46th ANNUAL MASSACHUSETTS MATHEMATICS OLYMPIAD. A High School Competition Conducted by. And Sponsored by FIRST-LEVEL EXAMINATION SOLUTIONS

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1 46th ANNUAL MASSACHUSETTS MATHEMATICS OLYMPIAD A High School Competition Conducted by THE MASSACHUSETTS ASSOCIATION OF MATHEMATICS LEAGUES (MAML) And Sponsored by THE ACTUARIES CLUB OF BOSTON FIRST-LEVEL EXAMINATION Thursday, October, 009 SOLUTIONS

2 The value of is 5 Solution: i9 = = = i5 5 The operation a b means Solution: ( ) ( ) a b b a Compute ( 4 ) ( ) The answer is 4 4 = = 48 6 = 6 = 44i6 6i = = 4 i i ( ) = 9 4 = 8 = 6 i i Tomás drives,000 miles a year as a sales rep for an auto parts company He wants to buy a hybrid car to get better gas mileage (48 mpg!), but the model he wants costs $400 more than its non-hybrid counterpart ( mpg) If gas costs $00 per gallon, about how long would it take Tomás to recoup the higher cost of the hybrid? The answer is 50 months Solution: The hybrid car will use 666 gallons per year, at a cost of $000 in gas, whereas the non-hybrid car will use 000 gallons per year, at a cost of $000 The difference in the cost of gas ($000 per year) will be made up in 4 years, which is about 50 months 4 Given: ( x + 5) varies directly as ( y + ) and inversely as ( ) and z = 7 Find x when y = and z = 6 The answer is z, and x = 5 when y = Solution: ( ) ( y ) ( z ) + 5 x + 5 = k 5 = k k = 5 ( ) ( y ) ( z ) + 4 x + 5 = ( x + 5) = x = x = 4

3 5 The product of 99 and an integer k is 50x8x, where x represents a digit in the five-digit product Find the integer k The answer is 5 Solution: If 50x8x is divisible by 99, it is divisible by 9, so its digits must sum to a multiple of 9 This means that x must equal = If 6( ) 6 x ( x+ 6) is 8 =, then x( ) ( x + 0 ) ( ) = pq( ) Solution: ( ) ( ) ( ) 0 Find the base 0 sum p + q The answer 6 = = = + 4 = 7 x x x x x x x+ 6 ( ) ( ) 7i 9 = 6 = 5 + = 5 p + q = The five-digit number ABCDE has only even digits, such that A cannot be zero and the other digits may repeat The four-digit number FGHJ has only odd digits, which may also repeat How many ways can these numbers be configured so that ABCDE is twice FGHJ? The answer is 0 Solution: The largest four-digit number with odd digits is 9999 Since twice 9999 is 9,998, any five-digit number ABCDE must have A = 8 In right ABC, AC BC, AC = 9, and BC = 56 PD is the perpendicular bisector of AB, where P AB, and D AC Compute PD The answer is 9 6 Solution: ABC is a triangle with hypotenuse 00 ADP is a similar triangle whose long leg has length PA = 00 (half the hypotenuse of ABC) The length of the short leg of ADP is therefore PD = ( ) = =

4 9 Find the sum of all the integers in the following list (which omits multiples of 4):,,, 5, 6, 7, 9,, 75, 77, 78, 79 The answer is 400 Solution: The sum of the first 79 positive integers is (after the method of Gauss) 9( 80) + 40 = 60 The sum of the first 9 multiples of 4 is ( ) Subtracting this sum from the first sum leaves = If P( x + ) = x + 7x + 4, and ( ) answer is (,, 8) Solution: Let y = x + Then, x = y Therefore, P x = ax + bx + c, find the ordered triple (a, b, c) The ( + ) = ( + ) = ( ) + 7( ) + 4 = + 8 Thus, ( ) P x P y y y y y equivalently, P( x) = x + x 8 ( a,b,c ) = (,, 8) P y = y + y 8, or, Given the following array, name the third number in the 0 th row (Note: is the second number in the third row) The answer is Solution: The number of numbers in each row follows the sequence,, 5, 7, In 9 rows, therefore, there are = 6 numbers, of which the last is 7 (the 6 st odd number) The next three odd numbers are 7, 75, and 77

5 In the figure below, AB =, and the area of ABC is 90 Find cot CBD The answer is 5 Solution: If AB = and the area of ABC is 90, then the height of ABC is 5 Thus, in right ACE below, CE = 5 and AE = 5, so that BE = Therefore, cot CBD = cot CBE = = 5 5 C A 0 E B D An octahedral die has eight faces, numbered through 8 Ayesha rolls a fair octahedral die three times What is the probability that the number on the third roll is the product of the numbers from the first two rolls? The answer is 5 8 Solution: Three rolls of a die produce 5 outcomes: (,, ), (,, ), (,, ), etc Of these outcomes, how many are such that the product of the first two rolls equals the third? The following table lists all the possible ordered triples that meet this criterion (,, ) (,, ) (,, ) (,, ) (,, ) (, 4, 4) (,, 4) (4,, 4) (, 5, 5) (5, 5, ) (, 6, 6) (,, 6) (,, 6) (6,, 6) (, 7, 7) (7,, ) (, 8, 8) (, 4, 8) (4,, 8) (8,, 8)

