MO-ARML Practice March 17, 2007 Relay Questions

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O-ARL Practice arch 17, 007 Relay Questions If a question depends on NYWR, the answer should be expressed in terms of the variable representing NYWR 1.

1 O-ARL Practice arch 17, 007 Relay Questions If a question depends on NYWR, the answer should be expressed in terms of the variable representing NYWR 1. Compute the number of distinct 11 letter palindromic permutations of the letters of ISSISSIPPI. he letter must lie in the middle position. o its left are permutations of SSIIP, and to its right are mirror images of the letters on its left. he number of distinct 5! palindromic permutations is thus =30.!! K. Let K = NYWR. If the regular octagon ABCDEFGH has sides equal to in length, 30 compute the value of AC AD. K Let x =. hen AC AD 30 x x 1 1 = x+ + = x = x ( + ) x AD = + ( +1) x = x. herefore, AC K K AD = ( + ) ( 1+ ) Let = NYWR. Let S n be a unit square subdivided into n congruent squares. he centers are all connected as shown in the diagram at the right for S. Compute the least value of n such that the sum of the lengths of the connecting line segments exceeds here are n subsquares in each row and column, and hence there are n horizontal 1 line segments of length 1. he sum of the vertical line segments equals one n 1 horizontal line segment which is 1. he total length of the connecting line n

2 segments thus equals to ( ) = likely, 4 n =. 1 1 n +1 1 n n. If n > 10 4, n > 4 n n. ost 4. Let ABCDEF be a regular hexagon. If the ratio of the area of region ABCE to the area of the hexagon in simplest terms is a, compute a + b. b Based on the diagram, the region consists of 4 triangles area of ABCE 4 congruent to AEF. hus, = =. area of ABCDEF 6 3 herefore, the sum a + b is Compute the number of sets of K = 5 consecutive integers in {1,, 3,., 50} such that the product of the elements of each set of K = 5 integers is divisible by both 1 and 1. A number is divisible by both 1 and 1 if and only if it is divisible by 3 7. For the product of a K- tuple to be divisible by 3 7, it must contain a multiple of 7. If K 3, the product is always divisible by 3. If K 4, the product is always divisible by. If K = 5, then for each multiple of 7 from 7 to 4, there are five 5-tuples in S containing this multiple of 7. For 49, there are only two such 5-tuples. hus, there are (6)(5) + = 3 sets of five integers whose product is divisible by both 1 and Let = NYWR. Given points D(0, ), O(0, 0), and E(, 0), equilateral triangle ABC is inscribed in DOE such that C is the midpoint of DE and AB is parallel to DE. he length of AB can be expressed in simplest form a( b c). Determine (a, b, c). Let x = OA = OB. hen, AB = x. Note that ACD and BCE are congruent, implying ACD = 60. Using the AC AD Law of Sines, = o o sin 45 sin 60 o ADsin 45 AD ( x) AC = = = = x. o sin Solving for x, we obtain answer is,6,. x = ( 3 1), making AB = AC = ( 6 ). he

3 7. In trapezoid ABCD, AB is parallel to CD and angles B and D are acute. If sin B = cos D, compute m A m B in degrees. Since B and D are acute, the property sin B = cos D implies that m B + m D = 90. But m A = 180 m D, we have m A m B = 180 m D m B = Let = NYWR and set K =. Square LNP is 15 inscribed in right triangle ABC as shown. If PN = K, compute the product (AP)(NC). Note that APL and NC are similar because m CN = m A. herefore, AP LP = N NC, and thus (AP)(NC) = (N)(LP) = K = Let = NYWR and set K = 6. here are K positive integers in an arithmetic progression with common difference 1. If the sum of the terms is 75, compute the value of the first term. If the first term is x, then the sum is x + (x + 1) + + (x + K 1) = implying x = +1 K. K 10. Let n be a positive integer. If the number of integers in the domain of y = 1 x x n equals n 6, compute n. (( )( )) ( x+ K 1) K = 75, he domain of the function y =( ( 1 x)( x n) ) is the interval (1, n) and thus the only integers in the domain are though n 1. Hence, n = n 6 and n = Let = NYWR. Shown is the graph of the first quadrant portion of a relation. he relation is symmetric over the x-axis and over the origin. Given A(0, ), 3 3 B,, and C(, 0), compute the area bounded by the graph.

