triangle

ya thats the answer.

We use the fact that in a triangle \tan A + \tan B + \tan C = \tan A \tan B \tan C

Lets say A=B.

Then tan A = tan B so that we have

2\tan A + \tan C = 300 = \tan^2 A \tan C

Eliminating tan C and simplifying gives us (setting t = tan A)

t + \frac{150}{t^2} = 150

Now, since A<\frac{\pi}{2}, we have t = tan A>0.

Lets prove that this equation has exactly two positive roots.

Let f(t) = t + \frac{150}{t^2}

f is continuous for t>0

f(1) = 151

Now,from AM-GM Inequality (or just taking derivative)

f(t) = \frac{t}{2} + \frac{t}{2} + \frac{150}{t^2} attains its least value of \frac {3 (150)^{\frac{1}{3}}}{2}

when t = 300^{\frac{1}{3}}

That means by IMV we have one root in the interval \left(1, 300^{\frac{1}{3}}\right)

Again f(151)>150

So one more root in the interval \left( 300^{\frac{1}{3}}, 151\right)

Now we write it as t^3-150t^2+150=0

This tells us that

(a) there are at most three roots and we have found two of them

(b) the product of roots is -ve. Since the two roots we have are +ve, this root is -ve and we can discard it as t>0

Hence we have two distinct triangles corresponding to these values of t