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
The area of a particular regular hexagon is x3 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
length s. We are told the area of the hexagon is x3 square units, so we can write
6 × (s2√3)/4 = x3 → (3√3/2)s2 = x3. 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)s2 = [(√3/2)s]3
→ (3√3/2)s2 = (3√3/8)s3 → 8(3√3/2)s2 = (3√3)s3 → [(12√3)/(3√3)]s2 = s3 → 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
