Two pegs on a square

$First\quad of\quad all\quad i\quad want\quad to\quad know\quad what\quad is\quad pegs..??\\ (BUT\quad I\quad solved\quad this\quad question\quad by\quad assuming\quad pegs\quad to\quad be\quad Point.)\\ SOL:\quad for\quad to\quad minimize\quad MA+MB+MC\quad \\ "M"\quad must\quad be\quad present\quad inside\quad the\quad triangle\quad ABC\\ and\quad makes\quad equal\quad angles\quad which\quad is\quad 120\quad degree.(THINKS\quad LOGICALLY)\\ now\quad drop\quad perpendicular\quad from\quad 'M'\quad to\quad side\quad AC=\sqrt { 2 } \\ and\quad let\quad Foot\quad of\quad perpendicular\quad is\quad 'L'\quad \\ Now\quad By\quad symmetry\quad \\ \Longrightarrow MN=2\times ML\\ Since\quad \angle MAL=30\quad degree\\ \Longrightarrow \tan { (30) } =\frac { 1 }{ \sqrt { 3 } } =\frac { ML }{ AL } \\ \Longrightarrow AL=\frac { AC }{ 2 } =\frac { \sqrt { 2 } }{ 2 } =\frac { 1 }{ \sqrt { 2 } } \\ \Longrightarrow ML=\frac { 1 }{ \sqrt { 6 } } \\ \Longrightarrow MN=\frac { \sqrt { 6 } }{ 3 } \\ \Longrightarrow a=6\quad \& \quad b=3\\ \Longrightarrow a+b=9$ .

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Think of distances MA, MB, MC whose sum is to be minimized as soap films (or rubber bands) meeting in M. Three soap film meet at 120° (Lami's theorem for 3 equal forces/tensions!)

Same happens at peg N

Note triangle MNA is equilateral! => AM = MN

Using Sine Rule in triangle AMB - $\frac {1}{\sin 120} = \frac {AM}{\sin 45}$ $AM = \frac{\sqrt 6}{3}$