Problem of the Week Archive
Some Pi to Celebrate St. Patrick’s Day! – March 17, 2014
Problems & Solutions
Originally posted 3/14/11
This four-leaf clover consists of four coplanar circles. Each circle is externally
tangent to two others, as shown. The two smaller circles are congruent, and the
two larger circles are congruent. A square is constructed such that each of its
four vertices is also the center of one of the four circles. Suppose the area of
the square is 6 ¼ in2, and the combined area of the four circular leaves of the
clover is 6 ½  π in2. What is the absolute difference between the radius of
a large circular leaf and the radius of a small circular leaf? Express your
answer as a common fraction.
The area of the square is 6 ¼ in2. If we let the side of the square be s, then s2 = 25/4 → s = 5/2. Let us denote the radius of
the large and small circular leaves as r1 and r2, respectively. We also note that s = r1 + r2 → r2 = s − r1 → r2 = 5/2 − r1. We are
told that the combined area of the four circular leaves is 6 ½  π in2. Since the two large circles are congruent, as are the two
small circles, we have 2π(r1 )2 + 2π(r2 )2 = (13/2)π → 2π( r1 )2 + 2π(5/2 − r1 )2 = (13/2)π → 2π[(r1 )2 + (5/2 − r1 )2] =
(13/2)π → (r1 )2 + (5/2 − r1 )2 = 13/4 → (r1 )2 + 25/4 − 5r1 + ( r1 )2 = 13/4 → 2(r1 )2 − 5r1 + 3 = 0. If we factor this quadratic
equation we get (2r1 − 3)(r1 – 1) = 0. So, r1 = 3/2 or r1 = 1. That means if r1 refers to the small circle its value is 1 inch. If,
as it does in our case, r1 refers to the large circle then its value is 3/2 inches. So, if r1 = 3/2 then r2 = 1. Therefore, the
absolute difference between the radius of a large circular leaf and the radius of a small circular leaf is 3/2 – 1 = 1/2 inch.
Another four-leaf clover also consists of 4 coplanar circles. The large
circular leaves are externally tangent to each other, as well as to each of
the smaller circular leaves, which are also congruent to one another.
The radius of the large and small circular leaves is 1 ½ inches and
1 inch, respectively. What is the area of a rhombus formed such that
each of its vertices is also the center of one of the four circular leaves?
One way we can determine the area of the rhombus is by taking half the product of its diagonals. The radius of each large
circular leaf is 3/2 inches. That makes the length of the shorter diagonal 3 inches. If we include
this diagonal in the figure we can see that two isosceles triangles are formed. From the figure
5
3
⁄2”
⁄2”
we can see that the length of the altitude of one of these triangles is equivalent to 1/2 of the
n
length of the longer diagonal. The length of the altitude we’ll call n. We can also see that this
altitude is the long leg of a right triangle whose hypotenuse is equal to 1 + 3/2 = 5/2 inches
3
⁄2”
and whose shorter leg is 3/2 inches. If we apply the Pythagorean Theorem we get (5/2)2 =
n2 + (3/2)2 → 25/4 = n2 + 9/4 → n2 = 16/4 → n2 = 4. So, n = 2, since n represents a
measurement and must be positive. The length of the longer diagonal is 2(2) = 4 inches. Now we can see that the area of the
rhombus is (1/2)(3)(4) = 6 in2.
Albert has a home screen printing business. He has designed a bracelet and T-shirt for
St. Patrick’s Day which both use a special ink, available for a limited time from his supplier.
Albert has already begun taking and fulfilling orders for merchandise. Albert noticed that
457.5 ml of the special ink was used to make 25 bracelets and 8 T-shirts, and he used
another 712.5 ml of ink to make 14 bracelets and 18 T-shirts.
Albert hopes he’ll have enough ink to fulfill an order for 42 bracelets and 21 T-shirts since
his printer indicates that he has 1 L of the special ink remaining. If Albert will have enough
ink, how many milliliters of ink will be left after fulfilling the order? If, on the other hand,
Albert does not have a sufficient amount of ink to fulfill the order, by how many milliliters is
he short? Express your answer as a decimal to the nearest hundredth.
Let x and y be the amount of ink used to make a T-shirt and a bracelet, respectively. Then 8x + 25y = 457.5 and 18x + 14y =
712.5. If we multiply the first equation by 9 and multiply the second equation by −4, we can add the two equations and solve
for y. Doing so yields (72x + 225y = 4117.5) + (–72x – 56y = −2850) → 169y = 1267.5 → y = 1267.5/169 → y = 7.5 ml.
Now we can substitute 7.5 for y in either of the two original equations and solve for x. The result when 7.5 is substituted into
the first equation is 8x + 25(7.5) = 457.5 → 8x + 187.5 = 457.5 → 8x = 270 → x = 33.75 ml. So, we know that 7.5 ml of ink
is used to make one bracelet and 33.75 ml of ink is used to make one T-shirt. For an order of 42 bracelets and 21 T-shirts,
Albert will use 42(7.5) + 21(33.75) = 1023.75 ml of ink. Since 1 L of ink is equivalent to 1000 ml, Albert will not have
enough ink to fulfill this order. In fact, Albert will be short by 1023.75 – 1000 = 23.75 ml of ink.