2017 AIME I Problems/Problem 8
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Problem 8
Two real numbers and are chosen independently and uniformly at random from the interval . Let and be two points on the plane with . Let and be on the same side of line such that the degree measures of and are and respectively, and and are both right angles. The probability that is equal to , where and are relatively prime positive integers. Find .
Solution 1
Noting that and are right angles, we realize that we can draw a semicircle with diameter and points and on the semicircle. Since the radius of the semicircle is , if , then must be less than or equal to .
This simplifies the problem greatly. Since the degree measure of an angle on a circle is simply half the degree measure of its subtended arc, the problem is simply asking:
Given such that , what is the probability that ? Through simple geometric probability, we get that .
The answer is
~IYN~ Note: The Geometric probability can be easily be found through graphic through the x-y plane.
Solution 2 (Trig Bash)
Put and with on the origin and the triangles on the quadrant. The coordinates of and is , . So = , which we want to be less then . So So we want , which is equivalent to or . The second inequality is impossible so we only consider what the first inequality does to our by box in the plane. This cuts off two isosceles right triangles from opposite corners with side lengths from the by box. Hence the probability is and the answer is
Solution by Leesisi
Solution 3 (Quicker Trig)
Let Since we are given many angles in the problem, we can compute the lengths of some of the lines in terms of trigonometric functions: Now observe that quadrilateral is a cyclic quadrilateral. Thus, we are able to apply Ptolemy's Theorem to it: We want (the absolute value comes from the fact that is not necessarily greater than so we cannot assume that is to the right of as in the diagram), so we substitute: By simple geometric probability (see Solution 2 for complete explanation), so
~burunduchok
Solution 4
Scale the circle down from radius 100 (diameter 200) to radius 6 (diameter 12). Then we want the probability that . Now note that all possible and lie on a interval on the circumference of the circle. But for , and must be less than apart on the circumference of the circle. Simple geometric probability gives us , so the answer is . (Professor-Mom)
Solution 5
Impose a coordinate system as follows:
Let the midpoint of be the origin, and let be the x-axis. We construct a circle with center at the origin with radius 100. Since and are both right angles, points and are on our circle. Place and in the first quadrant of the Cartesian Plane. Suppose we construct and such that they are clockwise rotations of and , respectively by an angle of degrees. Thus, we see that . We want this quantity to be less than . This happens when or when . The probability that the last inequality is satisfied is . Therefore, the probability that is less than is . Hence,
~MathIsFun286
slightly edited
~Txu
Solution 6
WLOG, let . It does not actually matter, but it is necessary for this particular setup. It should be apparent that . We write the equation
If we examine right triangle , we can see that . Also, we are given , so now we have
We want to be less than or equal to ; this is equivalent to We solve from there:
\begin{align*} \dfrac{QR}{200}&\le\dfrac12 \\ \sin(b-a)&\le\dfrac12 \\ \arcsin(\sin(b-a))&\le\arcsin\left(\dfrac12\right) \\ b-a&\le30^\circ. \\ \end{align*}
(Notice that if , then this would become As in Solution 1, we can write .) One can now proceed as in Solution 1, but let us tackle the geometric probability for completeness.
We now have transformed this problem into another problem asking for the probability of two uniformly, randomly, and independently chosen real numbers between and being no more than from each other.
If the first number (let this be ) is between and , then the other number can be from to - a range of . Thus, the probability that this contributes is .
If is between and or to (these two cases are equivalent), the chance is the same as that of the average value since the ranges are uniform. For (the average), the second number can be from to - a range of . The total range is . Thus, this case contributes .
Adding the two, we get for an answer of .
~~Technodoggo
See Also
2017 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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