Difference between revisions of "2023 AIME II Problems/Problem 6"
R00tsofunity (talk | contribs) (this is both a combinatorics and geometry problem) |
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | ==Solution 3 (Geometry)== | ||
+ | [[File:2023 AIME I 6.png|500px|right]] | ||
+ | The only configuration of two points that makes the midpoint outside of the <math>L-</math>shape is one point in the top square, and one in the right square with probability <math>\frac{2}{9}</math> (see Solution <math>1.)</math> | ||
+ | |||
+ | We use the coordinate system shown in diagram. Let the arbitrary point <math>A(x, -y), x \in [0,1], y \in [0,1]</math> be in right square. | ||
+ | It is clear that iff point <math>B</math> lies in yellow rectangle with sides <math>x</math> and <math>1 – y,</math> then the midpoint <math>AB</math> lies outside of the <math>L.</math> Probability of this is <math>p = x(1 – y).</math> | ||
+ | Consider the points <math>A_1(y,x-1), A_2(1-x,y-1), A_3(1-y,-x).</math> Similarly we find <math>p_1 = xy, p_2= (1-x)y, p_3 = (1-y)(1-x).</math> | ||
+ | The probability that the midpoint of one of the segments <math>A_1B, A_2B, A_3B</math> and <math>AB</math> is outside of the <math>L</math> shape is <math>\frac {p_1+p_2+p_3+p}{4} = \frac {1}{4}.</math> | ||
+ | If point <math>A</math> is in the right square and point <math>B</math> in the top square, then the probability that the midpoint of <math>\overline{AB}</math> lies outside <math>L-</math>shape is <math>\frac {1}{4} \implies </math> if points <math>A</math> and <math>B</math> are chosen independently and uniformly at random in <math>L-</math>shape, then the probability that the midpoint of <math>\overline{AB}</math> lies outside L-shape is <math>\frac {1}{4} \cdot \frac {2}{9} = \frac {1}{18}.</math> | ||
+ | Therefore the probability that the point is inside the <math>L-</math>shape is <math>1-\frac{1}{18}=\frac{17}{18} \implies \boxed{35}.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
== See also == | == See also == |
Revision as of 10:18, 6 March 2023
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 and point
in the top square, then 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 L-shape is
Therefore the probability that the point is inside the
shape is
vladimir.shelomovskii@gmail.com, vvsss
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 |
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