BS D Dear Participant, Shevat 5774 /January 2014 We congratulate you on your achievement in reaching the second round of the Ulpaniada Mathematics Competition and wish you continued success. Please fill in your personal details on this page before you start answering the questions. Name: Grade level: Email address: Name of High School: City: Country/State: This question paper is comprised of two parts. When you have completed the first part and encircled your answers, please copy them into this table (a,b,c,d or e) Question number 1 2 3 4 5 6 7 8 9 10 11 12 Your answer Please write your answers to the second part on the corresponding pages. Wishing you much hatzlacha The Ulpaniada Team Ulpaniada Math Department, Michlalah Jerusalem College P.O.B. 16078, Bayit Vegan, Jerusalem 91160, Israel Tel: +972-2-6750705, Fax: +972-73-2750665 http://ulpaniada.macam.ac.il/english ulpaniada@macam.ac.il 1
The following question paper consists of two parts. The first part contains 12 questions. Each question has 5 possible answers, only one of which is correct. Read the question carefully, solve it and encircle the correct answer. The second part contains 2 questions. Solve them, including all of your reasoning in the solution. Partial answers will also be accepted. You have three hours to complete the whole test. You may use a calculator. B hatzlacha! 1 Enter one of the digits 1,2,3,4,5 into each square of the following 5 5 board, so that all five numbers appear in each row, column or region surrounded by a bold line. Some of the squares already have numbers filled in. What number is supposed to be in the yellow square? a. 1 b. 2 c. 3 d. 4 e. 5 2 Naama chose a natural number. Efrat multiplied it either by 4 or by 5, Shira subtracted either 4 or 5 from Efrat s answer. Bilha added either 4 or 5 to Shira s answer and got 5774. What is the sum of the digits in the number that Naama chose? a. 10 b. 11 c. 12 d. 13 e. there is not enough information to calculate the sum 2
3 The following drawing is made up of five semicircles, the upper semicircle AB, and four smaller semicircles. 3 You can get from A to B via the upper arch (the upper path), or via the four short lower arches (the lower path). What is the ratio between the lengths of the two paths? a. The upper path is twice as long as the lower path b. The upper path is π/2 times as long as the lower path c. The two paths have equal lengths d. The lower path is π/2 times as long as the upper path e. There is not enough information to calculate the ratio 4 How many three digit numbers are there which are equal to 34 times the sum of their digits? a. 10 b. 8 c. 6 d. 4 e. 0 5 Ayala and Dina entered a room which had three bulbs, each connected to a different switch. The lights were all off, and Ayala chose one of the switches and turned it on. After that, Dina chose one of the three switches and changed its position (from on to off or from off to on). From then on, each girl in turn changed the position of one of the switches. After ten turns (five of Ayala s and five of Dina s), which of the following statements is necessarily true? a. All the lights are on b. All the lights are off c. There s at least one light on and one light off d. It can t be that all the lights are on e. It can t be that all the lights are off 3
6 The triangular plot of land near my house was divided up by the gardener into three sections (as shown in the diagram), with different plants in each section. Flowers were planted in the middle pink colored triangle, vegetables in the right triangle and trees in the left triangle. The above diagram is all that is left from the architect s plans, and from his comments it is clear that: ABD is a right angled triangle, DB=DG=GF=EB, and that the distance from A to F is twice the distance from G to B. The ratio between the area of the triangle of flowers and the area of the whole garden is: a. 1/5 b. 1/4 c. d. e. 7 A number N consists only of the digits 1,2,3,4,5-some of these digits may appear more than once, or not at all. If the sum of N s digits is 36, and the sum of the digits of the number 2N is also 36, how many times does N contain the digit 5? a. 0 b. 1 c. 2 d. 3 e. 4 8 An octahedron is a regular solid which is obtained from a cube by joining the centers of its six faces as shown in the diagram below. If each side of the cube has length one centimeter, the volume of the octahedron in cm 3 is: a. 1/8 b. 1/6 c. 1/4 d. 1/3 e. 1/2 4
9 The following multiplication exercise uses all 9 digits 1,2,3,...,9. The digits are encoded by asterisks. We are told that the numbers in each of the three rows are divisible by 9, and that the answer row consists of four consecutive numbers, not necessarily in order. The sum of the last digits of the three numbers (denoted by red asterisks in the illustration) is: a. 21 b. 20 c. 19 d. 18 e. 17 10 In the figure below, nine points on a circle form a regular nine sided polygon. All chords joining any two of the vertices of the polygon are shown. Many of these chords intersect inside the circle. The number of intersection points of chords inside the circle is: a. 252 b. 128 c. 126 d. 120 e. 112 5
11 On Mondays, Wednesdays and Fridays, Tom only speaks the truth, whilst on the other days of the week he only lies. Today Tom said four of the following sentences: 1. The number of my siblings who are older than me equals the number of my siblings who are younger than me. 2. Today is Tuesday 3. Three of my siblings got a 10 in mathematics. 4. 11011 is not a prime number. 5. I have a prime number of siblings. Which sentence did Tom NOT say today? a. 1 b. 2 c. 3 d. 4 e. 5 12 Enter the numbers 1,2,3,,9 into the circles of the W shape below, one in each circle, so that the sums of the three numbers in the circles of each of the four arms of the W are equal. If this equal sum is as large as possible, what is the number in the yellow circle? a. 9 b. 8 c. 7 d. 6 e. We cannot tell 6
Part Two: In this section there are two questions. Solve them, writing down all of your reasoning. Provide a proof where it is requested. Partial solutions will also be accepted. Write your answer on the question paper beneath each question. If there is not enough space, continue on the other side of the page. Please write all of your solutions clearly in black pen. 13 The midpoint polygon. a. Consider any triangle ABC. Join the midpoints of each side to obtain triangle DEF as shown below. Prove that the area of the midpoint triangle DEF is a quarter of the area of the original triangle ABC. b. Given a quadrilateral ABCD, join the midpoints of adjacent sides to form the midpoint quadrilateral EFGH, as shown. Prove that the midpoint quadrilateral is a parallelogram whose area is half the area of the quadrilateral ABCD. 7
c. Given a regular hexagon ABCDEF, form the midpoint hexagon as shown by joining midpoints of adjacent sides. Prove that the area of the midpoint hexagon GHIJKL is three quarters of the area of ABCDEF. d. Prove that for a regular n sided polygon, the ratio of the area of its midpoint polygon to that of the original polygon is e. We have shown that for a regular hexagon, the area of its midpoint hexagon is three quarters of the area of the original hexagon. Now consider any convex hexagon, not necessarily regular. (A convex hexagon is a hexagon all of whose interior angles are less than 180 0 ). Use the result from section b to prove that in every convex hexagon, the area of its midpoint hexagon is greater than half of the area of original hexagon. 8
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14 Food for thought... Consider a row of n squares of chocolate. Binyamin and Noam, two smart kids, play the following game: Binyamin starts by breaking the row along one of its lines into two (not necessarily equal) parts, each of which has a whole number of squares. Binyamin then eats the smaller of the two parts (if the two parts are equal, he eats one of them), and the game goes over to Noam. Noam proceeds in a similar fashion, breaking the remaining chocolate into two parts, and eating the smaller part (or one of the halves if they are equal), and so the game continues, alternating between the two boys. The first player to eat just one square of chocolate is the loser. If, at the beginning of the game, the row had 2 squares, Binyamin would be the loser, because after breaking the row into two parts, he d be forced to eat just one square of chocolate! a. Which of the players wins if, at the beginning of the game the row had 3 squares? b. Which of the players wins if, at the beginning of the game the row had 4 squares? Binyamin is said to be in a losing position, if, no matter how he plays his turn, he cannot prevent Noam from winning. Binyamin is said to be in a winning position, if (by playing cleverly) he can bring Noam to a losing position. If the game begins with two pieces of chocolate, Binyamin is in a losing position. Call the number 2 a losing position. c. Fill in the following table : L=losing position, W= winning position The position of the player whose turn is next n L 2 3 4 5 6 7 8 9 d. Prove that if n is a losing position, then n+2 is a winning position. e. Prove that if n, n+1 are losing positions, then n+2, n+3,...,2n+2 are winning positions. What is the next losing position? What is the losing position after that? f. Prove that apart from the first two losing positions, all the other losing positions are numbers ending with either 7 or 8. g. Derive a formula in terms of n, describing precisely all the losing positions of the game. 10
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