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Difference between revisions of "2025 AIME I Problems/Problem 11"
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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.
 
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.
  
−
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>.
+
The parabola <math>x = 34y^{2}</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>.
  
 
==Graph==
 
==Graph==
−
It may be helpful to graph certain parts of the graph to grasp a better understanding of what we need. I created an example diagram on Desmos here: https://www.desmos.com/calculator/ne8shyhyka
+
<asy>
 +
import graph;
 +
size(300);
 +
Label f;
 +
f.p=fontsize(6);
 +
yaxis(-2,2,Ticks(f, 2.0));
 +
xaxis(-6.5,6.5,Ticks(f, 2.0));
 +
draw((0, 0)..(1/4,sqrt(1/136))..(1/2,sqrt(1/68))..(0.75,sqrt(0.75/34))..(1, sqrt(1/34))..(2, sqrt(2/34))..(3, sqrt(3/34))..(4, sqrt(4/34))..(5, sqrt(5/34))..(6, sqrt(6/34))..(7, sqrt(7/34))..(8, sqrt(8/34)), red);
 +
draw((0, 0)..(1/4,-sqrt(1/136))..(0.5,-sqrt(1/68))..(0.75,-sqrt(0.75/34))..(1, -sqrt(1/34))..(2, -sqrt(2/34))..(3, -sqrt(3/34))..(4, -sqrt(4/34))..(5, -sqrt(5/34))..(6, -sqrt(6/34))..(7, -sqrt(7/34))..(8, -sqrt(8/34)), red);
 +
draw((-7,0)--(7,0), black+0.8bp);
 +
draw((0,-2.2)--(0,2.2), black+0.8bp);
  
−
~lprado
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draw((-6,-0.1)--(-6,0.1), black);
 +
draw((-4,-0.1)--(-4,0.1), black);
 +
draw((-2,-0.1)--(-2,0.1), black);
 +
draw((0,-0.1)--(0,0.1), black);
 +
draw((2,-0.1)--(2,0.1), black);
 +
draw((4,-0.1)--(4,0.1), black);
 +
draw((6,-0.1)--(6,0.1), black);
  
−
==Solution==
+
draw((-7,1)..(-5,-1), blue);
 +
draw((-5,-1)--(-3,1), blue);
 +
draw((-3,1)--(-1,-1), blue);
 +
draw((-1,-1)--(1,1), blue);
 +
draw((1,1)--(3,-1), blue);
 +
draw((3,-1)--(5,1), blue);
 +
draw((5,1)--(7,-1), blue);
 +
</asy>
 +
 
 +
~[[User:Mathkiddus|mathkiddus]]
 +
 
 +
==Solution 1==
 
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>
  
−
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>.
+
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>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
 +
~[[User:Mathkiddus|mathkiddus]]
 +
 
 +
==Solution 2==
 +
Drawing the graph, we can use the sawtooth graph provided so nicely by MAA and draw out the parabola <math>x = 34y^2</math>. We realize that the sawtooth graph is just a bunch of lines where the positive slope lines are <math>y = x, y = x + 4, y = x + 8,...</math>. The intersections of these lines, along with the parabola are just solving the system of equations: <math>x = 34y^2</math> and <math>y = x, y = x + 4, ...</math>. If we just take <math>y = x</math> and <math>x = 34y^2</math>, we see that the sum of all <math>y</math> by Vieta's is just <math>\frac{1}{34}</math>. Similarly, for <math>y = x + 4</math>, the sum of the roots by Vieta's is also <math>\frac{1}{34}</math>. So for all the positive slope lines intersecting with the parabola just gives the sum of all <math>y</math> to continuously be <math>\frac{1}{34}</math>. Okay, now let's look at the negative slope lines. These will have equations of <math>y = 2 - x, y = 6 - x, y = 10 - x, ..., y = 34 - x, ...</math>. Similar to what we did above, we just set each of these equations along with the parabola <math>x = 34y^2</math>. The sum of all <math>y</math> for each of these negative line intersections by Vieta's is <math>\frac{-1}{34}</math>. This keeps going for all of the lines until we reach <math>y = 34 - x</math>. 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 <math>34 - y = 34y^2</math> which means <math>34y^2 + y - 34 = 0</math>. Solving gives<cmath>y = \frac{-1 \pm \sqrt{68^2 + 1}}{68} = \frac{-1 + 5 \sqrt{185}}{68}.</cmath>So, the sums of the roots are <math>\frac{1}{34}</math> + <math>\frac{-1}{34}</math> + <math>\frac{1}{34}</math> + .... + <math>\frac{-1}{34}</math> + <math>\frac{1}{34}</math> + <math>\frac{-1 + 5 \sqrt{185}}{68}.</math> Nicely all the <math>\frac{1}{34}</math> terms cancel out leaving with only one <math>\frac{1}{34}</math> and <math>\frac{-1 + 5 \sqrt{185}}{68}.</math> So the sum of these two is <math>\frac{1 + 5 \sqrt{185}}{68}.</math> From there, the answer is <math>\boxed{259}</math>.
  
−
Solution credit: @EpicBird08
+
~ilikemath247365
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2025|n=I|num-b=10|num-a=12}}
 
{{AIME box|year=2025|n=I|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 00:34, 14 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 $x = 34y^{2}$ 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

[asy] import graph; size(300); Label f; f.p=fontsize(6); yaxis(-2,2,Ticks(f, 2.0)); xaxis(-6.5,6.5,Ticks(f, 2.0)); draw((0, 0)..(1/4,sqrt(1/136))..(1/2,sqrt(1/68))..(0.75,sqrt(0.75/34))..(1, sqrt(1/34))..(2, sqrt(2/34))..(3, sqrt(3/34))..(4, sqrt(4/34))..(5, sqrt(5/34))..(6, sqrt(6/34))..(7, sqrt(7/34))..(8, sqrt(8/34)), red); draw((0, 0)..(1/4,-sqrt(1/136))..(0.5,-sqrt(1/68))..(0.75,-sqrt(0.75/34))..(1, -sqrt(1/34))..(2, -sqrt(2/34))..(3, -sqrt(3/34))..(4, -sqrt(4/34))..(5, -sqrt(5/34))..(6, -sqrt(6/34))..(7, -sqrt(7/34))..(8, -sqrt(8/34)), red); draw((-7,0)--(7,0), black+0.8bp); draw((0,-2.2)--(0,2.2), black+0.8bp); draw((-6,-0.1)--(-6,0.1), black); draw((-4,-0.1)--(-4,0.1), black); draw((-2,-0.1)--(-2,0.1), black); draw((0,-0.1)--(0,0.1), black); draw((2,-0.1)--(2,0.1), black); draw((4,-0.1)--(4,0.1), black); draw((6,-0.1)--(6,0.1), black); draw((-7,1)..(-5,-1), blue); draw((-5,-1)--(-3,1), blue); draw((-3,1)--(-1,-1), blue); draw((-1,-1)--(1,1), blue); draw((1,1)--(3,-1), blue); draw((3,-1)--(5,1), blue); draw((5,1)--(7,-1), blue); [/asy]

~mathkiddus

Solution 1

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

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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