2 Geometry Solutions

Size: px

Start display at page:

Download "2 Geometry Solutions"

Harry Owen
5 years ago
Views:

1 2 Geometry Solutions Here is give problems and solutions in increasing order of difficulty. 2.1 Easier problems Problem 1. What is the minimum number of hyperplanar slices to make a d-dimensional Rubik s cube? Solution The key is to notice that 2d slices are sufficient, by construction, and that since the central cube requires 2d slices because it has 2d faces. Problem 2. [This was homework] Some squares can be non-trivially partitioned into squares all of whose side lengths are different (very hard). Are there any cubes which have a partition into cubes all of whose side lengths are different? Solution No: if there were a such a partition, then consider the cube with the smallest dimensions: it is impossible to fit the other cubes around it. Problem 3. [This was homework] Let δ be any positive real number. Determine lim n cos 4π 2 n 2 + δ. Solution Why this is under geometry I have no idea. Write cos 4π 2 n 2 + δ = cos( ( ) 4π 2 n 2 δ + δ 2πn) = cos. 4π 2 n 2 + δ So the limit is 1. Problem 4. Prove that if we choose more than 2 n+1 /n points in { 1, 1} n, then three of them form an equilateral triangle. Solution Let Y be a set of more than 2 n+1 /n points in X = { 1, 1} n. Let N(x) denote the set of points at distance two from x X (so these are points differing from x in one co-ordinate). Then any two elements of N(x) 1

2 are at distance 8 from each other, so it is enough to show that Y contains three points in N(x). We know that X max N(x) Y N(x) Y x X x X = n Y > 2 n+1 since there are n points at distance two from every x X. It follows that max N(x) Y > 2 x X which shows that N(x) contains three points in Y for some x X, as required. 2.2 Harder Problems Problem 5. [This was homework] Prove that if two of the angle bisectors of a triangle are equal in length, then the triangle is isosceles. Solution Let the triangle be abc, with angle 2α at a and 2β at b, with angle bisectors ad and be. Add lines dd and ee parallel to ab. We may assume, without loss of generality, that dd ee. Notice that add and bee are isosceles triangles with acute angles α and β respectively, since dd and ee are parallel. In these triangles, dd ee and ad = be, so cos α cos β and α β. However ad bd, so (exercise) ac bc, which means sin 2α sin 2β and α β. Therefore α = β, and abc is isosceles. c d e d e a α α β β b Figure 1 : Add parallel lines dd, ee. 2

3 Problem 6. Prove that if a finite number of points in the plane are not all collinear, then there is a line which passes through exactly two of the points. Can you arrange seven points such that only three lines pass through exactly two of the points? Solution Suppose such a finite configuration P of points exists. Let [p, q] denote the line through points p, q P. Choose three non-collinear points p, q, r P such that the distance from r to [p, q] is minimized. We can assume that r is the point (0, 1) and [p, q] is the x-axis. By assumption, [p, q] contains a third point s P, and we have found a point t P on the same side of the y-axis as p or on the same side of the y-axis as q. But then the distance from t to [p, r] or the distance from t to [q, r] is less than the distance from r to [p, q], contradicting our choice of p, q, r. The required arrangement is given below: for each pair of points, the line through them contains a third point, unless the pair is on the curve drawn in the picture below. Figure 2 : Many triples of collinear points. Problem 7. Prove that if we are given n distinct points in R 2, then there exists a point x in R 2 such that every line through x separates the n points into two groups of at least n/3 points each. Solution First we show that any four points in the plane can be partitioned into two sets such that the convex hull of the two sets has nonzero intersection. The points are affinely independent, so we know that 3

