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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?

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Areas and Ratios

Do you have enough information to work out the area of the shaded quadrilateral?

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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 $\sqrt{2}$?

Root Hunter

Age 16 to 18 Challenge Level:

$\mathbf{f(x) =\frac{1}{x-2} + \frac{1}{x-3}}$

For the argument to be valid, we must ensure that our functions are continuous in the range in question. For example, $f(1.9) \approx -10.9$ and $f(2.1) \approx 8.9$, but there is no zero between these values since the function in not continuous at $x = 2$.

However $f(2.9) \approx -8.9$, and $f$ is continuous for $2 < x < 3$ so there is a root between $2.1$ and $2.9$.

$\mathbf{f(x) = x^x - 1.5x}$

$f(1) =1^1 - 1.5 = -0.5$

and $f(2) = 2^2 - 1.5\times 2 = 4 - 3 = 1$

so there is a root between $1$ and $2$.

$\mathbf{f(x) = x^{1000000} + 1000000^x - 17}$

$f(-1) = (-1)^{1000000} + 1000000^{-1} - 17 = 1 + 10^{-6} - 17 = -15.999999$

and $f(1) = 1^{1000000} + 1000000^1 - 17 = 1 + 1000000 - 17 = 999984$

so there is a root between $-1$ and $1$.

$\mathbf{f(x) = \cos{(\sin{(\cos{x})})} -\sin{(\cos{(\sin{x})})}}$

$f(0) = \cos{(\sin{1})} - \sin{(\cos{0})} = \cos{0.84147 \cdots} -\sin{1} = 0.66637\cdots - 0.84147\cdots = -0.1751\cdots$

and $f(\frac{\pi}{2}) = \cos{(\sin{0})} - \sin{(\cos{1})} = \cos{0} - \sin{0.54030\cdots} = 1 - 0.5144\cdots = 0.4856\cdots$

So there is a root between $0$ and $\frac{\pi}{2}$.