Sunday Afternoon Maths VI
Triangle numbers
Let \(T_n\) be the \(n^\mathrm{th}\) triangle number. Find \(n\) such that: $$T_n+T_{n+1}+T_{n+2}+T_{n+3}=T_{n+4}+T_{n+5}$$
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\(T_{n} = \frac{1}{2}n(n+1)\), so:
$$T_{n}+T_{n+1} = \frac{1}{2}n(n+1) + \frac{1}{2}(n+1)(n+2)$$
$$= (n+1)^2$$
So, we are looking for \(n\) such that \((n+1)^2+(n+3)^2=(n+5)^2\). This is true when \(n=5\) (\(6^2+8^2=10^2\)).
Extension
Find \(n\) such that \(T_{n}+T_{n+1}+T_{n+1}+T_{n+2}=T_{n+2}+T_{n+3}\).
Ellipses
A piece of string 10cm long is tied to two pins 6cm apart.
The string is used to draw an ellipse. The pins are then moved 2cm further apart and a second ellipse is drawn. Which ellipse has the larger area?
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The area of an ellipse is \(\pi ab\) where \(a\) and \(b\) are the distances from the centre of the ellipse to the closest and furthest points on the ellipse.
In the first ellipse, \(a=5\mathrm{cm}\) and \(b=4\mathrm{cm}\), so the area is \(20\pi\mathrm{cm}^2\). In the second ellipse, \(a=5\mathrm{cm}\) and \(b=3\mathrm{cm}\), so the area is \(15\pi\mathrm{cm}^2\). Hence, the first ellipse has the larger area.
Extension
How far apart should the pins be placed to give the ellipse with the largest area?
