Bertie

Seventy two years.

13.8564i to 4 decimal places and where i is the square root of minus 1.

x = 2 implying y = 0 is ok. but x = -2 leads to a contradiction.

Also, the equations are symmetric in x and y, so it's possible to switch the x and y values to obtain a second solution corresponding to x=2, y=0.

An alternative method of solution would be to square the second equation and then combine that with the first leading to 2xy = 0, from which either x = 0, or y = 0.

Then substitute into the second equation, (not the first), to find the corresponding y or x value.

It depends on what information you have regarding it. The lengths of all three sides, the lengths of two sides and the included angle, two angles and the length of the included side, base length and height. Which ?

What do you mean by symmetrical ?

Solve the equations simultaneously.

Substitute the y from the first equation into the second and solve the resulting equation for x. Put that value back into the first (preferably), or second equation to find the corresponding value of y.

Then what ? We wait with bated breath.

On the assumption that alpha, beta, a and b are real,

$$\alpha -\imath \beta=\frac{1}{a-\imath b}=\frac{a+\imath b}{a^{2}+b^{2}}.$$

Taking the complex conjugate of both sides,

$$\alpha + \imath\beta=\frac{a-\imath b}{a^{2}+b^{2}},$$

and multiplying the equations together,

$$\alpha^{2}+\beta^{2}=\frac{a^{2}+b^{2}}{(a^{2}+b^{2})^{2}},$$

from which the result follows.

Hi Melody

The reason that you get the wrong answer for the 1-1-4 arrangement is that you count some of the possibilities twice.

Suppose that the two odd b***s (if I can call them that !) are labelled A and B, so that there are those two as singles and the others as a group of 4.

They can be placed in the three boxes in 3! = 6 different ways. A-B-4, B-A-4, A-4-B, B-4-A, 4-A-B and 4-B-A.

Now, how many A, B, 4 combinations are there ? The answer is 6C4 = 15, the number of ways in which the group of 4 can be taken from the original 6, (the two singles take care of themselves), not 6*5 = 30.

The problem with the 6*5=30, is that it takes order into account, that is, it sees A,B,4 and B,A,4 as being different and as such covers two of the 6 possibles listed earlier rather than just one.

To look at the 3-3-0 arrangement, suppose that a group of three is chosen and called 3(1), and that the remaining three are called 3(2).

Now suppose that they occupy the first two boxes, so that we have 3(1)-3(2)-0 or 3(2)-3(1)-0. How many possible combinations are there there ? It's tempting to say 6C3 for the first and 6C3 for the second making 2*6C3 = 2*20 = 40 in all. (That, in effect, is what is being done if you say that there are 3!=6 different 3,3,0 arrangements.) That's wrong though, it's actually just 6C3=20, because 6C3 includes both 3(1) and also 3(2).

The 0 can occupy three different positions, so there are 3*20 = 60 combinations in all.

It's easy if you look at it as

box 1    box 2    box 3

3            3           0

3            0           3

0            3           3

There are 6C3 possibilites for each row making 3*6C3 = 60 possibles in all.

The 3!=6, in effect, counts each row twice.

The people who say that maths is boring are usually the people who don't understand/can't do maths.

The boring response is a defense mechanism. They don't want to say that they don't understand the subject, so rather than admit that, they either try to belittle it or say that it's boring.