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?
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.
The illustration shows the graphs of twelve functions. Three of
them have equations y=x^2, x=y^2 and x=-y^2+2. Find the equations
of all the other graphs.
If we take any 8 consecutive numbers: $$n, n+1, n+2, n+3, n+4, n+5, n+6, n+7$$ then the sum of the squares of four of these numbers is equal to the sum of the squares of the other four.
This means that the terms in $x^2$ in $x$ and the constant term must be split equally.
If we sum the squares of each of the eight consecutive numbers and find the mean, this will equal the sum of each of the four terms needed. So adding all the squares we have: $$n^2 + (n+1)^2 + (n+2)^2 + (n+3)^2 + (n+4)^2 (n+5)^2 + (n+6)^2 + (n+7)^2 = 8n^2 + 56n + 140 $$ So the two sides of the equality must have the value $$4n^2 + 28n + 70$$ $$(n+1)^2 + (n+2)^2 + (n+4)^2 +(n+7)^2 = n^2 +
(n+3)^2 + (n+5)^2 + (n+6)^2$$