IMO 2012 Shortlist G8

Category: Geometry

Problem

Let ABC be a triangle with circumcircle ω and ℓ a line without common points with ω. Denote by P the foot of the perpendicular from the center of ω to ℓ. The side-lines BC,CA,AB intersect ℓ at the points X,Y,Z different from P. Prove that the circumcircles of the triangles AXP,BY P and CZP have a common point different from P or are mutually tangent at P. Solution 1. Let ωA,ωB,ωC and ω be the circumcircles of triangles AXP,BY P,CZP and ABC respectively. The strategy of the proof is to construct a point Q with the same power with respect to the four circles. Then each of P and Q has the same power with respect to ωA,ωB,ωC and hence the three circles are coaxial. In other words they have another common point P′ or the three of them are tangent at P. We first give a description of the point Q. Let A′ 6= A be the second intersection of ω and ωA; define B′ and C′ analogously. We claim that AA′ , BB′ and CC′ have a common point. Once this claim is established, the point just constructed will be on the radical axes of the three pairs of circles {ω,ωA},{ω,ωB},{ω,ωC}. Hence it will have the same power with respect to ω,ωA,ωB,ωC. ℓ ωA ωC ω ωB X Y Z P A B′ Q O C′ B A′ C Z′ Y ′ X′ P′ We proceed to prove that AA′ , BB′ and CC′ intersect at one point. Let r be the circumra- dius of triangle ABC. Define the points X′ ,Y ′ ,Z′ as the intersections of AA′ ,BB′ ,CC′ with ℓ. Observe that X′ ,Y ′ ,Z′ do exist. If AA′ is parallel to ℓ then ωA is tangent to ℓ; hence X = P which is a contradiction. Similarly, BB′ and CC′ are not parallel to ℓ. From the powers of the point X′ with respect to the circles ωA and ω we get X′ P · (X′ P + PX) = X′ P · X′ X = X′ A′ · X′ A = X′ O2 − r2 , hence X′ P · PX = X′ O2 − r2 − X′ P2 = OP2 − r2 . We argue analogously for the points Y ′ and Z′ , obtaining X′ P · PX = Y ′ P · PY = Z′ P · PZ = OP2 − r2 = k2 . (1) In these computations all segments are regarded as directed segments. We keep the same convention for the sequel. We prove that the lines AA′ ,BB′ ,CC′ intersect at one point by Ceva’s theorem. To avoid distracting remarks we interpret everything projectively, i. e. whenever two lines are parallel they meet at a point on the line at infinity.39 Let U,V,W be the intersections of AA′ ,BB′ ,CC′ with BC,CA,AB respectively. The idea is that although it is difficult to calculate the ratio BU CU , it is easier to deal with the cross-ratio BU CU /BX CX because we can send it to the line ℓ. With this in mind we apply Menelaus’ theorem to the triangle ABC and obtain BX CX · CY AY · AZ BZ = 1. Hence Ceva’s ratio can be expressed as BU CU · CV AV · AW BW

BU CU / BX CX · CV AV / CY AY · AW BW / AZ BZ . ℓ ω X Y P A V Q W U B C Z′ Z X′ Y ′ Project the line BC to ℓ from A. The cross-ratio between BC and UX equals the cross-ratio between ZY and X′ X. Repeating the same argument with the lines CA and AB gives BU CU · CV AV · AW BW

