Problems

Let’s prove the following statement: every graph without isolated vertices is connected.
Proof We use the induction on the number of vertices. Clearly the statement is true for graphs with \(2\) vertices. Now, assume we have proven the statement for graphs with up to \(n\) vertices.
Take a graph with \(n\) vertices by induction hypothesis it must be connected. Let’s add a non-isolated vertex to it. As this vertex is not isolated, it is connected to one of the other \(n\) vertices. But then the whole graph of \(n+1\) vertices is connected!

Let’s compute the infinite sum: \[1+2 + 4 + 8 + 16 + ... + 2^n + ... = c\] Observe that \(1+2+4+8+... = 1 + 2(1+2+4+8+16+...)\), namely \(c = 1+2c\), then it follows that \[c = 1+2+4+8+... = -1.\]

Let’s prove that any \(90^{\circ}\) angle is equal to any angle larger than \(90^{\circ}\). On the diagram

We have the angle \(\angle ABC = 90^{\circ}\) and angle \(\angle BCD> 90^{\circ}\). We can choose a point \(D\) in such a way that the segments \(AB\) and \(CD\) are equal. Now find middles \(E\) and \(G\) of the segments \(BC\) and \(AD\) respectively and draw lines \(EF\) and \(FG\) perpendicular to \(BC\) and \(AD\).
Since \(EF\) is the middle perpendicular to \(BC\) the triangles \(BEF\) and \(CEF\) are equal which implies the equality of segments \(BF\) and \(CF\) and of angles \(\angle EBF = \angle ECF\), the same about the segments \(AF=FD\). By condition we have \(AB=CD\), thus the triangles \(ABF\) and \(CDF\) are equal, thus \(\angle ABF = \angle DCF\). But then we have \[\angle ABE = \angle ABF + \angle FBE = \angle DCF + \angle FCE = \angle DCE.\]

Let’s prove that \(1=2\). Take a number \(a\) and suppose \(b=a\). After multiplying both sides we have \(a^2=ab\). Subtract \(b^2\) from both sides to get \(a^2-b^2=ab-b^2\). The left hand side is a difference of two squares so \((a-b)(a+b)=b(a-b)\). We can cancel out \(a-b\) and obtain that \(a+b=b\). But remember from the start that \(a=b\), so substituting \(a\) for \(b\) we see that \(2b=b\), dividing by \(b\) we see that \(2=1\).

Let’s prove that \(1\) is the smallest positive real number: Assume the contrary and let \(x\) be the smallest positive real number. If \(x>1\) then \(1\) is smaller, thus \(x\) is not the smallest. If \(x<1,\) then \(\frac{x}{2}<x\) so \(x\) can not be the smallest either. Then \(x\) can only be equal to \(1\).

Nick writes the numbers \(1,2,\dots,33\), each exactly once, at the vertices of a polygon with \(33\) sides, in some order.

For each side of the polygon, his little sister Hannah writes down the sum of the two numbers at its ends. In total she writes down \(33\) numbers, one for each side.

It turns out that these \(33\) sums are \(33\) consecutive whole numbers, written in some order. Can you find an arrangement of the numbers written by Nick that makes this happen?

Is it possible to arrange the numbers \(1,\, 2,\, ...,\, 50\) at the vertices and middles of the sides of a regular \(25\)-gon so that the sum of the three numbers at the ends and in the middle of each side is the same for all sides?

Draw a shape that can both be cut into 4 copies of the figure on the left or alternatively into 5 copies of the figure on the right. (the figures can be rotated).

A equilateral triangle made of paper bends in a straight line so that one of the vertices falls on the opposite side as shown on the picture. Show that the corresponding angles of the two white triangles are equal.

Sometimes proving a statement takes careful step-by-step reasoning. But other times, all you need is a single well-chosen example.

Indeed: many problems don’t ask you to prove that something always works. Instead, they ask whether something is possible at all: can an object exist, or can a situation happen, even if it sounds unlikely? In those cases, finding just one example that works is enough to solve the problem.

In this sheet, we’ll practise building examples and constructions. The goal is not to try lots of random cases and hope one works, but to think smartly and strategically about what an example should look like.

By the end of the session, you should feel more confident spotting when an example is enough, and how to construct one that does exactly what the problem asks for.