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Difference between revisions of "2025 AIME I Problems/Problem 13"

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~BS2012, eevee9406 ~hashbrown2009
 
~BS2012, eevee9406 ~hashbrown2009
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== Solution 2 (probably unrigorous) ==
 
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Ignore the fact that the segments must have endpoints in diametrically opposite quadrants.
 
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The key is to observe that the number of regions the disk is split into is equal to <math>K+S+1,</math> where <math>K</math> denotes the number of intersections† and <math>S</math> denotes the number of segments.
 
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We use linearity of expectation. Take two arbitrary segments on the disk, and they intersect with probability <math>\frac 12.</math>  In the problem, we are already given the fact that one pair of our <math>27</math> segments are guaranteed to intersect, and for the other <math>\binom{27}{2}-1=350</math> pairs of segments, since at least one segment is chosen randomly, these are all random and each of these pairs intersect with probability <math>\frac 12.</math> Thus, <math>175+1=176</math> intersections are expected.
 
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It is easy to see from here the answer is <math>176+27+1=\boxed{204}.</math>
 
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~pieMax2713
 
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†This only holds true if no three segments intersect at a common point, which happens with probability 0 in this scenario.
 
  
 
== Video Solution ==
 
== Video Solution ==

Latest revision as of 11:22, 19 August 2025

Problem

Alex divides a disk into four quadrants with two perpendicular diameters intersecting at the center of the disk. He draws $25$ more lines segments through the disk, drawing each segment by selecting two points at random on the perimeter of the disk in different quadrants and connecting these two points. Find the expected number of regions into which these $27$ line segments divide the disk.

Solution 1

First, we calculate the probability that two segments intersect each other. Let the quadrants be numbered $1$ through $4$ in the normal labeling of quadrants, let the two perpendicular diameters be labeled the $x$-axis and $y$-axis, and let the two segments be $A$ and $B.$

$\textbf{Case 1:}$ Segment $A$ has endpoints in two opposite quadrants.

[asy] pair A,B,C,D,E,F,O; A=(1,0);B=(0,1);C=(-1,0);D=(0,-1);E=(0.5,0.86602540);F=(-0.707106,-0.707106);O=(0,0); draw(A--C);draw(B--D);draw(circle((0,0),1));draw(E--F,blue);dot(E,blue);dot(F,blue); [/asy]

This happens with probability $\frac{1}{3}.$ WLOG let the two quadrants be $1$ and $3.$ We do cases in which quadrants segment $B$ lies in.

  • Quadrants $1$ and $2,$ $2$ and $3,$ $3$ and $4,$ and $4$ and $1$: These share one quadrant with $A,$ and it is clear that for any of them to intersect $A,$ the endpoint that shares a quadrant with an endpoint of $A$ on a certain side of that endpoint of $A$ For example, if it was quadrants $1$ and $2,$ then the point in quadrant $1$ must be closer to the $x$-axis than the endpoint of $A$ in quadrant $1.$ This happens with probability $\frac{1}{2}.$ Additionally, segment $B$ has a $\frac{1}{6}$ to have endpoints in any set of two quadrants, so this case contributes to the total probability

\[\dfrac{1}{3}\left(\dfrac{1}{6}\cdot\dfrac{1}{2}+\dfrac{1}{6}\cdot\dfrac{1}{2}+\dfrac{1}{6}\cdot\dfrac{1}{2}+\dfrac{1}{6}\cdot\dfrac{1}{2}\right)=\dfrac{1}{9}\]

  • Quadrants $2$ and $4.$ This always intersects segment $A,$ so this case contributes to the total probability

\[\dfrac{1}{3}\cdot\dfrac{1}{6}=\dfrac{1}{18}\]

  • Quadrants $1$ and $3.$ We will first choose the endpoints, and then choose the segments from the endpoints. Let the endpoints of the segments in quadrant $1$ be $R_1$ and $R_2,$ and the endpoints of the segments in quadrant $3$ be $S_1$ and $S_2$ such that $R_1,R_2,S_1,$ and $S_2$ are in clockwise order. Note that the probability that $A$ and $B$ intersect is the probability that $A_1$ is paired with $B_1,$ which is $\dfrac{1}{2}.$ Thus, this case contributes to the total probability

\[\dfrac{1}{3}\cdot\dfrac{1}{6}\cdot\dfrac{1}{2}=\dfrac{1}{36}.\]

$\textbf{Case 2:}$ Segment $A$ has endpoints in two adjacent quadrants.

[asy] pair A,B,C,D,E,F,O; A=(1,0);B=(0,1);C=(-1,0);D=(0,-1);E=(0.5,0.86602540);F=(-0.707106,0.707106);O=(0,0); draw(A--C);draw(B--D);draw(circle((0,0),1));draw(E--F,blue);dot(E,blue);dot(F,blue); [/asy]

This happens with probability $\frac{2}{3}.$ WLOG let the two quadrants be $1$ and $2.$ We do cases in which quadrants segment $B$ lies in.

  • Quadrants $1$ and $3,$ $1$ and $4,$ $2$ and $3,$ and $2$ and $4.$ This is similar to our first case above, so this contributes to the total probability

\[\dfrac{2}{3}\left(\dfrac{1}{6}\cdot\dfrac{1}{2}+\dfrac{1}{6}\cdot\dfrac{1}{2}+\dfrac{1}{6}\cdot\dfrac{1}{2}+\dfrac{1}{6}\cdot\dfrac{1}{2}\right)=\dfrac{2}{9}\]

  • Quadrants $3$ and $4.$ This cannot intersect segment $A.$
  • Quadrants $1$ and $2,$ Similar to our third case above, this intersects segment $A$ with probability $\frac{1}{2},$ so this case contributes to the total probability

\[\dfrac{2}{3}\cdot\dfrac{1}{6}\cdot\dfrac{1}{2}=\dfrac{1}{18}.\] Thus, the probability that two segments intersect is \[\dfrac{1}{9}+\dfrac{1}{18}+\dfrac{1}{36}+\dfrac{2}{9}+\dfrac{1}{18}=\dfrac{17}{36}.\] Next, we will compute the expected number of intersections of a segment with the axes. WLOG let a segment have an endpoint in quadrant $1.$ Then, it will intersect each axis with probability $\dfrac{2}{3}$ because two out of the three remaining quadrants let it intersect a specific axis, so the expected number of axes a segment intersects is $\frac{4}{3}.$

So, why do intersections matter? Because when adding a segment, it will pass through a number of regions, and for each region it passes through, it will split that region into two and add another region. The segment starts in a region, and for each intersection, it will enter another region, so the number of regions a segment passes through is $1$ more than the number of intersections with the axes and other segments. Thus, we have that by linearity of expectation the expected number of new regions created by adding a segment is \[\dfrac{17}{36}\cdot(\text{number of segments already added})+\dfrac{4}{3}+1,\] so the number of new regions added in total by $25$ segments again by linearity of expectation is \[\sum_{k=0}^{24}\left(\dfrac{17}{36}k+\dfrac{7}{3}\right)=\dfrac{17}{36}\cdot \dfrac{24\cdot 25}{2}+\dfrac{25\cdot 7}{3}\] which simplifies to $200$ as the expected number of new regions added by the $25$ segments. The axes create $4$ regions to begin with, so our answer is \[200+4=\boxed{204}.\]

~BS2012, eevee9406 ~hashbrown2009

Video Solution

2025 AIME I #13

MathProblemSolvingSkills.com


See Also

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

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