HMMT 二月 2013 · 团队赛 · 第 6 题
HMMT February 2013 — Team Round — Problem 6
题目详情
- [ 25 ] Let triangle ABC satisfy 2 BC = AB + AC and have incenter I and circumcircle ! . Let D be the intersection of AI and ! (with A, D distinct). Prove that I is the midpoint of AD .
解析
- [ 25 ] Let triangle ABC satisfy 2 BC = AB + AC and have incenter I and circumcircle ω . Let D be the 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 is well-known that B , I , and C lie on a circle centered at D . Thus BD = DC = DI . Applying Ptolemy’s theorem 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 AC that 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 AC bisects ∠ 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 perpendicu- lar bisectors of segments P E and EQ , respectively. Thus, the circumcenter of △ P QE is I , so the perpendicular 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 , whence the perpendicular bisector of BC meets the bisector of ∠ A at D . A homothety centered at A with factor 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 and circumradius of △ ABC , respectively, let E be the intersection of segments AD and BC , and let O be the 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 bc 2 theorem and the angle bisector theorem we can find AI = bc (1 − ) = . Thus, it a + b + c b + c 3 remains to show that 6 Rr = bc . abc Now, we use the combine the well-known formulas = R and K = rs to get abc = 4 Rrs , where K is 4 K 2 a +2( b + c ) 2 a +2(2 a ) 4 s the area of △ ABC and s is its semiperimeter. We have bc = Rr = Rr = Rr = 6 Rr , a a a as desired.