Difference between revisions of "2025 AIME I Problems/Problem 11"

Revision as of 19:56, 13 February 2025

Problem

A piecewise linear function is defined by $f(x) = \begin{cases} x & \operatorname{if} ~ -1 \leq x < 1 \\ 2 - x & \operatorname{if} ~ 1 \leq x < 3\end{cases}$ and $f(x + 4) = f(x)$ for all real numbers $x$ . The graph of $f(x)$ has the sawtooth pattern depicted below.

The parabola $x^{2} = 34y$ intersects the graph of $f(x)$ at finitely many points. The sum of the $y$ -coordinates of all these intersection points can be expressed in the form $\tfrac{a + b\sqrt{c}}{d}$ , where $a$ , $b$ , $c$ , and $d$ are positive integers such that $a$ , $b$ , $d$ have greatest common divisor equal to $1$ , and $c$ is not divisible by the square of any prime. Find $a + b + c + d$ .

Graph

It may be helpful to graph certain parts of the graph to grasp a better understanding of what we need and gain some intuition. I created an example diagram on Desmos here: https://www.desmos.com/calculator/ne8shyhyka

~lprado

Solution

Note that $f(x)$ consists of lines of the form $y = x - 4k$ and $y = 4k + 2 - x$ for integers $k$ . In the first case, we get $34y^{2} = y - 4k$ and the sum of the roots is $\tfrac{1}{34}$ by Vieta. In the second case, we similarly get a sum of $-\tfrac{1}{34}.$ Thus pairing $4k$ and $4k+2$ gives a $y$ -coordinate sum of $0.$

This process of pairing continues until we get to $k = 8$ . Then $y = x - 32$ behaves exactly as we expect, with a sum of $\tfrac{1}{34}$ .

However, $y = 34-x$ is where things start becoming fishy, since there is one root with absolute value less than $1$ and one with absolute value greater than $1$ . We get $34-34y^2 = y.$ Solving with the quadratic formula (clear to take the positive root) gives $y = \frac{-1 \pm \sqrt{68^2 + 1}}{68} = \frac{-1 + 5 \sqrt{185}}{68}.$ Adding our $\tfrac{1}{34}$ from earlier gives the answer $\frac{1 + 5 \sqrt{185}}{68} \implies \boxed{259}$ .

~ 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

@@ Line 14: / Line 14: @@
 This process of pairing continues until we get to <math>k = 8</math>. Then <math>y = x - 32</math> behaves exactly as we expect, with a sum of <math>\tfrac{1}{34}</math>.
-However, <math>y = 34-x</math> is where things start becoming fishy, since there is one root with absolute value less than <math>1</math> and one with absolute value greater than <math>1</math>. We get <cmath>34-34y^2 = y</cmath> and solving with the quadratic formula (clear to take the positive root) gives <cmath>y = \frac{-1 \pm \sqrt{68^2 + 1}}{68} = \frac{-1 + 5 \sqrt{185}}{68}.</cmath> Adding our <math>\tfrac{1}{34}</math> from earlier gives the answer <math>\frac{1 + 5 \sqrt{185}}{68} \implies \boxed{259}</math>.
+However, <math>y = 34-x</math> is where things start becoming fishy, since there is one root with absolute value less than <math>1</math> and one with absolute value greater than <math>1</math>. We get <cmath>34-34y^2 = y.</cmath>Solving with the quadratic formula (clear to take the positive root) gives <cmath>y = \frac{-1 \pm \sqrt{68^2 + 1}}{68} = \frac{-1 + 5 \sqrt{185}}{68}.</cmath> Adding our <math>\tfrac{1}{34}</math> from earlier gives the answer <math>\frac{1 + 5 \sqrt{185}}{68} \implies \boxed{259}</math>.
 ~ EpicBird08 ~mathkiddus

2025 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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