USA Aime 1983: Problems & Solutions 1 1 Problems 1. Let x,y, and z all exceed 1, and let w be a positive number such that log x w = 4, log y w = 40, and log xyz w = 1. Find log z w.. Let f(x) = x p + x 15 + x p 15, where p x 15. Determine the minimum value taken by f(x) by x in the interval 0 < p < 15. 3. What is the product of the real roots of the equation x + 18x + 30 = x + 18x + 45? 4. A machine shop cutting tool is in the shape of a notched circle, as shown. The radius of the circle is 50 cm, the length of AB is 6cm, and that of BC is cm. The angle ABC is a right angle. Find the square of the distance (in centimeters) from B to the center of the circle. 5. Suppose that the sum of the squares of two complex numbers x and y is 7 and the sum of the cubes is 10. What is the largest real value of x + y can have? 6. Let a n equal 6 n + 8 n. Determine the remainder upon dividing a 83 by 49. 7. Twenty five of King Arthur s knights are seated at their customary round table. Three of them are chosen - all choices being equally likely - and are sent of to slay a troublesome dragon. Let P be the probability that at least two of the three had been sitting next to each other. If P is written as a fraction in lowest terms, what is the sum of the numerator and the denominator? 8. What is the largest -digit prime factor of the integer ( 00 100)? 9. Find the minimum value of 9x sin x+4 x sin x for 0 < x < π. 10. The numbers 1447, 1005, and 131 have something in common. Each is a four-digit number beginning with 1 that has exactly two identical digits. How many such numbers are there? 11. The solid shown has a square base of side length s. The upper edge is parallel to the base and has length s. All edges have length s. Given that s = 6, what is the volume of the solid?
USA Aime 1983: Problems & Solutions 1. The length of diameter AB is a two digit integer. Reversing the digits gives the length of a perpendicular chord CD. The distance from their intersection point H to the center O is a positive rational number. Determine the length of AB. 13. For {1,, 3,..., n} and each of its non-empty subsets, an alternating sum is defined as follows. Arrange the number in the subset in decreasing order and then, beginning with the largest, alternately add and subtract succesive numbers. For example, the alternating sum for {1,, 3, 6, 9} is 9 6 + 3 + 1 = 6 and for {5} it is simply 5. Find the sum of all such alternating sums for n = 7. 14. In the adjoining figure, two circles with radii 6 and 8 are drawn with their centers 1 units apart. At P, one of the points of intersection, a line is drawn in sich a way that the chords QP and P R have equal length. (P is the midpoint of QR) Find the square of the length of QP. 15. The adjoining figure shows two intersecting chords in a circle, with B on minor arc AD. Suppose that the radius of the circle is 5, that BC = 6, and that AD is bisected by BC. Suppose further that AD is the only chord starting at A which is bisected by BC. It follows that the sine of the minor arc AB is a rational number. If this fraction is expressed as a fraction m n in lowest terms, what is the product mn?
USA Aime 1983: Problems & Solutions 3 Answers 1. 060. 015 3. 00 4. 06 5. 004 6. 035 7. 057 8. 061 9. 01 10. 43 11. 88 1. 065 13. 448 14. 130 15. 175
USA Aime 1983: Problems & Solutions 4 3 Solutions 1. Instead of applying logarithm rules, we exponentiate everything so that x 4 = w, y 40 = w, and (xyz) 1 = w. Now, we need to isolate z and w in order to find log z w. Notice that x 10 = w 5, y 10 = w 3, and (xyz) 10 = w 10. With a quick substitution, we get w 5 w 3 z 10 = w 10 and log z w = 060.. First, we eliminate the absolute value. With the conditions given, x p = x p, x 15 = 15 x, and x p 15 = 15+p x. Adding these together, the sum is equal to 30 x, of which the minimum value is attained when x = 15. Therefore, The answer is 015. 3. Instead of immediately squaring, we substitute y = x + 18x + 30 such that our expression becomes y = y + 15. Solving for y, y = 10 or y = 6. The second solution is extraneous, so we ll will go with the first. Substituting x + 18x + 30 back in for y and simplifying, x + 18x + 0 = 0. The product of our roots is therefore 0 1 = 00. 4. Because we are given a right angle, we look for ways to apply the Pythagorean Theorem. Extend a perpendicular from O to AB and label it D. Additionally, extend a perpendicular from O to the line BC, and label it E. Let OE = x and OD = y. We re trying to find x + y. Applying the Pythagorean Theorem, OA = OD + AD, and OC = EC + EO. Thus, ( 50) = y + (6 x), and ( 50) = x + (y + ). We solve this system to get x = 1 and y = 5, resulting in an answer of 1 + 5 = 06. 5. First, notice that x + y = (x + y) xy = 7 and x 3 + y 3 = (x + y)(x + y ) xy(x + y) = 7(x + y) xy(x + y) = (7 xy)(x + y) = 10. Now that we re left with x + y and xy, let a = x + y and b = xy. We get a b = 7 and a(7 b) = 10. Because we want the largest possible a, let s find an expression for b in terms of a. a 7 = b = b = a 7. Substituting, a 3 1a + 0 = 0. Factored, (a 1)(a + 5)(a 4) = 0 The largest possible solution is therefore 004. 6. Notice 6 83 + 8 83 = (7 1) 83 + (7 + 1) 83. Applying the Binomial Theorem, half of our terms cancel out and we are left with (7 83 + 3403 7 81 + + 83 7). Alsl of the terms in this big jumble of numbers are divisible by 49 except the final term. Therefore, our answer is (83)(7) (mod 49) = 035. 7. Some casework should do the trick. To position three knights adjacently, we can have the rightmost knight on ony of the 5 spots, so there are 5 ways. To position two knights adjacently, we once again have 5 positions for the rightmost knight, in addition to the 1 positions where we can place the remaining knight, so there are 5 1 ways. Dividing by the total number of ways to position the knights, ( 5 3 ), our probability is 5+5 1 ( 5 3 ) = 11 46, and the answer is 057.