4 he area of the first quadrant region is =. he symmetry over the x-axis and the origin implies that the total area is four times of the area of the first quadrant region, which is equal to Compute the largest integer n with 0 < n 96 (= NYWR) such that the triangle with sides n, n 3, and 3n 9 has integer area. he semiperimeter of the triangle is s = 3n 6. he area of the triangle is therefore equal to (3n- 6)(n- 6)( n- 3)(3) = 3( n- 3) ( n - ). his is an integer whenever n = m for some m. Equivalently, n = m - + or n he complete graph of x + y = 1 is shown on the viewing window of a graphing calculator. When the region enclosed by the equation is measured, its area is 10 square units. If the Xmin and Xmax of the viewing window are tripled, the Ymin and Ymax are doubled, and the new graph of the equation is measured, the new measured area of the region is K square units. Compute K. When the Xmin and Xmax of the viewing window are tripled, the measured length in the horizontal direction is reduced to 1/3 of the original length. Similarly, when the Ymin and Ymax are doubled, the measured length in the vertical direction is reduced to 1/ of its original length. herefore, the new measured area of the region is only 1/6 of the original area, which is equal to 10/6 = Let = NYWR. If K = 4 the following equation is true: ( ) ( K + x x ) (1) If x = 10, then ( ) ( K + x ) () If x = 1/10, then ( ) ( K + x ) odd integer. is an integer, compute the number of real values of x for which = 1. x = (1) (K + 1) = 1, so x = 1 is a solution. x = ( 1) (K + 1) = 1 K + 1 is an even integer K is an (3) If x 10, 1/10, then x ± 1, then ( ) ( K + x ) x = 1 K + x = 0 x = 10 K. herefore, there are two or three solutions depending on whether K is an even or odd integer.

5 15. Let = NYWR. Point N lies on y = x 1 and P lies on y = 1 x as shown. If the coordinates of are (, 0) and NPQ is a rectangle, compute the area of NPQ. Since 1 = N = PQ = 1 x, the coordinates of Q are (, 0). herefore, the area of the rectangle is ( )( 1). 16. Compute the number of integer values of n such that n 3 + n + n + 6 is divisible by n n + n + n+6 n ( n+1)+( n+1)+5 5 = = n +1+ n+1 n+1 n+1 n 3 + n + n + 6 is divisible by n is divisible by n + 1 n = 6,, 0, and 4, yielding four integer solutions. 17. Let = NYWR. In paralleram ABCD, EC is the extension of BC. If the area of ADF is 81 and the area of FCE =, compute the area of ABCD. Let a = DF, b = FC, h = the height of ADF, and k = the height of ECF. By similarity, 81 a a = = 9. Since the area of ABCD = h(a + b) = ha + hb = 16 + b b ha 81 = = 16 + = ha Let = NYWR and set K =. If y = x is tangent to y = x + Kx+ N, then compute N. At the tangency, x + Kx+ N = x has a unique solution. hat is, x + (K 1)x + N = 0 has a unique solution. herefore, (K 1) K -1 - = 4N. Equivalently, N = = 4

f(x + y) + f(x y) = 10.

f(x + y) + f(x y) = 10. Math Field Day 202 Mad Hatter A A Suppose that for all real numbers x and y, Then f(y x) =? f(x + y) + f(x y) = 0. A2 Find the sum + 2 + 4 + 5 + 7 + 8 + 0 + + +49 + 50 + 52 + 53 + 55 + 56 + 58 + 59. A3