ZX′ Y X′ / ZX Y X · XY ′ ZY ′ / XY ZY · Y Z′ XZ′ / Y Z XZ and hence BU CU · CV AV · AW BW = (−1) · ZX′ Y X′ · XY ′ ZY ′ · Y Z′ XZ′ . The equations (1) reduce the problem to a straightforward computation on the line ℓ. For instance, the transformation t 7→ −k2 /t preserves cross-ratio and interchanges the points X,Y,Z with the points X′ ,Y ′ ,Z′ . Then BU CU · CV AV · AW BW = (−1) · ZX′ Y X′ / ZZ′ Y Z′ · XY ′ ZY ′ / XZ′ ZZ′ = −1. We proved that Ceva’s ratio equals −1, so AA′ ,BB′ ,CC′ intersect at one point Q. Comment 1. There is a nice projective argument to prove that AX′,BY ′,CZ′ intersect at one point. Suppose that ℓ and ω intersect at a pair of complex conjugate points D and E. Consider a projective transformation that takes D and E to [i;1,0] and [−i,1,0]. Then ℓ is the line at infinity, and ω is a conic through the special points [i;1,0] and [−i,1,0], hence it is a circle. So one can assume that AX,BY,CZ are parallel to BC,CA,AB. The involution on ℓ taking X,Y,Z to X′,Y ′,Z′ and leaving D,E fixed is the involution changing each direction to its perpendicular one. Hence AX,BY,CZ are also perpendicular to AX′,BY ′,CZ′. It follows from the above that AX′,BY ′,CZ′ intersect at the orthocenter of triangle ABC. Comment 2. The restriction that the line ℓ does not intersect the circumcricle ω is unnecessary. The proof above works in general. In case ℓ intersects ω at D and E point P is the midpoint of DE, and some equations can be interpreted differently. For instance X′ P · X′ X = X′ A′ · X′ A = X′ D · X′ E, and hence the pairs X′X and DE are harmonic conjugates. This means that X′,Y ′,Z′ are the harmonic conjugates of X,Y,Z with respect to the segment DE.40 Solution 2. First we prove that there is an inversion in space that takes ℓ and ω to parallel circles on a sphere. Let QR be the diameter of ω whose extension beyond Q passes through P. Let Π be the plane carrying our objects. In space, choose a point O such that the line QO is perpendicular to Π and ∠POR = 90◦ , and apply an inversion with pole O (the radius of the inversion does not matter). For any object T denote by T ′ the image of T under this inversion. The inversion takes the plane Π to a sphere Π′ . The lines in Π are taken to circles through O, and the circles in Π also are taken to circles on Π′ . O ℓ P R Q Q′ R′ ω ℓ′ Π P′ Π′ ω′ Since the line ℓ and the circle ω are perpendicular to the plane OPQ, the circles ℓ′ and ω′ also are perpendicular to this plane. Hence, the planes of the circles ℓ′ and ω′ are parallel. Now consider the circles A′ X′ P′ , B′ Y ′ P′ and C′ Z′ P′ . We want to prove that either they have a common point (on Π′ ), different from P′ , or they are tangent to each other. H C′ O B1 X′ A′ W Y ′ P′ Z′ Π′ ℓ′ ω′ A1 C1 B′ The point X′ is the second intersection of the circles B′ C′ O and ℓ′ , other than O. Hence, the lines OX′ and B′ C′ are coplanar. Moreover, they lie in the parallel planes of ℓ′ and ω′ . Therefore, OX′ and B′ C′ are parallel. Analogously, OY ′ and OZ′ are parallel to A′ C′ and A′ B′ . Let A1 be the second intersection of the circles A′ X′ P′ and ω′ , other than A′ . The segments A′ A1 and P′ X′ are coplanar, and therefore parallel. Now we know that B′ C′ and A′ A1 are parallel to OX′ and X′ P′ respectively, but these two segments are perpendicular because OP′ is a diameter in ℓ′ . We found that A′ A1 and B′ C′ are perpendicular, hence A′ A1 is the altitude in the triangle A′ B′ C′ , starting from A. Analogously, let B1 and C1 be the second intersections of ω′ with the circles B′ P′ Y ′ and C′ P′ Z′ , other than B′ and C′ respectively. Then B′ B1 and C′ C1 are the other two al- titudes in the triangle A′ B′ C′ .41 Let H be the orthocenter of the triangle A′ B′ C′ . Let W be the second intersection of the line P′ H with the sphere Π′ , other than P′ . The point W lies on the sphere Π′ , in the plane of the circle A′ P′ X′ , so W lies on the circle A′ P′ X′ . Similarly, W lies on the circles B′ P′ Y ′ and C′ P′ Z′ as well; indeed W is the second common point of the three circles. If the line P′ H is tangent to the sphere then W coincides with P′ , and P′ H is the common tangent of the three circles.42