New York State Mathematics Association of Two-Year Colleges

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1 New York State Mathematics Association of Two-Year Colleges Math League Contest ~ Fall 06 Directions: You have one hour to take this test. Scrap paper is allowed. The use of calculators is NOT permitted, as well as computers, books, math tables, and notes of any kind. You are not epected to answer all the questions. However, do not spend too much time on any one problem. Four points are awarded for each correct answer, one point is deducted for each incorrect answer, and no points are awarded/deducted for blank responses. There is no partial credit. Unless otherwise indicated, answers must given in eact form, i.e. in terms of fractions, radicals,, etc. NOTE: NOTA = None Of These Answers. n. Let f ( ) denote the n th iteration of function f, i.e. f n ( ) f( f( f f ( ) )). For eample, f ( ) f( f( )) and f 06 ( ) when simplified? B) f ( ) f( f( f( ))). If C) D) NOTA n f( ), then which of the following represents. Suppose a regular octagon (i.e. an 8-sided polygon with congruent sides and interior angles) is to be constructed from a square by removing 4 triangular corners, as shown. What will be the length of each edge of the octagon?. How many real values for solve the equation 4? Note: is the absolute value of. B) C) D) 4 4. What is the remainder when the 06 digit number consisting of 0 groups of 06 followed by 0, is divided by? 0 B) C) D) , i.e. the number 0 groups of 06. If the positive integers are arranged in a triangular pattern, as shown, then "06" will eventually appear in the m th row and n th position from the left. What is m n? (e.g. "" appears in the th row and nd position from the left, thus m and n, making mn. ) Hint: k k( k )

2 6. The equation log 06( ) log (06) has two real solutions, and. What is the product,, of these two solutions? 06 B) 06 C) 06 D) A right triangle with a height of 4 and unknown base,, intersects a semicircle whose diameter coincides with the height of the triangle, as shown. A is the area outside the triangle, but inside the semicircle; while A is the area inside the triangle, but outside the semicircle. If A A, then what is? B) C) 0 D) 0 8. A lattice point in the y-plane is a point with integer coordinates. How many lattice points fall on the line y? none B) eactly one C) eactly two D) infinitely many 4 A A 9. If the line y is rotated counter-clockwise about the point (,), what will be the y-intercept of the new line? 0. A bo initially contains 0 marbles, red and blue. If 9 marbles are randomly selected and removed from the bo, what is the probability that the remaining marble is red?. arccos arcsin? B) C) D). If cos( ) cos(0 ) cos(0 ), then is? Hint: cos( AB) cos( cos( B) sin( sin( B) 0 B) 0 C) 0 D) 0. Which value for solves the equation 9 6 4? ln() ln() ln(9) B) C) ln() ln() ln() ln() ln() ln() D) ln(9) ln() Ant 4. A cube with edges of length, has an ant and a crumb both unit from the nearest corner on opposite edges, as shown. What is the shortest distance the ant can traverse to get to the crumb? Note: The ant cannot jump, and must remain in contact with the cube at all times. Crumb

3 4,0,8. As of September 06 the largest known prime number is. What are the last two 0 digits in its epansion? For eample, 04, so the last two digits are The Lambert W function is defined as the inverse function of f( ) e for, i.e. W( ) f ( ) for e. Thus, if y ye, then y W( ). If W B) W C) W D) W e 4, then equals. A circle of radius is placed in the first quadrant so that it is tangent to the line y and the -ais, as shown. What is the -coordinate of the point of tangency on the -ais? 8. If is the smallest positive angle that satisfies the equation: value of tan( )? 49 B) 49 4 C) 4 D) cos( ) sin( ), then what is the 9. The diagram shows a large circle of radius with four smaller circles of radius ½ that are tangent to the larger circle with centers along the coordinate aes. What is the area of the shaded region? 0. The seven cards shown have a letter on one side and a number on the other. How many of the cards must be turned over in order to verify the statement: If a card has a vowel on one side, then the other side must have an odd number.? A B E 0 6 B) 4 C) D) 6

4 New York State Mathematics Association of Two-Year Colleges Math League Contest ~ Fall 06 ~ Solutions f ( ) f( ) f ( ) f f( ) f. f ( ) f f ( ) f ( ) f ( ) f f ( ) f( ) f ( ). 4 n 4 k Thus, f ( ) cycles by : f ( ) f ( ) f ( )... f ( ), and f ( ) f ( )... f, 6 k 06 6 and f ( ) f ( )... f. Since 06 6, f ( ) f ( ). Answer: A. Let = the length of each edge. Since each corner is a triangle, with a hypotenuse of length, the leg length is. Thus, each side of the square has length. Solving for gives: Answer:.. Case I: 0 : 4 0, but 0. Thus, only is a solution in this case. Case II: 0 : 4 0, both satisfy 0. Therefore, there are eactly solutions. 4. Since 06 is a multiple of (06 = 88), and = = , the remainder must be. Answer: B. Notice that the last number in row k, called triangular numbers, is... k. For eample, the 6 in row equals and the 0 in row 4 equals 4. Thus, we seek the largest value of k such that... k 06 in order to identify the row preceding 06 (or the row if the sum = 06). Using the formula for the sum, we need to determine the largest k-value for which: kk ( ) 06 or ( ) 40. kk Since , we know k will be in the low 60's , 66 8, , and , giving a k-value of 6 with 06 being the last number of that row (since the sum = 06). Thus, 06 appears in the 6 rd row and the 6 rd position, making mn66 6. Answer: 6 k

