Obtuse Angle

Prove that among any 9 points in (3D) space, there are three which form an obtuse angle.

Let the points be labeled A1, A2, … , A9, and P be their convex hull. If we assume that all angles among the points are not obtuse, then the interiors of the bodies P + A1, P + A2, … , P + A9 should be all disjoint. That is because, for every Ai and Aj, P must be bound between the planes passing through Ai and Aj which are orthogonal to the segment AiAj. However, all of these 9 bodies have the same volume and are contained in the body P + P, which has 8 times larger volume. This is a contradiction, and therefore our assumption is wrong.

Riddled and Dismembered

The following is a riddle. Do not solve the riddle. Instead, explain the title.

My top is tilted to the light,
A trumpet, almost, in its form.
You court a woman? There I am.
I please, palliate, perfume and mourn.
Your grief I lament. Slay me and
I will you last amen adorn.

The riddle is about a (dismembered) flower. Its different parts are hidden within the words of the riddle:

My top is tilted to the light, -> pistil
A trumpetalmost in its form. -> petal
You court a womanThere I am. -> anther
I pleasepalliate, perfume and mourn. -> sepal
Your grief I lament. Slay me and -> filament
I will your last amen adorn. -> stamen

Source:

Puzzling StackExchange

Lost Boarding Pass

There are 100 passengers which are trying to get on a plane. The first passenger, however, has lost his boarding pass, so decides to sit on an arbitrary seat. Each successive passenger either sits on his own seat if it is empty or on an arbitrary other if it is taken. What is the chance that the last person will sit in his own seat?

The chance is 50%. Indeed, the last passenger will either have to sit in his own seat or the one which belongs to the first passenger. Since there hasn’t been any preference made by anyone at any time towards any of these two seats, the probability that either of them is left last is 1/2.

Mysterious Polynomial

You are given a polynomial P(x) of unknown degree with coefficients which are non-negative integers. You don’t know any of the coefficients, but you are allowed to evaluate the polynomial at any point you choose. What is the smallest number of evaluations you need to use, so that can find the degree and the coefficients of P(x)?

Just 2 evaluations are enough. First, get P(1). This will give you an upper bound on the number of digits of the largest coefficient of the polynomial, let’s say that is N. Now evaluate the polynomial at the point M = 10 to the power of N. This will make all of the coefficients of P appear in the number P(M), separated by strings of zeros.

Broken Clock

An old wall clock falls on the ground and breaks into 3 pieces. Describe the pieces, if you know that the sum of the numbers on each of them is the same.

The total of all numbers on the clock is 1 + 2 + … + 12 = 78. Therefore each piece must contain numbers with total sum 26. The only way for this to happen is if the pieces are broken via 3 parallel lines – {1, 2, 11, 12}, {3, 4, 9, 10}, {5, 6, 7, 8}.

Cover the Grid

You must cover a 7×7 grid with L-shaped triminos and S-shaped tetrominos, without overlapping (flipping and rotating is permitted). What is the minimum number of pieces you can use in order to do this?

Remark: All pieces must be placed entirely on the board.

Consider all cells in the grid which lie in an odd row and odd column – there are 16 of them. Since each of the two pieces can cover at most 1 of these cells, we need at least 16 pieces. Giving an example with 16 pieces is easy.