Problems

Look at the following diagram, depicting how to get an extra cell by reshaping triangle.

Can you find a mistake? Certainly the triangles have different area, so we cannot obtain one from the other one by reshaping.

This problem is often called "The infinite chocolate bar". Depicted below is a way to get one more piece of chocolate from the $5\times 6$ chocolate bar. Do you see where is it wrong?

Consider the following "proof" that any triangle is equilateral: Given a triangle $ABC$, we first prove that $AB=AC$. First let’s draw the bisector of the angle $\mathrm{\angle }A$. Now draw the perpendicular bisector of segment $BC$, denote by $D$ the middle of $BC$ and by $O$ the intersection of these lines. See the diagram

Draw the lines $OR$ perpendicular to $AB$ and $OQ$ perpendicular to $AC$. Draw lines $OB$ and $OC$. Then the triangles, $RAO$ and $QAO$ are equal, since we have equal angles $\mathrm{\angle }ORA=\mathrm{\angle }OQA=90°,$ and $\mathrm{\angle }RAO=\mathrm{\angle }QAO,$ and the common side $AO$. On the other hand the triangles $ROB$ and $QOC$ are also equal since the angles $\mathrm{\angle }BRO=\mathrm{\angle }CQO=90°$, the hypotenuses $BO=OC$ the legs $RO=OQ$. Thus, $AR=AQ,$ $RB=QC,$ and $AB=AR+RB=AQ+QC=AC.$ Q.E.D.

As a corollary, one can show that all the triangles are equilateral, by showing that $AB=BC$ in the same way.

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 $\mathrm{\angle }ABC={90}^{\circ }$ and angle $\mathrm{\angle }BCD>{90}^{\circ }$. 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 $\mathrm{\angle }EBF=\mathrm{\angle }ECF$, the same about the segments $AF=FD$. By condition we have $AB=CD$, thus the triangles $ABF$ and $CDF$ are equal, thus $\mathrm{\angle }ABF=\mathrm{\angle }DCF$. But then we have $$\mathrm{\angle }ABE=\mathrm{\angle }ABF+\mathrm{\angle }FBE=\mathrm{\angle }DCF+\mathrm{\angle }FCE=\mathrm{\angle }DCE.$$

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 $1$.

In how many ways can you read the word TRAIN from the picture below, starting from T and going either down or right at each step?

There are $100$ people in a room. Each person knows at least $67$ others. Show that there is a group of four people in this room that all know each other. We assume that if person $A$ knows person $B$ then person $B$ also knows person $A$.