2023 AIME II Problems/Problem 6
Contents
Problem
Consider the L-shaped region formed by three unit squares joined at their sides, as shown below. Two points and are chosen independently and uniformly at random from inside the region. The probability that the midpoint of also lies inside this L-shaped region can be expressed as where and are relatively prime positive integers. Find
Video Solution by MOP 2024
https://youtube.com/watch?v=0UViBt-LYTo&t=0s
Solution 1
We proceed by calculating the complement.
Note that the only configuration of the 2 points that makes the midpoint outside of the L shape is one point in the top square, and one in the right square. This occurs with probability.
Let the topmost coordinate have value of: , and rightmost value of: .
The midpoint of them is thus:
It is clear that are all between 0 and 1. For the midpoint to be outside the L-shape, both coordinates must be greater than 1, thus:
By symmetry this has probability . Also by symmetry, the probability the y-coordinate works as well is . Thus the probability that the midpoint is outside the L-shape is:
We want the probability that the point is inside the L-shape however, which is , yielding the answer of ~SAHANWIJETUNGA
Solution 2（Calculus)
We assume each box has side length 1. We index the upper left box, the bottom left box, the bottom right box as II, III, IV, respectively. We index the missing upper right box as I. We put the graph to a coordinate system by putting the intersecting point of four foxes at the origin, the positive direction of the axis at the intersecting line of boxes I and IV, and the positive direction of the -axis at the intersecting line of boxes I and II. We denote by the midpoint of .
Therefore,
We observe that a necessary for is either and , or and . In addition, by symmetry,
Thus, The second equality follows from the condition that the positions of and are independent. The sixth equality follows from the condition that for each point of and , the and coordinate are independent.
Therefore,
Therefore, the answer is .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 3 (Geometry)
The only configuration of two points that makes the midpoint outside of the shape is one point in the top square, and one in the right square with probability (see Solution
We use the coordinate system shown in diagram. Let the arbitrary point be in right square. It is clear that iff point lies in yellow rectangle with sides and then the midpoint lies outside of the
Probability of this is
Consider the points
Similarly we find
The probability that the midpoint of one of the segments and is outside of the shape is If point is in the right square, point in the top square, the probability that the midpoint of lies outside shape is
If points and are chosen independently and uniformly at random in shape, then the probability that the midpoint of lies outside shape is Therefore the probability that the point is inside the shape is
vladimir.shelomovskii@gmail.com, vvsss
Solution 4 (Coordinates, similar to Solution 1)
Consider this diagram:
First, the one of points must be in the uppermost box and the other in the rightmost box. This happens with probability 2/3*1/3=2/9.
We need the midpoints of the coordinates to be greater than but less than
We need the midpoints of the coordinates to be greater than but less than
Thus, we set up:
Using geometric probability, as shown below, we get that the probability of happening is 1/2. This is for both. So we have probability 1/2*1/2=1/4.
Thus, the desired is 2/9*1/4=1/18 => 1-1/18=17/18.
mathboy282
Video Solution by The Power of Logic
See also
2023 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.