IMO 2001 SL G5

Origin: GRE | Category: Geometry

Problem

Let ABC be an acute triangle. Let DAC, EAB, and FBC be isosceles triangles exterior to ABC, with DA = DC, EA = EB, and FB = FC such that \angleADC = 2\angleBAC, \angleBEA = 2\angleABC, \angleCFB = 2\angleACB. Let D′ be the intersection of lines DB and EF, let E′ be the intersection of EC and DF, and let F ′ be the intersection of FA and DE. Find, with proof, the value of the sum DB DD′ + EC EE′ + FA FF ′ .

Solution

It is easy to see that the hexagon AEBFCD is convex and \angleAEB + \angleBFC +\angleCDA = 360◦. Using this relation we obtain that the circles \omega1, \omega2, \omega3 with centers at D, E, F and radii DA, EB, FC respectively all pass through a common point O. Indeed, if \omega1 \cap\omega2 = {O}, then \angleAOB

180◦−\angleAEB/2 and \angleBOC = 180◦−\angleBFC/2; hence \angleCOA = 180◦−\angleCDA/2 as well, i.e., O \in\omega3. The point O is the re- A B C D E F D′ E′ F ′O flection of A with respect to DE. Similarly, it is also the reflection of B with respect to EF, and that of C with respect to FD. Hence DB DD′ = 1 + D′B DD′ = 1 + SEBF SEDF = 1 + SOEF SDEF . Analogously EC EE′ = 1+ SODF SDEF and F A F F ′ = 1+ SODE SDEF . Adding these relations gives us

DB DD′ + EC EE′ + FA FF ′ = 3 + SOEF + SODF + SODE SDEF = 4.