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                                 MATHCOUNTS  Problem of the Week Archive 
                                                Best of 2012 – January 7, 2013 
                                                                         
                 Problems & Solutions 
                 As we begin a new year, we decided to take a look back at some of our favorite problems from 2012… 
                 2012 School Sprint #30 
                                                               3
                 The area of a particular regular hexagon is x  square units, where x is the measure of the distance from 
                 the center of the hexagon to the midpoint of a side. What is the side length of the hexagon?
                                                                                                                  
                  
                 A regular hexagon can be divided into six congruent equilateral triangles, as shown. 
                 Therefore, the area of the hexagon is the sum of the areas of these six triangles, or 
                 6 × (s2√3)/4, where (s2√3)/4 represents the area of an equilateral triangle with side 
                                                                   3
                 length s. We are told the area of the hexagon is x  square units, so we can write 
                 6 × (s2        3           2   3
                       √3)/4 = x  → (3√3/2)s  = x . We know that the height of one of these triangles is 
                 the distance from the center of the hexagon to the midpoint of a side, which we are 
                 told is x. We can determine the side length of the hexagon using the properties of 30-60-90 right triangles. 
                 The drawing the altitude of the equilateral triangle creates two 30-60-90 right triangles, with hypotenuse of 
                 length s, shorter leg of length s/2, and longer leg of length (√3/2)s. So, we have x = (√3/2)s. We can 
                                                                                                                  2           3
                 substitute this expression for x in the equation above for the area of the hexagon to get (3√3/2)s  = [(√3/2)s]   
                            2          3            2        3                  2   3
                 → (3√3/2)s  = (3√3/8)s  → 8(3√3/2)s  = (3√3)s  → [(12√3)/(3√3)]s  = s  → s = 4 units. 
                  
                  
                 2012 Chapter Target #7 
                 A 5 × 5 × 5 wooden cube is painted on exactly five of its six faces and then cut into 125 unit cubes. One 
                 unit cube is randomly selected and rolled. What is the probability that the top face of the unit cube that 
                 is rolled is painted? Express your answer as a common fraction. 
                  
                 A 5 × 5 × 5 cube is painted on 5 of its 6 faces. It is then cut into 125 unit cubes. One unit 
                 cube is randomly selected and rolled. We are asked to find the probability that the top 
                 face of the cube that is rolled is painted. Assume that the top of the cube, as it sits on a 
                 surface, is the one face that is not painted. Then, of the 25 cubes on the top level, we have 
                 9 (in grey) that have no sides painted, 12 (in white) that have one side painted and 4 (in 
                 black) that have two sides painted as shown.  
                  
                 The same holds for the second, third and fourth level. On the fifth level, which is the bottom, all cubes have 
                 their bottom painted and some have some of their sides painted (but not their tops). In particular, using the 
                 same image above, 4 cubes have 3 sides painted (in black), 12 (in white) have 2 sides painted and 9 (in 
                 grey) have 1 side painted. Let’s total everything up. The number of cubes that have 0 sides painted is 
                 (9 ×
                      4) = 36. The number of cubes that have only 1 side painted is (4 × 12) + 9 = 57. The number of cubes 
                 that have 2 sides painted is (4 × 4) + 12 = 28. The number of cubes that have 3 sides painted is 4. Let’s just 
                 do a check here: 36 + 57 + 28 + 4 = 125. Okay, we’re good.  
                  
                 What is the probability that we choose a cube with no sides painted? 36/125  
                 What is the probability that we choose a cube with 1 side painted? 57/125  
                 What is the probability that we choose a cube with 2 sides painted? 28/125  
                 What is the probability that we choose a cube with 3 sides painted? 4/125  
                  
                     Now, let’s look at probabilities that a painted side will come up when the cube is rolled.  
                     What is the probability that a cube with no sides painted has a painted side on the top when rolled? 0 
                     What is the probability that a cube with 1 side painted has a painted side on the top when rolled? 1/6 
                     What is the probability that a cube with 2 sides painted has a painted side on the top when rolled? 2/6  
                     What is the probability that a cube with 3 sides painted has a painted side on the top when rolled? 3/6  
                      
                     Finally, let’s put it all together. The probability that one randomly chosen cube is rolled and the top face is 
                     painted is: (36/125) × 0 + (57/125) × (1/6) + (28/125) × (2/6) + (4/125) × (3/6) = (57 + 56 + 12)/(125 × 6) = 
                     125/(125 × 6) = 1/6. 
                      
                      
                     2012 State Sprint #28 
                     If the cost of a dozen eggs is reduced by x cents, a buyer will pay one cent less for x + 1 eggs than if the 
                     cost of a dozen eggs is increased by x cents. What is the value of x? 
                      
                     Let c be the cost of a dozen eggs in cents. Then the cost of one egg is c/12 cents. If we reduce that price 
                     by x cents, the cost of one egg is (c – x)/12. If we add x cents, then the cost of one egg is (c + x)/12. So, we 
                     have the following:  
                                −+
                             cx                 cx
                                   (xx++1)   1 =       ( +1)
                               12                 12
                         22 22
                      cx     c x       x         cx     c x      x
                           +−−+12 =                   +++
                                12                      12
                         22 22
                      cx     c x       x      =cx      c x       x  
                           +−−+12                    +++
                                   22
                                 x x          =x x
                                −−+12               +
                                 2
                              2xx+−2      12 =0
                               xx
                              ( +3)(2 −4)= 0
                     So, x + 3 = 0 and x = −3. But x can’t be negative. So, we have 2x – 4 = 0 → 2x = 4 → x = 2. 
                      
                      
                     2012 National Sprint #15 
                     The sum of the reciprocals of three consecutive positive integers is equal to 47 divided by the product of 
                     the integers. What is the smallest of the three integers? 
                      
                     Let the three integers be x – 1, x and x + 1. We have  
                                         1      1     1            47
                                             ++  = 
                                          −+−+
                                           1            1 (     1)( )(    1)
                                        x xx x xx
                       (  +1)+( +1)( −1)+( −1)
                      xx         x      x        xx  =             47
                               (  −+1)( )(    1)            (  −+1)( )(    1)
                                x xx                         x xx
                                 222  
                                   ++−1+−= 47
                                xxx xx
                                                    3x2 = 48
                                                       2    16
                                                      x=
                                                            4
                                                        x=
                     It follows that x – 1 = 4 – 1 = 3 is the smallest of the three integers.