1. (E) Suppose the two numbers are a and b. Then the desired sum is

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1 Solutions 00 5 th MC. (E) Suppose the two numbers are a and b. Then the desired sum is (a ) + (b + ) = (a + b) + = S +.. (E) Suppose N = 0a + b. Then 0a + b = ab + (a + b). It follows that 9a = ab, which implies that b = 9, since a 0.. () If Kristin s annual income is x 8,000 dollars, then p 00 8,000 + p + p (x 8,000) = x Multiplying by 00 and expanding yields 8,000p + px + x 8,000p 56,000 = px + 0.5x. So,.75x = 7 4x = 56,000 and x =, (D) Since the median is 5, we can write the three numbers as x, 5, and y, where (x y) = x + 0 and (x y) + 5 = y. If we add these equations, we get (x y) + 5 = x + y + 0 and solving for x+y gives x+y = 5. Hence the sum of the numbers x+y +5 = 0. Let m be the mean of the three numbers. Then the least of the numbers is m 0 and the greatest is m + 5. The middle of the three numbers is the median, 5. So ((m 0) (m + 5)) = m and m = 0. Hence, the sum of the three numbers is (0) = (D) Note that 9999 = = 0000! = 0000! ! 6. (E) The last four digits (GHIJ) are either 975 or 75, and the remaining odd digit (either or 9) is,, or C. Since + + C = 9, the odd digit among,, and C must be. Thus the sum of the two even digits in C is 8. The three digits in DEF are 864, 64, or 40, leaving the pairs and 0, 8 and 0, or 8 and 6, respectively, as the two even digits in C. Of those, only the pair 8 and 0 has sum 8, so C is 80, and the required first digit is 8. The only such telephone number is

2 Solutions 00 5 th MC 7. () Let n be the number of full-price tickets and p be the price of each in dollars. Then np + (40 n) p = 00, so p(n + 40) = 400. Thus n + 40 must be a factor of 400 = 9. Since 0 n 40, we have 40 n , and the only factor of 400 that is in the required range for n + 40 is 74 = 9. Therefore, n + 40 = 74, so n = 4 and p =. The money raised by the full-price tickets is 4 = 78 dollars. 8. (C) The slant height of the cone is 0, the radius of the sector. The circmference of the base of the cone is the same as the length of the secotr s arc. This is 5/60 = 7/0 of the circumference, 0π, of the circle from which the sector is cut. The base circumference of the cone is 4π, so its radius is (C) Note that For all positive x, f(600) = f( ) = f(500) = 6/5 6/5 = 5. f(x) = f( x) = f() x, so xf(x) is the constant f(). Therefore, 600f(600) = 500f(500) = 500() = 500, so f(600) = 500 given conditions. 600 = 5. Note. f(x) = 500 x is the unique function satisfying the 0. (D) The pattern shown at left is repeated in the plane. In fact, nine repetitions of it are shown in the statement of the problem. Note that four of the nine squres in the three-by-three square are not in the four pentagons that make up the three-by-three square. Therefore, the percentage of the plane that is enclosed by pentagons is 4 9 = 5 9 = %