4 α 1 x 1 + α 2 x 2 + α 3 x 3 + α 4 x 4 = 0 where α i s add up to zero. Let A = {x i : α i > 0} and B = {x i : α i < 0}. Then let x = i A x i α = x i α i B where α is the sum of α i s with x i A. So x is in the convex hull of A and of B. Now we proceed by induction on n, to prove that the intersection of any n convex sets is non-empty if any three of them have non-empty intersection. If n = 3, the result is obvious. Suppose n > 3. Let x i be a point which is in the intersection of convex sets A j where j i, where i = 1, 2,..., d + 2. Then there are disjoint sets A and B such that conv{x i : x i A} and conv{x i : x i B} intersect. Then a point in this intersection lies in all the sets A j. Problem 8. [This was homework] Prove that amongst 2 n + 1 points in the plane, some three of the points form an angle of at least π π/n. Solution Let X be the set of points. Each pair of points (x, y) defines a line segment [x, y] making an angle with the x-axis in [0, π). Partition the possible angles into the sets [iπ/n, (i + 1)π/n) for 0 i n 1. This partitions the set L = {[x, y] : x, y X} into n parts, say L j : j n. We claim that there is a closed walk consisting of an odd number of line segments in some L j. If not, then for all j we can partition X into two sets (U j, V j ) such that every [x, y] L j has x U j and y V j, or vice versa. For x X, define the binary string x by x j = 1 if x U j and x j = 0 otherwise, for j n. Note that x y by assumption. On the other hand, since X > 2 n, there exist x, y X such that x = y, a contradiction which proves the claim. To complete the proof, let [x 1, x 2 ], [x 2, x 3 ],..., [x 2k+1, x 1 ] be the line segments traversed in a closed walk in some L j. Then the angle between [x i, x i+1 ] and [x i 1, x i ] is either between 0 and π/n or between π π/n and π, for each i. It cannot be that the angle is always between 0 and π/n, since the walk would then have to have even length (it would zigzag). This completes the proof. Problem 9. Prove that for every positive integer n, there is a circle in the plane which contains n integer lattice points in its interior. Prove that there is also a circle containing n lattice points on its boundary. 4

5 Solution Given an open disc C and a finite set X of points in C, we can find an annulus of width ε separating X from C, the boundary of C, for some ε > 0. So it is enough to allow points to be on the boundary in this problem. For n = 1, the answer is obvious. Suppose we found a disc C such that C C contains a set X of n points. For some smallest δ > 0, (1 + δ)(c C) contains X and some new point/s, and all new points are on the boundary. If there is only one new point, we are done. If there are two or more new points, pick a circle contained in an annulus in (1 + δ)(c C)\C of width ε > 0 separating X from the new points, such that the circle contains exactly one new point. This circle is the required circle C for which C C contains n + 1 points. So the proof is complete by induction. For the second part, the key is to consider circles (2x 1) 2 + 4y 2 = 5 k (3x 1) 2 + 9y 2 = 5 k Prove that the first equation has 2k solutions whereas the second has 2k + 1 solutions. Problem 10. Draw a configuration of n straight lines in the plane. A monotone path in the configuration is a walk along lines such that the x- coordinates of points along the walk are always increasing and no points are visited more than once. A turning point of the walk is a point at which the walk moves fr om one straight line to another. Find a configuration which has a monotone path with at least cn 3/2 turning points for all n, where c is a positive constant. Can there be a configuration with a monotone path with at least n(n 1)/2 turning points for arbitrarily large n? Solution Consider the picture in the hint. The groups of long solid lines consist of k lines each, for a total of k 2 lines. The short solid lines have negative slope, and there are k(k 1) of them. The long dashed lines have positive slope and there are k 1 of them. So we have a total of n = k 2 + k(k 1) + k 1 = 2k 2 1 lines. A long x-monotone path is constructed as follows. In each k k subgrid of solid long lines, we can find an x-monotone walk of length 2k 2 from the leftmost to the rightmost point of the grid. Using the short solid lines and long dashed lines, we can clearly link these together to get an x-monotone path of length k 2 (2k 2) + k(k 1) + (k 1) = 2k 3 k

6 This is of order 2 1/2 n 3/2, as required. For the second part of the problem, if we consider five consecutive turns along an x-monotone path, then six lines are required to define these five turns. Two of the lines intersect at a point which cannot be a turn of the x-monotone path, so for every five consecutive turning points, we found a point which is not a turn. Since there are at most ( n 2) points which are intersections of lines, the number l of turning points of any x-monotone path satisfies l ( ) n l/ which gives l 5 6( n 2)

5 Graphs

5 Graphs jacques@ucsd.edu Some of the putnam problems are to do with graphs. They do not assume more than a basic familiarity with the definitions and terminology of graph theory. 5.1 Basic definitions