five line proofs Victor Ufnarovski, Frank Wikström 20 december 2016 Matematikcentrum, LTH

Transcription

1 five line proofs Victor Ufnarovski, Frank Wikström 20 december 2016 Matematikcentrum, LTH

2 projections The orthogonal projections of a body on two different planes are disks. Prove that they have the same radius R. If the planes are parallel, the conclusion is clear. Otherwise, consider the line of intersection between the planes and the projection on this line. This is a segment of length 2R. 2

3 common eigenvectors Let A, B be two square complex matrices and C = AB BA. Prove that if rank C = 1 then A and B have a common eigenvector. 3

4 mirror How tall does a mirror have to, be to ensure that you can see all of yourself in it? 4

5 no fixed points Let f (x) = ax 2 + bx + c. Show that if the equation f (x) = x has no real solutions then the equation f (f (x)) = x has no real solutions as well. Either f (x) > x for all x and f (f (x)) > f (x) > x or f (x) < x for all x and f (f (x)) < f (x) < x. The statement is valid for any continuous function f (x). 5

6 sylvester gallai s theorem In any configuration of n points in the plane, not all on a line, there is a line which contains exactly two of the points. Consider a pair consisting of a point P and a line l where the distance between P and l is minimal. The line must contain at least three other points. If P is the orthogonal projection of P on l, then at least two of those points, say X and Y must be on the same side relative P. Suppose that X is closer to P. What is the distance d between X and the line PY? We easily prove that d is less than PP and will obtain the desired contradiction. 6

7 cubic equation Solve the equation: x 3 + x 2 + x = 1 3. Aesthetically the equation 3x 3 + 3x 2 + 3x + 1 = 0 looks better. Something looks very familiar and we are almost forced to complete the picture that 3x 2 + 3x + 1 is painting for us. We get: 3x 3 + x 3 + 3x 2 + 3x + 1 x 3 = 2x 3 + (x + 1) 3. ( ) Now we can factorize the sum of two cubes 3 3 2x + (x + 1) 3. 7

8 equal unions Consider a set with n elements and n + 1 non-empty subsets S 1, S 2,..., S n+1. Show that we can always choose two non-empty disjoint sets of indices I and J such that i I S i = j J S j. Each S i can be encoded as a {0, 1}-vector v i of length n. The linear dependence between them in R n can be rewritten as α i v i = β j v j i I j J with positive α i, β j.the subsets of indices I, J are found! 8

9 chessboard There are k black squares on an n n board, the remaining ones are white. Each white square that has a common side with at least two black squares is painted black and this process is continued so long as it is possible to find such a white square. Prove that if at the end all the white squares become black then k n. 9

10 divisors (imo) The positive divisors of the integer n > 1 are d 1 < d 2 <... < d k, so that d 1 = 1, d k = n. Show that d 1 d 2 + d 2 d d k 1 d k < n 2. Consider the set of divisors { n d i }. We need n 2 d k d k n2 < n d 2 d 1 d 1 d 2 d 2 d 3 1 d k 1 d k < 1. Now compare with the differences ( 1 1 ) ( ) ( ) = 1 1 d 1 d 2 d 2 d 3 d k 1 d k n 10

11 rewriting things Let p > 3 be a prime. Show that if then p 2 m. Because 1 k + 1 p k = p 1 = m n p it follows that p m. k(p k) 11

12 rewriting things, solution If we cancel p, it remains to show that in Z p, But in Z p, p k = k and the sum is equal to p 1 2 k=1 1 k(p k) = 0. p 1 2 k=1 1 k 2 = 1 p 1 2 k=1 This is just the sum of all squares in Z p : 1 k (p 1) 2 = p(p + 1)(2p + 1), 6 which is obviously zero in Z p for p > 3. 12

13 nice rectangles A rectangle is nice if at least one of its side has integer length. Prove that every rectangle that can be cut into nice rectangles is itself nice. We need to understand what is nice in being nice. Consider a chessboard colouring, but where all the cells have size Then in every nice rectangle exactly half the area is white. The opposite is true as well. 13

14 measuring a brick We have several bricks of the same size and a tape measure. We need to measure the longest (three dimensional) diagonal. Of course, it can be done trivially: if a, b, c are the sizes then we can measure each of them and compute a 2 + b 2 + c 2 as the answer. The problem is that we are allowed to use only one measurement. Is it still possible? 14

15 rational powers There are two irrational numbers x and y such that x y is rational. Take x = y = 2. If 2 2 is rational, we are done. Otherwise and again, we are done. ( 2 2) 2 = ( 2) 2 = 2 15

16 guess the polynomial I m thinking of a polynomial p (of arbitrary degree) with non-negative integer coefficients. You may ask me the value p(n) for any integer n. Can you determine the polynomial with a finite number of questions? Ask me for p(1) and p(p(1) + 1). Cheating: Ask me for p(π). 16

17 100 prisoners problem 100 prisoners awaiting execution are given an opportunity to be pardoned. Each prisoner is assigned an integer between 1 and 100. In a neighbouring room, there are 100 boxes, each containing a piece of paper with an integer between 1 and 100. One at the time, the prisoners are let into the other room and are allowed to open up to 50 boxes. If every prisoner finds his/her own integer, all of them are freed. Otherwise they are all executed. (The boxes are closed again awaiting the next prisoner.) The prisoners may decide on a strategy, but once they start opening boxes, no communication is allowed. How big is their chance to survive? 17

