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The Ladder Problem
Introduction
My father does a lot of renovation work in the house and one
problem he had was bringing supplies up the stairs. More
specifically the ladder. The aim of this investigation would be
to find the longest ladder possible for a staircase. The width
of both the corridors would affect the large the ladder can be.
Moreover, the width of the ladder affects the length as well.
Figure 1
At first, I will find the general equation for the maximum
length of the ladder when both corridors are the same length. Here I will be assuming the
ladder is 2D. Afterwards, I will attempt to find the general equation for when the corridor
widths are different. Knowing these two equations I can manipulate them to take into account
the width of the ladder.
Exploration – 1st Attempt
When I first looked at this problem, I knew that it would involve trigonometry. The way I
imagined the problem was like coordinates.
Figure 2
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𝐴 – 𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑐𝑜𝑟𝑟𝑖𝑑𝑜𝑟
𝐵 – 𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑐𝑜𝑟𝑟𝑖𝑑𝑜𝑟
𝐶 − 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑎𝑑𝑑𝑒𝑟
𝐷 – 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑎𝑑𝑑𝑒𝑟
𝐸 – 𝑐𝑜𝑟𝑛𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑡𝑎𝑖𝑟𝑐𝑎𝑠𝑒
𝐿 – 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑎𝑑𝑑𝑒𝑟
Then because there is a right-angled triangle, Pythagoras Theorem could be used to find L in
terms of C and D.
Figure 3
This equation can be used to find the equation of a line, since we are thinking about the
problem within an axis and coordinates. Now, given that the ladder touches the corner, E, we
know that the coordinates are (A, B). So, when inputted into the equation of the line, it now
has 4 unknowns, A, B, L, and D. The equation can be differentiated to find the minima and
find an equation where D is written in terms of A and B. This is because the corridor can be
infinitely large in length, but the corridor are limits. So, that’s why we are finding a minima.
This can then be inputted into the equation of the line to get one in terms of A, B and L.
This attempt did not work because I had problems with simplifying the final equation.
2
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Exploration – Final Attempt – Same Corridor Length
First, we need to split the problem up into smaller parts so it’s easier to solve. Here, the ladder
is split into lengths depending on where it touches the corner of the staircase.
Figure 4
𝑎 − 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑐𝑜𝑟𝑟𝑖𝑑𝑜𝑟
𝐿! − 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑜𝑛𝑒 𝑠𝑖𝑑𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑎𝑑𝑑𝑒𝑟
𝐿" − 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑜𝑡ℎ𝑒𝑟 𝑠𝑖𝑑𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑎𝑑𝑑𝑒𝑟
After, trigonometry can be used to find an equation of 𝐿 in terms of 𝑎. However, we need to
find the angles needed to configure this equation.
Figure 5
3
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The Ladder Problem
Introduction
My father does a lot of renovation work in the house and one
problem he had was bringing supplies up the stairs. More
specifically the ladder. The aim of this investigation would be
to find the longest ladder possible for a staircase. The width
of both the corridors would affect the large the ladder can be.
Moreover, the width of the ladder affects the length as well.
Figure 1
At first, I will find the general equation for the maximum
length of the ladder when both corridors are the same length. Here I will be assuming the
ladder is 2D. Afterwards, I will attempt to find the general equation for when the corridor
widths are different. Knowing these two equations I can manipulate them to take into account
the width of the ladder.
Exploration – 1st Attempt
When I first looked at this problem, I knew that it would involve trigonometry. The way I
imagined the problem was like coordinates.
Figure 2
1
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𝐴 – 𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑐𝑜𝑟𝑟𝑖𝑑𝑜𝑟
𝐵 – 𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑐𝑜𝑟𝑟𝑖𝑑𝑜𝑟
𝐶 − 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑎𝑑𝑑𝑒𝑟
𝐷 – 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑎𝑑𝑑𝑒𝑟
𝐸 – 𝑐𝑜𝑟𝑛𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑡𝑎𝑖𝑟𝑐𝑎𝑠𝑒
𝐿 – 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑎𝑑𝑑𝑒𝑟
Then because there is a right-angled triangle, Pythagoras Theorem could be used to find L in
terms of C and D.
Figure 3
This equation can be used to find the equation of a line, since we are thinking about the
problem within an axis and coordinates. Now, given that the ladder touches the corner, E, we
know that the coordinates are (A, B). So, when inputted into the equation of the line, it now
has 4 unknowns, A, B, L, and D. The equation can be differentiated to find the minima and
find an equation where D is written in terms of A and B. This is because the corridor can be
infinitely large in length, but the corridor are limits. So, that’s why we are finding a minima.
This can then be inputted into the equation of the line to get one in terms of A, B and L.
This attempt did not work because I had problems with simplifying the final equation.
2
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Exploration – Final Attempt – Same Corridor Length
First, we need to split the problem up into smaller parts so it’s easier to solve. Here, the ladder
is split into lengths depending on where it touches the corner of the staircase.
Figure 4
𝑎 − 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑐𝑜𝑟𝑟𝑖𝑑𝑜𝑟
𝐿! − 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑜𝑛𝑒 𝑠𝑖𝑑𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑎𝑑𝑑𝑒𝑟
𝐿" − 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑜𝑡ℎ𝑒𝑟 𝑠𝑖𝑑𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑎𝑑𝑑𝑒𝑟
After, trigonometry can be used to find an equation of 𝐿 in terms of 𝑎. However, we need to
find the angles needed to configure this equation.
Figure 5
3
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