AIME 2025 I · 第 11 题

Solution 1

Note that the part of $f(x)$ in the first and fourth quadrants consists of lines of the form $y = x - 4k$ and $y = 4k + 2 - x$ for integers $k \geq 0$. Plugging in $x = 34y^2$ for $y = x - 4k$, we get $-34y^2 + y + 4k = 0,$ so the sum of the roots is $\tfrac{1}{34}$ by Vieta’s. Similarly, in the second equation, we get a sum of $-\tfrac{1}{34}.$

The top peaks of the graph form the line $y = 1$ and the bottom peaks form $y = -1.$ So the parabola will cross the positive line once it reaches the point $(34,1),$ and it will cross the negative line at $(34, -1).$ For $y = x - 4k,$ we find $k = \frac{33}{4}$ for the positive $y$ values and $k = \frac{35}{4}$ for the negative y values. Both round down to $8,$ so the parabola will stop intersecting $y = x - 4k$ entirely after $k = 8.$

Now, for $y = 4k + 2 - x,$ we find $k = \frac{33}{4}$ for the positive y values and $k = \frac{31}{4}$ for the negative y values. $\frac{33}{4}$ rounds down to $8,$ but $\frac{31}{4}$ rounds down to $7,$ so the parabola will stop intersecting the negative $y$ values of $y = 4k + 2 - x$ after $k = 7,$ but it will intersect the positive $y$ values until $k = 8.$

There are $8$ values of $k$ from $0 - 7,$ so the sum contributed by $y = 4k + 2 - x$ until $k = 7$ is $-\frac{8}{34}.$ There are $9$ values of $k$ from $0 - 8,$ so the sum contributed by $y = x - 4k$ until $k = 8$ is $\frac{9}{34}.$ Adding these together we get $\frac{1}{34}.$

Now all that's left is to figure out the positive $y$ value at $k = 8$ for the line $y = 4k + 2 - x.$ We get

$$34-34y^2 = y.$$ $$y = \frac{-1 \pm \sqrt{68^2 + 1}}{68} = \frac{-1 + 5 \sqrt{185}}{68}.$$ Adding our $\tfrac{1}{34}$ from earlier finally gives us $\frac{1 + 5 \sqrt{185}}{68} \implies \boxed{259}$.

~grogg007, EpicBird08, mathkiddus

Solution 2

Drawing the graph, we can use the sawtooth graph provided so nicely by MAA and draw out the parabola $x = 34y^2$. We realize that the sawtooth graph is just a bunch of lines where the positive slope lines are $y = x, y = x + 4, y = x + 8,...$. The intersections of these lines, along with the parabola are just solving the system of equations: $x = 34y^2$ and $y = x, y = x + 4, ...$. If we just take $y = x$ and $x = 34y^2$, we see that the sum of all $y$ by Vieta's is just $\frac{1}{34}$. Similarly, for $y = x + 4$, the sum of the roots by Vieta's is also $\frac{1}{34}$. So for all the positive slope lines intersecting with the parabola just gives the sum of all $y$ to continuously be $\frac{1}{34}$. Okay, now let's look at the negative slope lines. These will have equations of $y = 2 - x, y = 6 - x, y = 10 - x, ..., y = 34 - x, ...$. Similar to what we did above, we just set each of these equations along with the parabola $x = 34y^2$. The sum of all $y$ for each of these negative line intersections by Vieta's is $\frac{-1}{34}$. This keeps going for all of the lines until we reach $y = 34 - x$. Now, unfortunately, both solutions don't work as the negative solution is out of the range of $[1 , 3]$, $[5, 7]$ and so on. So we just need to take one solution for this and that being the positive one according to the graph. So we just need to solve $34 - y = 34y^2$ which means $34y^2 + y - 34 = 0$. Solving gives

$$y = \frac{-1 \pm \sqrt{68^2 + 1}}{68} = \frac{-1 + 5 \sqrt{185}}{68}.$$ So, the sums of the roots are $\frac{1}{34}$ + $\frac{-1}{34}$ + $\frac{1}{34}$ + .... + $\frac{-1}{34}$ + $\frac{1}{34}$ + $\frac{-1 + 5 \sqrt{185}}{68}.$ Nicely all the $\frac{1}{34}$ terms cancel out leaving with only one $\frac{1}{34}$ and $\frac{-1 + 5 \sqrt{185}}{68}.$ So the sum of these two is $\frac{1 + 5 \sqrt{185}}{68}.$ From there, the answer is $\boxed{259}$.

~ilikemath247365

Video Solution

2025 AIME I #11

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