[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

[MathsEd] The Secret to Mathematical Problem Solving

     Many students treat mathematics problem solving either as a mystery, or they like to shoot at random from a set of formulas or recipes that they have memorised and just see whether it works or not.  When this strategy does not work and after working for a few minutes, they just give up.  They wonder how all those maths geniuses get it.  Many do not know that even the professional mathematicians take many years or evencenturies to solve mathematics problems.  The mathematician George Polya has written a book “How to Solve It” in 1945, describing how mathematics problems are solved.  Since then many people have adapted and modified the steps slightly but they basically boil down to the following steps.

Step 1: Understanding

     Try to make sure you really understand the problem first.  If it is an examination question, read and take note of the given information.  Ask what is known, what can be known and what is to be found.

Step 2: Planning

     This is where you plan your strategy of attack.  Can you organise the information into a table or a diagram?  Heuristics(rules-of-thumb or guidelines) are usually useful to help you formulate yourstrategy, especially if you have not seen this type of question before.

Step 3: Execution

     Carry out your plan.  Make sure you are conscious of what you are doing.  Are you able to explain it to yourself, or younger brother/sister?

Step 4: Evaluation

     Check your calculations and logic.  Are there any careless mistakes?  Is your answer plausible or believable?  Are you on the right track?  Are you getting somewhere?  You also need good number sense.  For example, if you are calculating with a triangle with sides  4 cm,  5 cm  and  6 cm,  and you get an area like 1 000 cm ²,  does your answer even taste and smell right?  Do not continue doing the same wrong thing.  If you catch yourself making a mistake, go back to step 2 and change your strategy.  Try another approach.

Step 5: Reflection

     After solving the problem, think back at what lessons you have learnt by attempting / solving this problem.  How could you have done better?  Did you discover anything that can be applied in other problems?  What if the numbers are changed?  What if the conditions are changed?  You can test your own understanding by setting yourself a similar or modified question.  Can you generalise your results?  Can you link what you have learnt to daily life or to other subjects?

Suitable Levels

* all levels, all topics !!!

[Pri20150523WNCA] Trees arranged in a Hexagon Outline

Question

Introduction

     This is real eeeasy peasy lemon squeezy, isn’t it?  54 ¸ 6 = 9  Ta da!  The answer, right? Wrong!  You got tricked!  Ha!  Ha!

     Always tryto understand the question and do the planning first.  Never be in a hurry and jump to thecalculation stage.  So what went wrong?  Well, the tree at each vertex is counted twice.
     Huh?
     Sometimes to understand the situation, it may be easier to consider a simpler problem.  Let us say there are four trees per side.  This is how it looks like from above.

     You can see that the corner trees (coloured in orange instead of brown) are counted twice, because they each serve as an extreme marker of two of the sides of the hexagon.  There are 18 trees and if you divide by  6,  you get  3  and not  4.  One way to count properly is to start from one corner tree and count groups of three trees, either in a clockwise or anti-clockwise (American: counter-clockwise) direction.

     Notice that the number of trees on one edge of the hexagon is equal to the number of trees in one group plus one (the corner tree for the next group).  So for  18  trees, the correct calculation is  18 ¸ 6 + 1 = 3 + 1 = 4  for the number of trees along one edge.  We use the same procedure for  54  trees.

Solution

     number of trees on each side = 54 ¸ 6 + 1 = 9 + 1 = 10

Final Remarks

     You may want to generalise it into a formula

                    # trees on each side = total # trees ¸ #sides + 1

However, I do not recommend that you purposely memorise this formula.  Mathematics is not about memorisation.  It is about understanding.  Once you understand it, the formula comes out automatically.  You may test yourself or get a friend to test your understanding by setting a similar question but changing the number of trees and number of sides.

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H10. Simplify the problem

Suitable Levels

* Primary School Mathematics

* other syllabuses that involve whole numbers