#30 of Sept 2016 Grade 10 CSAT (Korean SAT) Mock Test

Let $B(0,~0)$ and $C(4,~0)$ and we see that $A(3,~3).$

The equation of line $AC$ is $y=-3x+12.$

Let $F(a,~b),$ and let $F_1$ and $F_2$ be the points of reflection over $\overline{AB}$ and $\overline{AC}.$

Then it is clear that $F_1(b,~a)$ and $F_2(a,~-b).$ Also, $b=-3a+12.$

The minimum of the perimeter of triangle $DEF$ is equal to the minimum of $\overline{F_1D}+\overline{DE}+\overline{EF_2}.$

The minimum would occur when $D,~E$ are on $\overline{F_1F_2}.$

$\begin{aligned} \overline{F_1F_2}&=\sqrt{(a-b)^2+(a+b)^2} \\ &=\sqrt{2a^2+2b^2} \\ &=\sqrt{20a^2-144a+288} \\ &=\sqrt{20\left(a-\dfrac{18}{5}\right)^2+\dfrac{144}{5}}. \end{aligned}$

Therefore, the minimum of the perimeter of triangle $DEF$ is $\dfrac{12}{\sqrt{5}}$ when $F\left(\dfrac{18}{5},~\dfrac{6}{5}\right).$

$\therefore~P^2=\dfrac{144}{5}=\dfrac{M}{N},\\ M+N=144+5=\boxed{149}.$

The in inscribed triangle is orthic because orthic triangle have the minimum perimeter.
Let the sides be $a' + b' + c' = aCosA + bCosB + cCosC = \dfrac 4 {\sqrt5} + \dfrac 5 {\sqrt5} + \dfrac 3 {\sqrt5} = \dfrac{12}{\sqrt5}=P.\\ P^2=\dfrac{144} 5=\dfrac M N\\ M+N=144+5=149.$
To find Cos, Cos Rule Is used.