6 4 Let f be a function defined for all real numbers with the property that ( ) ( ) f x = f + x Suppose that f has 6 roots Find the sum of the roots The answer is 8 Solution: If + x0 is a solution, then x0 is also a solution Therefore, the average value of the roots is And i 6 = 8 5 Find the positive integral values of x and y such that x + y = 0, and x + y is a multiple of 7 The answer is (8, 6) Solution: The positive integral solutions to x + y = 0 are (, 9), (4, 8), (6, 7), (8, 6), (0, 5), (, 4), (4, ), (6, ), and (8, ) The first of these ordered pairs, when substituted into x + y, produces 5, which is not a multiple of 7 Each successive ordered pair increases the sum x + y by 5 A multiple of 7 is achieved with the ordered pair (8, 6) 6 Find the numerical value of the st term of the geometric sequence whose first three terms are ( + i) 0, ( i) 6, ( ) 6 i The answer is Solution: ( ) ( ) [ ] + i = + i = + i + i 0 = i = i i = ( ) ( ) [ ] i = i = i + i = i = i i = ( ) i = i i = Thus, the terms are 0 8 6,,, The st term of this sequence is 0 = 0

7 7 Equilateral triangle ABC is inscribed in a circle D is a point on minor arc BC The length of chord BD is inches and the length of chord DC is 5 inches How long is AD (in inches)? The answer is 8 Solution: In the figure, the lengths of the sides of the triangle are given by s According to Ptolemy s theorem, A ( ) s + 5s = AD s, so that AD = 8 s s Alternatively, we may recognize that, as inscribed angles, m BDA = m CDA = 60, and then use the Law of Cosines with BCD and ACD B D s 5 C In BCD, s = + 5 5cos ( 0 ) = = 49, so that s = 7 Then, in ACD, we have s = 49 = 5 + ( AD) 5( AD) cos( 60 ), which, upon rearrangement, becomes the quadratic ( AD) ( AD) course, extraneous 5 4, which has roots and 8 The former is, of Ptolemy s theorem states that in a cyclic quadrilateral, the sum of the products of the lengths of the pairs of opposite sides equals the product of the lengths of the diagonals

8 8 Two values of x make the following statement true Find the absolute value of their difference The answer is 8 ( )( 5 ) log 5 log 9 + log x + log = 0 x Solution Using the change-of-base rules for logarithms: log5 log5 log 5 log 5 = = = log log log log 9 5 log log log 5 log 5 = = So ( log 5 )( log 9 ) 5 4 = log log x log x x = =, and log log log log = x log x So the original equation becomes: ( x) ( ) 4 log x log 4log log x + log + log + + = 0 = 0 log log x log log x The numerator of this last fraction is a quadratic that factors to ( log x log )( log x log ) + + Thus: log x = log x =, and 8 log x = log x = The positive difference of these two values of x is 8

9 9 The positive difference between the square of the arithmetic mean of two numbers and the square of their geometric mean is 6 Find the positive difference between the arithmetic mean of their squares and the square of their arithmetic mean The answer is 6 Solution Let x and y be the two numbers, and let x > y Then, the first of the above conditions produces ( ) x + y x + xy + y xy = xy = x + xy + y xy = to ( x y) = 44 x y =, because of the requirement that x > y This simplifies The second of the above conditions produces x + y x + xy + y x + y x xy y, which simplifies to ( x y ) Because x y =, this last expression equals 6 0 In ABC, A is at (, 4), B is at (8, ), and C is at (6, 6) Find the coordinates (x, y) of point K on the segment AB so that the ratio of the area of is :5 The answer is Solution 4 5, 5 5 BKC to the area of ABC ABC is an isosceles right triangle of leg length 0 and area 50 Therefore, we want K to be located on AB such that ( 8 ), 4 + ( 4 ) =, KB = AB The coordinates of K are given by 5

