Ohio Council of Teachers of Mathematics
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Ohio Council of Teachers of Mathematics Solutions to Thirty-Second Annual Contest February 26, 2005
1) 2005 2005 (2 0 05 ) 2005 (0 0) 2005 2) 5 Marie has one more brother (Donnie) and one fewer sister (herself) than Donnie, and thus she has 5 more brothers than sisters. Algebraically, if Donnie has x brothers and y sisters, then x y 3. Marie has x 1 brothers and y 1 sisters, and thus x 1 (y 1) x y 2 3 2 5 more brothers than sisters. 3) diagonal EXTRA PROBLEM: Find an equation for the number of diagonals an n- sided polygon has. E-mail [email protected] to check your answer. 4) 101 20.05x 20.05 2005 20.05(x 1) 20.05(100) x 1 100 x 101 5) B (,2) is equivalent to the other four, NOT (,2) 1 1 1 3 3 6) 7 3 1 7 2 45 2 3 3 7 7) Tuesday 2005 286 7 3 , so 2005 days is 286 weeks and 3 days, so just note 3 days from now is Tuesday (Note that 3 is the remainder when 2005 is divided by 7.) 8) A Note the domain of f is [1,2) (2,) . The range of f over [1,2) is (,0] and its range over (2,) is (0,) , giving the overall range of (,) . 10 1252 43 200 0 64 10 9) 46 27 or 27 or 46.370 27 5 2.9 (200.5) 27 40 3 1 46 27 10) A The whole numbers includes 0. min sec 11) 96 seconds 3% of an hour is .03hours(60 hour )(60 min ) 108 seconds. 20% of a minute sec is .2min(60 min ) 12 seconds. 108 12 96 seconds. 3 12) 45 units, 24 units If AB=AC, then 4x 8x 3 x 4 . The triangle would then have side lengths 3, 3, and 7.5. But such a triangle isn’t possible since the sum of the lengths of any two sides of a triangle is greater than the length of the third side, while 3 3 7.5 . If AB=AC, then 4x 2x 6 x 3. The triangle has legitimate side lengths of 12, 12, and 21 for a perimeter of 45. If AC=BC, then 2x 6 8x 3 9 6x x 1.5. The triangle has legitimate side lengths of 9, 9, 6 for a perimeter of 24. 13) (0,11) or 11 Plug x 0 into the equation and solve for y: y 5 (0 4)2 16 y 11. 2 14) 50 200 five 2 5 0 5 0 1 50 . 15) B 16) y 2sin(x 2 ) 1 Multiplying sine by 2 creates amplitude of 2; 1 for coefficient of x gives or y 2sin(x 2 ) 1 period of 2 , adding or subtracting 2 from x shifts left or right by 2 units, adding 1 translates up by 1 unit. 17) 8 years old Note the ten ages are 5, 6, 7, 8, 9, 10, 11, 12, 13, 14. First note 5 and 7 are the only two that add to 12 years old. Then 6 and 9 are the only two remaining that add to 15 years old. Then 8 and 11 are the only two remaining that add to 19 years old., so the 11 year old’s brother is 8. x 2 x 2 (x 2)(x 1) x 1 18) x 1 Note y , x 2 . Vertical asymptotes x 2 3x 2 (x 2)(x 1) x 1 correspond to where the denominator of this final term equals 0. 2 19) i. B (33 ) 2 36 39 33 ii. D Non-real complex numbers can’t be compared using less than or greater than signs. iii. A Supplement of X 180° mX 90° mX complement of X 20) Note 3 x 2 3 x 0 3 x 3.
