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Solutions to mathematical problems from the mathcounts problem of the week archive for the year 2012. The problems cover various topics such as geometry, probability, and algebra. Students and educators can use these solutions to check their understanding of the concepts and to prepare for exams.
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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 × (s^2 √ 3)/4, where (s^2 √ 3)/4 represents the area of an equilateral triangle with side length s. We are told the area of the hexagon is x 3 square units, so we can write 6 × (s^2 √ 3)/4 = x^3 → (3 √ 3/2)s^2 = x^3. 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 substitute this expression for x in the equation above for the area of the hexagon to get (3 √ 3/2)s^2 = [( √ 3/2)s] 3 → (3 √ 3/2)s^2 = (3 √ 3/8)s^3 → 8(3 √ 3/2)s^2 = (3 √ 3)s^3 → [(12 √ 3)/(3 √ 3)]s^2 = s^3 → s = 4 units.
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/ What is the probability that we choose a cube with 1 side painted? 57/ What is the probability that we choose a cube with 2 sides painted? 28/ What is the probability that we choose a cube with 3 sides painted? 4/
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/ What is the probability that a cube with 2 sides painted has a painted side on the top when rolled? 2/ What is the probability that a cube with 3 sides painted has a painted side on the top when rolled? 3/
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/.
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:
− (^) + + + +
2 2 2 2
2 2 2 2 2 2 2
c x (^) x c xx
cx c x x cx c x x
cx c x x = cx c x x x x = x x x x = x x
So, x + 3 = 0 and x = −3. But x can’t be negative. So , we have 2x – 4 = 0 → 2x = 4 → x = 2.
Let the three integers be x – 1, x and x + 1. We have
− + − +
x x x x x x x x x x x x x x x x x x x x x x x x x = x = It follows that x – 1 = 4 – 1 = 3 is the smallest of the three integers.