[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