Problems
On the diagram below \(AD\) is the bisector of the triangle \(ABC\). The point \(E\) lies on the side \(AB\), with \(AE = ED\). Prove that the lines \(AC\) and \(DE\) are parallel.
On the diagram below the line \(BD\) is the bisector of the angle \(\angle ABC\) in the triangle \(ABC\). A line through the vertex \(C\) parallel to the line \(BD\) intersects the continuation of the side \(AB\) at the point \(E\). Find the angles of the triangle \(BCE\) triangle if \(\angle ABC = 110^{\circ}\).
We want to wrap \(12\) Christmas presents in different coloured paper. We have \(6\) different patterns of paper and we want to use each one exactly twice. In how many ways can we do this?
Mr Roberts wants to place his little stone sculptures of vegetables on the different shelves around the house. He has \(17\) sculptures in total and three shelves that can fit \(7\), \(8\) and \(2\) sculptures respectively. In how many ways can he do this?
The order of sculptures on the shelf does not matter.
Scrooge McDuck has \(100\) golden coins on his office table. He wants to distribute them into \(10\) piles so that no two piles contain the same amount of coins. Moreover, no matter how you divide any of the piles into two smaller piles, among the resulting \(11\) piles there will be two with the same amount of coins. Find an example of how he could do that.
In a certain state, there are three types of citizens:
A fool considers everyone a fool and themselves smart;
A modest clever person knows truth about everyone’s intellectual abilities and consider themselves a fool;
A confident clever person knows about everyone intellectual abilities correctly and consider themselves smart.
There are \(200\) deputies in the High Government. The Prime Minister conducted an anonymous survey of High Government members, asking how many smart people are there in the High Government. After reading everyone’s response he could not find out the number of smart people. But then the only member who did not participate in the survey returned from the trip. They filled out a questionnaire about the entire Government including themselves and after reading it the Prime Minister understood everything. How many smart could there be in the High Government (including the traveller)?
The dragon locked six dwarves in the cave and said, "I have seven caps of the seven colors of the rainbow. Tomorrow morning I will blindfold you and put a cap on each of you, and hide one cap. Then I’ll take off the blindfolds, and you can see the caps on the heads of others, but not your own and I won’t let you talk any more. After that, everyone will secretly tell me the color of the hidden cap. If at least three of you guess right, I’ll let you all go. If less than three guess correctly, I’ll eat you all for lunch." How can dwarves agree in advance to act in order to be saved?
Find the mistake in the sequence of equalities: \(-1=(-1)^{\frac{2}{2}}=((-1)^2)^{\frac{1}{2}}=1^{\frac{1}{2}}=1\).
Theorem: If we mark \(n\) points on a circle and connect each point to every other point by a straight line, the lines divide the interior of the circle is into is \(2n-1\) regions.
"Proof": First, let’s have a look at the smallest natural numbers.
When \(n=1\) there is one region (the whole disc).
When \(n=2\) there are two regions (two half-discs).
When \(n=3\) there are \(4\) regions (three lune-like regions and one triangle in the middle).
When \(n=4\) there are \(8\) regions, and if you’re still not convinced then try \(n=5\) and you’ll find \(16\) regions if you count carefully.
Our proof in general will be by induction on \(n\). Assuming the theorem is true for \(n\) points, consider a circle with \(n+1\) points on it. Connecting \(n\) of them together in pairs produces \(2n-1\) regions in the disc, and then connecting the remaining point to all the others will divide the previous regions into two parts, thereby giving us \(2\times (2n-1)=2n\) regions.
Let’s "prove" that the number \(1\) is a multiple of \(3\). We will use the symbol \(\equiv\) to denote "congruent modulo \(3\)". Thus, what we need to prove is that \(1\equiv 0\) modulo \(3\). Let’s see: \(1\equiv 4\) modulo \(3\) means that \(2^1\equiv 2^4\) modulo \(3\), thus \(2\equiv 16\) modulo \(3\), however \(16\) gives the remainder \(1\) after division by \(3\), thus we get \(2\equiv 1\) modulo \(3\), next \(2-1\equiv 1-1\) modulo \(3\), and thus \(1\equiv 0\) modulo \(3\). Which means that \(1\) is divisible by \(3\).