[S1_20151107PAXA] Angle between two Aussie Fifty Cent Coins

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 n-sided 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 half-angle is in fact an exterior angle of the 50 cent coin (a 12-sided 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 50-cent 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 ten-year-old kid can do this!

Heuristics Used

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 year-old kids

[OlymPri20150527AGSQ] Three Angles and Three Squares

Question

Introduction

     This question, from a Facebook forum, is essentially the same problem as the one in a previous article.  However, here we are not allowed to use any advanced mathematics like arctangent (inverse tangent).  It must be a solution that a primary school pupil can come up with, or at least understand.

Solution 1

     Refer to the article “Angles with nice tans add up nicely”.  Ðb + Ðc  = Ða = 45°.  So  Ða + Ðb + Ðc  = 2 Ða = 90°.  But using this method is “cheating”.

Solution 2a

     I happened to be watching this YouTube video that solves this problem.  What a coincidence!  The presenter is Professor Zvezdelina Stankova from the University of California, Berkeley.  I liked the way she used the “Act it out” heuristic by using a pair of scissors to cut out the angles on paper, and then putting them together to see if  they actually add up to  90°.  The trouble with this approach is: How do you know that they add up exactly to 90°?  So Stankova proceeded to give an elementary but rigorous proof.  Her proof is elegant, but I think the demonstration of the fact that  ÐEHD  is a right angle can be tightened using the idea of rotation.

Solution 2b

     I am going to follow Stankova’s naming of the vertices, except that our problem is the mirror image of the one that is shown in the video.  Ðc  is easily seen to be  45°.  This is because  DBAE  is an isosceles triangle with  ÐBAE = 90°.  Ðc = (180° – 90°) ¸ 2 = 45°.  We extend the grid to a  2  by  3  grid, and construct  HD  and  HE. 

Obviously  HD  = HE = CE,  because they are all hypothenuses (that is the correct plural form, not “hypotheni”.  I checked)  of right-angled triangles with some length of  1  unit and another length of  2  units including the right angle in between.  Another way to think about it is that they are all diagonals of a  2  by  1  rectangle of some orientation.

Notice that  ÐIDH = ÐFHE = ÐECA = Ðb,  for the same reason.  Observe also that the rectangle  HCDI  is rectangle  HFEK  rotated by  90°  clockwise.  Every point and every line of the original rectangle is rotated by the same amount.  In particular,  HE  is rotated by  90°  to get  HD.  Hence  ÐEHD = 90°,  and since  DHDE  is an isosceles triangle,  ÐHDE = 45° = Ðc.   Ða + Ðb + Ðc = ÐEDA + ÐIDH + ÐHDE = ÐJDC = 90°.  ©

Solution 3

     One of the viewers, Ian Agol, contributed a solution, which I think is the most elegant solution.  It is a proof without words.  If you just stare at the picture and think, you should “get it” without explanation.  The diagram below is an adaptation of his solution.

If you want an explanation:  Ðc = ÐJDK  = 45°, obviously.  The red grid is constructed at a  45°  angle to the black grid, but has a bigger length.  Nevertheless,  ÐEDQ = ÐECA = Ðb,  because  DQDE  is just an enlarged version of  DACE.  [They are similar triangles.  The enlargment factor is Ö2,  but we do not need this.]  By looking at  ÐJDC,  you will see that  Ða + Ðb + Ðc  = 90°.

Remarks

     Unlike the usual examination-based school “mathematical” practices, real mathematics is not just about getting the answer.  It is about creativity, seeking understanding and seeking elegance.  We prefer short and sweet solutions to long and complicated solutions.  Sometimes when we finish solving a problem, another solution or a few other solutions pop-up.  Having different approaches to a mathematical problem allows us to understand the problem and the inter-relationships better.

Heuristics Used

H01. Act it out

H02. Use a diagram / model

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

Suitable Levels
* Primary School Mathematics Olympiad
* other syllabuses that involve angles

[OlymPri20150526] The Smallest Angle in a Special Trapezium

Question

Introduction

     This looks challenging because we do not seem to be given much information.  For example, we do not know the individual angles of the trapezium (American: trapezoid).  Does this mean that this puzzle cannot be solved?  Are we trapped by this trapezium?  What is the secret key that unlocks the problem?

Solution

     First sketch trapezium  ABCD.  Introduce point  F,  the mid-point of  AD.  All contructed lines and points are shown in grey (American: gray).  Then  DF = FA = BC. 

Now draw a line parallel to  BC  passing through  A  and intersecting  DC  at  E, say.  Note that ÐAED = ÐBCD  (corresponding angles).  Then  ÐADE + ÐAED = ÐADC + ÐBCD = 120°.  Hence the remaining angle  in  DDAE,  ÐDAE = 180° – 120° = 60°.  We can deduce this, even though we do not (currently) know the individual angles  ÐADE  and  ÐAED.  This is the key step.

From here, things get easier.  DAFE  is an isosceles triangle with  ÐAFE = ÐAEF = (180° – 60°) ¸ 2 = 60°.  So  DAFE  is in fact an equilateral triangle.  That means  FE = FA = FD.  So  DFDE  is an isosceles triangle.

Note that  ÐAFE  is an exterior angle of the triangle  DFDE.  If you know that exterior angle of a triangle is the sum of the interior opposite angles,  then from  ÐAFE = ÐFED + ÐFDE,  we easily see that  ÐADE = ÐFDE = 60° ¸ 2 = 30°.  If you do not know the theorem about exterior angles, you can still quickly work out that  ÐDFE = 120°,  and then use  ÐADE = (180° – 120°) ¸ 2  and arrive at the same conclusion.  ©

Remarks

     We can now see that actually  ÐBCD  = ÐAED  = 90°,  but we do not need to rely on that (or on accurate drawing) to deduce the answer.  From solving this question, we learn that even though we do not know the individual angles, by construction and using the sum of angles in a triangle, it is possible to solve for an important angle, namely  ÐAFE.  The rest of the solving uses isosceles triangles and equilateral triangles, which are part of the common repertoire of tactics.

HeuristicsUsed

H02. Use a diagram / model

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

H12* Think of a related problem   (Draw construction lines)

Suitable Levels
* Primary School Mathematics

* other syllabuses that involve area of triangles and circles