STEP III 2007 question 6 solution

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Firstly,

and similarly for q.

Thus we have

and the first result follows.

Then, note that if and are perpendicular,

or

Either way, we have

and thus

from which it follows that

as required.

For the next part, we have in the case : (let represent in the Argand plane, with the circle centred in the origin and with radius a)

Multiplying the first and the third equations, and then dividing by the second, we get

there are obviously two distinct that satisfy this, as required.

For , it's instead

In this case, any can satisfy this, becayse if you choose a , you can always choose a so that the first equation is satisfied, and then a so that the second one is satisfied, and then a so that the third is satisfied, and then, and here's the catch, equation 4 follows from the other three: (multiply the first and third equations and divide by the fourth) and so the fourth equation will automatically satified.

For larger n, we have exactly the same sitution as in n = 3 when n is odd (you can use the same reasoning, just alternatingly multiply and divide all the equations together) and as in n = 4 when n is even (again the same reasoning, the last equation can always be obtained by alternatingly multiplying and dividing the first ones together.

Solution by ukgea.