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

(→‎Solution 1)
(→‎Solution 1)
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==Solution 1==
 
==Solution 1==
 
First, we calculate the probability that two segments intersect each other. Let the quadrants be numbered <math>1</math> through <math>4</math> in the normal labeling of quadrants, let the two perpendicular diameters be labeled the <math>x</math>-axis and <math>y</math>-axis, and let the two segments be <math>A</math> and <math>B.</math>
 
First, we calculate the probability that two segments intersect each other. Let the quadrants be numbered <math>1</math> through <math>4</math> in the normal labeling of quadrants, let the two perpendicular diameters be labeled the <math>x</math>-axis and <math>y</math>-axis, and let the two segments be <math>A</math> and <math>B.</math>
−
[b]Case 1:[/b] Segment <math>A</math> has endpoints in two opposite quadrants. This happens with probability <math>\frac{1}{3}.</math> WLOG let the two quadrants be <math>1</math> and <math>3.</math> We do cases in which quadrants segment <math>B</math> lies in.
+
<cmath></cmath>
 +
<math>\textbf{Case 1:}</math> Segment <math>A</math> has endpoints in two opposite quadrants. This happens with probability <math>\frac{1}{3}.</math> WLOG let the two quadrants be <math>1</math> and <math>3.</math> We do cases in which quadrants segment <math>B</math> lies in.
 
* Quadrants <math>1</math> and <math>2,</math> <math>2</math> and <math>3,</math> <math>3</math> and <math>4,</math> and <math>4</math> and <math>1</math>: These share one quadrant with <math>A,</math> and it is clear that for any of them to intersect <math>A,</math> it must be on a certain side of <math>A.</math> For example, if it was quadrants <math>1</math> and <math>2,</math> then the point in quadrant <math>1</math> must be closer to the <math>x</math>-axis than the endpoint of <math>A</math> in quadrant <math>1.</math> This happens with probability <math>\frac{1}{2}.</math> Additionally, segment <math>B</math> has a <math>\frac{1}{6}</math> to have endpoints in any set of two quadrants, so this case contributes to the total probability
 
* Quadrants <math>1</math> and <math>2,</math> <math>2</math> and <math>3,</math> <math>3</math> and <math>4,</math> and <math>4</math> and <math>1</math>: These share one quadrant with <math>A,</math> and it is clear that for any of them to intersect <math>A,</math> it must be on a certain side of <math>A.</math> For example, if it was quadrants <math>1</math> and <math>2,</math> then the point in quadrant <math>1</math> must be closer to the <math>x</math>-axis than the endpoint of <math>A</math> in quadrant <math>1.</math> This happens with probability <math>\frac{1}{2}.</math> Additionally, segment <math>B</math> has a <math>\frac{1}{6}</math> to have endpoints in any set of two quadrants, so this case contributes to the total probability
 
<cmath>\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}</cmath>
 
<cmath>\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}</cmath>
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* Quadrants <math>1</math> and <math>3.</math> We will first choose the endpoints, and then choose the segments from the endpoints. Let the endpoints of the segments in quadrant <math>1</math> be <math>R_1</math> and <math>R_2,</math> and the endpoints of the segments in quadrant <math>3</math> be <math>S_1</math> and <math>S_2</math> such that <math>R_1,R_2,S_1,</math> and <math>S_2</math> are in clockwise order. Note that the probability that <math>A</math> and <math>B</math> intersect is the probability that <math>A_1</math> is paired with <math>B_1,</math> which is <math>\dfrac{1}{2}.</math> Thus, this case contributes to the total probability
 
* Quadrants <math>1</math> and <math>3.</math> We will first choose the endpoints, and then choose the segments from the endpoints. Let the endpoints of the segments in quadrant <math>1</math> be <math>R_1</math> and <math>R_2,</math> and the endpoints of the segments in quadrant <math>3</math> be <math>S_1</math> and <math>S_2</math> such that <math>R_1,R_2,S_1,</math> and <math>S_2</math> are in clockwise order. Note that the probability that <math>A</math> and <math>B</math> intersect is the probability that <math>A_1</math> is paired with <math>B_1,</math> which is <math>\dfrac{1}{2}.</math> Thus, this case contributes to the total probability
 
<cmath>\dfrac{1}{3}\cdot\dfrac{1}{6}\cdot\dfrac{1}{2}=\dfrac{1}{36}.</cmath>
 
<cmath>\dfrac{1}{3}\cdot\dfrac{1}{6}\cdot\dfrac{1}{2}=\dfrac{1}{36}.</cmath>
−
In total, case <math>1</math> contributes
+
<cmath></cmath>
−
<cmath>\dfrac{1}{9}+\dfrac{1}{18}+\dfrac{1}{36}=\dfrac{7}{36}</cmath>
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<math>\textbf{Case 2:}</math>
−
to the total probability that two segments intersect.
+
Segment <math>A</math> has endpoints in two adjacent quadrants. This happens with probability <math>\frac{2}{3}.</math> WLOG let the two quadrants be <math>1</math> and <math>2.</math> We do cases in which quadrants segment <math>B</math> lies in.
 +
* Quadrants <math>1</math> and <math>2,</math> <math>3</math> and <math>4,</math> <math>1</math> and <math>3,</math> and <math>2</math> and <math>4.</math> This is similar to our first case above, so this contributes to the total probability
 +
<cmath>\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}</cmath>
 +
* Quadrants <math>2</math> and <math>3.</math> This cannot intersect segment <math>A.</math>
 +
* Quadrants <math>1</math> and <math>4,</math> Similar to our third case above, this intersects segment <math>A</math> with probability <math>\frac{1}{2},</math> so this case contributes to the total probability
 +
<math></math>\dfrac{2}{3}\cdot\dfrac{1}{6}\cdot\dfrac{1}{2}=\dfrac{1}{18}.$
 +
Thus, the probability that two segments intersect is
 +
<cmath>\dfrac{1}{9}+\dfrac{1}{18}+\dfrac{1}{36}+\dfrac{2}{9}+\dfrac{1}{18}=\dfrac{17}{36}.</cmath>
 
(Work in Progress)
 
(Work in Progress)
  

Revision as of 21:12, 13 February 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. 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 $A,$ it must be on a certain side 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. 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 $2,$ $3$ and $4,$ $1$ 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 $2$ and $3.$ This cannot intersect segment $A.$
  • Quadrants $1$ and $4,$ Similar to our third case above, this intersects segment $A$ with probability $\frac{1}{2},$ so this case contributes to the total probability

$$ (Error compiling LaTeX. Unknown error_msg)\dfrac{2}{3}\cdot\dfrac{1}{6}\cdot\dfrac{1}{2}=\dfrac{1}{18}.$ Thus, the probability that two segments intersect is \[\dfrac{1}{9}+\dfrac{1}{18}+\dfrac{1}{36}+\dfrac{2}{9}+\dfrac{1}{18}=\dfrac{17}{36}.\] (Work in Progress)

See also

2025 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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