I Have Seen Sine, Cosine, What About Tangent?

Relevant wiki: Solving Triangles - Problem Solving - Hard

By Vieta's formula : $\begin{cases} a+b+c = 24 \\ ab+bc+ca = 180 \\ abc = 420 \end{cases}$

By sine rule : $\dfrac {\sin A}a = \dfrac {\sin B}b = \dfrac {\sin C}c = k \implies \begin{cases} \sin A = ka \\ \sin B = kb \\ \sin C = kc \end{cases}$

By Heron's formula :

$\begin{aligned} A_\triangle & = \sqrt{\color{#3D99F6}{s}(\color{#3D99F6}{s}-a)(\color{#3D99F6}{s}-b)(\color{#3D99F6}{s}-c)} & \small \color{#3D99F6}{\text{where }s = \frac {a+b+c}2 = 12} \\ & = \sqrt{12\color{#3D99F6}{(12^3-24(12)^2+ 180(12)-420)}} & \small \color{#3D99F6}{\text{Note that } (x-a)(x-b)(x-c) = x^3-24x^2+ 180x-420} \\ & = \sqrt{12\color{#3D99F6}{(12)}} \\ & = 12 \end{aligned}$

But $A_\triangle = \frac 12 ab \sin C = \frac 12 abck \implies abck = 24$

By cosine rule :

$\begin{aligned} a^2 & = b^2 + c^2 - 2bc \cos A \\ \implies \cos A & = \frac {b^2 + c^2 - a^2}{2bc} \\ & = \frac {\color{#3D99F6}{a^2 + b^2 + c^2} - 2a^2}{2bc} & \small \color{#3D99F6}{\text{See Note}} \\ & = \frac {\color{#3D99F6}{216} - 2a^2}{2bc} \\ & = \frac {108 - a^2}{bc} \end{aligned}$

Now, we have:

$\begin{aligned} \tan A + \tan B + \tan C & = \tan A \tan B \tan C & \small \color{#3D99F6}{A+B+C = 180^\circ} \\ \implies \tan A \tan B \tan C & = - \frac pq \\ \implies - \frac qp & = \cot A \cot B \cot C \\ & = \frac {\cos A \cos B \cos C}{\sin A \sin B \sin C} \\ & = \frac {108 - a^2}{bc\cdot ka} \cdot \frac {108 - b^2}{ca\cdot kb} \cdot \frac {108 - c^2}{ab\cdot kc} \\ & = \frac {108^3-108^2(a^2+b^2+c^2)+ 108 (\color{#3D99F6}{(ab)^2 + (bc)^2 +(ca)^2})-(abc)^2}{(abck)^3} & \small \color{#3D99F6}{\text{See Note}} \\ & = - \frac {114192}{13824} = - \frac {793}{96} \end{aligned}$

$\implies p+q = 96 + 793 = \boxed{889}$

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$\color{#3D99F6}{\text{Note:}}$

By Newton's sums (identities) :

$\begin{aligned} a^2+b^2+c^2 & = (a+b+c)^2 - 2(ab+bc+ca) \\ & = 24^2 - 2(180) \\ & = \color{#3D99F6}{216} \\ (ab)^2 + (bc)^2 +(ca)^2 & = (ab+bc+ca)^2 - 2(a^2bc+ab^2c+abc^2) \\ & = (ab+bc+ca)^2 - 2abc(a+b+c) \\ & = 180^2 - 2(420)(24) \\ & = \color{#3D99F6}{12240} \end{aligned}$

The polynomial $f(X) = X^3 - 24X^2 + 180X - 420$ has roots $a,b,c$ . Thus the triangle has semiperimeter $s = 12$ and hence area $\Delta = \sqrt{sf(s)} = 12$ and hence inradius $r = \tfrac{\Delta}{s} = 1$ . Thus $\tan\tfrac12A \; = \; \frac{r}{s-a} \; = \; \frac{1}{12-a} \qquad \tan\tfrac12B \; = \; \frac{1}{12-b} \qquad \tan\tfrac12C \; = \; \frac{1}{12-c}$ Using the substitution $X = 12 - Y^{-1}$ , the monic cubic with roots $\tan\tfrac12A$ , $\tan\tfrac12B$ and $\tan\tfrac12C$ is $g(Y) \; = \; Y^3 - 3Y^2 + Y - \tfrac{1}{12}$ Using the substitution $Y = \sqrt{Z}$ , the monic cubic with roots $\tan^2\tfrac12A$ , $\tan^2\tfrac12B$ and $\tan^2\tfrac12C$ is $h(Z) \; = \; Z^3 - 7Z^2 + \tfrac12Z - \tfrac{1}{144}$ Thus $\begin{array}{rcl} \tan A + \tan B + \tan B & = & \tan A \tan B \tan C \; =\; \frac{8\tan\frac12A \tan\frac12 B \tan\frac12C}{(1 -\tan^2\frac12A)(1 - \tan^2\frac12B)(1 - \tan^2\frac12C)} \\ & = & -\frac{8g(0)}{h(1)} \; = \; -\frac{96}{793} \end{array}$ making the answer $96 + 793 = \boxed{889}$ .

$\displaystyle Using Vieta'sFormula\ \ \sum_{cyc}a=24,\ \ \ \ \ \sum_{cyc}ab=180,\ \ \ \ \ abc=420. \\ \displaystyle Semi \ perimeter \ \ =s=\frac 1 2 *\sum_{cyc}a=12\\ After\ normal\ manupulations,\\ \displaystyle we\ get\ \sum_{cyc}a^2=216,\ \ \ \ \ and\ \ \ \ \ \sum_{cyc}a^2b^2=180^2-2*420*24=12240.\\ \text{ Now using Heron's formula for finding area:-}\\ \because\ \ f(x)=x^3-24x^2+120x-420=(x-a)(x-b)(x-c)\\ \therefore\ \ [ABC]=\sqrt{s*f(s)}= \sqrt{12\{\underbrace{(12-a)(12-b)(12-c)\}}_{\color{#3D99F6}{f(12)}}}\\ \therefore\ \ [ABC]=\sqrt{12f(12)}=\sqrt{12\cdot 12}=12\\ SinA=\dfrac {2[ABC]}{bc}=\dfrac {24}{bc},\ \ \ \ \ CosA =\sqrt{1 - Sin^2A}\\ \therefore\ \ TanA =\dfrac{24}{\sqrt{b^2c^2 - 24^2}}\\ TanA+TanB+TanC=TanA*TanB*TanC\\ =\dfrac{24}{\sqrt{b^2c^2 - 24^2}}*\dfrac{24}{\sqrt{c^2a^2 - 24^2}}*\dfrac{24}{\sqrt{a^2b^2 - 24^2}}\\$ $\displaystyle =\dfrac{24^3} {\sqrt{ a^4b^4c^4 - 24^2*\sum_{cyc}a^2*(a^2b^2c^2 )+ 24^4*\sum_{cyc}a^2b^2 - 24^6 } } \\ =\dfrac {24^3} {\sqrt{ 420^4 - 24^2*216*420^2+ 24^4*12240 - 24^6 } } \\ =\dfrac {96} {-793} =- \dfrac p q.\\ p+q=\Huge\ \ \ \color{#D61F06}{889}$

Note:- Answered required was negative, so on checking one of the angles is 138 degrees, so I have taken negative sign for the square root.