-4 -3 -2 -1 0 1 2 3 4
21) $24060 Let P be the amount invested at 8%. Then 0.08P 0.04(P 22055) 2005 .12P 882.2 2005 3V V 22) h or h s 2 (1/ 3)s 2 23) 1037,1716,2005 A Note 2005 5 401, both of which are prime factors. Start or 1357,1476,2005 B with the well-known Pythagorean triple 3,4,5 and multiply all or 2005,4812,5213 C entries by 401 to get answer D. Start with the Pythagorean or 1203,1604,2005 D triple 5,12,13 and multiply each entry by 401 to get answer C. or 200,1995,2005 E Alternatively, let a (or b) equal 2005. Then or 2005,80388,80413 F 20052 c 2 a 2 (c a)(c a) . We consider ways of or 2005,402000,402005 G 2 factoring 2005 4020025 into two distinct factors with the or 2005,2010012,2010013 H. smallest factor listed first: 1 4020025 , 5804005 , 25160801, 40110025 . For each of these factorizations, let c equal 0.5 times the sum of the two numbers and a (or b) equal 0.5 times the difference of the two numbers. This gives answers H, G, F, and C respectively (all the answers where 2005 is a or b). 24) a) (200, -5) Reflecting across the x-axis changes the sign of y. b) (5, 200) Reflecting across the line y x interchanges the x and y coordinates. n!(n 1) 1 n(n 1)!(n 1) 25) 1 1 (n 1)! n 2 n (n 1)!n(n 1) 26) 5 3 5 10 cm. The horizontal and vertical pieces of the hexagon have length or 14.8705 cm. 2+3=5. There are 3 slanted pieces, that can be seen as diagonals of a 1x2 rectangle have length 5 . The one slanted piece that can be seen as a diagonal of a 1x3 rectangle has length 10 . 27) 27.2614○, 332.7386○ cos x tan 2005 sin 2005 x arccos(tan 2005 sin 2005) 27.2614 This gives the angle in the 1st quadrant. Cosine is also positive in the 4th quadrant, and the solution in the 4th quadrant is x 360 27.2614 332.7386 . 1 28) k 6 or –0.1667 or 0.16 The second line’s equation can be written as y 2x 5, and hence the 2nd line has a slope of 2. We need the first line to have a 1 1 1 slope of 2 , so 3k 2 k 6 . 2 29) 13 or 40.8407 cm r Consider the 2-dimensional view of the sphere and
6 cm . intersecting plane. Using the Pythagorean Theorem,
7cm . 7cm . we have the radius r of the circle of intersection is the solution to r 2 49 36 13 cm2. Then the area of this circle is r 2 13 cm2.
30) –17 or –16.5 There are 15 scores, so the median score is the 8th score which is 83. The lower quartile is the median of the scores below 83, or the 4th score, so x 72. The upper quartile is the median of the scores above 83, or the 12th score, so y 89 . Thus, x y 17 . Some textbooks include the median when finding the lower quartile and upper quartile. If we do this, then x 72 and y 88.5 , giving x y 16.5 .
log 2 2 4 31) 2, 16 5 4log x 2 log 2 x 4 log 2 x log 2 x . Let y log 2 x , log 2 x log 2 x 4 so 5 y 5y 4 y 2 y 2 5y 4 0 (y 4)(y 1) 0 y
y 4 or y 1 log 2 x 4 or log 2 x 1 x 16 or x 2 . 32) –4, 1 a a 2 a 2 4a 4 a 2 3a 4 0 (a 4)(a 1) 0 a 4 or a 1 2005 2 3 1 3 2 3 2 33) 4 2 2(2 ) 2005 586.8127 cm Let r be the radius of the sphere. Then the cylinder has bases that are circles of radius r and a height of 2r. Using the formula for the volume of a cylinder, 2005 r 2 2r 2r 3 . Thus, 2005 1 3 r 2 . Using the formula for the surface area of a sphere, we get a 2 2005 2 / 3 surface area of 4r 4 2 . 34) 12 Completing the square for y, 0 y 2 12x 2y 25 (y 1) 2 112x 25 (y 1) 2 12(x 2) . Thus, (y 1)2 12(x 2) , so the length of the latus rectum is 12. 2 35) 3 Solution #1 – There are 333 27 triples of colors that the teachers may have selected from the bag, each of which is equally likely to occur since there are an equal number of marbles of each color. 3 of these have all the same color and 3! 6 have all different colors. Thus, 27 3 6 18 triples have exactly two marbles of the same color, and the probability this 18 2 occurs is 27 3 . Solution #2 – After RT pulls a marble out, the probability both JVS and 1 1 1 AN pull out the same colored marble is 3 3 9 . After RT pull a marble out, the probability JVS pulls out a marble of a different color and then 2 1 2 AN pulls out a marble of the unselected color is 3 3 9 . Thus, the 1 2 6 2 probability exactly two marbles are the same color is 1 9 9 9 3 . 1 2 2 36) 5 2.2361 x 2 x2 x 4 1 x 5 x 5 , but we only want the positive value. 37) B Show the conjecture holds for the base case (n 1) . 38) catenary or hyperbolic cosine The word “catenary” is derived from the Latin word for chain and was first used by Huygens in a letter to Leibniz in 1690. It has also been called the Alysoid and the Chainette. 5m3n 39) 8 Note a b c d e 5m and f g h 3n . Thus, the mean of all abcd e f gh 5m3n five numbers is 8 8 . 2 40) 56 3 mph 56.6667mph Note the total distance Rosie travels is 110 215 170 495 miles. Let t be the amount of time in hours that it takes her to drive from Austintown to Ayersville. Then the total time Rosie travels (in hours) is 1 2 2 3 3 3 t 6 t hours. To average 55mph for the entire trip, we must 495 have 6t 55 495 330 55t 165 55t t 3 hours. Thus, Rosie 170 2 must drive at a rate of 3 56 3 mph.
The names of the students who qualify for the OHMIO competition will be posted on the OCTM website (www.ohioctm.org) by noon on Monday, March 7.
Solutions provided by: Dr. Gordon Swain, Ashland University Dr. Christopher Swanson, Ashland University