Pretty Narrow Alley

Number Theory Level 5

Two ladders of different lengths cross in a narrow alley (see graphic below picture).

The rungs of the ladders are spaced a foot apart, from end to end, and the alley is an integer number of feet wide. The ladders cross exactly where there are rungs, i.e., integer number of feet from intersection point to ends of ladders. If no ladder is greater than $30$ feet in length, how wide is the alley?

The answer is 8.

1 solution

Michael Mendrin
Oct 7, 2014

The ladders are $12$ and $28$ feet long, the only solution possible.

Edit: Let $a,b,c,d$ be the integer distances from the ends of the ladders to the point of intersection, where $c+b$ is the length of the longer ladder and $a+d$ is the length of the shorter ladder. Given integers $a,b,c,d$ , then the square of the width of the alley is

${ w }^{ 2 }=\left( { a }^{ 2 }-{ b }^{ 2 } \right) \dfrac { a+d }{ a-d }$

yielding an integer $w$ only for $a,b,c,d$ equal to $9,7,21,3$ , and $w=8$

To reduce the search, we limit $\ a,b,c,d$ to those that satisfy $ab=cd$ , $a+d<31$ and $c+b<31$

question

Guiseppi Butel - 6 years, 8 months ago

I reviewed my notes, and found a bug in it. But, fortunately, the solution is still unique. Thank goodness for that, because it seems like this is now one of my more popular problems. I've revised my answer.

Michael Mendrin - 6 years, 8 months ago

How do you arrive at the answer?

Niranjan Khanderia - 6 years, 8 months ago

First, I'll give others the chance to offer their method of arriving at the answer.

Michael Mendrin - 6 years, 8 months ago

OK. Thank you.

Niranjan Khanderia - 6 years, 8 months ago

Hi. Michael. I, too, am quite intrigued by this problem. Would you please explain how you formulated the equation for finding the width? I have a list of at least 80 sets of 4 integers that satisfy the primary conditions set on the ladders and several that come to integral values for the width of the alley however the walls are not quite vertical. Your solution appears to be unique.

Guiseppi Butel - 6 years, 8 months ago

Let $x+y=w$ be the width of the alley, where $x,y$ are the bases of the right triangles that have $a,b$ as the hypotenuse. Because the right triangles share a common altitude, we have

${ a }^{ 2 }-{ x }^{ 2 }={ b }^{ 2 }-{ y }^{ 2 }$

${ a }^{ 2 }-{ b }^{ 2 }={ x }^{ 2 }-{ y }^{ 2 }=w(x-y)$

But we also know that $y=\dfrac { d }{ a } x$ , so that $w=\left( 1+\dfrac { d }{ a } \right) x$ , and so we end up with

${ a }^{ 2 }-{ b }^{ 2 }={ w }^{ 2 }{ \left( 1+\frac { d }{ a } \right) }^{ -1 }\left( 1-\frac { d }{ a } \right)$

which leads to the expression used to filter out solutions.

Michael Mendrin - 6 years, 8 months ago

Thanks, Michael.

Guiseppi Butel - 6 years, 8 months ago

I have some sets as well and I can't understand why they're not accettable

Paolo Quadri - 6 years, 7 months ago

Wonder why there aren't any answers for my adaption of this problem? 'Take Off on "Pretty Narrow Alley"'

Guiseppi Butel - 6 years, 7 months ago

Pretty Narrow Alley

The answer is 8.

1 solution

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