Celebrating International Pi Day 2014
Worksheet for primary students
The Australian Mathematics Trust wishes every student a happy International Pi Day.
Please enjoy these activities in your classroom and celebrate the fun of mathematics
with the world.
International Pi Day is celebrated on March 14th (3/14)
and Pi (Greek letter “π”) is the symbol used in mathematics
to represent a constant — the ratio of the circumference
of a circle to its diameter — which is approximately 3.14159.
As Pi is a constant number, meaning that for all circles of
any size, Pi will be the same. The number Pi is extremely
useful when solving geometry problems involving circles.
radius
diameter
Originating in the US, International Pi Day started out as a
celebration of one of the most recognisable mathematical
symbols but has grown the world over as a day to embrace,
share and enjoy mathematics.
area
circumference
History of Pi
By measuring circular objects, it has always turned out that a circle is a little more
than three times its width around. The mathematician Archimedes used polygons with
many sides to approximate circles and determined that Pi was approximately 22/7. In
Australia, Pi Approximation Day is celebrated due to our date format of day/month. The
symbol (Greek letter “π”) was first used in 1706 by William Jones. A ‘p’ was chosen for
‘perimeter’ of circles, and the use of π became popular after it was adopted by the Swiss
mathematician Leonhard Euler in 1737.
Remembering
Most of the time, Pi is recognised as 3.14 but for additional accuracy 3.14159 is used. A
good way to remember this longer number is to count the number of letters in each word of
this phrase.
May I have a large container of butter today
3 1 4 1 5
9
2 6 5
For more information on International Pi Day visit www.piday.org
©2014 Australian Mathematics Trust www.amt.edu.au
1
About the Australian Mathematics Trust
Australian Mathematics Trust is a national non-profit organisation whose purpose is to
enrich the teaching and learning of mathematics and informatics for all students. The
vision of the AMT is to challenge and encourage Australians in the understanding of
mathematics and informatics through competitions and original enrichment programs.
The best known activity of the Trust is the Australian Mathematics Competition (AMC)
sponsored by Commonwealth Bank which is the original mathematics competition in
Australia. The AMC began in 1978, and attracts hundreds of thousands of entries annually,
from Australia and overseas. Many of Australia’s leading mathematicians aged below 40
were identified and developed as a result of taking part in the competition. Entries are now
open for the 2014 Australian Mathematics Competition held on Thursday 7 August.
For more information about the AMT visit www.amt.edu.au.
©2014 Australian Mathematics Trust www.amt.edu.au
2
Pi Problems
Whilst questions involving Pi are not in the primary curriculum, they relate to area,
perimeter, spatial thinking and circles. Here are some questions on these topics which
you might like to try. They involve some careful thinking and some may be best done
by practical experiments, perhaps working in pairs or small groups.
1. Which shape has the largest area?
(A)
(B)
(D)
(C)
(E)
2. A large square contains two small squares with areas 25
square centimetres and 4 square centimetres. What is
the perimeter, in centimetres, of the shaded area?
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3. The perimeter of this rectangular paddock is 700 m. It is subdivided into six
identical paddocks as shown.
The perimeter, in metres, of each of the six smaller paddocks is
(A) 116
1
3
(B) 300
©2014 Australian Mathematics Trust www.amt.edu.au
(C) 200
(D) 150
(E) 600
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4. A square piece of paper is folded along its centre line and then folded again as
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A cut is made along the dotted line in the third diagram. When the paper is
unfolded, it now has a hole in the centre. Which of the following could be the
shape of this hole?
(A)
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(E)
5. A rectangular sheet of paper is folded in
half and then folded in half again. A piece
is cut out of the folded paper as shown.
The sheet is then smoothed out to its
original size again.
Which one of the following could it be?
(A)
(B)
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©2014 Australian Mathematics Trust www.amt.edu.au
(C)
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6. A square piece of paper is folded along its centre line and then folded again as
shown.
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A cut is made along the dotted line in the third diagram. Which of the following
could be the shape of the paper when it is unfolded?
