Converse

Clearly if a, b and c are the lengths of the sides of a triangle and the triangle is equilateral then a^2 + b^2 + c^2 = ab + bc + ca. Is the converse true, and if so can you prove it? That is if a^2 + b^2 + c^2 = ab + bc + ca is the triangle with side lengths a, b and c necessarily equilateral?

Consecutive Squares

The squares of any 8 consecutive numbers can be arranged into two sets of four numbers with the same sum. True of false?

Parabolic Patterns

The illustration shows the graphs of fifteen functions. Two of them have equations y=x^2 and y=-(x-4)^2. Find the equations of all the other graphs.

Stage: 4 Challenge Level:

Elliot from Wilson's School sent us the following observations:

The first two functions are almost identical, as they both square the number, add on the original number and then subtract six.
However, function $g$ also adds two at the end, causing function $f$ to always be two less than function $g$, when $x$ is the same number. Because of this, their graphs are almost identical curves, apart from function $f$'s graph being two below function $g$'s graph.

The graph of function $f$ dips to $-6.25$ and function $g$ dips down to $-4.25$.

Krystof from Uhelny Trh, Prague, describes this by saying that the second function "shifts" the first function upwards by 2 units.
Elliot goes on to explore the second pair:

For the second pair of functions, function $f$ remains the same, but function $g$ now adds on two before the number is squared, instead of at the end. This changes the relationship considerably, as before, the output of $f$ would always be two less than function $g$'s output, but if the number is squared after two is added, the difference between the outputs is very different. The two graphs are again very similar to each other, however, this time function $f$ produces a curve slightly to the right of function $g$'s curve with them both dipping down to $-2.5$.

In fact, function $g$'s curve is two units to the right of function $f$'s curve - can you see why?

George, who is also from Wilson's School, sent us a very clear explanation of the first part of the problem, which you can read by clicking here.