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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 10: Solve Me!

Stage: 5 Short Challenge Level:

Any sensible numerical method will lead to a solution $-16.3(2)$.

In particular, an interval-halving method is efficient and simple to implement. You can make it a little quicker by choosing a sensible starting point: note that any solution would have to be negative, since all of the coefficients are positive; another moment of inspection will also show that the solution must lie between $-10$ and $-100$, giving you a sensible starting point for a computation.

To determine whether there are any other solutions, note that the expression is a cubic and will therefore have either $1$ or $3$ real solutions.

To explore the properties of the cubic $y=2x^3+34x^2+567x+8901$, look at the turning points. Differentiating, we find that
$$\frac{dy}{dx} = 6x^2+68x+567$$
The discriminiant of this quadratic is $68^2-4\times 6 \times 567 = -8984$. Since this is negative, there are no turning points and the cubic consequently only has $1$ real solution.