WYSE ACADEMIC CHALLENGE State Math Exam 2009 Solution Set. 2. Ans E: Function f(x) is an infinite geometric series with the ratio r = :

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1 WYSE ACADEMIC CHALLENGE State Math Eam 009 Solution Set 40. Ans A: ( C( 40,8 ) * C( 3,8 ) * C( 4,8 ) * C( 6,8 ) * C( 8,8 )) / 5 = Ans E: Function f() is an infinite geometric series with the ratio r = : + k k = 0 + f( ) = = = + +. Therefore, ( ) = f and f () =. 3. Ans D: We either have two A s and one other letter, or three different letters. This gives us C(3,) * 6 + P (7,3) = 8 possible rearrangements. 4. Ans A: If f ( ) = a + b + c, then f ( ) = a + b and f ( ) = a. Since f () =, a =. Therefore, f ( ) = + b, so f ( 3) = ( 3) + b = 0, so b = -6. Thus, the first derivative of f() equals f ( ) = 6, so f () = 6 = 8. ( ), then for the set of / n 5. Ans D: If we use mean= and s.d.= n n 0, 00=, =, 000 / =, and = For the set of, = 00 and = 000. This makes the 000 ( 00 ) / standard deviation s.d.= = If we use the sample standard deviation formula, we end up with , but this also rounds to Ans A: = nn ( + ) n n+. Therefore, 49 n= 4 = = = 0.3 nn ( + ) kt 7. Ans C: Using A= Pe and 50=00e k 300, we get k = If we let t = , we end up with A=00e = Don t round too much on k! log3 8. Ans E: log 3 > > > 0 > 0. Let log log log log =. Then > 0 0< < 0< log3 < < <3.

2 9. Ans C: The left side has a domain of b < 0. log( b) + log( ) = ( ) log ( b)( ) = + b = 0 + b 0 = 0. This has real solutions if b 40 > 0. Due to the original domains, we onl have a real solution when b 6.3, so we have no real solution when b > Ans E: Let us multipl the first equation b i and add it to the second equation: i( i ) = i i = i i + ( + i) = i + 0= i +. This equation never holds true, so the sstem has no solution.. Ans A: Split the figure up into a pentagon and five attached isosceles triangles. The outer five triangles have angles of 7, 7, and 36 and sides of 0, 0 and (use law of sines to find the third side). If we use as the base, the height is (tan 7 = base/half the height). Each of these triangles therefore has an area of 0.5 * * = We can then split the pentagon up into five more isosceles triangles. The triangles from the pentagon have angles of 7, 54, and 54 and sides of 6.803, 5.573, and (use law of sines again to find the remaining two lengths). Use base of and tan 54 = base/half the height to get a base of We then get that the five triangles in the pentagon have areas of 0.5 * * 4.53 = We then find the overall area of the star b taking 5 * * =.66.. Ans D: cosθ = r cosθ = ( r 0 ) r (cos θ sin θ) = r ( r cosθ) ( r sinθ) = =, which represents the equation of a hperbola in rectangular coordinates. 3. Ans D: First draw lines AP, PB, PD, DQ, BQ, and CQ each from particular points to the center of the circles. We have tan ABP = 0, so angle ABP is degrees. Angle PBE is also.80 degrees, so angles DBQ and CBQ are degrees. Now tan68.99 = 0, which makes BC 4 inches. BC Ans B: π + π 6 = Ans B: a b = + ( ) + ( 3) 0, b c = + + ( 3) ( ) 0. a c = + ( ) + ( ) = 0, 6. Ans B: The equation of the parabola can be represented as + = + ( ) = ( + ). Therefore, the coordinates of the verte are (-, ) and hence its distance from the origin is.

3 7. Ans A: (Note: all angles are listed in degrees) Let = the distance from the bo s original position to the point at the bottom of segment h. We get the h following two equations: tan39 = and = h tan37. First solve for to + 00 get 00 tan37 = tan39 tan37 = Solve for h to get h = tan39 = Ans D: /6 /6 / lim = lim = lim + /6 lim + = e /6 = /6 6 lim + = 9. Ans D: Let be the distance from the light to the wall, be the height of the girl, and z be the height of the shadow. = 5 and 6 = z based on equal tangents. B equation, = 60. B equation, z = /6 = 60/6 = 0 feet Ans C: Finding the determinant of the matri A+ B = gives us 3 6+ det( A+ B) = (+ 3 )(6 + ) ( 3 )( + ) = = 0. Factor and solve to get (5 + 8) = 0, = 0, -8/5. Ans D: Bill must be taking the tires to Detroit, since statement IV keeps him from being either of the people going to Chicago. B statement II, Stan went to Chicago, so he must be the person taking the seats. This leaves Ted as the other person in Detroit. Since statement III implies the mirrors were taken to Detroit, we conclude that Ted took the mirrors to Detroit. We can also conclude that Pete must have taken the transmissions to Chicago.. Ans E: N = 0N = 50 N = Ans C: If we let = old old and = old + old, then substitution and solving gives us old = and old = Put these into old + old = 5 to get ( ) + ( ) = 5. So, = 5, = 5, + = 5. The equation is a circle, and it has a radius of 5 =.80. Alternativel, we could transform the center and one point on the circle and find the distance between the points, but doing so assumes the result is a circle.

