Problems

This is a famous problem, called Monty Hall problem after a popular TV show in America.
In the problem, you are on a game show, being asked to choose between three doors. Behind each door, there is either a car or a goat. You choose a door. The host, Monty Hall, picks one of the other doors, which he knows has a goat behind it, and opens it, showing you the goat. (You know, by the rules of the game, that Monty will always reveal a goat.) Monty then asks whether you would like to switch your choice of door to the other remaining door. Assuming you prefer having a car more than having a goat, do you choose to switch or not to switch?

Alice the fox and Basilio the cat have grown \(20\) counterfeit bills on a money tree and now write seven-digit numbers on them. Each bill has \(7\) empty cells for numbers. Basilio calls out one digit "1" or "2" (he doesn’t know the others), and Alice writes the number into any empty cell of any bill and shows the result to Basilio. When all the cells are filled, Basilio takes as many bills with different numbers as possible (out of several with the same number, he takes only one), and the rest is taken by Alice. What is the largest number of bills Basilio can get, regardless of Alice’s actions?

Cut a \(7\times 7\) square into \(9\) rectangles, out of which you can construct any rectangle whose sidelengths are less than \(7\). Show how to construct the rectangles.

Can three points with integer coordinates be the vertices of an equilateral triangle?

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 \[\angle ABE = \angle ABF + \angle FBE = \angle DCF + \angle FCE = \angle DCE.\]

Jess and Tess are playing a game colouring points on a blank plane. Jess is moving first, she picks a non-colored point on a plane and colours it red. Then Tess makes a move, she picks \(2022\) colourless points on the plane and colours them all green. Jess then moves again, and they take turns. Jess wins if she manages to create a red equilateral triangle on the plane, Tess is trying to prevent that from happening. Will Jess always eventually win?

Can you cover a \(10 \times 10\) board using only \(T\)-shaped tetrominos?

Can you cover a \(10 \times 10\) square with \(1 \times 4\) rectangles?

Two opposite corners were removed from an \(8 \times 8\) chessboard. Is it possible to cover this chessboard with \(1 \times 2\) rectangular blocks?