IMO 1995 SL G4

Origin: UKR | Category: Geometry

Problem

An acute triangle ABC is given. Points A1 and A2 are taken on the side BC (with A2 between A1 and C), B1 and B2 on the side AC (with B2 between B1 and A), and C1 and C2 on the side AB (with C2 between C1 and B) such that \angleAA1A2 = \angleAA2A1 = \angleBB1B2 = \angleBB2B1 = \angleCC1C2 = \angleCC2C1. The lines AA1, BB1, and CC1 form a triangle, and the lines AA2, BB2, and CC2 form a second triangle. Prove that all six vertices of these two triangles lie on a single circle.

Solution

Let the two triangles be X1Y1Z1, X2Y2Z2, with X1 = BB1 \capCC1, Y1 = CC1 \capAA1, Z1 =AA1 \capBB1, X2 = BB2 \capCC2, Y2 = CC2 \cap AA2, Z2 = AA2 \capBB2. First, we observe that \angleABB2 = \angleACC1 and \angleABB1

\angleACC2. Conse- quently \angleBZ1A1

\angleBAA1 + \angleABB1 = \angleBCC2 + \angleC2CA = \angleC and similarly \angleAZ2B2 = \angleC, \angleAY1C1

\angleCY2A2

\angleB. Also, \triangleABB2 ∼\triangleACC1; hence AC1/AC = AB2/AB. A B C A1 A2 B1 B2 C1 C2 Y1 Z1 X1 Z2 X2 Y2 From the sine formula, we obtain AZ1 sin \angleABZ1

AB sin \angleAZ1B = AB sin \angleC = AC sin \angleB = AC sin \angleAY2C

AY2 sin \angleACY2 =⇒ AZ1 = AY2. Analogously, BX1 = BZ2 and CY1 = CX2. Furthermore, again from the sine formula,

AY1 sin \angleAC1Y1

AC1 sin \angleAY1C1 = AC1 AC AC sin \angleB = AB2 AB AB sin \angleC = AB2 sin \angleAZ2B2

AZ2 sin \angleAB2Z2 . Hence, AY1 = AZ2 and, analogously, BZ1 = BX2 and CX1 = CY2. We deduce that Y1Z2 \parallelBC and Z2X1 \parallelAC, which gives us \angleY1Z2X1 = 180◦−\angleC = 180◦−\angleY1Z1X1. It follows that Z2 lies on the circle cir- cumscribed about \triangleX1Y1Z1. Similarly, so do X2 and Y2. Second solution. Let H be the orthocenter of \triangleABC. Triangles AHB, BHC, CHA, ABC have the same circumradius R. Additionally, \angleHAAi = \angleHBBi = \angleHCCi = \theta (i = 1, 2). Since \angleHBX1 = \angleHCX1 = \theta, BCX1H is concyclic and therefore HX1 = 2R sin \theta. The same holds for HY1, HZ1, HX2, HY2, HZ2. Hence Xi, Yi, Zi (i = 1, 2) lie on a circle centered at H.