Thank you for viewing my question.

Please view my diagram at this page: 
http://www10.brinkster.com/divet/Challenges/triangle.htm

You know that the side AB = 7
You know that DE = CF = CD = EF = 1

You know that angle theta in the figure has this formula...
theta = (a/b) * arcsin((c + d*sqrt(e)) / f)
(also shown on diagram)

Due to the fact that there could be more than one combination of
values it is also known that f is a perfect square and d=5e.  The
terms are the smallest possible values.

** PLEASE NOTE THAT THE VALUES OF a,b,c,d,e,f IN THE EQUATION ARE NOT
RELATED TO THE VERTICES A,B,C,D,E,F IN THE FIGURE **

The question is to find a,b,c,d,e,f such that they are all intergers
and the angle theta in the drawing equals theta in the equation.  THE
ANSWER MUST BE IN LOWEST POSSIBLE VALUES OR THE ANSWER IS INVALID.

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Request for Question Clarification by mathtalk-ga on 10 Jul 2005 14:30 PDT

Hi, aaronfarr-ga:

The data provided for the diagram, apparently of a right-angled
triangle ABC, is symmetric with regard to exchanging the base BC and
altitude AC.  As drawn the figure suggests that theta should be taken
opposite to the shorter side, i.e. that length(BC) < length(AC).

Would that be your interpretation as well?

regards, mathtalk-ga

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Clarification of Question by aaronfarr-ga on 10 Jul 2005 18:14 PDT

Hello mathtalk-ga,
That is a good question.  Since the square is inscribed within the
triangle and the hypotenuse is fixed that means theta must also be
fixed.  As you pointed out theta could take on 2 possible values
depending on the ratio between the adjacent and opposite sides.  I
believe it is safe to assume from the diagram that the adjacent side >
opposite side.
Thanks for your interest and good luck all!
aaronfarr-ga

Calculating for AC through the principle of similar triangles, we get
a fourth order polynomial

AC^4 - 2AC^3 + AC^2 = 48 AC^2 - 98 AC + 49

Solving gives us AC = 6.90162, 1.16944, 0.87413, or -6.94521.  The
latter two are discarded as we know AC > 1.  But, it can be either of
the first two, hence the two possible values for theta.

Let's assume as you said, aaronfarr, though, that due to the relative
sizes of the image, that AC is the "longer" leg of the triangle,
meaning AC = 6.90162.

From this, we get a theta of 0.167 radians or approximately 9.62 degrees.

Creating a simple program with a lot of for loops within for loops,
sweeping from 1 to 50 for each value of a,b,c,d,e, and f, with the
constraints that d = 5e and that f = (integer)^2, and also taking into
account that there will be some rounding errors within the computer, I
had the computer display anything within 5% of the desired value of
0.167 radians.

And the answer is...drumroll please...

a = b = c = e = 1
d = 5
f = 36

This results in theta = 0.167 radians

Mathtalk, I admire your work on thse types of math problems, so please
double check the work!

The challenging part of this exercise is to produce an exact
expression for the angle theta in the form specified in the Question:

  theta = (a/b) * arcsin((c + d*sqrt(e)) / f)

subject to f being a perfect square, d = 5e, and all variables a thru
f are as integers "the smallest possible values".

So, toufaroo-ga, you and I have some work left to do!

regards, mathtalk-ga

Hello toufaroo and mathtalk,
Unfortunately the answer that toufaroo came up with is incorrect.  The
approach was creative but the precision was lost when working
numerically.  According to my calculations your answer would have
equated 0.1678506156 to 0.1674480792.
So I will have to agree with mathtalk here and say that we are looking
for a symbolic expression of the same form as the a...f term.

Part of the frustration for me is the number of methods available to
solve for theta.  Good luck and thanks for your work so far!
-aaronfarr

I am happy to say that I have successfully solved this problem and
therefore dont require a solution anymore.  I would like to thank
toufaroo and mathtalk for their efforts.  I would recommend completing
the question if it has sparked your interest since it is rewarding.

It can be solved on pen & paper but stay away from pythagorean theorem
and all of methods which result in 4th order equations.
If can also be solved on a computer using any initial method but whats
the fun in that??

Thanks again
aaronfarr-ga