Prove that one can not tile the whole plane with regular pentagons.
Draw the plane tiling using convex hexagons with parallel and equal opposite sides:
Draw how to tile the whole plane with figures, consisting of squares \(1\times 1\), \(2\times 2\), \(3\times 3\), \(4\times 4\), \(5\times 5\), and \(6\times 6\), where each square appears equal amount of times in the design of the figure. Can you think of two essentially different ways to do this?
Find a non-regular octagon which you can use to tile the whole plane and show how to do that.
Observe that \(14\) isn’t a square number but \(144=12^2\) and \(1444=38^2\) are both square numbers. Let \(k_1^2=\overline{a_n...a_1a_0}\) the decimal representation of a square number.
Is it possible that \(\overline{a_n...a_1a_0a_0}\) and \(\overline{a_n...a_1a_0a_0a_0}\) are also both square numbers?
Let \(ABCDEF\) be a regular hexagon. Points \(G\) and \(H\) lie on \(EF\) and \(DE\) respectively such that \(|EG|=|EH|\). Furthermore, the area of quadrilateral \(ABGF\) is equal to the area of quadrilateral \(BGEH\), which are both equal to the area of \(BCDH\). What’s the ratio \(\frac{|EG|}{|EF|}\)?