Celebrating International Pi Day 2014
Worksheet for secondary students
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.
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. 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 problems
involving circles, but also appears in many other applications of mathematics.
radius
diameter
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.
area
circumference
About Pi
Pi is an infinite non-recurring decimal which means Pi has infinitely many numbers to the right of the
decimal point and no repeating pattern. As a number that cannot be written as a repeating decimal or a finite
decimal (you can never get to the end of it) pi is irrational. Pi shows up in some unexpected places including
the 'famous five' equation connecting the five most important numbers in mathematics, 0, 1, e, π, and i,
known in mathematics, as Euler's identity eiπ + 1 = 0.
History of Pi
By measuring circular objects, it has been known for a long time that the circumference of a circle is a
little more than 3 times its diameter. 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
31415
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
The 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 the
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
38
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8
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P
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(D) a22
(E) a
...............................................
(D) a
(E) 2
............... ..................... .....
.........................................................
....... . . ..........................
2
......................................
.....
Questions geometry — circles
73
Questions geometry — circles
69
8
2000 I.28 (9%), S.27 (9%)
Questions geometry — circles
69
8
2000 I.28 (9%), S.27 (9%)
5
◦
Each of the four chords in the diagram
Each of the four chords in the diagram
cuts the area of the outer circle in the
cuts the area of the outer circle in the
ratio 1 : 3. The points of intersection of
ratio 1 : 3. The points of intersection of
these chords are the vertices of a square.
these chords are the vertices of a square.
The chords cut the circle into 9 regions.
The chords cut the circle into 9 regions.
The ratio of the area of region P to the
The ratio of the area of region P to the
area of the circle passing through the
area of the circle passing through the
vertices of the square is
vertices of the square is
(A) 1 : 4
(A) 1 : 4
√
(B) 1 : √2
(B) 1 : 2
(C) 1 : 2
(C) 1 : 2
©2014 Australian Mathematics Trust www.amt.edu.au
◦
...........................................
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P
P
(D) 1 : π
(D) 1 : π
(E) 1 : 2π
(E) 1 : 2π
3
Solutions
and Teachers Notes
Alternative 2
Q,Competitions.
S
These
questionsofhave
previous
AustralianTMathematics
The first should be
The position
S iscome
not from
specified
in the
4
.
.
......
...
.....
.
.
..
.
.
.
question
so may
be assumed
to learnt
coin- the area .....formula for
....a circle. The next two are a little more
.
.
quite
easy and
for most
students
who have
.
.
....
.
....
......
...
.....
cide with Q.but
Thus
the area
of �RQT
equals
..
...
..
......
challenging
require
no
algebra.
The
last
two
are
difficult,
requiring some spatial insight into how
.
.
.
.
...quite
..
1
1
.....
..
6
2
.....
...
.....
..
T
Q
×
QP
=
×
4
×
6
=
12cm
,
.
..
.
.
.
.
.
.. is not complicated, but careful thought is required to
shapes
can be added
or subtracted. The algebra involved
......
...
2
2
..
.... ..........
...
.. ...
(A)
... .......
........
establish the combination of areashence
required.
......
.
...
R
P
CIRCLES
11
It is clear that the width of the rectangle is twice the radius of the circle,
or 2cm, and the length is four times the radius, or 4cm. The shaded
4
2000 J.16 (60%)
area is the area of the rectangle minus 2 times the area of the circle, i.e.
Alternative
(4 × 2 − 2 ×1π × 12 )cm2 = (8 − 2π)cm2
, x x x ... .....
...
....
...... ....
.......
....
The figure is a parallelogram
hence (B)
....
..
.. ......
....
....
.
.
..
....
...
with base 2x and height 12.
..
..
....
....
.
.
....
.
.
.
....
....
...
.. 12
..
....
....
Its
..
.... ...
....
2 area is then 2x × 12 = 24x.
...
.... ..
...
....
.... ......
.
.
.
.
.
.
...
.... ..
So
24x
=
408
and
x
=
17,
.. ..
P
After making the constructions shown
...
...
hence (D).
...
and denoting the circle radius as r we
..
.
C
IRCLES
4 ...... Alternative
2 QOS, QOT are congruhave
that �s
.
............... .
.......... .... ........
..
......
... x,......with
..... V
The (RHS),
figure consists
of two
12 and.........base
area
ent
therefore
QT triangles,
= QS = height
4.
.
.......
.
..1998
T ............. ....... ...... ..
1
I.11 (33%)
2Similarly,
...
1
...
.
.
.
.
.
P
V
=
P
T
=
4
and
V
R
=
O
.
..
