A blog about Mathematics and Mathematics Education in Singapore, as well as Mathematics Education in general. Written for students, parents, educators and other stakeholders in Singapore, and around the world.
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Wednesday, May 27, 2015
[OlymPri20150527AGSQ] Three Angles and Three Squares
question, from a Facebook forum, is essentially the same problem as the one in a
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.
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.
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.
HD = HE
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.
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
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
you will see that Ða + Ðb + Ðc =
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