3 Solutions 00 5 th MC 4. (D) Think of continuing the drawing until all five chips are removed form the box. There are ten possible orderings of the colors: RRRWW, RRWRW, RWRRW, WRRRW, RRWWR, RWRWR, WRRWR, RWWRR, WRWRR, and WWRRR. The six orderings that end in R represent drawings that would have ended when the second white chip was drawn. Imagine drawing until only one chip remains. If the remaining chip is red, then that draw would have ended when the second white chip was removed. The last chip will be red with probability /5.. () For integers not exceeding 00, there are 00/ = 667 multiples of and 00/4 = 500 multiples of 4. The total, 67, counts the 00/ = 66 multiples of twice, so there are = 00 multiples of or 4. From these we exclude the 00/5 = multiples of 5 and the 00/0 = 00 multiples of 0, since these are multiples of 5. However, this excludes the 00/60 = multiples of 60 twice, so we must re-include these. The number of integers satisfying the conditions is = 80.. (E) The equation of the first parabola can be written in the form y = a(x h) + k = ax axh + ah + k, and the equation for the second (having the same shape and vertex, but opening in the opposite direction) can be written in the form Hence, y = a(x h) + k = ax + axh ah + k. a + b + c + d + e + f = a + ( ah) + (ah + k) + ( a) + (ah) + ( ah + k) = k. The reflection of a point (x, y) about the line y = k is (x, k y). Thus, the equation of the reflected parabola is k y = ax + bx + c, or equivalently, y = k (ax + bx + c). Hence a + b + c + d + e + f = k. 4. (D) Each of the ( 9 ) 9C = 6 pairs of vertices determines two equilateral triangles, for a total of 7 triangles. However, the three triangles 4 7, 5 8, and 6 9 are each counted times, resulting in an overcount of 6. Thus, there are 66 distinct equilateral triangles. 5. () Unfold the tetrahedron onto a plane. The two opposite-edge midpoints become the midpoints of opposite sides of a rhombus with sides of length, so are now unit apart. Folding back to a tetrahedron does not change the distance and it remains minimal.

4 Solutions 00 5 th MC 5 6. (D) Number the spider s legs from through 8, and let a k and b k denote the sock and shoe that will go on leg k. possible arrangement of the socks and shoes is a permutation of the sixteen symbols a, b,... a 8, b 8, in which a k precedes b k for k 8. There are 6! permutations of the sixteen symbols, and a precedes b in exactly half of these, or 6!/ permutations. Similarly, a precedes b in exactly half of those, or 6!/ permutations. Continuing, we can conclude that a k precedes b k for k 8 in exactly 6!/ 8 permutations. 7. (C) Since P = 90 if and only if P lies on the semicircle with center (, ) and radius 5, the angel is obtuse if and only if the point P lies inside this semicircle. The semicircle lies entirely inside the pentagon, since the distance,, from (, ) to DE is greater than the radius of the circle. Thus the probability that the nagle is obtuse is the ratio of the area of the semicircle to the area of the pentagon. Let O = (0, 0), = (0, ), = (4, 0), C = (π +, 0), D = (π +, 4), and E = (0, 4). Then the area of the pentagon is [CDE] = [OCDE] [O] = 4 (π + ) ( 4) = 8π, and the area of the semicircle is π( 5) = 5 π. The probability is 5 π 8π = 5 6. E O D C 8. (D) Let C be the intersection of the horizontal line through and the vertical line through. In right triangle C, we have C = and = 5, so C = 4. Let x be the radius of the third circle, and D be the center. Let E and F be the points of intersection of the horizontal line through D with the vertical lines through and, respectively, as shown. In ED we have D = 4 + x and E = 4 x, so DE = (4 + x) (4 x) = 6x, and DE = 4 x. In DF we have D = + x and F = x, so and F D = x. Hence, and x =, which implies x = 4 9. F D = ( + x) ( x) = 4x, 4 = C = F D + DE = x + 4 x = 6 x F D C E

5 Solutions 00 5 th MC 6 9. () The sum and product of the zeros of P (x) are a and c, respectively. Therefore, a = c = + a + b + c. Since c = P (0) is the y-intercept of y = P (x), it follows that c =. Thus a = 6 and b =. 0. (C) Let the midpoints of sides, C, CD, and D be denoted M,N,P,and Q, respectively. Then M = (, 5) and N = (, ). Since MN has slope, the slope of MQ must be /, and MQ = MN = 0. n equation for the line containing MQ is thus y 5 = (x ), or y = (x+)/. So Q has coordinates of the form (a, (a + )). Since MQ = 0, we have (a ) + ( a + 5) = 0 (a ) + ( a ) = (a ) = 0 (a ) = 9 a = ± Since Q is in the first quadrant, a = 5 and Q = (5, 6). Since Q is the midpoint of D and = (, 9), we have D = (7, ), and 7 + = 0. Use translation vectors. s before, M = (, 5) and N = (, ). So NM =,. The vector MQ must have the same length as MN and be perpendicular to it, so MQ =,. Thus, Q = (5, 6). s before, D = (7, ), and the answer is 0. Each pair of opposite sides of the square are parallel to a diagonal of CD, so the diagonals of CD are perpendicular. Similarly, each pair of opposite sides of the square has length half that of a diagonal, so the diagonals of CD are congruent. Since the slope of C is - and C is perpendicular to D, we have b a =, so a = (b ). Since C = D, 40 = (a ) + (b ) = 9(b ) + (b ) = 0(b ), and since b is positive, b = and a = + (b ) = 7. So the answer is 0.