.... So
.. = 408 and
2 × × x × 12 = 12x, and a rectangle with area.... 12x.
..
................. 24x
.
.
.
.
.
.
.
.
.
.
. .
.
2radius
The=
of the circular
arcs is 4is units
...... .. .....of
........ the
.. straight
SR
7. Therefore
the perimeter
30, and the
...... ..
....... .... ......... length
......
........
x = 17,
.......
......
edges
is 4 units.
....... hence (A)
.................................... r .... .............hence
.......
.......
....... (D).
.
.......
....
.......
.......................................
There are 8 straight edges and 3 full circular........arcs,
so
4
7
Q
S
R
the perimeter = (8 × 4) + (3 × π × 8) = 24π + 32,
2000 I.8 (37%)
5
hence (B).
3
The shaded area is the difference between two right-angled
.................triangles
.............
..............
.
.
........
.
.
.
.
.
......
.....
......
..... (37%)
Let
and about
z be the
as shown
the24 and .....5.............and
with
theangles
right angle
of 10inand
12 1997
respectively.
2 ysides
I.13
....
...
.
..
...
.
◦
.
1
...
diagram.
Since
y
is
subtended
from
the
......
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.............
....... units,
.
.
.
.
.......
.
.
The shaded area is then (10 × 24 − ◦5 × 12) = ..90
.
.
.
.
.
.
.
... .......... square
.
...◦
............... .....
...........y
.......◦
.
.
.
x
......
... ...... ....
2subtends 44
.
The track
one semi-circle
of rasame
chordconsists
as thatofwhich
,y=
.
.
.
.
.
....
.
.
.... ..
.. ........
....
.
.... ..
.....
.....
....
..
.....
hence
(E).
......
.....
.....
......
......
dius
300, andz three
radius
44. Clearly,
= 180semi-circles
− 96 = 84.ofThus
x = ..................
..
..... z ◦ ..........
...... .....
....
... ...
............
.. ..
....
.
◦
.. ....
.......... 96
. ...
100.
180 − 44 − 84 = 52,
.
.
.
.
.
.
.
.
..........................
..
.
.
.
.. ....
.
.
.
.
.
........... I.13 (22%)
..
... ...
3 total length of the track is hence (C) ..... ......
6
2000
..
.........
..
..... ....
..
The
......
..... ..
..
...
...
...... .....
.....
..
.
.
.
.
.
.
.
.....
.
.
.
.
...
.. ........ ....
....
..◦
... ..........
.
...
.
.. ............ ..
. ..
....... 44
300π
+
3
×
100π
=
600π,
.
.
...
.
...
.
.
.
.
.....
.. .... . .......
.......... ..
..
......
The path has the shape of a rectangle with quarter
...
.
.
.
.
.
.
.
........
... ..
...... ..... ..
.
...
.
.
.
.
.
........... .. .. ..
.... .. .. ...
........
.
hence
(E).
.
...
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.....................................
.....
....
circles, as shown in the diagram.
....
....
........
.....................
........
.....................
The area of the path is then
20
3
1997 I.22 (21%),
50 S.19 (26%)
(2 × 50 × 2) + (2 × 20 × 2) + (π × 22 )
.... .. .. ...
... .. ....
... ..
... ....
.
...... .............R
.......................
...........
......
.
.
.
.
.
= 280 in
+ 4π
.
.
Draw
the construction lines as shown, where
.
.
.
.
.
.
.
.
........
.....
.
.
.
.
.
.
.
.
.
...
...
Solutions
geometry
—
circles
205
...
.
.
...
L is the centre of the circular arc and LR is the
M
30 ....... N (A).
...
...
hence
...
...
.
.
...
vertical radius.
x .......
...
...
...
...
...
...
Then LR = LN = 50.
60 cm
... ..... 50
... ..
......
Let LM = x.
L
In the right angled ΔLM N
60 cm
LN 2
©2014 Australian Mathematics Trust
=
LM 2 + M N 2
502 = x2 + 302
x2 = 402
x = 40
www.amt.edu.au
M R is then 10 cm, so the maximum height is 60 + 10 = 70 cm,
hence (A).
Solutions geometry — area
213
4
477
2000 I.19 (5%)
2000 I.19 (5%)
Alternative 11
Alternative
The area
area of
of the
the four
four lunes
lunes is
is the
the area
area
The
of
four
semi-circles,
diameter
a,
less
that
of four semi-circles, diameter a, less that
portion of
of the
the area
area of
of the
the larger
larger circle
circle
portion
which
is
outside
the
square.
which is outside the square.
The area
area of
of the
the four
four semi-circles
semi-circles is
is
The
2
πa22
11 × π × aa2 = πa
4
×
4 × 2 × π × 4 = 2 ..
