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Math Contest 6 Calulators Are Not Allowed Math Contest 6 Calulators are not allowed. Name: Area 1: First Period Instructor: 1. Leah and Nicole are riding a Ferris wheel. This particular Ferris wheel has radius 20 feet and revolves at a constant rate of one revolution per minute. How many seconds does it take Leah and Nicole to travel from the bottom of the wheel to a point 10 vertical feet from the bottom? 1. Solution: In the figure below, the center of the wheel is at O,andtheriders(Leah and Nicole) travel from A to B.SinceAC =10andOB = OA =20,thepointC is the midpoint of OA.Intheright OCB,wehaveOC half of the length of the hy- potenuse OB,som∠COB =60◦. Since the wheel turns through an angle of 360◦ in 60 60 seconds, the time required to turn through an angle of 60◦ is 60( 360 )=10seconds. 2. Awindowhas9panesintheformofa3x3grid.Inhowmanywayscanonecolor6of these panes yellow so that the window looks the same from the inside and outside the house? 2. Solution: If we first color all the panes yellow and then color the remaining 3 panes white (to represent unpainted) we can solve an easier problem. The solution has to be mirror symmetric, if we reflect along the vertical middle line. In the central column there could be either 3 or 1 white panes. For 3, there is only one possibility. For 1 white pane in the middle there are 3 3 = 9 possibilities: 3 for the middle one, and 3 for the white panel in the left column;∗ then we have to color a pane in the similar position on the right. Thus the total number is 1 + 9 = 10. 3. Consider all three digit numbers. How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits? 3. Solution: The first and last digits must be both odd or both even from their average to be an integer. There are 5 5=25odd-oddcombinationsforthefirstandlast digits. There are 4*5=20 even-even∗ combinations that do not use zero as the first digit. Hence there are 45. 4. For this problem we will consider happy numbers. A happy number is defined by the following process. Starting with any positive integer replace the number by the sum of the squares of its digits and repeat. If this process ends in 1 we call the number happy. For example: 7 72 =49 42 +92 =97 . Find all two digit happy numbers that reach 1 in exactly→ 2 steps. → ··· 4. Solution: We can represent our two digit number as xy where x is the first digit and y is the second digit. We also know that 0 x 9and0 y 9. This means that the largest possible number attained from≤ our first≤ is 162≤ = 92≤+92.Fromthis we know in order to reach 1 in exactly two steps that the sum of the digits must equal either 10 or 100. Since 10 becomes 12 +02 =1and12 +02 +02 =1.Weneed only pairs of perfect squares that sum to 10 and 100. We get that 10 = 1 + 9 and 100 = 64 + 36 thus the answers are 86, 68, 31, 13. 5. The sides of a triangle are consecutive integers and the area of the triangle is also an integer. What is the smallest perimeter possible such that the triangle is not a right triangle? 5. Page 2 Solution: The answer can be found using and alternative formula for the area of a triangle know as Heron’s formula. Heron’s formula is s(s a)(s b)(s c)where s = a+b+c is the semiperimeter of the triangle. − − − 2 Let the sides be b 1,b,b+1,sotheareais 3b 3b − 3b 3b 3b b+2 b b 2 K = ( (b 1))( (b))( (b +1))= ( )( )( − ). This simplifies 2 2 − − 2 − 2 − 2 2 2 2 b further to 4 3(b 2)(b +2), so it looks like b has to be a multiple of 4 or one of b 2orb +2 must− be a perfect square of 3 times a perfect square. If b =4,then the− area is an integer, (6), but the triangle is then a 3 4 5trianglebutisaright triangle. The next value that works, is b =14,withthearea− − 14 14 14 4 3(14 2)(14 + 2) = 4 3(12)(16) = 4 (6)(4) = 84. Thus the answer is 13 + 14 + 15 =− 42. 6. We consider the equation x2 + kx +1 = 0. A single fair die is rolled to determine the value of the middle coefficient, k. The value for k is the number of dots on the upper face of the die. What is the probability that the equation will have real, unequal root? 6. b √b2 4ac Solution: Using the quadratic formula, − ± 2a − ,wecanwriteallsolutionstothe given equation. We now need to determine when this is real and produces unequal roots. This happens when b2 4ac > 0. Using the given equation we then get that k2 4 1 1 > 0, thus k must− be greater than 2. So k can equal 3, 4, 5, or 6. Thus − ∗ ∗ 4 2 the probability is 6 = 3 Page 3.
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