6 Solutions 00 5 th MC 7. (D) Note that and (a + )(b + ) = ab + a + b + = 54 + = 55 = 5 7, (b + )(c + ) = bc + b + c + = 46 + = 47 = 7. Since (a + )(b + ) is a multiple of 5 and (b + )(c + ) is not a multiple of 5, it follows that a + must be a multiple of 5. Since a + divides 55, a is one of 4, 74, 74, or 54. mong these only 4 is a divisor of 8!, so a = 4. This implies that b + =, and b = 0. From this it follows that c + = 7 and c = 6. Finally, (c + )(d + ) = 05 = 5 7, so d + = 5 and d = 4. Therefore, a d = 4 4 = 0.. (C) The area of triangle EF G is (/6)(70) = 5/. Triangles F H and CEH are similar, so / = EC/F = EH/HF and EH/EF = /5. Triangles GJ and CEJ are similar, so /4 = EC/G = EJ/JG and EJ/EG = /7. D E C J H F G Since the areas of the triangles that have a common altitude are proportional to their bases, the ratio of the area of EHJ to the area of EHG is /7, and the ratio of the area of EHG to that of EF G is /5. Therefore, the ratio of the area of EHJ to the area of EF G is (/5)(/7) = 9/5. Thus, the area of EHJ is (9/5)(5/) =.. () If r and s are the integer zeros, the polynomial can be written in the form P (x) = (x r)(x s)(x + αx + β). The coefficient of x, α (r +s), is an integer, so α is an integer. The coefficient of x, β α(r + s) + rs, is an integer, so β is also an integer. pplying the quadratic formula gives the remaining zeros as ( α ± α 4β) = α 4β α ± i. nswer choices (), (), (C), and (E) require that α =, which implies that the imaginary parts of the remaining zeros have the form ± 4β /. This is true only for choice (). Note that choice (D) is not possible since this choice requires α =, which produces an imaginary part of the form β, which cannot be.

7 Solutions 00 5 th MC 8 4. (D) Let E be a point on D such that CE is perpendicular to D, and draw E. Since DC is an exterior angle of D it follows that DC = D + D = = 60. D E C Thus, CDE is a triangle and DE = CD = D. Hence, DE is isosceles and ED = ED = 0. ut EC is also equal to 0 and therefore EC is isosceles with E = EC. On the other hand, E = D ED = 45 0 = 5 = E. Thus, E is isosceles with E = E. Hence E = E = EC. The right triangle EC is also isosceles with EC = EC = 45. Hence, C = EC + ECD = = (D) If a, b, and c are three consecutive terms of such a sequence, then ac = b, which can be rewritten as c = ( + b)/a. pplying this rule recursively and simplifying yields..., a, b, + b a, + a + b ab, + a, a, b,... b This shows that at most five different terms can appear in such a sequence. Moreover, the value of a is determined once the value 000 is assigned to b and the value 00 is assigned to another of the first five terms. Thus, there are four such sequences that contain 00 s a term, namely 00, 000,, 000, , 00,...,, 000, 00, , 000,,..., , 000, , 00, , , 000, ,..., and , 00, 00, ,..., respectively. The four values of x are 00,, , and