IMO 1997 SL 16

Origin: BLR

Problem

In an acute-angled triangle ABC, let AD, BE be altitudes and AP, BQ internal bisectors. Denote by I and O the incenter and the cir- cumcenter of the triangle, respectively. Prove that the points D, E, and I are collinear if and only if the points P, Q, and O are collinear.

Solution

Let da(X), db(X), dc(X) denote the distances of a point X interior to \triangleABC from the lines BC, CA, AB respectively. We claim that X \inPQ if and only if da(X) + db(X) = dc(X). Indeed, if X \inPQ and PX = kPQ then da(X) = kda(Q), db(X) = (1 −k)db(P), and dc(X) = (1 − k)dc(P)+kdc(Q), and simple substitution yields da(X)+db(X) = dc(X). The converse follows easily. In particular, O \inPQ if and only if da(O) + db(O) = dc(O), i.e., cos \alpha + cos \beta = cos \gamma. We shall now show that I \inDE if and only if AE + BD = DE. Let K be the point on the segment DE such that AE = EK. Then \angleEKA = 2\angleDEC = 1 2\angleCBA = \angleIBA; hence the points A, B, I, K are concyclic. The point I lies on DE if and only if \angleBKD = \angleBAI = 1 2\angleBAC = 2\angleCDE = \angleDBK, which is equivalent to KD = BD, i.e., to AE+BD = DE. But since AE = AB cos \alpha, BD = AB cos \beta, and DE = AB cos \gamma, we have that I \inDE ⇔cos \alpha + cos \beta = cos \gamma. The conditions for O \inPQ and I \inDE are thus equivalent. Second solution. We know that three points X, Y, Z are collinear if and only if for some \lambda, µ \inR with sum 1, we have \lambda−−\to CX + µ−−\to CY = −\to CZ. Specially, if −−\to CX = p−\to CA and −−\to CY = q−−\to CB for some p, q, and −\to CZ = k−\to CA + l−−\to CB, then Z lies on XY if and only if kq + lp = pq. Using known relations in a triangle we directly obtain −−\to CP = sin \beta sin \beta + sin \gamma −−\to CB, −−\to CQ = sin \alpha sin \alpha + sin \gamma −\to CA, −−\to CO = sin 2\alpha \cdot −\to CA + sin 2\beta \cdot −−\to CB sin 2\alpha + sin 2\beta + sin 2\gamma ; −−\to CD = tan \beta tan \beta + tan \gamma −−\to CB, −−\to CE = tan \beta tan \beta + tan \gamma −\to CA, −\to CI = sin \alpha \cdot −\to CA + sin \beta \cdot −−\to CB sin \alpha + sin \beta + sin \gamma .

Now by the above considerations we get that the conditions (1) P, Q, O are collinear and (2) D, E, I are collinear are both equivalent to cos \alpha+cos \beta = cos \gamma.