Question:
Let $a,$ $b,$ $c,$ $d$ be real numbers such that $a + b + c + d = 17$ and $ab + bc + cd + da = 46.$  Find the minimum possible value of $a^2 + b^2 + c^2 + d^2.$

Answer:
Note that $ab + bc + cd + da = 46$ factors as $(a + c)(b + d).$  So, let $r = a + c$ and $s = b + d.$  Then $r + s = 17$ and $rs = 46,$ so by Vieta's formulas, $r$ and $s$ are the roots of $x^2 - 17x + 46 = 0.$  Thus, $r$ and $s$ are equal to
\[\frac{17 \pm \sqrt{105}}{2},\]in some order.

We can let $a = \frac{r}{2} + t,$ $c = \frac{r}{2} - t,$ $b = \frac{s}{2} + u,$ and $d = \frac{s}{2} - u.$  Then
\[a^2 + b^2 + c^2 + d^2 = \frac{r^2}{2} + 2t^2  +\frac{s^2}{2} + 2u^2 \ge \frac{r^2 + s^2}{2} = \frac{197}{2}.\]Equality occurs when $a = c = \frac{r}{2}$ and $b = d = \frac{s}{2},$ so the minimum value is $\boxed{\frac{197}{2}}.$