HMMT 二月 2011 · TEAM1 赛 · 第 4 题

英文解析

4. [ 15 ] Let a , b , and c be complex numbers such that | a | = | b | = | c | = | a + b + c | = 1. If | a − b | = | a − c |
   and b 6 = c , prove that | a + b || a + c | = 2.
   Solution:
   b c
   First Solution. Since | a | = 1, a cannot be 0. Let u = and v = . Dividing the given equations by a a
   | a | = 1 gives | u | = | v | = | 1 + u + v | = 1 and | 1 − u | = | 1 − v | . The goal is to prove that | 1 + u || 1 + v | = 2.
   By squaring | 1 − u | = | 1 − v | , we get (1 − u )(1 − u ) = (1 − v )(1 − v ), and thus 1 − u − u ¯ + | u | =2
   1 − v − v ¯ + | v | , or u + ¯ u = v + ¯ v . This implies Re( u ) = Re( v ). Since u and v are on the unit circle in 2
   the complex plane, u is equal to either v or ¯ v . However, b 6 = c implies u 6 = v , so u = ¯ v .
   Therefore, 1 = | 1 + u + ¯ u | = | 1 + 2 Re( u ) | . Since Re( u ) is real, we either have Re( u ) = 0 or Re( u ) = − 1.
   The first case gives u = ± i and | 1 + u || 1 + v | = | 1 + i || 1 − i | = 2, as desired. It remains only to notethat Re( u ) = − 1 is in fact impossible because u is of norm 1 and u = − 1 would imply u = ¯ u = v .
   Remark: by the rotational symmetry of the circle, it is acceptable to skip the first step of this solutionand assume a = 1 without loss of generality.
   Second Solution. Let a , b , and c , be the vertices of a triangle inscribed in the unit circle in thecomplex plane. Since the complex coordinate of the circumcenter is 0 and the complex coordinate ofa + b + cthe centroid is , it follows from well-known facts about the Euler line that the complex coordinateof the orthocenter is a + b + c . Hence the orthocenter lies on the unit circle as well. Is it not possible for 3
   the orthocenter not to be among the three vertices of the triangle, for, if it were, two opposite angles ofthe convex cyclic quadrilateral formed by the three vertices and the orthocenter would each measuregreater than 90 degrees. It follows that the triangle is right. However, since | a − b | = | a − c | , the rightangle cannot occur at b or c , so it must occur at a , and the desired conclusion follows immediately.