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

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Note that <math>f(x)</math> consists of lines of the form <math>y = x - 4k</math> and <math>y = 4k + 2 - x</math> for integers <math>k</math>. In the first case, we get <math>34y^{2} = y - 4k</math> and the sum of the roots is <math>\tfrac{1}{34}</math> by Vieta. In the second case, we similarly get a sum of <math>-\tfrac{1}{34}.</math> Thus pairing <math>4k</math> and <math>4k+2</math> gives a <math>y</math>-coordinate sum of <math>0.</math>
 
Note that <math>f(x)</math> consists of lines of the form <math>y = x - 4k</math> and <math>y = 4k + 2 - x</math> for integers <math>k</math>. In the first case, we get <math>34y^{2} = y - 4k</math> and the sum of the roots is <math>\tfrac{1}{34}</math> by Vieta. In the second case, we similarly get a sum of <math>-\tfrac{1}{34}.</math> Thus pairing <math>4k</math> and <math>4k+2</math> gives a <math>y</math>-coordinate sum of <math>0.</math>
  
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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>.
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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>.  
  
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Solution credit: @EpicBird08
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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>.
  
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~ EpicBird08 ~mathkiddus
  
 
==Solution 2==
 
==Solution 2==

Revision as of 19:55, 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 \[34-34y^2 = y\] and 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

See Also

2025 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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