18 100 prisoners problem Number the boxes from 1 to 100. Every prisoner starts opening his own box, and they continue to open the box whose number is the one just revealed. This strategy succeds if the corresponding permutation has no cycle with length > 50. The probability that the permutation has a cycle of length l > 50 is (there can only be one such cycle) ( ) 100 (l 1)! (100 l)! = 100!. l l Hence, the probability of survival is ! 100 l=51 100! l = 1 (H 100 H 50 ) Letting each prisoner look in 61 boxes, the probability goes above 50 %. 18

19 pascal s theorem Pascal s theorem (also known as the Hexagrammum Mysticum Theorem) states that if six arbitrary points are chosen on a conic (i.e., ellipse, parabola or hyperbola) and joined by line segments in any order to form a hexagon, then the three pairs of opposite sides of the hexagon (extended if necessary) meet in three points which lie on a straight line, called the Pascal line of the hexagon. 19

20 pascal s theorem 20

21 pascal s theorem Let f be the cubic polynomial vanishing on the lines through AB, CD and EF, and let g the the cubic polynomial vanishing on the lines through BC, DE and FA, respectively. Take any point P on the cubic, and choose the constant λ such that h = f + λg = 0 at P. Then the cubic h vanishes at seven points on the conic. By Bezout s theorem, h is reducible: one component of h = 0 must be the conic, and the other a line through the intersection points, i.e. the Pascal line. 21

22 lower bound for ramsey numbers Suppose we have a complete graph on n vertices. We wish to show (for small enough values of n) that it is possible to color the edges of the graph in two colors (say red and blue) so that there is no complete subgraph on r vertices which is monochromatic (every edge colored the same color). 22

23 lower bound for ramsey numbers Color each edge independently with probability 1 2 red or blue. For any set S of r vertices, let X(S) = 1 if every edge between the r vertices have the same color, otherwise 0.The expectation of X(S) is E(X(S)) = 2 2 r(r 1)/2. Hence E = S E(X(S)) = 2 2 r(r 1)/2 = ( ) n 2 1 r(r 1)/2 r is the expected number of monochromatic complete r-subgraphs.if E < 1, there must exist a coloring with no monochromatic complete r-subgraph.this happens for example with n = 5, r = 4. Hence R(4, 4) > 5. 23

24 lower bound for ramsey numbers Similarly, the above shows that R(r, r) grows at least exponentially in r. For example, almost every coloring of K n where n = (1.1) r contains no monochromatic complete r-subgraph. (But there is no explicit construction known!) 24

25 a theorem by wiener Assume f is a continuous zero-free function on the circle with absolutely convergent Fourier series. Then so is 1/f. This was proved by Wiener in 1932 with a long technical argument with delicate estimates. 25

26 a theorem by wiener, gelfand s solution (1939) The set of functions on the circle with absolutely convergent Fourier series is the image of the Gelfand transform Γ : l 1 (Z) C(S 1 ). Recall some Banach algebra theory If Γ is the Gelfand transform from any commutative Banach algebra B to the ring of continuous functions on its maximal ideal space M, then x is invertible in B iff Γ(x) is invertible in C(M). So, by assumption f = Γ(x) for some invertible x l 1 (Z). The Gelfand transform is an algebra homomorphism, so 1/f = Γ(x 1 ). 26

27 the fundamental theorem of algebra Every (non-constant) polynomial p C[z] has a zero. Assume p is zero-free. Then by Cauchy s integral formula 1 zp(z) dz = 2πi 2p(0). z =r Let r. The left hand side tends to zero (why?) which is a contradiction. 27

28 the fundamental theorem of algebra, revisited Every (non-constant) polynomial p C[z] has a zero. Extend p to a continuous function p : CP 1 CP 1. This is a continuous mapping from a compact space to a Hausdorff space, so the image is closed. The mapping is also holomorphic and non-constant, so the image is open. Finally CP 1 is connected, so the only clopen sets are and CP 1 itself. 28

29 problems to take home

30 common eigenvectors Let A, B be two square complex matrices and C = AB BA. Prove that if rank C = 1 then A and B have a common eigenvector. 30

31 lots and lots of gangsters There is an infinite number of gangsters and every one of them tries to shoot one of his colleagues. Prove that it is possible to choose an infinite subset of gangsters not shooting one another. 31

32 integer (?) sequence The sequence a n is defined by a 0 = a 1 = 1; a n+2 = 1 + a2 n+1 a n. Prove that all terms in the sequence are integers. 32

33 three equal parts Divide a segment on a plane in three equal parts using compasses only. 33

34 the windmill problem (imo) Let S be a finite set containing at least two points in the plane. Assume that no three points of S are collinear. A windmill is a process that starts with a line l going through a single point P S. The line rotates clockwise about the pivot P until the first time that the line meets some other point in S. This point, Q, takes over as the new pivot, and the line now rotates clockwise about Q, until it next meets a point in S. This process continues indefinitely. Show that we can choose a point P S and a line l going through P such that the resulting windmill uses each point of S as a pivot infinitely many times. 34

35 inscribed circles Let AA 1, BB 1 and CC 1 intersect each other in the point O inside the triangle ABC (A 1 belongs to BC, B 1 to AC and C 1 to AB.) Prove that if OB 1 + AC 1 = AB 1 + OC 1 and OC 1 + BA 1 = OA 1 + BC 1, then OB 1 + CA 1 = OA 1 + CB 1. 35

36 sums divisble by n Prove that given any 2n 1 natural numbers it is possible to choose n of them such that their sum will be divisible by n. 36