HMMT 十一月 2012 · 冲刺赛 · 第 18 题
HMMT November 2012 — Guts Round — Problem 18
题目详情
- [ 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
英文原题
[ 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
英文解析
- [ 10 ]
√
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 3
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
AB + AC = 1 + .3 3