**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”. Ð

*b*+ Ð*c*= Ð*a*= 45°. So Ð*a*+ Ð*b*+ Ð*c*= 2 Ð*a*= 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 Ð

*EHD*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. Ð

*c*is easily seen to be 45°. This is because D*BAE*is an isosceles triangle with Ð*BAE*= 90°. Ð*c*= (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 Ð

*IDH*= Ð*FHE*= Ð*ECA*= Ð*b*, 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 Ð*EHD*= 90°, and since D*HDE*is an isosceles triangle, Ð*HDE*= 45° = Ð*c*. Ð*a*+ Ð*b*+ Ð*c*= Ð*EDA*+ Ð*IDH*+ Ð*HDE*= Ð*JDC*= 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: Ð

*c*= Ð*JDK*= 45°, obviously. The red grid is constructed at a 45° angle to the black grid, but has a bigger length. Nevertheless, Ð*EDQ*= Ð*ECA*= Ð*b*, because D*QDE*is just an enlarged version of D*ACE*. [They are*similar triangles*. The enlargment factor is Ö2, but we do not need this.] By looking at Ð*JDC*, you will see that Ð*a*+ Ð*b*+ Ð*c*= 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.

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

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