HMMT 二月 2013 · 团队赛 · 第 6 题

英文解析

6. [ 25 ] Let triangle ABC satisfy 2 BC = AB + AC and have incenter I and circumcircle ω . Let D bethe intersection of AI and ω (with A, D distinct). Prove that I is the midpoint of AD .
   Answer: N/A Since AD is an angle bisector, D is the midpoint of arc BC opposite A on ω . It iswell-known that B , I , and C lie on a circle centered at D . Thus BD = DC = DI . Applying Ptolemy’stheorem to cyclic quadrilateral ABDC , we get
   AB · DC + AC · BD = AD · BC = AD · ( AB + AC ) / 2
   Team Round
   Using BD = DC we have immediately that AD = 2 BD = 2 DI so I is the midpoint of AD as desired.
   Solution 2: Let P and Q be the midpoints of AB and AC , and take the point E on segment BC such
   AB + AC AB ACthat BE = BP . Note that CE = AB − BE = AB − BP = − = = CQ , so triangles
   2 2 2
   AB/ 2
   BE AB
   BP E and CQE are isosceles. In addition, = = , so by the angle bisector theorem, AE
   EC AC/ 2 ACbisects ∠ CAB , whence E must lie on the bisector of ∠ A .
   Since triangles BP E and CQE are isosceles, the bisectors of angles B and C are the perpendicular bisectors of segments P E and EQ , respectively. Thus, the circumcenter of △ P QE is I , so theperpendicular bisector of P Q meets the bisector of ∠ A at I .
   ̂ ̂
   Furthermore, since ∠ DAB = ∠ CAD , arcs BD and DC have the same measure, so BD = DC , whencethe perpendicular bisector of BC meets the bisector of ∠ A at D . A homothety centered at A withfactor 1/2 maps BC to P Q , and so maps D to I . Thus, D is the midpoint of AI .
   Solution 3: Let a = BC , b = CA , and c = AB , let r and R denote the lengths of the inradius andcircumradius of △ ABC , respectively, let E be the intersection of segments AD and BC , and let O bethe circumcenter of △ ABC . I is the midpoint of chord AD if and only if OI ⊥ AD , which is true
   2 2 2 2
   if and only if OA = OI + AI . By Euler’s distance formula, OI = R ( R − 2 r ), and by Stewart’s
   ( ) ( )
   2 2
   b + c a bctheorem and the angle bisector theorem we can find AI = bc (1 − ) = . Thus, it 2
   a + b + c b + c 3
   remains to show that 6 Rr = bc .
   Now, we use the combine the well-known formulas = R and K = rs to get abc = 4 Rrs , where K isabc
   4 K
   2 a +2( b + c ) 2 a +2(2 a )
   4 sthe area of △ ABC and s is its semiperimeter. We have bc = Rr = Rr = Rr = 6 Rr ,
   a a aas desired.