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Given a rectangle $ABCD$ on a cell sheet. Here $AB = 3$; $AD = 4$. Two point $M$ and $N$ are chosen inside the rectangle. The distance from $M$ to $AB$ is equal to $1$, the distance from $M$ to $AD$ is equal to $2$. The distance from $N$ to $CD$ equals $2$; from $N$ to $AD$ equals $1$. Points $Q$, $R$, $S$, $T$ were chosen on the sides $AB$, $BC$, $CD$, $AD$. At what positions of the points $P$, $Q$, $S$, $T$ is the sum $MQ + MR + MS + MT + NQ + NR + NS + NT$ minimal?

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From the conditions we know that we cannon change the positions of points $ M $ and $ N $. Also I think I know a theorem that might help here.

Theorem. Given two points $ A $ and $ B $ and a line $ l $, find a point $ C $ on the line $ l $ if we know that $ AC + BC $ is minimum. (In the case when the points lie on different sides of the line, we must only draw a line, and if not, we have to draw a point $ D $ which is symmetrical to $ B $, and then connect $ A $ and $ D $.)

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    Hint: minimize each partial sum $MQ+NQ$ independently. – 2017-02-24
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    So I have to do MQ and NQ, MR and NR, MS and NS..., right? – 2017-02-24
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    Right. You can choose the $4$ points independently, since the position of $Q$ only affects $MQ+NQ\,$, and similar for the others. – 2017-02-24
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    There is a symmetry to the figure such that $MQ+NQ = MT + NT = MR+NR$ – 2017-02-24
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    As others have pointed out, there are four independent sub-problems solved by finding the optimal position of $Q$, $R$, $S$ and $T$. So consider just the $Q$ location: Consider Fermat's principle of least time applied to light passing from a point $M$ to a mirror ($AB$) to $N$: It does so along the shortest path... *which leads to the law of reflection*: that the angle of incidence equals the angle of reflection. Application of this law will solve your problem. – 2017-02-24

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As others have pointed out, there are four independent sub-problems solved by finding the optimal position of $Q$, $R$, $S$ and $T$. So consider just the $Q$ location: Consider Fermat's principle of least time applied to light passing from a point $M$ to a mirror ($AB$) to $N$: It does so along the shortest path... which leads to the law of reflection: that the angle of incidence equals the angle of reflection. Application of this law will solve your problem. enter image description here

If you're really sophisticated, you'll use the principle of virtual images to find the location of $Q$: The total distance $MQ + QN$ is the same as $MQ + QN^\prime$, where $N^\prime$ is the "virtual" image of $N$ on the other side of the mirror $AB$, i.e., at location $(-2,1)$.