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Previous Puzzles of the Month + Solutions

 
December 2006 - January 2007  
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thinking manDifficulty level: bulbbulbbulb, basic trigonometry knowledge.
Puzzle # 109
 Italiano italiano Français francais
  Print and cut out the seven right-triangle shapes below and join them together to form a square (no overlappings!). If the hypotenuse of the largest right triangle is 10 units, what is the area of the whole square?
puzzle 109

solution

It is not hard to solve this problem using trigonometry. The basic trigonometric functions are functions of an angle; they are important when studying triangles and are commonly defined as ratios of two sides of a right triangle containing the angle.
We use the following names for the sides of the right triangle:
- The hypotenuse is the longest side of a right triangle.
- The opposite side is the side opposite to the angle we are interested in.
- The adjacent side is the side that is in contact with the angle we are interested in.

right triangle trigo

Some fundamental trigonometric functions
1) The sine of an angle is the ratio of the opposite side to the hypotenuse: sin(A) = a / h, and h·sin(A) = a
2) The cosine of an angle is the ratio of the adjacent side to the hypotenuse: cos(A) = b / h, and h·cos(A) = b
3) The tangent of an angle is the ratio of the opposite side to the adjacent side: tan(A) = a / b
(The sine, cosine and tangent ratios can be remembered by SOH, CAH, TOA -- Sine-Opposite-Hypotenuse, Cosine-Adjacent-Hypotenuse, Tangent-Opposite-Adjacent)
Some fundamental trigonometric identities
4) sin(A) / cos(A) = tan(A)
5) sin2(A) + cos2(A) = 1


The right triangles of the puzzle 109 seem to be similar, that is, they all have at least two angles equal: 90° and x.

similar triangles

By rotation we can fan the triangles out around their vertex (as shown below), making the adjacent side to the angle x of some triangles merge with the hypothenuse of their adjacent right triangle. Now, using the basic trigonometric functions seen above, we can give a value to each side of the triangles as follows:

right triangles fan

But the problem is that we do not know the exact value of x. We can only empirically guess that x is approximately Pi/6 and must be 29° < x < 31°, as the drawing suggests:

right triangles fan 2

However, with those triangles it is possible to form the following square:

triangles to square

The sides of the square must have the same length, thus we obtain the following equation:
10cos(x) = 10cos7(x) + 10sin(x), simplified in
1 = cos6(x) + tan(x)
and also:
10cos(x) = 10sin(x)[cos2(x) + cos4(x) + cos6(x)], simplified in
1 = tan(x)[cos2(x) + cos4(x) + cos6(x)]

Thanks to the above equations and to the following exact trigonometric constants cos(30) = square root3/2 and tan(30) = square root3/3, we can already prove that x CANNOT be equal to 30°, in fact:
1 = square root3/3[(square root3/2)2 + (square root3/2)4 + (square root3/2)6], and then
1 ≠ square root3/3[(3/4) + (9/16) + (27/64)] = 111square root3/192
(an irrational number multiplied by a rational number can never be equal to an integer!)

With some patience and the help of trigonometric tables (or a scientific calculator), we can however approach the real value of x which is ± 29.68. Knowing x it is now easy to find the area of the square:
[10cos(29.68)]2 = 75.482 units square

To conclude, puzzles don't always have precise solutions or can be systematically solved by defined mathematical processes. This puzzle shows you that the trial-and-error process may also be important to puzzle solving.

The winner of the puzzle of the month is: Thierry LEBORDAIS, France. Congratulations Thierry!

© 2002 G. Sarcone, www.archimedes-lab.org
You can re-use content from Archimedes’ Lab on the ONLY condition that you provide credit to the authors (© G. Sarcone and/or M.-J. Waeber) and a link back to our site. You CANNOT reproduce the content of this page for commercial purposes.
You're encouraged to expand and/or improve this article. Send your comments, feedbacks or suggestions to Gianni A. Sarcone.


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