How to prove that there exist exactly two non-similar isosceles triangles ABC such that tanA+tanB+tanC=100?

Question

I actually have a doubt about the solution of this question given in my book. It uses the equations tan 2A = - tan C (from A=B, A+B+C = 180 degrees) and 2 tan A + tan C = 100, thereby formulating the cubic equation $x^3 - 50x^2 + 50=0$. The discriminant is $3x^2- 100x$. Since $f(0) f(100/3) <0 $, it is found there are three distinct real roots. After this my book simply states that one of these roots is obtuse so the other two values are taken. Why is one of the values of A obtuse when we have used 2A + C = 180 degrees and an obtuse value of A would not satisfy this equation? Would someone please help?

Answer 1

Since $A$ is the repeated angle, $A$ is acute so $x:=\tan A>0$. We must thus count the positive roots of $x^3-50x^2+50=0$. The roots have a negative product, $-50$, so there's one negative root plus two positive ones**. (They can't all be negative because their sum is $50$).

** Or a conjugate pair of non-real complex roots, which would result in no solutions to the original problem. But we can verify from a sign change on $[1, 2]$ that $x^3-50x^2+50$ has at least one positive root, so in fact it has two.

Answer 2

Because $$2\alpha+\gamma=180^{\circ}$$ or $$\alpha=90^{\circ}-\frac{\gamma}{2}<90^{\circ},$$ which says that $\alpha$ is an acute angle and we need to take $x>0$ only.

Easy to see that the equation, which you got has two positive roots.

One on $(1,2)$ and the second on $(49,50)$.

There is also a negative root, which is not valid.