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Previous Puzzles of the Month + Solutions

 
February-March 2003  
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Solutions are at the bottom of this page
 
Puzzle 1
Send us your comments on puzzle 1

puzzle no. 1

Puzzle 2
Send us your comments on puzzle 2

puzzle no. 2


solution
::::::::::::::Solutions::::::::::::::

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Solution
Puzzle 1

Solution
Puzzle 2

solution

A reaction from one of our visitors:

visitor solutionI 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 = square root3 / 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

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 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.

angle problem
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 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

© 2002 G. Sarcone, www.archimedes-lab.org

You're encouraged to expand and/or improve this article. Send your comments, feedbacks or suggestions to Gianni A. Sarcone.


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