Fun with triangles and polynomials

Relevant wiki: Vieta's Formula Problem Solving - Intermediate

Calling the expression $\small{\mathcal I\left(=\displaystyle\sum_{\text{cyc}}\sin A\right)}$ and using sine rule, $\small{\color{teal}{\sin A=\dfrac{a}{2R}}}$ .

$\mathcal I=\displaystyle\sum_{\text{cyc}}\dfrac{a}{\color{#D61F06}{2R}}=\dfrac{\displaystyle\sum_{\text{cyc}} a}{\color{#D61F06}{2R}}=\dfrac{\displaystyle\sum_{\text{cyc}}a}{\color{#D61F06}{\dfrac{abc}{2\Delta}}}$

Now using Vieta's formula, $\small{\color{teal}{\displaystyle\sum_{\text{cyc}}a=24,abc=420}}$ so that:

$\mathcal I=\dfrac{4\Delta}{35}$

Semi perimeter $=s=\dfrac{\displaystyle\sum_{\text{cyc}}a}{2}=12$ . Now using Heron's formula for finding area:-

$\Delta=\sqrt{12\{\underbrace{(12-a)(12-b)(12-c)\}}_{\color{#20A900}{f(12)}}}$

$(\because \small{f(x)=x^3-24x^2+120x-420=(x-a)(x-b)(x-c)})$

$=\sqrt{12f(12)}=\sqrt{12\cdot 12}=12$

Hence, $\mathcal I=\dfrac{48}{35}$ .

$\large 48+35-10=\boxed{73}$

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

$\begin{aligned} \implies \sin A + \sin B + \sin C & = k(\color{#3D99F6}{a+b+c}) & \small \color{#3D99F6}{\text{Vieta's formula: }a+b+c = 24} \\ & = \color{#3D99F6}{24}k \\ \implies \frac pq & = 24k \end{aligned}$

Area of triangle $A_\triangle = \frac 12 ab \sin C = \frac 12 abck$ . By Vieta's formula , we have $abc = 420$ $\implies A_\triangle = 210k$ .

By Heron's formula ,

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

$\implies 210k = 12 \implies k = \dfrac 2{35} \implies \dfrac pq = \dfrac {48}{35} \implies p+q-10 = \boxed{73}$

$x^3-24x^2+180x-420=0.\ \ \ \ by\ Vieta's\ formula,\\ \displaystyle \sum_{cyc}a=24,\ \ \ \ \ \sum_{cyc}ab=180,\ \ \ \ \ abc=420. \ \ \ After\ normal\ manupulations,\\ \displaystyle we\ get\ \sum_{cyc}a^2=216,\ \ \ \ \ and\ \ \ \ \ \sum_{cyc}a^2b^2=180^2-2*420*24=12240.\\ Using\ Cos\ Law,\ SinA=\sqrt{1-Cos^2A}=\dfrac 1 {2bc}*\sqrt{4b^2c^2- ( - a^2+b^2+c^2)^2}.\\ \displaystyle SinA=\dfrac a {2abc} * \sqrt{4 \sum_{cyc}a^2b^2\ \ -\ \ \ \left \{ \sum_{cyc}a^2 \right \}^2.} \\ \displaystyle \sum_{cyc}SinA= \dfrac{\displaystyle \sum_{cyc}a} {2abc}* \sqrt{ 4\sum_{cyc}a^2b^2\ \ -\ \ \ \left \{ \sum_{cyc}a^2 \right \}^2.}\\ \displaystyle \sum_{cyc}SinA= \dfrac {24}{2*420}*\sqrt{4* 12240\ -\ 216^2}\\ =\dfrac{24*48}{2*420}=\dfrac{48}{35}=\dfrac p q.\\ p+q-10= \huge \color{#D61F06}{73}. \\. \\$

This is how I arrived at the solution. But a straight solution below.

Thanks to @Chew-Seong Cheong and @Rishabh Cool for the method below.

$\displaystyle Using Vieta'sFormula\ \ \sum_{cyc}a=24,\ \ \ \ \ \sum_{cyc}ab=180,\ \ \ \ \ abc=420. \\ \displaystyle Semi \ perimeter \ \ =s=\frac 1 2 *\sum_{cyc}a=12\\ \text{ Now using Heron's formula for finding area:-}\\ \because\ \ f(x)=x^3-24x^2+120x-420=(x-a)(x-b)(x-c)\\ \therefore\ \ [ABC]=\sqrt{s*f(s)}= \sqrt{12\{\underbrace{(12-a)(12-b)(12-c)\}}_{\color{#3D99F6}{f(12)}}}\\ \therefore\ \ [ABC]=\sqrt{12f(12)}=\sqrt{12\cdot 12}=12\\ SinA=\dfrac {2[ABC]}{bc} =\dfrac{2a[ABC]}{abc}.\\ \displaystyle\ \sum_{cyc}SinA=\sum_{cyc}\dfrac{2a[ABC]}{abc}\\ \dfrac{2*24*12}{420}=\dfrac{48}{35}=\dfrac p q.\\ p+q-10= \huge \color{#D61F06}{73}.$