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I seek to express $DE$ and $UI$ and $IE$ in terms of $\theta$:

$$\sec { \theta } =DE+IE+UI \tag{1.1}\label{myeqone}$$ $$\frac { 1 }{ 4 } =UI(UI+IE) \tag{2.1}\label{myeqtwo}$$ $$(\frac { 1 }{ 2 } -\tan { \theta } )^{ 2 }=DE(DE+IE) \tag{3.1}\label{myeqthree}$$

I rewrote $(2)$ and $(3)$ as

$$\frac { 1 }{ 4 } =UI(sec\theta -DE) \tag{2.2}\label{myeqtwo_one}$$ $$(\frac { 1 }{ 2 } -{ tan }^{ 2 }\theta )=DE(sec\theta -UI) \tag{3.2}\label{myeqthree_one}$$

This is where I'm stuck. I tried solving the system by replacing $DE$ in $(2.2)$ with $$\frac { (\frac { 1 }{ 2 } -tan^{ 2 }\theta ) }{ (sec\theta -UI) } $$but to no avail. I've also tried multiplying, adding, and subtracting the two systems in the hopes of something symmetric or meaningful, but once again to no avail. I even replaced $UI$, $DE$, and $IE$ with $x$,$y$, and $z$ in the hopes of simplifying some notation and hopefully solving the equation, but to no avail.

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    $\begingroup$ The left-hand sides of $(3.1)$ and $(3.2)$ don't match. Which is correct? $\endgroup$ – Blue Oct 10 '19 at 0:53
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Running this through Mathematica, I get ...

$$\begin{align} IE &= \pm\sqrt{\sin 2\theta} \\[4pt] UI &= \phantom{\sec\theta}-\frac12 \left(\pm\sqrt{\sin2\theta}-(\cos\theta+\sin\theta)\right)\\[4pt] DE &= \sec\theta-\frac12\left(\pm \sqrt{\sin2\theta}+(\cos\theta+\sin\theta)\right) \end{align}$$ where the same "$\pm$" is used throughout.

The algebra for deriving the above looks a bit messy; if you really want to see it, I can take a shot at writing it down. Personally, I'd like to see how the equations arose; there may be a straightforward geometric argument that avoids most of the algebra.


Here's what I did in Mathematica ...

Mathematica doesn't always simplify Sec[] and Tan[] the way I like, so I tend to write equations in terms of cost and sint, and then do a separate simplification pass. In this case, the crux of the work is done via:

Solve[{ 1/cost == DE + IE + UI, 1/4 == UI (UI + IE), (1/2 - sint/cost)^2 == DE (DE + IE)}, {DE, IE, UI}]

Easy-peasy!

This gave two solutions (differing only in sign, hence the "$\pm$"), with this expression needing some attention:

Sqrt[cost^2 (-1 + sint^2) (-1 + cost^2 - 2 cost sint + sint^2)]

Without too much trouble, this simplifies to cost^2 Sqrt[2 sint cost]; that is, $\cos^2\theta \sqrt{\sin 2\theta}$. From there, a little clean-up gives the result as shown above.

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  • $\begingroup$ Wow! How exactly did you run the equations through Mathematica? Can I see a screenshot so that I can try using it in the future? Thanks. In the future, I may work backwords and try to arrive at some geometric interpretation of these equations. $\endgroup$ – DarkRunner Oct 10 '19 at 0:35
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For notational brevity, let $x=DE$, $y=IE$ and $z=UI$ and work with the set below,

$$x+y+z = \sec \theta \tag{1}$$ $$z(z+y) =\frac 14\tag{2}$$ $$x(x+y)=\left(\frac 12 -\tan\theta \right)^2\tag{3}$$

Take (3)-(2) and use (1),

$$x-z=\sin\theta(\tan\theta-1)\tag{4}$$

From (3)+(2),

$$x^2+z^2+y(x+z)=\left(\frac 12 -\tan\theta \right)^2+\frac14\tag{5}$$

Square (1) and rearrange, $$y^2-(x-z)^2+2[x^2+z^2+y(x+z)]= \sec^2\theta\tag{6}$$

Substitute (4) and (5) into (6) to get the equation for $y$,

$$y^2-\sin^2\theta(\tan\theta-1)^2+2\left(\frac 12 -\tan\theta \right)^2+\frac12=\sec^2\theta$$

Then, simplify and obtain the solution for $y$,

$$y^2=\sin 2\theta,\>\>\>\>\>y=\pm\sqrt{\sin 2\theta}$$

Then, the solutions for $x$ and $z$ follow,

$$\begin{align} x&= -\frac12\left(y+\cos\theta+\sin\theta\right) +\sec\theta\\ z&= -\frac12 \left(y-\cos\theta-\sin\theta\right) \end{align}$$

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