Triangular Tan Tutu

Let $D$ be the common point of the incircle and side $a$ and $I$ the incircle. Denote the inradius by $r$ and the semiperimeter of the triangle by $s$ . Set $t=\tan \dfrac{C}{2}$ . Then, $CD=s-c=\frac{1}{2}\left( a+b+c \right)-c=\frac{1}{2}\left( a+b-c \right)=\frac{1}{2}\cdot 2=1$ and

$\begin{aligned} \tan C=-2 & \Rightarrow \frac{2\tan \frac{C}{2}}{1-{{\left( \tan \frac{C}{2} \right)}^{2}}}=-2 \\ & \Rightarrow \frac{t}{1-{{t}^{2}}}=-1 \\ & \Rightarrow {{t}^{2}}-t-1=0 \ \ \ \ \ (1)\\ \end{aligned}$ Moreover, $\tan \frac{C}{2}=\frac{ID}{CD}\Rightarrow t=\frac{r}{1}\Rightarrow t=r$ Combining with $(1)$ we get ${{r}^{2}}-r-1=0$ Thus, $m=n=-1$ and the answer is $m+n=\boxed{-2}$ .

The trick here will be to avoid calculating $a,b$ explicitly, and just use symmetric functions. We already have $a+b=c+2$ from the question.

From $\tan C=-2$ we get $\cos C=-\frac{1}{\sqrt5}$ and $\sin C=\frac{2}{\sqrt5}$ .

Using the cosine rule, $\begin{aligned} c^2 &=a^2+b^2-2ab \cos C \\ c^2 &= a^2+b^2+\frac{2ab}{\sqrt5} \\ c^2 &= (c+2)^2-2ab+\frac{2ab}{\sqrt5} \\ c^2 &= c^2+4c+4-2ab+\frac{2ab}{\sqrt5} \\ ab &=\frac{2(c+1)}{1-\frac{1}{\sqrt5}} = \frac12 \left(5+\sqrt5\right)(c+1) \end{aligned}$

The area of the triangle is $\Delta=\frac12 ab \sin C= \frac14 \left(5+\sqrt5\right)(c+1) \frac{2}{\sqrt5}=\frac12 \left(1+\sqrt5\right)(c+1)$

and the semiperimeter is $s=\frac{a+b+c}{2}=c+1$

so the inradius is $r=\frac{\Delta}{s}=\frac12 \left(1+\sqrt5\right)$

which is the golden ratio and is a root of $x^2-x-1=0$ , giving the answer $m+n=-1-1=\boxed{-2}$ .

Given that $\tan C = -2$ , $\implies \sin C = \dfrac 2{\sqrt 5}$ and $\cos C = - \dfrac 1{\sqrt 5}$ . By cosine rule ,

$\begin{aligned} c^2 & = a^2 + b^2 - 2 ab \cos C \\ c^2 & = (\blue{a+b})^2 - 2ab + \frac 2{\sqrt 5}ab & \small \blue{\text{Since }a+b - c = 2} \\ c^2 & = (\blue{c+2})^2 - 2\left(1 - \frac 1{\sqrt 5}\right) ab \\ c^2 & = c^2 + 4c + 4 - 2\left(1 - \frac 1{\sqrt 5}\right) ab \\ \implies 2c + 2 & = \left(1 - \frac 1{\sqrt 5}\right) ab \end{aligned}$

The inradius $r$ is given by:

$\begin{aligned} r & = \frac \blue \Delta{\frac 12(a+b+c)} = \frac {\frac 12 ab \sin C}{\frac 12(a+b+c)} = \frac {\frac 2{\sqrt 5}ab}{a+b+c} & \small \blue{\text{where }\Delta \text{ is the area of }\triangle ABC} \\ & = \frac {\frac 2{\sqrt 5}ab}{2c+2} = \frac {\frac 2{\sqrt 5}ab}{\left(1-\frac 1{\sqrt 5}\right)ab} = \frac {1+\sqrt 5}2 = \blue \varphi & \small \blue{\text{where }\varphi \text{ denotes the gold ratio.}} \end{aligned}$

Therefore $r = \varphi$ is the root of $f(x) = x^2 - x -1$ . The required answer is $m+n = -1 -1 = \boxed {-2}$ .

$BC=a\;\;\;CA=b\;\;\;AB=c\;\;\;OD=OE=OF=r\;\;\;CE=CF=\frac{a+b-c}{2}=\frac{2}{2}=1\;\;\;\tan C=-2\;\;\;\Rightarrow\;\;\;90^\circ<C<180^\circ\;\;\;and\;\;\; GE=2GC\\GC^2+(2GC)^2=1^2\;\;\;\Rightarrow\;\;\;GC=\frac{1}{\sqrt{5}}\;\;\;\bigtriangleup CEG\sim\bigtriangleup OEH\;\;\;\Rightarrow\;\;\;\frac{GE}{1}=\frac{HE}{r}\;\;\;\Rightarrow\;\;\;\frac{2}{\sqrt{5}}=\frac{1+\frac{1}{\sqrt{5}}}{r}\;\;\;\Rightarrow\;\;\;r=\frac{1+\sqrt{5}}{2}\\x^2+mx+n=0\;\;\;\Rightarrow\;\;\; \left(\frac{1+\sqrt{5}}{2}\right)^2+m\frac{1+\sqrt{5}}{2}+n\equiv 0\;\;\;\Rightarrow\;\;\;\frac{1+2\sqrt{5}+5}{4}+\frac{m+m\sqrt{5}}{2}\equiv -n\;\;\;\Rightarrow\;\;\;(m+1)\sqrt{5}+m+3\equiv -2n\;\;\;\Rightarrow\\(m+1)=0\;\;\;\Rightarrow\;\;\;m=-1\;\;\;\Rightarrow\;\;\; -1+3=-2n\;\;\;\Rightarrow\;\;\;n=-1\;\;\;\Rightarrow\;\;\;m+n=\boxed{-2}$