BMOS MENTORING SCHEME (Senior Level) February 2011 (Sheet 5) Solutions

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1 MOS MENTORING SCHEME (Senior Level) February 011 (Sheet 5) Solutions These solutions represent one way of solving the questions. They are not the only way, and just because your solution is different does not mean that it is worse or wrong. You will get most from the solutions if you have at least attempted the questions. Only then will you appreciate the difficulties that a problem posed and how they were addressed by the solution. Relevant theorems have been identified where possible, and references to further information on the Web have been given. If you are still unclear, ask your mentor. A. Rzym & J. Cranch c UKMT Let f(n) denote the average number of circles in the bowl having started with n lengths of string. Starting with n lengths of string and having selected the first end, the second end is selected from one of the n 1 remaining ends. There are then two possibilities: (a) In one case (i.e. with probability 1/(n 1)) the second end was the other end of the same string. In this case, we have created a circle and have n 1 strings left (yielding f(n 1) circles on average), i.e. a total of 1 + f(n 1) circles on average. (b) In the remaining n cases (i.e. with probability (n )/(n 1)), the second end is of a string distinct from the first. On joining, we are left with n 1 strings, i.e. a total of f(n 1) circles on average. Probability weighting these possibilities (and given that f(1)=1) gives and hence f(n) = 1 n (1 + f(n 1)) + n 1 n 1 f(n 1) = 1 + f(n 1) n 1 f(6) = 1/11 + 1/9 + 1/7 + 1/5 + 1/ If α, β are the roots of x Ax = 0 then αβ = and α + β = A. Defining S k = α k + β k, we have S 0 =, S 1 = A and S k = (α k + β k ) = (α k 1 + β k 1 )(α + β) αβ(α k + β k ) = AS k 1 + S k S k = A + (1) S k 1 S k 1 /S k Using (1) for k = n, and substituting (1) for k = n 1 we obtain S n S n 1 = A +. A + S n /S n 3 Repeating the process and observing that S 1 /S 0 = A/ gives the desired result. Supported by the Man Group plc Charitable Trust

2 3. (a) Repeatedly squaring the given equation we obtain x + x 1 + x x 1 = (x + x 1) + (x x 1) + x x + 1 = (x 1) = 1 x 1 = 1. We had hoped that squaring the original equation would give us a finite number of possible values for x (even if not all are actually solutions), but this has not happened. Instead, we will now consider whether the argument above can guide us to start with an identity and derive the required equation. For x 1 we have 1 x = 1 x and therefore (1 x) = 1 x x + x x + 1 =. Providing x 1/, x 1 exists and so (x + x 1) + (x x 1) + x x + 1 =. Since 1/ x 1, x + x 1 0 and x x 1 0, therefore the above equation implies ( x + ( x 1) + x x 1) + (x + x 1)(x x 1) =. Taking the square root of both sides yields the given equation. In other words, the given equation is satisfied if and only if 1/ x 1. (b) y squaring the given equation repeatedly we obtain x + x 1 + x x 1 = 1 (x + x 1) + (x x 1) + x x + 1 = 1 x + x x + 1 = 1/ 1 x = 1/ x. However there are no solutions to this equation, and therefore we conclude that there are no solutions to the given equation either. (c) y squaring the given equation repeatedly we obtain x + x 1 + x x 1 = (x + x 1) + (x x 1) + x x + 1 = 4 x x + 1 = x x = 3. Thus the only possible solution is x = 3/, and on substituting in the given equation we see that it is, indeed, a solution.

3 4. We will describe a sum of non-negative powers of without regard to order and where no power can be used more than twice as an a-partition, and denote the number of a-partitions of a positive integer n by a(n). We seek the value of a(011). We will describe a sum of positive powers of without regard to order and where no power can be used more than twice as a b-partition, and denote the number of b-partitions of a positive integer n by b(n). For every b-partition of an even integer, k, we can halve each term to give an a-partition of the integer k and vice-versa. This correspondence implies that b(k) = a(k). For every a-partition of an odd number, k + 1, there will be exactly one 0 in the sum. Removing this gives a b-partition of k and vice-versa. This correspondence implies a(k + 1) = b(k) = a(k). Every a-partition of an even integer, k, contains either no powers of 0 (in which case it is also a b-partition of k) or two powers of 0 (in which case, by dropping these, we have a b-partition of k. We conclude that a(k) = b(k) + b(k ) = a(k) + a(k 1). To summarise: a(k + 1) = a(k), a(k) = a(k) + a(k 1). Observing that a(1) = 1, a() = we then have a(011) = a(1005) = a(50) = a(51) + a(50) = a(15) + a(14) = 3a(6) + a(61) = 3a(31) + 4a(30) = 7a(15) + 4a(14) = 11a(7) + 4a(6) = 15a(3) + 4a() = 15a(1) + 4a() = Denote the centre of the circle of radius r by O. Denote the area of a triangle P QR by P QR. y the law of sines, Also, we have a sin A = b sin = c sin C = R AC = AO + OC + COA = 1 cr + 1 ar + 1 br = 1 r(a + b + c) = rs ut we also have AC = 1 bc sin A, which from the law of sines above can be written as AC = 1 bc sin A = 1 ( a ) bc = abc R 4R Equating these two expressions for AC we have AC = abc 4R = rs abc = 4srR 3

