MHB What Are the Possible Values of (ab+cd)/(ad+bc)?

[anemone]

Here is this week's POTW:

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Given positive real numbers $a,\,b,\,c,\,d$ satisfy the equalities

$a^2-ad+d^2=b^2+bc+c^2$ and $a^2+b^2=c^2+d^2$,

find all possible values of the expression $\dfrac{ab+cd}{ad+bc}$.

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[anemone]

Congratulations to Opalg for his correct solution:), which you can find below:

Spoiler

Everything in this problem is homogeneous of degree 2, so we can multiply all four numbers $a,b,c,d$ by the same positive constant and the problem will be unchanged. Therefore, choosing a suitable constant, we can assume that $a^2+b^2 = c^2 + d^2 = 1.$ Then there must exist $\theta,\ \phi$ in $(0,\pi/2)$ such that $$a = \cos\theta, \quad b = \sin\theta, \quad c = \cos\phi,
\quad d = \sin\phi.$$ The equation $a^2 - ad + d^2 = b^2 + bc + c^2$ becomes $$\cos^2\theta - \cos\theta\sin\phi + \sin^2\phi = \sin^2\theta + \sin\theta \cos\phi + \cos^2\phi,$$ so that $$\sin\theta \cos\phi + \cos\theta\sin\phi = \cos^2\theta - \sin^2\theta - \cos^2\phi + \sin^2\phi,$$ $$\sin(\theta + \phi) = \cos(2\theta) - \cos(2\phi) = 2\sin(\theta + \phi)\sin(\theta - \phi).$$ Therefore $\sin(\theta - \phi) = \frac12$, which implies that $\cos(\theta - \phi) = \frac{\sqrt3}2.$

Next, $$\begin{aligned} \frac{ab+cd}{ad+bc} &= \frac{\cos\theta\sin\theta + \cos\phi\sin\phi}{\cos\theta\sin\phi + \sin\theta\cos\phi} \\ &= \frac{\sin(2\theta) + \sin(2\phi)}{2\sin(\theta + \phi)} \\ &= \frac{2\sin(\theta + \phi)\cos(\theta - \phi)}{2\sin(\theta + \phi)} \\ &= \cos(\theta - \phi) = \frac{\sqrt3}2. \end{aligned}$$

Therefore $$\frac{ab+cd}{ad+bc} = \frac{\sqrt3}2.$$ So that is the only possible value of the expression.