University of Melbourne Schools Mathematics Competition 2004 Senior Solutions

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1 University of Melbourne Schools Mathematics Competition 004 Senior Solutions 1. Jane and her friends are standing in a circle. It turns out that both neighbours of each child are the same gender. If there are six boys in the circle, how many girls are there? Solution: Suppose that two boys are standing next to each other. Then for the neighbours of each child to be of the same gender, the next child in the circle must also be a boy. Similarly, all children in the circle must be boys, contradicting the fact that Jane, who is a girl, is standing in the circle. The same argument shows that no two girls can stand next to each other. Therefore, the circle contains alternating boys and girls. We conclude that there are equally many girls as boys so the circle contains six girls in total.. Suppose that it takes Mary and Jane two hours to do a certain job, it takes Mary and Sue three hours to do the job and that it takes Jane and Sue four hours to do the same job. How long would it take all three girls working together to do the same job? (Of course you should assume that each girl has an individual work rate that remains constant.) Solution: Mary and Jane do 1 a job in one hour, Mary and Sue do 1 a job in one hour and Sue and Jane do 1 4 a job in one hour. So if it were possible to gather two Marys, two Janes and two Sues together, then they would do = 1 1 jobs in one hour. This means that one Mary, one Jane and one Sue would manage to do = 1 4 jobs in one hour. Hence, the time taken for all three girls working together to do one job would be = 4 1 hours.. The elements of a set of n consecutive integers sum to n. The elements of another set, with n consecutive integers, sum to n. The difference between the largest elements of the two sets is 99. Find n. Solution: Let the n consecutive integers be a, a + 1, a +,..., a + n 1 and let the n consecutive integers be b, b + 1, b +,..., b + n 1. By considering the sums of these two sets, we have the following two equations. a + (a + 1) + (a + ) + + (a + n 1) = n n(a+n 1) = n a + n 1 = 4 a = 5 n b + (b + 1) + (b + ) + + (b + n 1) = n n(b+n 1) = n b + n 1 = 1 b = 1 n 1

2 Since the difference between the largest elements of the two sets is 99, we also have the equation (a + n 1) (b + n 1) = ±99 a b n = ±99 Now we can substitute the values for a and b determined earlier. 5 n (1 n) n = ±99 (5 n) (1 n) n = ±198 n = ±198 n = ± 198 n = 195 or 01 Since there are n numbers in the first set, n cannot be negative and we deduce that n = Tamref s theorem says that n a + n b = n c has no solutions for n, a, b, c positive integers and n >. Prove it. Solution 1: Suppose that there do exist positive integers n, a, b, c which satisfy the equation n a + n b = n c. For this to occur, it is clear that c > a and c > b. Without loss of generality, we may assume that a b. Now we have the inequality n b n a + n b = n c n c b. Since c b is positive and n >, it is not possible for this inequality to hold. Hence, we have the desired contradiction. Solution : (Provided by Stephen Muirhead, Scotch College) We can consider the equation modulo n 1 n a + n b n c (mod n 1) 1 a + 1 b 1 c (mod n 1) 1(mod n 1) This cannot hold for n >, so Tamref s Theorem is proven. 5. A partition of a positive integer n is a sum of positive integers that add to n. Thus n = p 1 + p + p p k. For example, possible partitions of 4 include: 4 = 1 +, 4 = 4, 4 = Consider partitions into non-decreasing parts, p 1 p p... p k such that p k = p 1 or p k = p For example, under these restrictions we exclude the partition 4 = 1+. How many such restricted partitions of n are there? Solution: We can calculate the number of restricted partitions for the first few values of n.

