Difference between revisions of "2015 AMC 12A Problems/Problem 23"
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<cmath>2(\int_{0}^{\frac{1}{2}} 3.5dn -\int_{0}^{\frac{1}{2}}ndn -\int_{0}^{\frac{1}{2}}\sqrt{\frac{1}{4}-n^2}dn) </cmath> | <cmath>2(\int_{0}^{\frac{1}{2}} 3.5dn -\int_{0}^{\frac{1}{2}}ndn -\int_{0}^{\frac{1}{2}}\sqrt{\frac{1}{4}-n^2}dn) </cmath> | ||
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==Solution 4 (Extension, CALCULUS)== | ==Solution 4 (Extension, CALCULUS)== |
Revision as of 17:53, 9 September 2019
Contents
Problem
Let be a square of side length 1. Two points are chosen independently at random on the sides of . The probability that the straight-line distance between the points is at least is , where and are positive integers and . What is ?
Solution
Divide the boundary of the square into halves, thereby forming 8 segments. Without loss of generality, let the first point be in the bottom-left segment. Then, it is easy to see that any point in the 5 segments not bordering the bottom-left segment will be distance at least apart from . Now, consider choosing the second point on the bottom-right segment. The probability for it to be distance at least 0.5 apart from is because of linearity of the given probability. (Alternatively, one can set up a coordinate system and use geometric probability.)
If the second point is on the left-bottom segment, then if is distance away from the left-bottom vertex, then must be at least away from that same vertex. Thus, using an averaging argument we find that the probability in this case is
(Alternatively, one can equate the problem to finding all valid with such that , i.e. (x, y) is outside the unit circle with radius 0.5.)
Thus, averaging the probabilities gives
Our answer is .
Case Solution
Fix one of the points on a SIDE. There are three cases: the other point is on the same side, a peripheral side, or the opposite side, with probability , respectively.
Opposite side: Probability is obviously , no matter what.
Same side: Pretend the points are on a line with coordinates and . If , drawing a graph will give probability .
Peripheral side: superimpose a coordinate system over the points; call them and . WLOG set and . We need , and drawing the coordinate system with bounds gives probability that the distance between the points is .
Adding these up and finding the fraction gives us for an answer of .
Solution 3 (Average Function Value/Faux Integration)
WLOG, let the first point be on the bottom side of the square. The points where the second point could exist are outside a circle of radius 0.5 centered on the first point. The parts of the square that lie in this circle are the distance from the point to the closest side of the square , the distance from the point to the outside of the circle (the radius ), and any portion of the nearest side that lies within the circle as represented by the Pythagorean Theorem . Note that the perimeter of the square is 4. Thus, the total length the second point can exist in can be represented by . Distributing, .
Then, we can find the average of this function through calculus (wow more calc?). This formula is as follows,
For this case, the limits of integration are and (). Then, we have,
$$ (Error compiling LaTeX. ! Missing $ inserted.)
Working On it Right NOW as this is written
Solution 4 (Extension, CALCULUS)
Set the problem up similarly to in solution 1, where we split the square into 8 segments. Notice that each segment is the same, so WLOG use any one of them. For the purposes of this solution, I will assume the segment we use starts at (0, 0) and ends at (0.5, 0). The square I use will have vertices of (0,0), (1,0), (0,1), and (1,1).
A way we can figure out when the distance is at least 0.5 is if we figure out when it isn't 0.5. Let's pick a point on our segment and denote it with (x, 0). Then, there are three ways a second point can be within the boundaries of this first point. Either, it is to the left of it (if possible), to the right of it, or it is on the segment that forms a right angle with it.
Obviously, there is x distance if the point is to the left of the point x. Since we denoted this segment to be from 0 to 0.5, then there will always be 0.5 distance to the right of the point x (as that's the maximum that we are trying to figure out).
The difficult part is finding the total length on the segment that is perpendicular to our segment. However, since the square has a right angle, we can first find that the segment should have a length of (by the Pythagorean theorem, with hypotenuse 0.5 and one leg being x).
Now that we have our three distances, all we need to do is find the average value of them. We can best do this with "Integral/Interval", and so we take the integral from 0 to 0.5 of and then divide it by 0.5 (the interval). To integrate by hand, we want to pull out the , and then apply u-sub and our integration rules to find the answer.
We get 0.75 + times 2. Simplify this into .
Now we are on our last stage. Proceed to make this equivalent to what the question is asking, as we have found the probability that the second point is within a distance of 0.5, whereas the question asks for at least a distance of 0.5 (so more than). We can do this simply by doing (as 4 is the total amount of length). This equates to and then we divide by 4 as that is the total amount of length (remember this is probability).
Thus, we get for our probability, and so the answer is 26 + 1 + 32 = (A) 59.
Some notes: I tried to explain everything but it's quite difficult to explain - there is a way of non-calculus (like there always is) that I think was mentioned above, something with circles (since the thing under the square root is just x^2+y^2 = 1/4, so finding the average value of that isn't difficult). IronicNinja~
See Also
2015 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
2015 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |