**Visualising Algebra Through Geometry**

This picture above is a fantastic example of how we can use geometry to visualise an algebraic expression. It’s taken from Brilliant – which is a fantastic new forum for sharing maths puzzles. This particular puzzle was created and uploaded by Arron Kau. The question is, which of the following mathematical identities does this image represent?

See if you can work it out! I will put the answer in white text at the bottom of the post – highlight it to reveal the solution.

Another example of the power of geometry in representing mathematical problems is provided by Ian Stewart’s Cabinet of Mathematical Curiosities. The puzzle itself is pretty famous:

*A farmer wants to cross a river and take with him a wolf, a goat, and a cabbage. There is a boat that can fit himself plus either the wolf, the goat, or the cabbage. If the wolf and the goat are alone on one shore, the wolf will eat the goat. If the goat and the cabbage are alone on the shore, the goat will eat the cabbage. How can the farmer bring the wolf, the goat, and the cabbage across the river?*

The standard way of solving it is trial and error with some logic thrown in. However, as Ian Stewart points out, we can actually utilise 3 dimensional geometry to solve the puzzle. We start with a 3D wolf-goat-cabbage (w,g,c) space (shown in the diagram). All 3 start at (0,0,0). 0 represents this side of the bank, and 1 represents the far side of the bank. The target is to get therefore to (1,1,1). In (w,g,c) space , the x direction represents the wolf’s movements, the y direction the goat and z the cabbage. Therefore the 8 possible triplet combinations are represented by the 8 vertices on a cube.

We can now cross out the 4 paths:

(0,0,0) to (1,00) as this leaves the goat with the cabbages

(0,0,0) to (0,0,1) as this leaves the wolf with the goat

(0,1,1) to (1,1,1) as the farmer would leave the goat and cabbage alone

(1,1,0) to (1,1,1) as the farmer would leave the wolf and goat alone.

which reduces the puzzle to a geometric problem – where we travel along the remaining edges – and the 2 solutions are immediately evident.

(eg. (0,0,0) – (0,1,0) – (1,1,0) – (1,0,0) – (1,0,1)- (1,1,1) )

What’s really nice about this solution is that it shows how problems seemingly unrelated to mathematics can be “translated” in mathematics – and also it shows how geometrical space can be used for problem solving.

Solution to the initial puzzle, highlight to reveal: The answer is the third option – 1^{3} + 2^{3}…. = (1+2+….)^{2}. This is quite a surprising identity. You can see it by seeing that there are (for example) 2 squares of length 2 – this gives you a total area of 2x2x2 = 2^{3}. Adding all the squares will give you the same area as a square of sides (1+2+3….)(1+2+3….) – hence the result.

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October 2, 2016 at 3:00 pm

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