3 n Restricted Partitions Number 1 1 1, 1+1, 1+, , +, 1+1+, , +, 1++, , From this evidence, it seems reasonable to conjecture that there are n restricted partitions of n. In fact, the table seems to indicate that for every k which satisfies 1 k n, there is a unique restricted partition of n with k parts. This is exactly what we will try to prove. Consider a positive integer n and an integer k which satisfies 1 k n. By the division algorithm, we can uniquely write n = ka + b where 0 b < k. From this, we can construct a restricted partition of n consisting of k b occurrences of the number a followed by b occurrences of the number a + 1. This clearly has (k b)+b = k parts and has the sum (k b)a+b(a+1) = ka ba+ab+b = ka+b = n. This shows that there exists a restricted partition of n with k parts. We still have to show that the restricted partition of n with k parts is unique. To do this, suppose that we have the two restricted partitions of n with k parts. They must both have the following form, where we can impose the condition 0 x, y < k. p,..., p, p + 1,..., p + 1 }{{}}{{} k x x q,..., q, q + 1,..., q + 1 }{{}}{{} k y y Now for these two partitions to have the same sum, the following equality must hold. x(p + 1) + (k x)p = y(q + 1) + (k y)q kp + x = kq + y But the division algorithm states that there is a unique way to write an integer in the form kp + x where 0 x < k, so we conclude that p = q, x = y and the two restricted partitions are actually the same. So we have proven that for every positive integer n and 1 k n, there exists a restricted partition of n with k parts. Hence, we can conclude that the number of restricted partitions of n altogether is simply n students took an exam comprising 6 questions. Every question was correctly answered by at least 10 students. Show that there must be two students such that every question was correctly answered by at least one of them. Solution: There were at least 6 10 = 70 correct solutions submitted in the exam. By the pigeonhole principle, we deduce that at least one student, whom

we ll call X, correctly solved at least 4 of the questions. Suppose that student X solved questions 1,, and 4. Note that at least 10 students solved each of questions 5 and 6.

4 we ll call X, correctly solved at least 4 of the questions. Suppose that student X solved questions 1,, and 4. Note that at least 10 students solved each of questions 5 and 6. But since there are only 00 students altogether, at least = 40 students must have solved both questions 5 and 6. If we call one of these students Y, then it is clear that students X and Y together solved all of the six questions between them. 7. ABC is an equilateral triangle with the circumcentre O. P is a point inside the circumcircle. Show that there is a triangle with side lengths P A, P B, P C and that its area depends only on P O. Solution 1: Let AB = x, P A = a, P B = b, P C = c and consider a rotation of 60 anticlockwise about the point A. This takes B to C and P to a point which we will call P A. This rotation takes the segment P B to the segment P A C so we have P A C = P B = b. The rotation also takes the segment P A to the segment P A A so we have P A A = P A = a. Since we also know that P AP A = 60, the triangle P AP A must be equilateral. In particular, P P A = a. Therefore, the triangle P P A C has side lengths P P A = a, P A C = b and P C = c, as required. Let the rotation of 60 anticlockwise about the points B and C take the point P to P B and P C, respectively. By a similar argument to the one above, we find that the shaded triangles P P A C, P P B A, P P C B are all congruent to each other, since they have sides of length a, b and c. Denote the area of these triangles by T. We also note that the unshaded triangles P AP A, P BP B and P CP C are all equilateral, with sides of length a, b and c, respectively. The rotation about A takes triangle AP B to AP A C, so the two triangles are congruent. Similarly, triangle BP C is congruent to triangle BP B A and triangle CP A is congruent to triangle CP C B. In particular, this means that the area of 4

5 the hexagon AP B BP C CP A is twice the area of triangle ABC. area(ap B BP C CP A ) = area(ap B B + BP C C + CP A A + AP B + BP C + CP A) = area(ap B + BP C + CP A) = area(abc) = x We can calculate the area of the hexagon in another way as follows. area(ap B BP C CP A ) = area(p P A C + P P B A + P P C B + P AP A + P BP B + P CP C ) = T + 4 (a + b + c ) Equating these two expressions gives us an expression for the area of T. T + 4 (a + b + c ) = T = x 1 x 1 4 (a + b + c ) Since x is constant, we need to show now that a + b + c depends only on P O. We do this using vectors by setting O to be the origin. a + b + c = PA + PB + PC = (PO + OA) + (PO + OB) + (PO + OC) = PO + PO OA + OA + PO + PO OB + OB + PO + PO OC + OC = P O + PO (OA + OB + OC) + OA = P O + OA This is true since OA + OB + OC = 0 and the lengths OA = OB = OC are constant. Hence, we conclude that the triangle with side lengths a, b and c has an area that depends only on P O. 5

( 1 ) Show that P ( a, b + c ), Q ( b, c + a ) and R ( c, a + b ) are collinear.

Problems 01 - POINT Page 1 ( 1 ) Show that P ( a, b + c ), Q ( b, c + a ) and R ( c, a + b ) are collinear. ( ) Prove that the two lines joining the mid-points of the pairs of opposite sides and the line