### Weekly Challenge 43: A Close Match

Can you massage the parameters of these curves to make them match as closely as possible?

### Weekly Challenge 44: Prime Counter

A weekly challenge concerning prime numbers.

### Weekly Challenge 28: the Right Volume

Can you rotate a curve to make a volume of 1?

# Weekly Challenge 7: Gradient Match

##### Stage: 5 Short Challenge Level:

There are various intuitive ways to think about such results but creating really clear arguments is somewhat more difficult. The descriptions given here do not constitute proofs as such, but do introduce advanced analytical ways of thinking which are refined at university.

Consider the first case; the others are similar.

Clearly the special case of the curve joining the two points $(0, 0)$ and $(8, 8)$ with a straight line has gradient $1$ everywhere. Any other curve between these two points can be considered as a deformed version of this line. Wherever the curve is deformed outwards point bulges will necessarily occur. Imagine dragging the line $y=x$ up or down. It will cut through each bulge but eventually pass out of each bulge. As it passes out of each bulge it will touch each bulge at a single point. These are the points with gradient 1.

Analytical proofs proceed along these sorts of lines:

Sketch: Imagine the curve being sketched starting from the origin. Imagine that the gradient of your curve is always less than or equal to some number $M$ which satisfies $M< 1$. Then in $8$ units of $x$ the $y$ value of the curve can increase by at most $8M$, which is less than $8$. Thus, it could not pass through the point $(8, 8)$; therefore the maximum achieved gradient cannot be less than $1$. Similarly the minimum achieved gradient cannot be greater than $1$. It is thus intuitively clear that a gradient of $1$ is achieved somewhere.

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