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(B) regular polygon with $n$ sides can tile the plane if and only if its interior angle is a divisor of $360^{\circ}$.The interior angle of a regular

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exactly three of the four shapes

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(B) regular polygon with $n$ sides can tile the plane if and only if its interior angle is a divisor of $360^{\circ}$.The interior angle of a regular $n$-gon is given by $\frac{(n-2) 	imes 180^{\circ}}{n}$.$(1)$ For an equilateral triangle $(n=3)$: Interior angle = $\frac{(3-2) 	imes 180^{\circ}}{3} = 60^{\circ}$. Since $360^{\circ} / 60^{\circ} = 6$,it can tile the plane.$(2)$ For a square $(n=4)$: Interior angle = $\frac{(4-2) 	imes 180^{\circ}}{4} = 90^{\circ}$. Since $360^{\circ} / 90^{\circ} = 4$,it can tile the plane.$(3)$ For a regular pentagon $(n=5)$: Interior angle = $\frac{(5-2) 	imes 180^{\circ}}{5} = 108^{\circ}$. Since $360^{\circ} / 108^{\circ} = 3.33$ (not an integer),it cannot tile the plane.$(4)$ For a regular hexagon $(n=6)$: Interior angle = $\frac{(6-2) 	imes 180^{\circ}}{6} = 120^{\circ}$. Since $360^{\circ} / 120^{\circ} = 3$,it can tile the plane.Thus,the shapes that can tile the plane are the equilateral triangle,the square,and the regular hexagon. There are exactly three such shapes.

Suppose we have to cover the $XY$-plane with identical tiles such that no two tiles overlap and no gap is left between the tiles. Suppose that we can choose tiles of the following shapes: equilateral triangle,square,regular pentagon,regular hexagon. Then,the tiling can be done with tiles of

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