### 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?

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.
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.