10 The country of Halfway issues coins in denominations of 8, 9, and 0 halfmarks, the unit of currency What is the largest number of halfmarks that cannot be expressed with these coins? The answer is Solution We want to find a string of eight consecutive numbers of halfmarks that can be expressed with these denominations Once we have that string, we can express the next eight by increasing by adding one halfmark coin to each of the first eight numbers Each new string of eight can be expressed in a similar manner Obviously, we can express 8, 9, and 0 halfmarks with these coins We can also express 6, 8, and 0 by using two of each coin, but also 7 (by using an 8 and a 9) and 9 (by using a 9 and a 0) Likewise, we can express 4 0 halfmarks notice the pattern? and then 40 This last is actually a string of nine, so is the largest unexpressable number Let AB 9 and CD 9 be non-negative two-digit integers in base 9 (A and C may be zero), such that ( AB) ( CD) ( AB) ( CD) = For a particular solution (,,, ) 68 9 let k = A + B + C + D Find the sum (in base 9) of all k i The answer is 44 Solution: Adding to both sides of the equation and factoring, we get (( AB) )( CD ) = 60 9 In base 0, this is A B C D, 6(9 ) + (9) = 504 = 7, which has 4()() = 4 positive integer factors, or factor pairs Since one of the factors is AB, we are looking for a factor that is less than a perfect square The number 504 has five such factors:, 8, 4, 6, and 68 Hence, in base 0, if ( AB) =, 8, 4, 6, or 68, AB =,, 5, 8, or Of these, we discover that is too small a value for our needs The remaining numbers, in base 9, are 0, 05, 08, and 4 Correspondingly, if CD = 68, 6,, 8, or, then CD = 69, 64,, 9, or 4 Of these, we discover that 69 is too large a value for our needs Again, the remaining numbers, in base 9, are 7, 4, 0, and 04 Thus, (A, B, C, D) = (0,, 7, ), (0, 5,, 4), (0, 8,, 0) and (, 4, 0, 4), and the respective values of k, in base 9, are,, 0, and 0, which add to 44 9

11 Find the polynomial of least positive degree that, when divided by x x, x x, or x x + + +, leaves a remainder of 5 Write your answer as a polynomial of least degree with integer coefficients The answer is x + x - 5x + 7 Solution If P(x) denotes the polynomial of least degree that we seek, we have ( ) P x ( x x )( x a) ( x x )( x b) = (Any pair of divisors would have worked here) Equating the coefficients of x, we have a + = b a b = Equating the coefficients of x, we have a = b + a + b = Solving this system, we have (a, b) = (, ), and ( ) ( )( ) P x x x x x x x = = We can verify that x + x is a divisor of this polynomial

12 4 Let ABC and A' B ' C ' be equilateral triangles with the same center such that the distance between corresponding sides is and each side of A'B'C'= Let D, E, F, D ', E ', and F ' be midpoints of the sides as pictured A mosquito flies along a straight path from D to E to F to D to E ' to F ' to D ' to E ' Find the length of the mosquito s flight The answer is 6 + Solution In trapezoid BB ' C ' C, m B ' BC = m C ' CB = 60, so the lengths of the sides of ABC are Thus, the paths represented by DE, EF, and FD are midsegments whose combined length is 9 E ' F ', F ' D ', and D ' E ' are midsegments whose combined length is These six parts of the path thus have a combined length of 6 This leaves DE ' In the figure, D is at the origin, and the other points are as shown Applying the distance formula, we have DE ' = Likewise, the paths represented by

13 5 Consider the set S of all positive fractions whose denominator is 4 and whose numerator is less than 6 and relatively prime with 4 How many nonempty subsets of this set have the property that the sum of all its elements is a reduced fraction? The answer is Solution: S is the 9-element set ,,,,,,,, N S has 9 = 5 subsets, but those with an even number of elements will sum to, 4 where N will be an even number and this is a reducible fraction Therefore, we restrict our attention to subsets with an odd number of elements All -element subsets satisfy the requirement We must look closely at, 5, 7 and 9-element subsets N must be a sum of an odd number of elements from {, 5, 7,,, 7, 9,, 5} which is relatively prime with 4 Five of these numbers are of the form k +, that is, they are more than a multiple of Call this set A = {, 7,, 9, 5} Then, let B = {5,, 7, }, the set of numerators that are more than a multiple of -element subsets: Eliminate any that contain three elements from A (which add to a multiple of ) or three elements from B (which add to a multiple of 6) = = 70 5-element subsets: Eliminate any that contain four elements from A and one element from B (which add to a multiple of 6) and any that contain one element from A and four elements from B (which add to a multiple of 9) = = element subsets: Eliminate any that contain five elements from A and two elements from B (which add to a multiple of 9) = 6 6 = element subsets: There is only one, namely S itself This set produces a reduced fraction The total is =

46th ANNUAL MASSACHUSETTS MATHEMATICS OLYMPIAD. A High School Competition Conducted by. And Sponsored by FIRST-LEVEL EXAMINATION

46th ANNUAL MASSACHUSETTS MATHEMATICS OLYMPIAD 009 00 A High School Competition Conducted by THE MASSACHUSETTS ASSOCIATION OF MATHEMATICS LEAGUES (MAML) And Sponsored by THE ACTUARIES CLUB OF BOSTON FIRST-LEVEL