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 From the same authors of the article Curiopticals, ISBN 1847322298 What are you looking at, ISBN 1847321836 Eyetricks, ISBN 1844427773 Make your own 3D illusions, ISBN 1844426327 Big Book of Optical Illusions, ISBN 0764135201 New Optical Illusions, ISBN 1844423271 Fantastic Optical Illusions, ISBN 184442295X Puzzillusions, ISBN 1844420647 MateMagica, ISBN 8889197560 L'Almanach du mathématicien en herbe, ISBN 2844690254

# Previous Puzzles of the Month + Solutions

February-March 2003
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Puzzle 1

Puzzle 2

 ::::::::::::::Solutions:::::::::::::: Roll over the references below to see the solutions Solution Puzzle 2

A reaction from one of our visitors:

I noticed that the answer to February-March 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 =
Black2 + Blue2 – 2 x Black x Blue x Cos 10 = 0.467911...
Red angle = 30 degrees.

- Posted by Omar A. Alrefaie

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 step-by-step 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.

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 step-by-step. It was really a complete and well-structured 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 archimedes-lab.org for your care and nice attitude. I will keep in touch. Best regards.
- Posted by Omar A. Alrefaie

Previous puzzles of the month...
contents + solutions

 August 98: the irritating 9-piece puzzle September 98: the impossible squarings October 98: the multi-purpose hexagon November 98: the incredible Pythagora's theorem December 98: the cunning areas January 99: less is more, a square root problem February 99: another square root problem... March 99: permutation problem... April 99: minimal dissections July 99: jigsaw puzzle August 99: logic? Schmlogic... September 99: hexagon to disc... Oct-Nov 99: curved shapes to square... Dec-Jan 00: rhombus puzzle... February 00: Cheeta tessellating puzzle... March 00: triangular differences... Apr-May 00: 3 smart discs in 1... July 00: Funny tetrahedrons... August 00: Drawned by numbers... September 00: Leonardo's puzzle... Oct-Nov 00: Syntemachion puzzle... Dec-Jan 01: how many squares... February 01: some path problems... March 01: 4D diagonal... April 01: visual proof... May 01: question of recflection... June 01: slice the square cake... July 01: every dog has 3 tails... Aug 01: closed or open... Sept 01: a cup of T... Oct 01: crank calculator... Nov 01: binary art... Dec 01-Jan 02: egyptian architecture... Feb 02: true or false... March 02: enigmatic solids... Apr 02: just numbers... May 02: labyrinthine ways... June 02: rectangle to cross... July-Aug 02: shaved or not... Sept 02: Kangaroo cutting... Oct 02: Improbable solid... Dec-Jan 03: Hands-on geometry

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