Alan

See the following:

It seems fairly clear from the following diagram that it is not always true.  The specified angle bisectors clearly won't intersect on AD. 

The proposition may well be true for a non-intersecting diagram.

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1.4E+07 usually means 1.4*10⁷ = 14 000 000, or 14 million.  

As follows:

Actually, there is no need for my equation (1).  Equation (2) shows that u_n is strictly less than 2 - √n for all n greater than or equal to zero. 

"why do we reflect across the x axis????"

Let \(z=p+iq\) , then \(\overline{z}=p-iq\)

So, if z is the resultant obtained by adding two complex numbers, the real part of the complex conjugate of z has the same magnitude and the same sign, but the imaginary part of its complex conjugate has the same magnitude but opposite sign i.e. its complex conjugate is just a reflection in the real (or x) axis.

Of course the problem could be tackled by finding both com plex conjugates of a pair and then simply adding them, but the result would be the same.

"Here's a list of pairwise sums of the conjugates of these complex numbers:

\(\overline{z}_1+\overline{z}_2, \overline{z}_1 + \overline{z}_3, \overline{z}_1 + \overline{z}_4, \overline{z}_2 + \overline{z}_3, \overline{z}_2 + \overline{z}_4, \overline{z}_3 + \overline{z}_4\)

Find the number of the quadrant each of these pairwise sums is in, and answer with the ordered list, such that your first number corresponds to the quadrant that \(\overline{z}_1+\overline{z}_2\)  is in, your second number corresponds to the quadrant that \(\overline{z}_1+\overline{z}_3\)

is in, etc."

Add two of the given complex numbers, \(z_1+z_2\) say, and reflect the result in the x-axis (i.e. the Real axis).

So, for example, it looks like  \(z_1+z_2\) when added will result in a complex number in the 2nd quadrant.  Reflect this in the x-axis and the result will be in quadrant 3. i.e. \(\overline{z}_1+\overline{z}_2\)will be in quadrant 3.  

Repeat this process for all the specified pairs.

Yours is a much simpler approach Rom, and, yes, I totally forgot about the "integer coefficients" requirement.

On the assumption that a) is indeed  \(\frac{1}{\alpha},\frac{1}{\beta},\frac{1}{\gamma}\) we have:

L(I)=10log⁡(I/10⁻¹²)

L(I) = 125, so  125 = 10log⁡(I/10⁻¹²)

Divide both sides by 10

12.5 = log⁡(I/10⁻¹²)

Antilog both sides:

10^12.5 = I/10⁻¹²

Multiply both sides by 10^-12 

10^-1210^12.5 = I

Using the fact that 10^a10^b = 10^a+b we have:

I = 10^0.5   or  I = √10

Perhaps these graphs will help (note, I've rearranged your equality to create a function f that is zero when the equality is true):