Question
Introduction
This
question appeared as Question 7 of this year's Australian VCE Further Mathematics
Examination. It has caused an uproar
among some students who sat for the examination, hurling abuse at “50 cent”. For a moment I was wondering: why should the
eponymous American rapper face the music for an exam question he never set? Hmmm...
Anyway, in Singapore , this
type of problem belongs to just the plain old Lower Secondary Mathematics
syllabus for the topic of angles and polygons, usually taught in secondary 1 (equivalent
to about grade 7 or 8). There is no “additional”
or “further” mathematics in our lower secondary curriculum. Students who do well can opt to take
Additional Mathematics. In the
Integrated Programme schools, they may call their mathematics curricula by
different names.
I would say
this question would be of intermediate difficulty level in Singapore .
It is not so straightforward, but there is a quick solution, given the right
insight. How can we obtain the right insights? This can be done by making observations and
splitting a problem into smaller problems!
[Heuristics H04, H10, H11]
Important Principles
Let us do
some recap. A polygon
is any closed shape consisting of a number (at least three) of straight edges. An exterior angle
of a polygon is obtained by producing (extending) an edge in one direction – it is the
angle between this extended edge and the next nearby edge. It is a fact that
the sum of all exterior angles of any polygon is
always 360°

This can be proven in various ways, but I think the
best way to see this intuitively is to imagine that you are an ant on the
polygon. Starting from any vertex, go
along the edges and every time you walk on a new edge, you turn by an amount
equivalent to the exterior angle. By the
time you have come back round to your starting point, you would have turned a
total of 360°. [H01]
By its
construction, the exterior angle and its interior angle always add up to 180°. For a regular polygon with n sides, all its sides (edges) are equal and all
its interior angles are equal and so all its exterior angles are also equal. Then it is immediately obvious that
each exterior angle of a regular nsided polygon is 360° ¸ n

Solution
We construct
a vertical line segment in the middle and focus on, say, the right half of the
angle (shown highlighted in magenta). [H09, draw an “imaginary” line]
This halfangle
is in fact an exterior angle of the 50 cent coin (a 12sided regular polygon)
and so it measures 360° ¸ 12 = 30°. Thus the
value of q is double
that, i.e. 60°. [H09] Done!
Remarks
If you have three such 50cent coins,
you can actually put them together and you see a triangular hole in the centre.
[H01]
This is actually an equilateral triangle
and each angle would be 180° ¸ 3 = 60°. I am sure even
a tenyearold kid can do this!
H01. Act it out
H02. Use a
diagram / model
H04. Look for
pattern(s)
H05. Work
backwards
H09. Restate
the problem in another way
H11. Solve part
of the problem
Suitable Levels
* Singapore Lower Secondary Mathematics (Sec 1 » Grade 7/8)
* Victorian Certificate of Education Further
Mathematics (Australia )
* other syllabuses that involve polygons
and angles
* whoever is interested, even 10 yearold
kids
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