4 6. Defining p = d/a, q = e/b, r = f/c, dividing the inequality to be proved by abc then cubing it, we see that the inequality to be proved follows providing that we can prove (1 + p 3 )(1 + q 3 )(1 + r 3 ) (1 + pqr) (p 3 + q 3 + r 3 ) + (p 3 q 3 + q 3 r 3 + r 3 p 3 ) + (pqr) pqr + 3(pqr) + (pqr) 3. Our proof therefore proceeds as follows: Applying AM-GM to the collection {p 3, q 3, r 3 } we have p 3 + q 3 + r p 3 q 3 r 3 = 3pqr. () Applying AM-GM to the collection {p 3 q 3, q 3 r 3, r 3 p 3 } gives p 3 q 3 + q 3 r 3 + r 3 p p 6 q 6 r 6 = 3(pqr). (3) Using () and (3) we have 1 + (p 3 + q 3 + r 3 ) + (p 3 q 3 + q 3 r 3 + r 3 p 3 ) + (pqr) pqr + 3(pqr) + (pqr) 3 from which the result follows. (1 + p 3 )(1 + q 3 )(1 + r 3 ) (1 + pqr) 3 7. Let F denote the function mapping the quadruple q = (a, b, c, d) onto the quadruple ( a b, b c, c d, d a ). We will refer to the maximum value of the quadruple q as meaning the maximum value of the four integers a, b, c, d. We will prove that a quadruple q cannot exist such that repeated application of F does not eventually yield (0, 0, 0, 0) by assuming the converse and deriving a contradiction. Suppose such a q does, in fact, exist. Then all elements of F (q) are positive and therefore a q must exist with all elements positive. If q = (a, b, c, d) and a, b, c, d 0 then the maximum value of q is finite and non-negative. y assumption, not all of a, b, c, d are zero, and therefore the maximum value of q is positive. Furthermore, if repeated application of F on q does not yield (0, 0, 0, 0) it must be the case that after a certain number iterations, the maximum value of the quadruple is positive and does not decrease with further applications of F. Thus there exists q = (a, b, c, d) with the property that the maximum of q and F (q) = ( a b, b c, c d, d a ) are equal. Without loss of generality we can take a to be the maximum value of q. If a b takes the (same) maximum value, then b = 0. If b c takes the maximum value, a, then b = a and c = 0 etc. Thus one of the elements of q which takes the maximum value of q must be followed by a zero. Without loss of generality assume that these are a and b respectively. Thus we now have q = (a, 0, c, d), with a > 0, c, d 0 and a equal to the maximum value of q. Then q and F (F (q)) = ( a c, c c d, c d d a, d a a ) have the same maximum value. There are then four cases to consider: 4

5 If the first element of F (F (q)) is equal to the maximum value of q (a), this implies c = 0, i.e. the quadruple q is of the form (a, 0, 0, d); If the second element of F (F (q)) is equal to the maximum value of q (a), then c = a, d = 0, i.e. the quadruple q is of the form (a, 0, a, 0); If the third element of F (F (q)) is equal to the maximum value of q (a), then either d = 0, c = 0 or d = a, c = 0, i.e. the quadruple is of the form q = (a, 0, 0, d) or (a, 0, 0, a); If the fourth element of F (F (q)) is equal to the maximum value of q (a), then d = a, i.e. the quadruple q is of the form q = (a, 0, c, a). We now dispose of these possibilities. For example, if q = (a, 0, 0, d) then F (q) = (a, 0, d, d a ) F (F (q)) = (a, d, d d a, d a a ) F (F (F (q))) = ( a d, d d d a, d d a d a a, a d a a ). Each of the final terms can only equal a if d is one of 0, a. ut if q = (a, 0, 0, 0) then F (F (F (F (q)))) = (0, 0, 0, 0). Similarly, if q = (a, 0, 0, a) then F (F (F (q))) = (0, 0, 0, 0). The other cases can be similarly disposed of, so we conclude that there is no q for which repeated application of F does not lead to (0, 0, 0, 0). References Vieta s Fundamental Theorem of Algebra s formulas Law of sines: of sines AM-GM inequality: of arithmetic and geometric means 5

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