What is the value of the integers a and b where sqrt(8-4sqrt3) = sqrt a - sqrt b?
If a is the radius of the axle, b the radius of each ball-bearing, and c the radius of the hub, why does the number of ball bearings n determine the ratio c/a? Find a formula for c/a in terms of n.
The incircles of 3, 4, 5 and of 5, 12, 13 right angled triangles have radii 1 and 2 units respectively. What about triangles with an inradius of 3, 4 or 5 or ...?
Alex from Stoke-on-Trent Sixth Form College, Dapeng Wang from Claremont Fan Court School, Chuyi Yang from Loughborough High School, Manuele Cavalli-Sforza from the British School of Manila, Feline Angel from Wootton Upper School and Chong Ching Tong, Chan Hei Leong, Chen Wei Jian and Ng Yan Shun from River Valley High School, Singapore all sent in excellent solutions. The first part of the solution came from the Singapore group and the second part, with the diagrams, from Alex.
If $a + b + c = 4$, $ab + bc + ca = 6$ and $abc = 3$, then
$$\frac {1}{a} +\frac{1}{b} + \frac{1}{c} = \frac {(bc + ac +ab)}{ abc}= \frac {6}{3} = 2.$$
Also $$\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}=\frac{( c + a + b )}{abc}= \frac{4}{3}.$$
In the image below, each side of the coloured squares has been assigned the lengths, and the area of each rectangle is written inside the rectangle. The total area of the diagram is the sum of these areas, and is equivalent to $(a+b+c)^2$ showing that this expands to: $a^2 + b^2 + c^2 + 2ab + 2ac + 2bc$.
Squaring both sides of the first equation gives $(a+b+c)^2 = 16$. Multiplying each side of the second equation by 2 gives $2ab + 2ac + 2bc = 12$ . Subtracting these last two equations leaves the sum of the squares so we have $a^2 + b^2 + c^2 = 4$.
Adding the volumes we get
$(a+b+c)^3 = a^3 + b^3 + c^3+ 3a^2b+3a^2c+3ab^2+3ac^2 +3b^2c+3bc^2+6abc$.