STEP III 2007 question 6 solution

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STEP III 2007 question 6 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP III 2007 question 1 solution

Firstly,

pp^* = |p|^2 = a^2

and similarly for q.

Thus we have

pq(p^* - q^*) = pp^*q - qq^*p = a^2(q-p)

and the first result follows.

Then, note that if and are perpendicular,

$\displaystyle \frac{p-q}{s-r} = \frac{|p-q|}{|s-r|}i$

$\displaystyle \frac{p-q}{s-r} = -\frac{|p-q|}{|s-r|}i$

Either way, we have

$\displaystyle \frac{p^* - q^*}{s^* - r^*} = \left(\frac{p-q}{s-r}\right)^*$

$\displaystyle \frac{p^* - q^*}{s^* - r^*} = -\frac{p-q}{s-r}$

and thus

$\displaystyle \frac{s-r}{s^* - r^*} = - \frac{p-q}{p^*-q^*}$

$\displaystyle \frac{s-r}{s^* - r^*} + \frac{p-q}{p^*-q^*} = 0$

from which it follows that

pq + rs

$\displaystyle = -a^2 \left(\frac{p-q}{p*-q*} + \frac{s-r}{s^* - r^*}\right)$

= 0

as required.

For the next part, we have in the case n=3 : (let a_i, b_i represent in the Argand plane, with the circle centred in the origin and with radius a)

$\displaystyle \left\{\begin{array} .a_1a_2 = -b_1b_2 \\ a_2a_3 = - b_2b_3 \\ a_3a_1 = -b_3b_1 \end{array}\right.$

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

$\displaystyle a_1^2 = -b_1^2$

there are obviously two distinct b_1 that satisfy this, as required.

For n = 4 , it's instead

$\displaystyle \left\{\begin{array} .a_1a_2 = -b_1b_2 \\ a_2a_3 = - b_2b_3 \\ a_3a_4 = -b_3b_4 \\ a_4a_1 = -b_4b_1 \end{array}\right.$

In this case, any b_1 can satisfy this, becayse if you choose a b_1 , you can always choose a b_2 so that the first equation is satisfied, and then a b_3 so that the second one is satisfied, and then a b_4 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.

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