2
4
2
By Pythagoras,
Pythagoras, the
the diameter
diameter of
of the
the larger
larger
By
2
√
πa
2
√
circle is
is aa 22 and
and its
its area
area is
is πa ..
circle
22
..........................
........ ....................
..........................................................
............................................................
....................................................................................................
.
...............................................................................
.................................. ..............................
..............................................................................................
........ ..........
......................
.
.
.
.
.
.
.
.
.
........... ............
.
.
.
.
.
.
........................
......... . .....
... .......................
....... ....................
........................
.................................
.............................
.......................................
.
............................
...........................
.
.........................
. . ..... .
............................
...........................................
.........................
.................................
............................
.........................................
.........................
...............................
.............................
...........................
.................................
.............................................
.
.
.
.
.
.
.
. .
.
.. .........................
......................................
.........................
.
...................
.........................
.........................................
.................... ...
.
.
.
..... ......................
......... . .. .....
..........................
................ .....
.......... ..............
....... ..........
.........................
............... ......................................................................
.................................................................................
.. ........................................................................
.................................................... ...
......................................................................
...........................................................
.......................................
................................
πa22
πa22
2
2
πa
+
a
The
required
area
is
then
2 − πa = a2 ,
The required area is then 2 + a − 2 = a ,
2
2
hence (D).
(D).
hence
Alternative
2
Alternative 2
The total
total area
area is
is the
the central
central square
square (a
(a22)) plus
plus four
four semicircles
semicircles
The
1
a 2
1
1
2
2
(4
×
)
=
πa
),
that
is
a
(1
+
π).
12 π( a
1
1
2
2
2
2 π).
(4 × 2 π( 22 ) = 22 πa ), that is a√(1+
a√ 2 22 2 1 2
=
It
is
also
the
inner
circle
(π
a 22
1 πa2 ) plus four lunes, so the area
It is also the inner circle
2 (π 1 2 1 = 222πa ) 2plus four lunes, so the area
(1 +
+ 12 π)
π) −
− 12 πa
πa2 =
= aa2 ,,
of the
the four
four lunes
lunes is
is aa2 (1
of
2
2
hence (D).
(D).
hence
8
2000 I.28 (9%), S.27 (9%)
5
8
2000 I.28 (9%), S.27 (9%)
Let the
the areas
areas of
of the
the regions
regions be
be P
P ,, Q
Q and
and S
S as
as shown
shown and
and the
the area
area of
of the
the
Let
larger
circle
be
A.
Since
each
chord
divides
the
area
of
the
circle
in
the
larger circle be A. Since each chord divides the area of the circle in the
ratio 11 :: 3,
3, we
we get
get
ratio
1
............................................
.......
.........
4P + 2Q =
A
(1)
.......
.....
....
.....
2
....
.....
.
.
.
....
.
Q
...
...
.
...
1
.... .... .... .
.. P
.
.
.
.
.
... . P
.
...
..
.
.
.
.
...
2Q + S =
A
(2)
..
. ...
...
...
.
..
.. .
..
...
2
.
...
.
.
...
.
..
r
.
.
.
...
..
...
.
...
.
...
..
.
(1)−(2)
4P − S = 0
...
.
.
...
...
..
....
x
..
...
.
Q
S
Q
.
.
....
..
...
..
.
.
.
.
.
1
..
...
.
...
.
.
.
.
r ..... ...
...
..
thus P =
S
...
..
..
....
...
...
..
..
....
4
..
...
.... .... .... .... ...
..
...
.
.
....
1 2
P...........
....P
....
Q
.....
..
=
x ,
........
......
.
.
.
.
.
.
.
.
.
.
............
4
..................................
214 x is the side of the square S.
geometry —
— area
area Solutions
Solutions
where
214
geometry
Now, if r is the radius of the smaller (dashed) circle, then 2r 2 = x2 .
π
The area of the smaller circle is then πr 2 = x2 .
2
The ratio of the area P to the area of the smaller circle is then
1 2 π 2
x : x = 1 : 2π,
4
2
hence (E).
9
2000 S.11 (31%)
©2014 Australian Mathematics Trust www.amt.edu.au
Alternative 1
Since two non-shaded squares are added for every
shaded square at every step, the ratio of the shaded
area to the non-shaded area will be 13 ,
hence (A).
.
......
.............. ..
..................
.........
................................................
......................................
......................................
......................................
...................
..................................................................................................
..............................................................................
..............................................................................
..............................................................................
..............................................................................
..............................................................................
.. ... ... .... ... ... .
..................................................................................................
..........................................................
5