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HMMT 十一月 2012 · 团队赛 · 第 8 题

HMMT November 2012 — Team Round — Problem 8

专题
Discrete Math / 离散数学
难度
L3
来源
HMMT

题目详情

  1. [ 4 ] ABC is a triangle with AB = 15, BC = 14, and CA = 13. The altitude from A to BC is extended to meet the circumcircle of ABC at D . Find AD .
解析
  1. [ 4 ] ABC is a triangle with AB = 15, BC = 14, and CA = 13. The altitude from A to BC is extended to meet the circumcircle of ABC at D . Find AD . 63 Answer: Let the altitude from A to BC meet BC at E . The altitude AE has length 12; one 4 way to see this is that it splits the triangle ABC into a 9 − 12 − 15 right triangle and a 5 − 12 − 13 right triangle; from this, we also know that BE = 9 and CE = 5. Now, by Power of a Point, AE · DE = BE · CE , so DE = ( BE · CE ) /AE = (9 · 5) / (12) = 15 / 4. It then follows that AD = AE + DE = 63 / 4.