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        riddles >> medium >> William the Logger
        
(Message started by: rloginunix on Apr 1^st, 2015, 6:43pm)

Title: William the Logger
Post by rloginunix on Apr 1^st, 2015, 6:43pm

William the Logger.

Down the Math River that has parallel banks and a width a Willy Wu is floating logs which can be approximated as absolutely rigid line segments. At some point the river makes a ninety-degree turn and while its banks remain parallel its width is now b:

http://www.ocf.berkeley.edu/~wwu/YaBBAttachments/rlu_wtl.png

For a given a and b what is the longest log L can William safely float around the bend in such a way that the log does not get bent, broken, stuck, submerged below or popped above water (when only 2-dimensional transformations are allowed)?

Put it another way. If William knows a, b and L ahead of time when should he make the trip?

Title: Re: William the Logger
Post by jollytall on Apr 1^st, 2015, 9:45pm

First idea:
Make the log in alpha degree compared to the horizontal line, when it is almost stuck. Then the "corner" splits it into two parts, x across river a and y across river b.
x*cos(alpha) = a
y*sin(alpha) = b
The log is a/cos(alpha)+b/sin(alpha)
The minimum of it is when the derivative is 0, i.e.
-a*cos(alpha)/sin2(alpha)+b*sin(alpha)/cos2(alpha)=0.
tg(alpha) = (a/b)^(1/3)
From alpha x+y = L can be calculated.

Title: Re: William the Logger
Post by pex on Apr 1^st, 2015, 10:06pm

on 04/01/15 at 21:45:31, jollytall wrote:

That's pretty much what I did. The result is rather cute: L = [hide](a^2/3 + b^2/3)^3/2[/hide].

Title: Re: William the Logger
Post by rloginunix on Apr 2^nd, 2015, 9:28am

You guys are both correct and what is rather amazing is that I took exactly the same route. Interesting.

There is also a more analytic (academic) approach, if you will: [hide]the orthogonal banks are actually the XOY axes. The log must slide along both playing a role of a tangent to some curve. So the question is - find the curve the length of tangent to which trapped between the axes is constant at all times. If the (a,b) corner is inside that curve - no go. Pythagoras and an equation of a straight line lead to a differential equation[/hide] in this case.

Turns out that the equation given by pex describes a [hide]curve known as astroid[/hide].

Well done.

(I interpreted jollytall's last equation like so: [hide]tg(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif) is almost a derivative, for it to be a real one the angle must be http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif- http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif and then we get a trivial differential equation dy/dx = -(y/x)^1/3[/hide] ...)

PS
This particular puzzle interconnects with [hide]A Cat on a Ladder[/hide] and [hide]Infinite Fast Ladder[/hide] in the easy section.

[edit]
Fixed two spelling errors.
[/edit]