A geometry problem by Omkar Kulkarni

It can be hard to deal with such an "unbalanced" equation. As such, we use the half angle tangent substitution .
Let $\tan \frac{ \theta }{ 2} = t$ . Then, we have to solve:

$( m + 2 ) (2t ) + (2m-1 ) ( 1-t^2 ) = (2m+ 1 ) ( 1+t^2 )$

Expanding and simplifying, we obtain $4mt^2 - 2mt - 4t + 2 = 0$ . Viewing this as a quadratic in $t$ , we obtain $2 ( 2t - 1) ( mt - 1) = 0$ .

Hence, $t = \frac{1}{2}$ or $t = \frac{1}{m}$ . Have we introduced extraneous solutions? The only change we've done is to multiply throughout by $1 + t^2 > 0$ , so we didn't introduce extraneous solutions.

Case 1: $t = \frac{1}{2}$ .
Then $\tan \theta = \frac{ 2 t } { 1 - t^ 2 } = \frac{4}{3}$ .
Note that for this specifc value, we have $\sin \theta = \frac{4}{5}$ and $\cos \theta = \frac{3}{5}$ , and thus the given equation is satisfied for all $m$ .

Case 2: $t = \frac{1}{m}$ .
Then, $\tan \theta = \frac{ 2t} { 1 - t^2 } = \frac{ 2m } { m^2 - 1 }$ .

Thus, we have 2 possible solutions.

I would like to offer this as a solution.

First, we make a trig substitution:

(m + 2) = R sin (phi), (2 m - 1) = R cos (phi), with R^2 = 5 (m^2 + 1)

Then, R sin (phi) sin (theta) + R cos (phi) cos (theta) = 2 m + 1,

and cos (theta - phi) = (2 m + 1) / R

Then sin^2 (theta - phi) = 1 - cos^2 (theta - phi) = [R^2 - (2 m + 1)^2] / R^2 =

[5 (m^2 + 1) - (4 m^2 + 4 m + 1)] / R^2 = (m^2 - 4 m + 4) / R^2 = (m - 2)^2 / R^2

Depending on which quadrant (theta - phi) falls in, the sin and cos, and also the tan, may be either positive or negative.

So, | tan (theta - phi) | = | sin (theta - phi) / cos (theta - phi) | = | [(m - 2) / R] / [(2 m + 1) / R)] | = | (m - 2) / (2 m + 1) |

For m >> 0 or m << 0, the RHS argument is positive. If tan (theta - phi) is also positive, we can proceed without the bars.

Then, tan (theta - phi) = [tan (theta) - tan (phi)] / [1 + tan (theta) tan (phi)] =

[tan (theta) - (m + 2) / (2 m -1)] / [1 + tan (theta) (m + 2) / (2 m - 1)] = (m - 2) / (2 m + 1)

tan (theta) - (m + 2) / (2 m -1) = [(m - 2) / (2 m + 1)] [1 + tan (theta) (m + 2) / (2 m - 1)]

tan (theta) [1 - (m + 2) (m - 2) / (2 m - 1) (2 m + 1)] = [(m - 2) / (2 m + 1)] + [(m + 2) / (2 m -1)] =

tan (theta) [(2 m - 1) (2 m + 1) - (m + 2) (m - 2)] = [(m - 2) (2 m -1) + (m + 2) (2 m + 1)]

tan (theta) = 4 (m^2 + 1) / 3 (m^2 +1) = 4/3

Close, but no cigar.

However, for -1/2 < m < 2, the quantity in the RHS absolute value bars is negative.

Then, tan (theta - phi) = [tan (theta) - tan (phi)] / [1 + tan (theta) tan (phi)] =

[tan (theta) - (m + 2) / (2 m -1)] / [1 + tan (theta) (m + 2) / (2 m - 1)] = - (m - 2) / (2 m + 1)

tan (theta) - (m + 2) / (2 m -1) = [- (m - 2) / (2 m + 1)] [1 + tan (theta) (m + 2) / (2 m - 1)]

tan (theta) [1 + (m + 2) (m - 2) / (2 m - 1) (2 m + 1)] = [- (m - 2) / (2 m + 1)] + [(m + 2) / (2 m -1)] =

tan (theta) [(2 m - 1) (2 m + 1) + (m + 2) (m - 2)] = [- (m - 2) (2 m -1) + (m + 2) (2 m + 1)]

tan (theta) = 10 m / 5 (m^2 - 1) = 2 m / (m^2 - 1)

For m = 2 or m = -1/2, this evaluates to 4/3, as above.

These solutions follow from the selection of the positive square root of sin^2 (theta - phi). Had we used the negative square root

instead, the same solutions result, but for -1/2 < m < 2, the tan (theta) is 4/3, and outside this range, it is 2 m / (m^2 - 1).