Problem of The Month From Math Central Uregina

MP12: August 2001 You are given five points in the plane and are told that they are the midpoints of the edges of some pentagon. Find the pentagon. Given the midpoints of the sides of an n-gon in the plane, is it always possible to determine the n-gon? Solution to August 2001 problem MP11: July 2001 An old edition of Ripley's Believe It Or Not states that the number N = 526315789473684210 is a persistent number: if N is multiplied by any positive integer, the resulting product always contains all ten digits 0, 1, 2, ..., 9 (with repetitions, of course). Prove or disprove Ripley's statement. Are there any persistent numbers that are smaller than N? Solution to July 2001 problem MP10: June 2001 We found this month's problem in Crux Mathematicorum 27:3 (April 2001) pages 204-205 - it is problem 3 from the Ninth Annual Konhauser Problemfest (Carleton College, prepared by David Savitt and Russell Mann. a. Begin with a string of 10 A's, B's, and C's, as for example A B C C B A B C B A. Underneath, form a new row of length 1 less as follows: between two consecutive letters that are different you write the third letter, and between two letters that are the same, you write the same letter again. Repeat this process until you have only one letter in the new row. For example, for the string given above you will have A B C C B A B C B A C A C A C C A A C B B B B C B A B B B B A A C C B B C A B C B A B C A C C A B C B C A A A Prove that the letters at the corners of the resulting triangle are always either all the same or all different. b. For which numbers n (besides 10) is the result from part (a) true for all strings of n A's, B's, and C's? Solution to June 2001 problem MP9: May 2001 This month's problem has been taken from ALGEBRA (1572), http://www-groups.dcs.st-and.ac.uk:80/~history/Bookpages/Bombelli4.gif whose author, Rafael Bombelli, lived from 1526 to 1573. Thanks Raf. Find the side length of a square DEFG inscribed in triangle ABC with D on AB, EF along BC, and G on AC, given AB = 13, BC = 14, and CA = 15. Solution to May 2001 problem MP8: April 2001 The square with vertices (0, 0), (1, 0), (1, 1), (0, 1) has two properties that are of interest here: all its vertices have integer coordinates (so we call them lattice points), except for the vertices, no point on the boundary or in the interior is a lattice point. This is not the case, for example, for the square with vertices (0, 0), (2, 0), (2, 2), (0, 2); although its vertices are lattice points, there are several lattice points (such as (0, 1)) on the boundary , and even an interior lattice point (1, 1). On the other hand, it is easy to find convex quadrangles with vertices at lattice points that contain no lattice points -- it just has to be long and thin, such as (0, 0), (1, 2), (2, 5), (1, 3). (We call a polygon convex if its boundary has no indentations; more precisely, line segments that join two of its vertices must contain no exterior points.)Is there a convex pentagon with lattice points as vertices that contains no other lattice points either inside or on the boundary? Solution to April 2001 problem MP7: March 2001 The diagram shows a unit circle with the boundary of the shaded portion consisting of three circular arcs of radius 1 whose centres are equally spaced on the surrounding circle. (If you imagine the diagram to be a clock's face, the centres are at 12:00, 4:00, and 8:00.) The problem is to dissect the unshaded portion of the clock's face into pieces that can be reassembled to form a rectangle. Solution to March 2001 problem MP6: February 2001 This month we have a variation on a well-known problem: It is easy to cover all the squares of an ordinary (8 by 8) chessboard using 32 dominoes; however, if we remove the squares from two opposite corners of the chessboard, the resulting grid cannot be covered with 31 dominoes. If you have not seen this classic problem before, you may wish to try solving it as a warm-up exercise to our problem.Consider a tetromino consisting of four squares arranged in a T as follows: It is easy to cover the 8 by 8 chessboard with these tetrominoes. Is it also possible to cover a 10 by 10 board with t-shaped tetrominoes? Either show how to do it, or prove that it cannot be done. Solution to February 2001 problem   MP5: January 2001 An equilateral triangle is inscribed in, and has a common vertex with, a rectangle. If A, B, C are the areas of the complementary right triangles (as shown in the figure), prove that A + B = C. Solution to January 2001 problem MP4: December 2000 The counting numbers are partitioned into the groups (1), (2, 3), (4, 5, 6), (7, 8, 9, 10), (11, 12, 13, 14, 15), ... . Find the sum of the numbers in the nth group. Solution to December 2000 problem MP3: November 2000 How many faces can a polyhedron have if all those faces are triangles? Some background information. A polyhedron is a solid figure with flat sides (called faces). In general each of those sides would be a polygon, but for our problem they are triangles. Most of you are probably familiar with the three examples whose faces are equilateral triangles, namely the tetrahedron (4 faces), octahedron (8 faces), and icosahedron (20 faces). In this problem there is no requirement that the triangular faces be equilateral.    Solution to November 2000 problem MP2: October 2000 Here is something you can do with a crossword puzzle even when you don't know any words. You start with a grid consisting mostly of white squares, but with some of them black. We say that two squares are neighbours if they share an entire edge. Start filling in the white squares that have at least two neighbours already black.For example, with we would fill in the square labelled X, because in each case it has two black neighbours. The new black squares allow you to fill in other squares; after a while you either fill in the whole grid, or you end up with a bunch of white squares that are adjacent to at most one black square. Our problem deals with the possibility of making an entire 8 by 8 grid all black. For instance, suppose that your original configuration is an 8 by 8 grid with the black squares down the main diagonal: leading to and eventually to Of course, there are many initial configurations of eight black squares that lead to a completely filled in 8 by 8 grid, such as Does there exist a configuration of 7 black squares that will allow one to completely fill in an 8 by 8 grid, or does one always need at least 8 black squares in the original configuration to do that? Solution to October 2000 problem MP1: September 2000  Eight nickels are touching as in the figure. If the upper right nickel (fully drawn) is rolled, without slipping, around the other seven, how many complete somersaults will the beaver make as it rolls through one full circuit of the array? Solution to September 2000 problem