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Shades of Fermat's Last Theorem

The familiar Pythagorean 3-4-5 triple gives one solution to (x-1)^n + x^n = (x+1)^n so what about other solutions for x an integer and n= 2, 3, 4 or 5?

Two Cubes

Two cubes, each with integral side lengths, have a combined volume equal to the total of the lengths of their edges. How big are the cubes? [If you find a result by 'trial and error' you'll need to prove you have found all possible solutions.]


Find the positive integer solutions of the equation (1+1/a)(1+1/b)(1+1/c) = 2

Approximating Pi

Age 14 to 18 Challenge Level:

A method is to calculate the perimeters or areas of the inscribed and circumscribed polygons to find an upper and lower limit for Pi. I have included a solution from Andrei below based on the perimeter of the two polygons. You could then find the average of the two limits to find an approximation to Pi. As Michael points out Pi is irrational and therefore cannot be calculated exactly. We can still obtain very good approximations to Pi using this method by increasing the number of sides. No one really addressed the issue of the problems with this method, especially for Archimedes as he did not have a calculator or tables to look up roots of numbers. For Archimedes methos of calculating square roots see the problem Archimedes and Numerical Roots.  
Andrei's Solution
The length of the side of the circumscribed square is $2r$, where $r$ is the radius of the circle. So, its perimeter is $8r$.

Now, I calculate the length of the inscribed square, using the Pythagoras Theorem, in a right-angled isosceles triangle, with the right angle in the centre of the circle, the two congruent sides - radii of the circle, and the a vertex of the hypotenuses - the side of the inscribed square:

$2r^2 = l^2$

$l = r\sqrt{2}$

So, the perimeter of the inscribed square is $4r\sqrt{2}$.

Knowing that the circumference of a circle is $2{\pi}r$, I can find a minimum and a maximum limit for $p$ from the condition that the circumference of the circle is lower than the perimeter of the circumscribed square and higher than the circumference of the inscribed square:

$4r\sqrt{2} < 2{\pi}r < 8r$

$4\sqrt{2} < 2\pi < 8$

$2\sqrt{2} < \pi < 4$

and approximating $\sqrt{2}$:

$2.8284 < \pi < 4$

Now, I use the same method for the hexagon. First I calculate the side of the circumscribed hexagon in a right-angled triangle with one side (the hypotenuses) the line connecting the centre of the circle with the vertex of the circumscribed hexagon, a side - half the side of the circumscribed hexagon, and the other - the radius of the circle. The side of the hexagon is perpendicular on the radius in the tangency point. The angle between the radius and the hypotenuses is $30^o$ (Here I use the Pythagoras Theorem and also the theorem that says than in a right-angled triangle with one angle of $30^o$, the side opposite to the angle of $30^o$ is 2 times smaller than the hypotenuses): $${ \left({L \over2}\right)^2 + r^2 = L^2 }$$ $${ {{L^2} \over 4} + r^2 = L^2 }$$ or $${ {3L^2 \over{4}} = r^2 }$$ so, $${ L = {2r \over{\sqrt{3}}}= {{2 \sqrt{3}r \over{3}}} }$$ The side of the inscribed hexagon is $r$. So, I found the following inequalities:

$6r < 2\pi r < 6\times (2r\sqrt{3})/3$

$6r < 2\pi r < 4\sqrt{3}r$

$6 < 2\pi < 4\sqrt{3}$

$3 < \pi < 2\sqrt{3}$

and approximating $\sqrt{3}$:

$3 < \pi < 3.4641$