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Geometry Level 4

If $A+B+C=\pi$ , $\sin A \sin B \sin C=p$ and $\cos A \cos B \cos C=q$ , then which among the following options shows the equation whose roots are $\tan A,\tan B,\tan C$ ?

pqx^{3}-qx^{2}-qx-p^2=0

pq^2x^{3}-px^{2}-qx-p^2q=0

px^{3}-qx^{2}-x-q=0

qx^{3}-px^{2} + (1 + q)x-p=0

2 solutions

Reineir Duran
Feb 3, 2016

I really don't know how to arrive at $\displaystyle \tan A \tan B + \tan B \tan C + \tan C \tan A = -\frac{1}{q} = -\frac{1}{\cos A \cos B \cos C}.$ But since $\displaystyle \sin A \sin B \sin C = p$ and $\displaystyle \cos A \cos B \cos C = q$ , with $\displaystyle A + B + C = \pi$ , then $\displaystyle \tan A \tan B \tan C = \tan A \tan B \tan C = \frac{p}{q}$ . So obviously, the first choice must be right ( $\displaystyle qx^3 - px^2 - x - p = 0$ ).

I agree this is not a complete solution, so tell me if you know how.

Reineir Duran - 5 years, 4 months ago

@Reineir Duran The (first) equality stated above is actually wrong (as seen by taking all the angles to be acute (the LHS is positive whereas the RHS is negative)). See the report, there I've posted the actual relation with its proof.

Deeparaj Bhat - 5 years, 3 months ago

On a scale of 1 to 10 sir... ,the wrongness of your solution is an inevitable 11.

Joe Bobby - 4 years, 7 months ago

Sorry, that was a bit too much... I overreacted

Joe Bobby - 4 years, 7 months ago

Joe Bobby
Oct 23, 2016

Tan(a+b+c) = 0

=> tan a + tan b + tanc = Πtana

Also Σtana.tanb ÷ (Πtana) = Σcota

Cos(a+b+c) = -1

Cos(a+b+c) = Πcosa - Πsina(Σcota)

Now we know:

Σtana
Σtanatanb
Πtana

End of story lads....

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