2025 AIME I Problem 11

On the parabola $x = 34y^2,$ intersections are the solutions of $y = f(34y^2)$ with $-1 \le y \le 1,$ so $0 \le x \le 34$

Each linear piece turns $y = f(34y^2)$ into a quadratic; add the roots with Vieta’s formulas, but check which roots actually land in the piece’s interval

Since $f$ only takes values in $[-1, 1],$ any intersection has $-1 \le y \le 1$ and hence $x = 34y^2 \in [0, 34].$ On the rising pieces, $x \in [4k - 1, 4k + 1)$ with $f(x) = x - 4k,$ so $y = f(34y^2)$ becomes $34y^2 - y - 4k = 0;$ on the falling pieces, $x \in [4k + 1, 4k + 3)$ with $f(x) = 4k + 2 - x,$ giving $34y^2 + y - (4k + 2) = 0.$ In each case a root is valid exactly when it lies in $[-1, 1)$ (rising) or $(-1, 1]$ (falling), since then $x = 34y^2$ automatically falls in the correct interval.

For the rising pieces the roots are $\frac{1 \pm \sqrt{1 + 544k}}{68},$ and both are valid exactly when $\sqrt{1 + 544k} \le 67,$ i.e. for $k = 0, 1, \ldots, 8:$ nine quadratics, each contributing root sum $\frac{1}{34}$ by Vieta. For the falling pieces the roots are $\frac{-1 \pm \sqrt{544k + 273}}{68}.$ The root with the minus sign requires $\sqrt{544k + 273} \lt 67,$ which holds for $k = 0, \ldots, 7;$ those eight quadratics each contribute $-\frac{1}{34}.$ For $k = 8$ only the positive root $\frac{-1 + \sqrt{4625}}{68} = \frac{-1 + 5\sqrt{185}}{68}$ is valid.

The total is $$\begin{aligned} &\frac{9}{34} - \frac{8}{34} + \frac{-1 + 5\sqrt{185}}{68} \\ &\quad = \frac{1 + 5\sqrt{185}}{68}, \end{aligned}$$ and $185 = 5 \cdot 37$ is squarefree, so $a + b + c + d$ $= 1 + 5 + 185 + 68$ $= 259.$