(A)
(B)
T
(C)
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(E)
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7. An apartment block has a number of square apartments and a number of square
gardens. Apartments must have at least one window, either to the outside or to a
garden. In figure 1, one apartment has a window to an internal garden G, and ten
have windows to the outside.
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figure 1
figure 2
What is the smallest number of gardens needed for an apartment block built on a
6 × 6 square, as in figure 2, so that each apartment has a window to the outside
or to an internal garden?
©2014 Australian Mathematics Trust www.amt.edu.au
5
4
8. In a circle dance, everyone is evenly spaced around a circle and has a number in
the order 1, 2, 3, 4, 5, . . . , and so on. The dancer with number 15 is directly
opposite dancer number 3. How many dancers are in the circle?
(A) 18
(B) 20
(C) 22
(D) 24
(E) 26
9. Two identical coins are placed one inside and one outside a thin circular wire,
which has twice the diameter of each coin. Each coin is rolled, without slipping,
around the wire until it returns to its starting point.
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⇑
If the direction of rotation does not matter, what is the difference, in degrees,
between the angle through which the outer coin rotates and the angle through
which the inner coin rotates?
10. Four 10 c coins lie on a table as shown. Keeping in contact with the other three
coins, the shaded coin is rolled around the other three coins until it returns to its
starting place.
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Through what angle does the shaded coin turn, on its axis, in rolling once around
the other three coins?
(A) 360◦
(B) 540◦
©2014 Australian Mathematics Trust www.amt.edu.au
(C) 720◦
(D) 900◦
(E) 1080◦
6
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Solutionsand
and Teachers’
Notes
Solutions
Teachers’
Notes
These questions vary in difficulty but have been chosen because they have the potential to
lead to some class discussion about the underlying ideas. As well as giving the solutions,
I have made some suggestions about possible discussion which might arise from each
question or group of questions.
1. The shapes have areas 4, 4, 5, 5 and 6 respectively.
Although this is a very simple question, it highlights that the largest area and
the largest perimeter do not necessarily come from the same shape. Working with
polyominoes like these can lead to some interesting investigations, such as which
polyominoes of a certain size have the largest perimeter or the reverse question, for
a given perimeter, what is the largest area you can get. Students can work with
squares (I often use carpet tiles) to explore these questions.
2. Since the large square contains two small squares with areas 25 square centimetres
and 4 square centimetres, the sides of these squares are 5 cm and 2 cm as shown.
5
2
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5
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7
This means that the side of the largest square is 7 cm and we can place the other
measurements, in centimetres, as shown.
The perimeter of the shaded region is then 2 + 5 + 7 + 2 + 5 + 3 = 24 cm.
Another straightforward question but it inovlves several simple steps. At first it
may appear that you do not have enough information, so persistence is rewarded.
The next question is similar.
3. Call the length of each of the smaller rectangles l and the width b. We can see
from the diagram on the right, that the length of the smaller rectangles is twice
the width.
perimeter = 700
b
l
b
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l
b
b
l
l
The perimeter of the large rectangle is then equivalent to 14 lots of the width of
the smaller rectangles. As the perimeter of the larger rectangle is 700 m, 14 times
©2014 Australian Mathematics Trust www.amt.edu.au
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the breadth of the smaller rectangles is 700 m, and so the breadth of the smaller
rectangles is 700 ÷ 14 = 50 m.
The perimeter of each of the smaller rectangles is then 50 + 100 + 50 + 100 = 300 m.
The next three questions involve folding and cutting. This can be quite hard for
students (and teachers!)to visualise and these questions are often not done well.
The rules of the AMC allow primary students to use any implements which are
part of the normal classroom, so why not get students to actually do the folding
and cutting to see what happens? This will undoubtedly convince them more than
the solutions bvelow. such activties are terrific for developing spatial awareness
and appreciation of symmetry.
4. When we look at the cut, it is clear that as the paper has 4 layers in the fold, so
there will be 4 cuts of equal length.
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The only shape shown with four equal sides is (D).
