Can you tile the plane with regular octagons?
Sam the magician shuffles his hand of six cards: joker, ace (\(A\)), ten, jack (\(J\)), queen (\(Q\)) and king (\(K\)). After his shuffle, the relative order
of joker, \(A\) and \(10\) is now \(A\), \(10\), joker. Also, the relative order of
\(J\), \(Q\) and \(K\) is now \(Q\), \(K\)
and \(J\).
For example, he could have \(A\), \(Q\), \(10\), joker, \(K\), \(J\)
- but not \(A\), \(Q\), \(10\), joker, \(J\), \(K\).
How many choices does Sam has for his shuffle?
Draw how to tile the whole plane with figures, composed from squares \(1\times 1\), \(2\times 2\), \(3\times 3\), \(4\times 4\), and \(5\times 5\) where squares of all sizes are used the same amount of times in the design of the figure.
The perimeter of the triangle \(\triangle ABC\) is \(10\). Let \(D,E,F\) be the midpoints of the segments \(AB,BC,AC\) respectively. What is the perimeter of \(\triangle DEF\)?
Let \(\triangle ABC\) be a triangle and \(D\) be a point on the edge \(BC\) so that the segment \(AD\) bisects the angle \(\angle BAC\). Show that \(\frac{|AB|}{|BD|}=\frac{|AC|}{|CD|}\).
Show that if \(1+2+\dots+n = \frac{n(n+1)}{2}\), then \(1+2+\dots+(n+1) = \frac{(n+1)((n+1)+1)}{2}\).
Show that \(1+2+\dots+n = \frac{n(n+1)}{2}\) for every natural number \(n\).
Show that if \(1+2^1+2^2+\dots+2^{10} = 2^{11} - 1\), then \(1+2^1+2^2+\dots+2^{11} = 2^{12} - 1\).
Show that \(1+2^1+2^2+\dots+2^n = 2^{n+1} - 1\) for every natural number \(n\).
What is wrong with the following proof that “all rulers have the same length" using induction?
Base case: suppose that we have one ruler, then clearly it clearly has the same length as itself.
Assume that any \(n\) rulers have the same length for the induction hypothesis. If we have \(n+1\) rulers, the first \(n\) ruler have the same length by the induction hypothesis, and the last \(n\) rulers have the same length also by induction hypothesis. The last ruler has the same length as the middle \(n-1\) rulers, so it also has the same length as the first ruler. This means all \(n+1\) rulers have the same length.
By the principle of mathematical induction, all rulers have the same length.