5 log( ) 6. Since log 06( ) and log(06) log(06) log (06) log( ), let y log 06 ( ), then y y log (06). Now we solve y or y y 0. We know the sum of the two solutions of the quadratic is. Thus, y y or log 06( ) log 06( ), / which gives log 06( ), and finally 06.. Let the unshaded region (common to both the triangle and semicircle) be A. Thus, the area of the semicircle is A AA, and the area of the S triangle is AT AA. Since A A, AS AT, which gives or. Answer: B 8. y y, which has no integer solutions. Answer: A 0 9. The rotation of the line y (solid) makes the new line (dashed) form angles of 0 (4 + ) and 60 with the -ais, as shown. Thus, we see two triangles formed. The smaller one has a base (dotted) of length and height. Therefore, the y-intercept of the new line is. Answer: Alternate Solution: The slope of a line equals tan( ), where is the angle of inclination, 0 in this case. Thus, the slope is tan(0 ) tan(60 ), and the equation is: y ( ) or y. Hence, the y-intercept is. 0. The probability the remaining marble is red, is the same as the probability of selecting a red one from the 0 marbles. Thus, the probability is 0. Answer: 0. Letting arccos and y arcsin sin( y)., gives cos( ) and A right triangle corresponding to these values is shown. Thus, y.. cos( AB) cos( cos( B) sin( sin( B) cos( AB) cos( cos( B) sin( sin( B) cos( AB) cos( cos( B) sin( sin( B), since cosine is an even function, and sine is odd. Now let A 60 and B 0, so we get: cos(600 ) cos(60 )cos(0 ) sin(60 )sin(0 ) and cos(600 ) cos(60 )cos(0 ) sin(60 )sin(0 ). Adding these two results gives: cos(0 ) cos(0 ) cos(60 )cos(0 ) cos(0 ) cos(0 ). Thus, 0. Answer: B y 0 60 (,)

6 Dividing by 4 gives: 9 6 or Now let giving: y y y or ( y)( y) 0. So, y or.. Solving for : ln() ln(). ln() ln() ln. y But, 4. Unfolding the cube as shown, reveals a shorter distance than the obvious walk directly down the wall and across the floor (giving a distance of 6). Thus, the shortest path is the hypotenuse of the right triangle with legs of length 4 (i.e. +). Making the shortest distance 4 (we know this is less than 6 since.).. Using 0 04 Answer: 4 0, we can look for a pattern , y, y which is positive, thus Answer: A , and so we see a pattern after only iterations. We see that positive odd powers of 04 end with 4,0,8 4,0,80 4, and positive even powers of 04 end with 6. Hence, 0,40,8,40,8 04 (...6) (...)... Answer: 6. Taking the square root of both sides gives: sides by to get e 4 e, for 0. Now divide both e, which now fits the form for the Lambert W function. Thus, W () or W().. Let a be the -value we seek, with a 0. Drawing a line from a on the -ais through the center of the circle to the line y forms two triangles; a large one whose legs have length a, and a small one with legs of length (the radius of the circle). The small triangle has hypotenuse of length, making the legs (a) of the larger triangle. Answer: Alternate Solution: Simple right triangle trigonometry, on the triangle whose hypotenuse is from the origin to the center of the circle, tells us tan(. ). Using the half-angle formula for tangent, cos( ) tan (again), a. sin( ), we get Ant a a tan(. ) cos(4) sin(4 ) tan(. ). So, we get Crumb a

7 8. Squaring both sides gives: 49 cos( ) sin( ) cos ( ) cos( )sin( ) sin ( ), 49 4 which reduces to: cos( )sin( ) and then: sin( ), since sin( ) cos( )sin( ). Now form a right triangle with this angle, which yields 4 tan( ). 9. Let's determine the shaded area in the first quadrant, then we'll multiply by 4 to obtain the entire area. A is the area of the segment cut off by the line y (dashed), while A is the area of the region within the sector (bound by the circle of radius, y, and the -ais) but A outside the lower semi-circle of radius ½. area of the ¼ - circle area of the right triangle with A A of radius ½ a base and height of ½ Notice the area of the each small semi-circle is, and the area of the sector in 8 question is. Since the two areas are equal, A 8 A. Thus, the area of the shaded 8 region in the first quadrant is AA 4A 4. Therefore, the total shaded area is 4. Answer: 4 0. Five cards must be turned over, which are: A, E,, 0, and 6. (This is a version of the Wason Selection Task.) 4