2011 AIME II Problems/Problem 15
Problem
Let . A real number is chosen at random from the interval . The probability that is equal to , where , , , , and are positive integers. Find .
Solution 1
Table of values of :
In order for to hold, must be an integer and hence must be a perfect square. This limits to or or since, from the table above, those are the only values of for which is an perfect square. However, in order for to be rounded down to , must be less than the next perfect square after (for the said intervals). Now, we consider the three cases:
Case :
must be less than the first perfect square after , which is , i.e.:
(because implies )
Since is increasing for , we just need to find the value where , which will give us the working range .
So in this case, the only values that will work are .
Case :
must be less than the first perfect square after , which is .
So in this case, the only values that will work are .
Case :
must be less than the first perfect square after , which is .
So in this case, the only values that will work are .
Now, we find the length of the working intervals and divide it by the length of the total interval, :
Thus, the answer is .
P.S. You don't need to calculate all the values of P(x) calculated by the above solution. Some very simple modular arithmetic eliminates a large portion of the numbers. The time saved is not that much if you are already at your mathcounts prime.
Solution 2
Make the substitution , so We're looking for solutions to with the new bounds . Since the right side is an integer, it must be that is a perfect square. For simplicity, write and Since , it must be that , which gives solutions , respectively. But this gives us three cases to check:
Case 1: .
In this case, we have Case 2: .
In this case, we have Case 3:
In this case, we have To finish, the total length of the interval from which we choose is . The total length of the success intervals is which means the probability is AIMEifying the answer gives .
Solution 3 (Graphing)
It's definitely possible to conceptualize this problem visually, if that's your thing, since it is a geometric probability problem. Let and . The graph of and will look like this, with having only integral y-values and having only integral x-values:
As both and consist of a bunch of line segments, the probability that is the "length" of the overlap between the segments of and divided by the total length of the segments of .
Looking at the graph, we see that and will overlap only when is an integer. Specifically, each region of overlap will begin when has solutions for integral in the range of , which consists of the integers , and end when jumps up to its next y-value.
Using the quadratic formula, we see that the start point of each of these overlapping segments will be the integral values of for in the specified range, meaning must be a perfect square. Plugging in all the possible values of , we get , corresponding to start points of . As already stated, the endpoints will occur when jumps up to the next integer at each of these segments, at which point the x-value will be . On the graph, the overlapping segments of and would be represented by the highlighted green segments below:
Taking the difference between this second x-value and the start point for each of our start points and summing them will give us the total length of these green segments. We can then divide this value by ten (the total length of the segments of ) to give us the probability of overlap between and .
Doing so gives us:
.
~ anellipticcurveoverq
Solution 4
Note that all the "bounds" have to be less than the number+1, otherwise it wouldn't fit the answer format. Therefore, the answer is
~Lcz
See also
2011 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
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