HMMT 十一月 2012 · 冲刺赛 · 第 18 题

HMMT November 2012 — Guts Round — Problem 18

[ 10 ] Let △ ABC be a triangle with AC = 1 and ∠ ABC obtuse. Let D and E be points on AC such ◦ that ∠ DBC = ∠ ABE = 90 . If AD = DE = EC , find AB + AC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT NOVEMBER 2012, 10 NOVEMBER 2012 — GUTS ROUND

解析

[ 10 ] √ 3 Answer: 1 + DBC is a right triangle with hypotenuse DC. Since DE = EC, E is the midpoint 3 of this right triangle’s hypotenuse, and it follows that E is the circumcenter of the triangle. It follows that BE = DE = CE, as these are all radii of the same circle. A similar argument shows that BD = DE = AD. ◦ Thus, BD = DE = DE, and triangle BDE is equilateral. So, ∠ DBE = ∠ BED = ∠ EDB = 60 . We ◦ ◦ ◦ have ∠ BEC = 180 − ∠ BED = 120 . Because BE = CE , triangle BEC is isosceles and ∠ ECB = 30 . ◦ ◦ ◦ Therefore, DBC is a right triangle with ∠ DBC = 90 , ∠ BCD = 30 , and ∠ CDB = 60 . This √ √ 2 2 3 3 √ means that CD = BC . Combined iwth CD = , we have BC = . Similarly, AB = , so 3 3 3 3 √ 3 AB + AC = 1 + . 3