You are currently browsing the tag archive for the ‘ladder’ tag.

**Can you solve Oxford University’s Interview Question?**

The excellent Youtube channel Mind Your Decisions is a gold mine for potential IB maths exploration topics. I’m going to follow through my own approach to problem posed in the video. The problem is to be able to trace the movement of the midpoint of a ladder as it slides down a wall. This has apparently been used as an Oxford interview question to test the ability to investigate novel problems.

It’s normally a good idea to start with a specific case with some nice numbers, to see what happens. So, I’ll choose a 3,4,5 triangle, where the ladder has a fixed length of 5 and has endpoints with coordinates (0,4) and (3,0). The midpoint is given by ((0+3)/2, (4+0)/2) = (1.5, 2).

Next I imagine what would happen to the point (3,0) if the ladder slipped down the wall. (3,0) would become (3+t,0) where t is a parameter. Given that the length is fixed as 5, I can now find the new height of the ladder up the wall using Pythagoras:

The new height is given by:

Therefore the new midpoint is given by:

We can now define our curve parametrically:

Therefore we can make t the subject in the first equation to get an equation just in terms of x and y.

Therefore we can rearrange to get:

This is the equation of a circle centred at (0,0) with radius 2.5:

This graph therefore traces the movement of the midpoint of the ladder (note that when the ladder was vertical against the wall the midpoint would be 2.5 high hence the graph starts at (0,2.5).

**The general case**

Now we have worked through the maths for a specific case, the general case isn’t too much extra work. For a triangle with base a and height b we would have the following midpoint coordinates:

This would lead to the following equation:

Which would rearrange to give the equation of the circle:

This is a circle centered at (0,0) with radius:

**Another approach**

This method is an alternative to the version above – this time using trigonometry. We start with the triangle below:

and then let the ladder slide to get the following (as the angle will get smaller t will be negative):

We can then define the midpoint coordinates as:

We can then rearrange and square both sides to get the following:

We can then use the trig identity for cosine squared theta + sine squared theta = 1:

Which rearranges to give the same result as before:

So, there we go – we’ve passed an Oxford interview question with a couple of different methods! The approach of first exploring the topic with a simple case is often a good starting point for these sorts of problems – as it allows you to gain an understanding of what is happening without getting too bogged down with variables. You can watch the video for a quicker solution – are there any other ways of approaching this problem you can find? How could this problem be modified?