An easier version of this puzzle is
here.
A large panel needs to be moved through a corridor, the panel is tall as the corridor. The corridor is A feet wide before a right angle turn, after the turn, it is B feet wide.
What is the maximum length of the panel that can pass through this corner.
Overhead view of the hallway:
+------------+---
| / |
| / |B ft
| / |
| /+------
| / |
| / |
| / |
| / |
| / |
| / |
| / |
|/ |
+<-A ft-->|
Let's call the portion of the panel from the inside corner to the wall of corridor A x, and the portion from the corner to the wall of B y. Then x + y = Length. ¥è is the angle formed between the panel and the side wall of A, and also between the panel and a line parallel to width B.
sin ¥è = A/x cos¥è = B/y
x = A/sin¥è y = B/cos¥è
L = x + y = A/sin¥è + B/cos¥è
To find the minimum length (or the max allowable), differentiate the above equation to get:
- Acos¥è/sin©÷¥è + Bsin¥è/cos©÷¥è = 0
or Acos¥è/sin©÷¥è = Bsin¥è/cos©÷¥è
rewrite to A/B = tan©ø¥è or tan¥è = ©ø¡îA / ©ø¡îB
So picture a right triangle with sides of cube root of A and cube root of B. The hypotenuse H equals the square root of the sum of the cube root of A squared and the cube root of B squared.
sin¥è = ©ø¡îA / H cos¥è = ©ø¡îB / H
L = A/sin¥è + B/cos¥è = A/©ø¡îA / H + B/©ø¡îB / H
Simplifying, L = the sum of A to the 2/3 power and B to the 2/3 power, the sum raised to the 3/2 power.
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Posted by lake
on 2004-10-21 15:20:19 |