5. Since the paper is folded twice with the cut not going to any edge, the resulting
cut-out must be two identical figures. Flattening the given figure out gives the
Σ-like shape, so the result is two of these,
hence (D).
Comment: If the two folds are at right angles, the result is a Σ-like shape with its
mirror image either side, which was not an alternative.
©2014 Australian Mathematics Trust www.amt.edu.au
8
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6. As the cut contains corner S and not corner T, the shape cannot be alternatives
(C), (D) or (E).
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Now, since the right-angled cut goes through the folded paper, there
(A)
(B)
T
(C)
T
(E)
(D)
S
S
must be two right angles in the cut when unfolded, so when unfolded will be (B).
7. Consider the 6 by 6 block as shown. All the outside apartments have exterior
views, so we need only deal with the 16 interior squares.
figure 1
8
Putting in one garden will remove one apartment and bring light to (at most) four
other apartments.
G
©2014 Australian Mathematics Trust www.amt.edu.au
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other apartments.
G
This complying arrangement will deal with at most five of the internal squares and
so, as 3 × 5 < 16, at least four gardens are required.
It is relatively easy to find such a solution with 4 gardens, as shown in figure 2.
G
G
G
G
figure 2
So the smallest number of gardens required is 4
A trial and error approach will convince many students of the answer but it is
interesting to get students who arrive at this answer to prove that it is a minimum.
This is very hard for young students and should provide some interesting discussion.
8. As dancer number 3 is directly opposite dancer 15, the line joining those two is a
diameter of the circle.
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4•
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This means that dancers numbered 4 to 14 are above the diameter and by symmetry
there will be the same number below the diameter.
The number above the diameter is 14 − 4 + 1 = 11. So there are the same number
below plus dancer 3 and 15, so there are 11 + 11 + 2 = 24 dancers altogether.
©2014 Australian Mathematics Trust www.amt.edu.au
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Students often have difficulty with questions involving counting when they need to
deal with end-points which need to be inlcuded or excluded. A diagram explaining
their thinking will certainly help.
The last two questions involve circles rotating round each other. They have both
proved to be very hard for many students, though some students of all ages can
visualise what happens quite easily. The best way to demonstrate the solutions is
by a prctaical experiment. If you have spirograph cogs of appopriate size, this will
be a great help. With coins or other circles, care must be taken in not allowing
them to slip as you rotate them.
9. As the diameter of each of the coins is half the diameter of the wire circle, the
circumference of the wire circle is twice that of each coin.
Consider the point P common to all three circles in the initial position. As we roll
each of the coins around the larger circle, the points P will then coincide at the
bottom of the larger circle as shown, as the the circumference of each of the coins
is half that of the larger wire circle.
As we rotate the outer coin, consider the point Q on the opposite side of its diameter
to P . This point Q will meet the larger circle halfway down as shown, and at this
stage this coin will have rotated 270◦ to get to this position, and so will rotate 540◦
to get to the position shown at the bottom. So when it completes the rotation to
get back to its starting position it will have rotated 2 × 540 = 1080◦ .
Q
•
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Q
Consider the point Q opposite P on the inner coin. When we rotate this coin around
the inside of the larger circle, the point C will meet the larger circle halfway down.
At this stage the inner coin will have rotated 90◦ , so when it gets to the bottom it
will have rotated 180◦ and so will rotate 360◦ to get back to its starting position.
The difference, in degrees, between the rotations of the two coins is then 1080 −
360 = 720.
©2014 Australian Mathematics Trust www.amt.edu.au
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10. In rolling coin 1 around the coin to its left from position 1 to position 2, the point
Y on coin 1 rolls around the circle to go to the point Y in position 2, and X goes
to the point X in 2.
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X
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So coin 1 rotates through an angle of 360◦ when it rolls from position 1 to position
2. Similarly, it will rotate another 360◦ in rolling from position 2 to position 3, and
another 360◦ in rolling from position 3 to position 1.
This means that it rolls through an angle of 3 × 360◦ = 1080◦ to get back to its
original position (and it will also have the same orientation).
©2014 Australian Mathematics Trust www.amt.edu.au
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