### Circles Ad Infinitum

A circle is inscribed in an equilateral triangle. Smaller circles touch it and the sides of the triangle, the process continuing indefinitely. What is the sum of the areas of all the circles?

### Climbing Powers

$2\wedge 3\wedge 4$ could be $(2^3)^4$ or $2^{(3^4)}$. Does it make any difference? For both definitions, which is bigger: $r\wedge r\wedge r\wedge r\dots$ where the powers of $r$ go on for ever, or $(r^r)^r$, where $r$ is $\sqrt{2}$?

A quadrilateral changes shape with the edge lengths constant. Show the scalar product of the diagonals is constant. If the diagonals are perpendicular in one position are they always perpendicular?

# Weekly Challenge 6: AP Train

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

This problem is based on the formula for the sum of the first $n$ natural numbers
$$1+2+\dots + (n-1) + n =\frac{1}{2}n(n+1)$$
We have
Its easier in a problem like this to introduce some notation. Let's write $S(n)$ to mean the sum of the first $n$ natural numbers.

Since the two sums are equal we have

$$S(1111+N-1)-S(1110) = S(1111+N -1 +D) - S(1111 + D)$$
Let's put the formula into each of these and cancel each factor of a half.
$$(1110+N)(1111+N)-(1110)(1111)=(1110+N+D)(1111+N+D)-(1111+D)(1112+D)$$
Before leaping in we can see that many parts cancel, so we can put
$$2221N + N^2 = 2221(N+D) + (N+D)^2 - 2223D - D^2- 2222$$
Collecting things together gives
$$D(N-1) = 1111$$
Now for a bit of number theory: $1111 = 11\times 101$. Thus, for solutions we require
$$D = 11, N=102\quad\mbox{ or } D=101, N=12$$

You can see these two sums in action on this spreadsheet screenshot