The
right triangles of the puzzle 109 seem to be similar,
that is, they all have at least two angles equal:
90° and x.
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:
But
the problem is that we do not know the exact value
of x. We can only empirically guess that x is approximately /6
and must be 29° < x < 31°, as the drawing
suggests:
However,
with those triangles it is possible to form the following
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) = 3/2 and tan(30)
= 3/3,
we can already prove that x CANNOT be equal to 30°,
in fact:
1
= 3/3[(3/2)2 +
(3/2)4 +
(3/2)6],
and then
1 ≠ 3/3[(3/4)
+ (9/16) + (27/64)] = 1113/192
(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!
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