I've drawn 2 triangles with one inscribed in the other, i.e. with the vertices of the smaller triangle lying on different sides of the large triangle. Given that the large triangle has side lengths $3\sqrt{2},\, 4,$ and $\sqrt{10},$ as shown in the figure, let $P$ denote the minimum possible perimeter of the smaller triangle.

If $P^2$ can be expressed as $\frac MN,$ where $M$ and $N$ are coprime positive integers, find $M+N.$

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This problem is a part of
<Grade 10 CSAT Mock Test> series
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The answer is 149.

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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 $F\left(\dfrac{18}{5},~\dfrac{6}{5}\right).$

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