A
reaction from one of our visitors:
I
noticed that the answer to FebruaryMarch 2003
puzzle is wrong (using only elementary geometry
find angle a). Your answer
is 20 degrees while the right answer is 30 degrees.
You can double check the answer by using trigonometry
sine and cosine laws to solve the triangles. In
addition, your answer is not based on theory and
lacks logical reasoning. Thanks!
Assuming
that a = 1 then,
Black = 2 Sin 80 = 1.969615...
Blue = 3
/ 2(Sin 40) = 1.347296...
Red =
Black^{2} + Blue^{2} – 2 x
Black x Blue x Cos 10 = 0.467911...
Red angle = 30 degrees.

Posted by Omar A. Alrefaie
Our
answer...
Dear Omar, you're absolutely right, our answer is
not complete and lacks logical reasoning. Actually,
this puzzle is one of the hardest puzzles to solve
with elementary geometry you may find on Internet!
The answer is really 20 degrees,
to understand why, please follow the stepbystep
explanation below and the relative images shown further
below.
Fig
a)
Let MLN be a triangle, then...
.1) A straight angle = 180°; so, angle PMN =
180°  100°  10° = 70°.
.2) Knowing that in a triangle the angles always
add up to 180°; then, angle MLN = 180° 
10°  70°  60°  20° = 20°.
.3) Angle LMN = angle MNL (10° + 70° = 60° +
20°), then triangle MLN is isosceles.
.4) Notice that angle MPN = 180°  70° 
60°  20° = 30°.
.5) Notice that triangle LMQ, having angles LMQ and
MLQ equal (20°), is isosceles.
Fig. b)
.6) Draw segment MQ (mirror of segment ON).
.7) Join Q to O with a segment, OQ is parallel to
MN by symmetry.
Fig. c)
.8) Knowing that pairs of alternate interior angles,
formed by a transversal intersecting 2 parallels
(here, segment OQ and segment MN), are equal; then,
triangle MNR is similar to triangle ORQ (that means
they are proportional to each other).
.9) Notice that triangle ORQ is an equilateral triangle
(with 3 sides equal).
.10) From L, join R with a segment. By symmetry,
the segment LR bisects (divides into 2 equal angles)
angle MLN.
Fig. d)
.11) Triangle MQP triangle LRQ, because they have
1 side and 2 adjoining angles equal: segment LQ =
segment QM (by .5); angle PMQ = angle RLQ; and angle
MQL is shared. So, segment PQ = segment QR.
.12) PQ = OQ (by .9 and .11), then triangle OQP is
isosceles and angle OPQ = (180°  80°) /
2 = 50°.
.13) Angle OPM = Angle OPQ  Angle MPN = 50° 
30° = 20°.
Q.E.D.
Reply of the visitor
I checked your site last Thursday (April 19, 2007),
and I found that you have posted my solution (which
was wrong anyway) and I also went through your solution
stepbystep. It was really a complete and wellstructured
solution. When I rechecked my answer, I realized
that I made a mistake calculating the red segment;
I did not take the square root of the answer (0.4679…..)
which resulted in my wrong answer. So you were absolutely
right, the answer is 20 degrees.
I really thank you and all colleagues at archimedeslab.org
for your care and nice attitude. I will keep in touch.
Best regards. 
Posted by Omar A. Alrefaie