HMMT 二月 2000 · CALC 赛 · 第 6 题

HMMT February 2000 — CALC Round — Problem 6

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 b e arried horizon tally around this orner.

解析

Assume the pip e barely ts around the orner (i.e. it is in on ta t with the orner). p 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 on 0 the lo w er w all it hits at the tigh test sp ot. Giv en an x , the longest a pip e ould b e with 0 r p p p 2 36 5 2 one end at x and leaning against the (6 ; 6 5 ) orner is x + (6 5 + ) . W e w an t 0 0 x 6 0 the minim um of all of these "longest pip es", b e ause the pip e needs to t at all angles 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, w e need to solv e x 6 x + 36 x 1296 = 0. W e an qui kly nd 0 0 0 p that x = 12 is the only go o d solution, so the maxim um length is 12 6 . 0 dy