IMO 2003 SL G7

Origin: SAF | Category: Geometry

Problem

Let ABC be a triangle with semiperimeter s and inradius r. The semicircles with diameters BC, CA, AB are drawn outside of the triangle ABC. The circle tangent to all three semicircles has radius t. Prove that s 2 < t \leqs 2 + 1 − \sqrt r.

Solution

Let D, E, F be the midpoints of BC, CA, AB, respectively. We construct smaller semicircles \Gammad, \Gammae, \Gammaf inside \triangleABC with centers D, E, F and radii d = s−a 2 , e = s−b 2 , f = s−c respectively. Since DE = d + e, DF = d + f, and EF = e + f, we deduce that \Gammad, \Gammae, and \Gammaf touch each other at the points D1, E1, F1 of tangency of the incircle \gamma of \triangleDEF with its sides (D1 \inEF, etc.). Consider the circle \Gammag with center O and radius g that lies inside \triangleDEF and tangents \Gammad, \Gammae, \Gammaf.

Now let OD, OE, OF meet the semicircles \Gammad, \Gammae, \Gammaf at D′, E′, F ′ respectively. We have OD′ = OD + DD′ = g + d + a 2 = g + s 2 and sim- ilarly OE′ = OF ′ = g + s 2. It fol- lows that the circle with center O and radius g + s 2 touches all three semicircles, and consequently t = g + s 2 > s 2. Now set the coordinate system such that we have the points D1(0, 0), E(−e, 0), F(f, 0) and such A B C D E F D1 E1 F1 D′ E′ F ′ \gammad \gammae \gammaf \Gammad \Gammae \Gammaf that the y coordinate of D is positive. Apply the inversion with center D1 and unit radius. This inversion maps the circles \Gammae and \Gammaf to the lines @ \Gammae / x = −1 2e and @ \Gammae  x = 2f  respec- tively, and the circle \gamma goes to the line <\gamma / y = 1 r . The images @ \Gammad and @ \Gammag of \Gammad, \Gammag are the circles that touch the lines @ \Gammae and @ \Gammaf. Since @ \Gammad, @ \Gammag are perpendicular to \gamma, they have radii equal to R = 4e + 4f and centers at  −1 4e + 4f , 1 r

and  −1 4e + 4f , 1 r + 2R

respectively. Let p and P be the distances from D1(0, 0) to the centers of \Gammag and @ \Gammag respectively. We have that P 2 =  4e − 4f +
1 r + 2R 2, and that the circles \Gammag and @ \Gammag are homothetic with center of homothety D1; hence p/P = g/R. On the other hand, @ \Gammag is the image of \Gammag under inversion; hence the product of the tangents from D1 to these two circles is equal to 1. In other words, we obtain

p2 −g2 \cdot \sqrt P 2 −R2 = 1. Using the relation p/P = g/R we get g = R P 2−R2 . The inequality we have to prove is equivalent to (4 + 2 \sqrt 3)g \leqr. This can be proved as follows: r −(4 + 2 \sqrt 3)g = r(P 2 −R2 −(4 + 2 \sqrt 3)R/r) P 2 −R2

r 
1 r + 2R 2 +  4e − 4f −R2 −(4 + 2 \sqrt 3) R r  P 2 −R2

r P 2 −R2

 R \sqrt 3 −1 r 2 +  1 4e −1 4f 2 \geq0. Remark. One can obtain a symmetric formula for g: 2g = s −a + s −b + s −c + 2 r .