rendered due to their size.
Instances For
Configuration properties that are obvious from the diagram, and construction of the #
ABC as a Triangle.
Instances For
A₂ is the second point of intersection of the ray AA₁ with the circumcircle of ABC.
Equations
- cfg.A₂ = (Affine.Simplex.circumsphere cfg.triangleABC).secondInter cfg.A (cfg.A₁ -ᵥ cfg.A)
Instances For
B₂ is the second point of intersection of the ray BB₁ with the circumcircle of ABC.
Equations
- cfg.B₂ = (Affine.Simplex.circumsphere cfg.triangleABC).secondInter cfg.B (cfg.B₁ -ᵥ cfg.B)
Instances For
Relate the orientations of different angles in the configuration #
More obvious configuration properties #
The first equality in the first angle chase in the solution #
More obvious configuration properties #
QPA₂ as a Triangle.
Instances For
PQB₂ as a Triangle.
Instances For
ω is the circle containing Q, P and A₂, which will be shown also to contain B₂,
P₁ and Q₁.
Equations
- cfg.ω = Affine.Simplex.circumsphere cfg.triangleQPA₂