4 cos( α) cos α sin α cosα sinα 4. Ans C: cot( α ) = = = sin( α) sinα cosα sinα cosα = c tanα = tanα c c = c a b c 5. Ans A: ( + ) + ( + ) = ( + ), a ( + ) + b= c, + ( + ) c a a + a + b = c, a + b = c a, ( a+ b) = c a, = a + b 6. Ans B: Since the right side of the equation is nonnegative, must be also nonnegative. Therefore, given 0, the original equation is equivalent two equations: c = and c =, which can be rewritten as + c = 0 ± + and c = 0. The solutions to these two equations are: = ± + and =. If c < 0.5, then the equations have no solution. If c = 0.5, then there is onl one solution = /. If 0.5 < c 0, then +, so the equations have eactl two nonnegative solutions: ± + =. If c > 0, then + >, so there are also two nonnegative ± + + solutions: =. Therefore, the original equation has eactl two solutions when c > Ans D: ( ) ( ( )) = 8. 0, = Alternativel, we end up with /3 left after the car, / left after the plane, and /5 left after the cand. So /5 = Ans C: + = 4, + = 4 = + = = 5. Let + 4 us square the equation + = 4 : + + = 764. Let us subtract the equation + = 5 from the last equation: 3 = 73. Thus = Ans E: If t is the minutes after 9:00 AM, t + ( t 5) = 00 and t = Round this to 9 hours, 4 minutes, or 6:4 PM 30. Ans B: Let =. Then =, f( ) = ( ) + ( ) = + +. Therefore, f( ) = 3 +, so f( ) = Ans E. The -intercepts are at -0.5 and 3, so -0.5 * 3 = -.5

5 3. Ans E: () If 0, then + > < 0 ( )( + ) < 0. The last inequalit holds true if < <. Taking into account that 0, the solution is 0 <. () If < 0, then the original equalit is true as long as the square root is defined: + 0. Therefore, <0. Combining these two solutions, we get <. 33. Ans B: The following triangle shows the situation t seconds before the car crosses the finish line: t*308 ft θ 50 ft The distance from the car to the finish line is 308t, and the distance from the track to the spectator is 50 ft. This sets up the equation tanθ = 308 t. If we 50 dθ 308 implicitl take the derivative, we end up with sec θ =. At the moment dt 50 dθ 308 the car crosses the finish line, θ = 0, so = = 6.6. dt Ans A: α = arcsin α = arcsin sin( α ) =. Therefore, cosα sinα cos α + sin α cotα + tanα = + = = = = sinα cosα sinαcosα sin( α) / sin( α) 35. Ans E: ( + )( ) log = log = log( + ) + log( ) logz z z 36. Ans C: The surface area of a clinder of radius r and height h equals S = π rh+ π r. Therefore πrh + πr = πh rh + r = 6h r + rh 6h =0. The last h± h 4 ( 6 h ) h± 5h equation has two solutions: r = =, but the onl h+ 5h 3 positive solution is r = = h. ThereforeV = π r h = 4π h.

6 ( n ) Ans E. Each interior angle of a regular n-gon equals = 74 n 80n 360 = 74n 6n = 360 n = 60. Hence, p =.5 60 = 75 (inches). 38. Ans D: Knowing BC gives us SAS, which determines the triangle. The other two do not give enough information. On a side note, knowing the measure of angle BAC would have determined the triangle, but it was not an option Ans C: π π ( ) 3 V =.5t = r h = d d = π d. Rearrange to get The derivative w.r.t. t is 3 30t 30 d' = 3 π π. When t = 60, d' = t d = π. 40. Ans B: Let and be the lengths of the two diagonals. We need to find + = +. Each side of the rhombus equals 5. According to the Pthagorean theorem, + = 5. On the other hand, the area of the rhombus equals + = = 39. Adding these two equations, we get + + = 5+ 39= 64. Therefore, ( + ) = 64 and + =

Academic Challenge 2009 Regional Mathematics Solution Set. #2 Ans. C. Let a be the side of the cube. Then its surface area equals 6a = 10, so

Academic Challenge 2009 Regional Mathematics Solution Set. #2 Ans. C. Let a be the side of the cube. Then its surface area equals 6a = 10, so Academic Challenge 009 Regional Mathematics Solution Set #1 Ans. C: x 4 = x 9 = -5 # Ans. C. Let a be the side of the cube. Then its surface area equals 6a = 10, so a = 10 / 6 and volume V = a = ( 10 /