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 ...?
Hazel of Madras College, St Andrew's, Fife and
David of Reading School, Berkshire both sent in good proofs of the
first result. Hazel worked out some examples of her own and then
made the conjecture that
$2(x^2 + y^2) = (x + y)^2 + (x - y)^2$
Hazel's proof was as follows:
$(x + y)^2 + (x - y)^2 = (x + y)(x + y) + (x - y)(x - y)$
$= x^2 + 2xy + y^2 + x^2 - 2xy + y^2$
$= 2x^2 + 2y^2$
$= 2(x^2 + y^2)$
So double the sum of two squares is always equal to the sum of
Robert of Newcastle-Under-Lyme School provided
an excellent solution and managed to crack the second toughnut part
of the problem .
Robert sensibly looked at some special cases
with low numbers before making a conjecture that
$3(x^2+y^2+z^2) = (x+y+z)^2 +(z-x)^2 +(z-y)^2 +(y-x)^2$
His proof of the result was:
L.H side = $3x^2 + 3y^2 + 3z^2$
R.H side = $(x + y + z)^2 + (z^2 -2xz + x^2) + (z^2 - 2yz +
y^2)+ (y^2 -2xy + x^2)$
$= (x + y + z)^2+2z^2 + 2y^2 + 2x^2 - 2xz - 2yz -2xy $
$ =(x^2 + y^2 +z^2 +2xy +2yz + 2xz)+ 2z^2 + 2y^2 + 2x^2 -2xz-2yz
- 2xy $
$ = 3x^2 + 3y^2 + 3z^2 $
Finally, Simon of Elizabeth College,
Guernsey managed to prove an extended general version of the result
incorporating the sum of n squares. This uses a lot of summations
and notations that you would usually only encounter at A-level, so
this part is for 16 and above!