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) = 10cos^{7}(x) + 10sin(x), simplified
in
1 = cos^{6}(x) + tan(x)
and also:
10cos(x) = 10sin(x)[cos^{2}(x) + cos^{4}(x)
+ cos^{6}(x)], simplified in
1 = tan(x)[cos^{2}(x) + cos^{4}(x)
+ cos^{6}(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 trialanderror
process may also be important to puzzle solving.
The
winner of the puzzle of the month is: Thierry
LEBORDAIS, France. Congratulations Thierry!
