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        riddles >> medium >> Ladder and Crate (Part 2)
        
(Message started by: ThudanBlunder on Jul 18th, 2008, 5:27pm)
Title: Ladder and Crate (Part 2)
Post by ThudanBlunder on Jul 18th, 2008, 5:27pm
A headmaster composed the following problem for his group of four high-flying maths students:

A long fireman's ladder of length d" rests squarely, and at an angle of less than 45o, against a vertical wall with its foot on the horizontal ground, a distance b" from the wall. A cubical crate fits flush into the angle of the wall and the ground, and just touches the ladder. What is the length of the side of the crate?

After some thought he then replaced the constants b and d with integers, such that the answer would also be an integer. However, to each student he gave a different integer value (< 1800) for d, the length of the ladder. The answers submitted by the four students were all the same. And they were all correct! So what was the side length of the crate?
Title: Re: Ladder and Crate (Part 2)
Post by Eigenray on Jul 18th, 2008, 6:27pm
High-flying, eh?  ;)

But wouldn't d http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif 2008 be more appropriate?
Title: Re: Ladder and Crate (Part 2)
Post by ThudanBlunder on Jul 18th, 2008, 7:17pm

on 07/18/08 at 18:27:21, Eigenray wrote:
High-flying, eh?  ;)

Yeah, it's a Brit primary school. :P But I am prepared to concede that you know the meaning of that word better than I do. ;)

I can't' see why 2008 would be more appropriate.
Unless one is a numerologist (perish the thought), 1800 rules out the next possible solution just as well as 2008.
Or are you just letting me know that you have already solved it? :) Perhaps I should have put it in Easy. ::)

Title: Re: Ladder and Crate (Part 2)
Post by Eigenray on Jul 18th, 2008, 8:49pm

on 07/18/08 at 19:17:06, ThudanBlunder wrote:
Yeah, it's a Brit primary school. :P But I am prepared to concede that you know the meaning of that word better than I do. ;)

I thought it was a reference to the answer. ???


Quote:
I can't' see why 2008 would be more appropriate.
Unless one is a numerologist (perish the thought), 1800 rules out the next possible solution just as well as 2008.

It is the largest bound for which there is a unique solution, not to mention the current year.


Quote:
Or are you just letting me know that you have already solved it? :)

What, me solve a math problem?  You must be thinking of someone else.


Quote:
Perhaps I should have put it in Easy. ::)

Depends, is it meant to be done by hand?
Title: Re: Ladder and Crate (Part 2)
Post by ThudanBlunder on Jul 18th, 2008, 9:24pm

on 07/18/08 at 20:49:26, Eigenray wrote:
It is the largest bound for which there is a unique solution, not to mention the current year.

True; but I thought a 150' ladder was already pushing it.


Quote:
What, me solve a math problem?  You must be thinking of someone else.

Someone else? Well, I reckon anyone whose surname begins with a trigonometric function must be pretty good at maths. :P


Quote:
Depends, is it meant to be done by hand?

Even if not, one still needs to derive the formulae.
Title: Re: Ladder and Crate (Part 2)
Post by Eigenray on Jul 19th, 2008, 12:52pm
Say the height of the ladder is a, so a2+b2=d2.  Then
[hideb]c = ab/(a+b), and a = bc/(b-c) is rational.
Since b,c are integers, a is an integer.  So
a = k(m2-n2)
b = k(2mn)
d = k(m2+n2),
where n<m are relatively prime of opposite parity, and
c = ab/(a+b) = 2kmn(m2-n2)/[m2+2mn-n2]
is an integer.
But m2+2mn-n2 is relatively prime to 2mn(m2-n2), so
k = u(m2+2mn-n2)
for some integer u.  Now
d = u(m2+2mn-n2)(m2+n2) < 1800.
Since m2+2mn-n2 = 2m2-(m-n)2 >= m2+2m-1,
m4-1 < (m2+2m-1)(m2+1) < 1800,
so m <= 6.  So there are not very many possibilities.
For each pair (m,n), c = 2mn(m+n)(m-n)*u, where
u < 1800/[(m2+2mn-n2)(m2+n2)].
Here are the possibilities for (m,n), and c:
(2,1): c=12*{1,...,51}
(3,2): c=60*{1,...,8}
(4,1): c=120*{1,2,3,4}
(4,3): c=168*{1,2}
(5,2): c=420
(6,1): c=420
If c appears 4 times, it must be on one of the last 3 lists.
168 and 336 are only on 2 of the lists, so c=420.[/hideb]
Title: Re: Ladder and Crate (Part 2)
Post by ThudanBlunder on Jul 23rd, 2008, 6:50pm
To sum up,
a = Max or Min {n(p2 - q2)(p2 - q2 + 2pq), 2npq(p2 - q2 + 2pq)}
b = Max or Min {n(p2 - q2)(p2 - q2 + 2pq), 2npq(p2 - q2 + 2pq)}
c = 2npq((p2 - q2)
d = n(p2 + q2)(p2 - q2 + 2pq)
for integers n,p,q


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