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

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==Problem==
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A piecewise linear function is defined by <cmath>f(x) = \begin{cases} x & \operatorname{if} ~ -1 \leq x < 1 \\ 2 - x & \operatorname{if} ~ 1 \leq x < 3\end{cases}</cmath> and <math>f(x + 4) = f(x)</math> for all real numbers <math>x</math>. The graph of <math>f(x)</math> has the sawtooth pattern depicted below.
  
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The parabola <math>x^{2} = 34y</math> intersects the graph of <math>f(x)</math> at finitely many points. The sum of the <math>y</math>-coordinates of all these intersection points can be expressed in the form <math>\tfrac{a + b\sqrt{c}}{d}</math>, where <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are positive integers such that <math>a</math>, <math>b</math>, <math>d</math> have greatest common divisor equal to <math>1</math>, and <math>c</math> is not divisible by the square of any prime. Find <math>a + b + c + d</math>.
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==Solution==
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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>
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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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Solution credit: @EpicBird08
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==See Also==
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{{AIME box|year=2025|n=I|num-b=10|num-a=12}}
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{{MAA Notice}}

Revision as of 18:40, 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 $a + b + c + d$.

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 $\frac{1 + 5 \sqrt{185}}{68} \implies \boxed{259}$.

Solution credit: @EpicBird08

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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