Estonian Math Competitions

Denote the vertices of the dodecagon as on Fig. 21. Since $QI=AS$ and $IS=QA$, we have $$DQ+QI+IS+SG=DQ+AS+QA+SG=DA+AG.$$ Similarly we obtain $ER+RK+KP+PB=EA+AB$. Therefore $$\begin{array}{rl}& CD+DQ+QI+IS+SG+GF+FE+ER+RK+KP+PB+BC\\ & =CD+DA+AG+GF+FE+EA+AB+BC\\ & =AB+BC+CD+DA+AE+EF+FG+GA.\end{array}$$ So the sum of the perimeters of the squares $ABCD$ and $AEFG$ is equal to the perimeter of the dodecagon (72 cm).



Since $QA=IS$, $AP=RK$, $SA=IQ$ and $AR=PK$, we have $$\begin{array}{rl}& AP+PH+HQ+QA+AR+RJ+JS+SA\\ & =RK+PH+HQ+IS+PK+RJ+JS+IQ\\ & =HI+IJ+JK+KH.\end{array}$$ So the perimeter of the square $HIJK$ is equal to the sum of the perimeters of the rectangles $APHQ$ and $ARJS$. However we are given that the sum of the perimeters of $APHQ$ and $ARJS$ is equal to one fifth of the sum of the perimeter of $ABCD$, the perimeter of $AEFG$ and twice the perimeter of $HIJK$. Denoting the side length of $HIJK$ by $x$ cm, its perimeter will be $4x$ cm. Then we can combine the relations of the previous two paragraphs into the equation $$4x=\frac{1}{5}(72+2\cdot 4x).$$ Solving it, we obtain $4x=24$, which gives $x=6$. So the side length of the square $HIJK$ is 6 cm.