USA Aime 1983: Problems & Solutions 5 8. Expanding, we get ( ) 00 100 = 00! 100!100!. Thus, our two digit prime p must satisfy 3p < 00. The largest such prime is 061, which is our answer. 9. First off, substitute y = x sin x so our expression becomes 9y +4 y = 9y + 4 y. Since x > 0 and sin x > 0, we have y > 0. Now we apply AM-GM to obtain 9y + 4 y 9y 4 y = 1. 10. We start with the case where there are two 1 s and the other two digits are do not repeat. Letting x denote the two unknown digits. There are 9(8) = 7 unknown digits that can fill in the spaces. Additionally, there are 3 ways to arrange the 1 s, so we get 3(9)(8) = 16 possibilities from this case. Next, we count the number of numbers in which 1 doesn t repeat. Similarly, we get 16 numbers. In total, our answer is 43. 11. First, draw the altitude from E to ABCD, and let the point of intersection be P. Additionally, draw the altitude from E to side AD, which equals 3 6 by the Pythagorean Theorem. We apply the Pythagorean Theorem once more to find that EP = (3 6) (3 ) = 6. Next, we complete the figure into a triangular prism. The volume is the base times the height, 6 1 6 = 43. An accurate diagram helps us see that the original figure is only 3 as large as the trianglular prism we just formed, so our answer is 43( 3 ) = 88. 1. Let AB = 10x + y and CD = 10y + x. It follows that CO = AB = 10x+y and CH = CD = 10y+x. Applying the Pythagorean Theorem on CO and CH, ( ) ( ) OH = 10x+y 10y+x = 9 4 11(x + y)(x y) = 3 11(x + y)(x y). Because OH is a positive rational number, the quantity 11(x + y)(x y) cannot contain any square roots. Therefore, x + y must equal eleven and x y must be a perfect square (since x + y > x y). The only pair (x, y) that satisfies this condition is (6, 5), so our answer is 65. 13. Let S be a non-empty subset of {1,, 3, 4, 5, 6}. Notice that the alternating sum of S plus the alternating sum of S with 7 included is 7. For ( the sake of convenience, let s consider {} and {7} a pair. There are 6 ) ( 0 + 6 ) ( 1 + + 6 ) 6 = 6 = 16 different sets and 64 pairs. Therefore, our answer is 64 7 = 448.
USA Aime 1983: Problems & Solutions 6 14. In the diagram below, we let QP = x, AX = a, and BY = b. From right triangle AZB, we see that 1 = 4x + (a b). From right triangle AXP, 64 = a + x and 36 = b + x. Solving these three equations, we find that QP = 4x = 130. 15. Let A be any fixed point on circle O and let AD be a chord of circle O. The locus of midpoints N of the chord AD is a circle P, with diameter AD. Generally, the circle P can intersect the chord BC at two points, one point, or they may not have a point of intersection. By the problem condition, however, the circle P is tangent to BC at point N. Let M be the midpoint of the chord BC such that BC = 3. From right angle triangle OMB, OM = OB BM = 4. Thus, tan BOM = BM OM = 3 4. Notice that the distance OM equals P N + P O cos AOM = r(1 + cos AOM) (Where r is the radius of circle P). Evaluating this, cos AOM = OM r 1 = OM R 1 = 8 5 1 = 3 5. From cos AOM, we see 1 cos that tan AOM = AOM 5 = 3 = 4 cos AOM 3 3 Next, notice that AOB = AOM BOM. We can therefore apply the tangent subtraction formula 4 tan AOM tan BOM to obtain, tan AOM = 1 + tan AOM tan AOM = 3 3 4 1 + 4 3 3 = 7. It follows that sin AOM = 4 4 7 7 + 4 = 7, resulting in an answer of 7 5 = 175. 5