HMMT 二月 2000 · CALC 赛 · 第 6 题
HMMT February 2000 — CALC Round — Problem 6
题目详情
英文原题
- A hallw a y of width 6 feet meets a hallw a y of width 6 5 feet at righ t angles. Find the length of the longest pip e that an be arried horizon tally around this orner.
解析
英文解析
- Assume the pip e barely ts around the orner (i.e. it is in on ta t with the orner).
The lo w er orner is at (0 ; 0) and the upp er orner is at (6 ; 6 5). Call x the p oin t onpthe lo w er w all it hits at the tigh test sp ot. Giv en an x , the longest a pip e ould be with 0
r 0
p pp
36 52
one end at x and leaning against the (6 ; 6 5 ) orner is x + (6 5 + ) . W e w an t 2
x 600
the minim um of all of these "longest pip es", be ause the pip e needs to t at all angles 0
around the orner. T aking the deriv ativ e (without the square ro ot for simpli it y) and
3 2
setting it equal to 0, we need to solv e x 6 x + 36 x 1296 = 0. W e an qui kly nd
0 00
that x = 12 is the only go o d solution, so the maxim um length is 12 6 .pdy 0