2017 AMC 10A Problems/Problem 17
Problem
Distinct points , , , lie on the circle and have integer coordinates. The distances and are irrational numbers. What is the greatest possible value of the ratio ?
Solution 1
Because , , , and are lattice points, there are only a few coordinates that actually satisfy the equation. The coordinates are and We want to maximize and minimize They also have to be the square root of something, because they are both irrational. The greatest value of happens when and are almost directly across from each other and are in different quadrants. For example, the endpoints of the segment could be and because the two points are almost across from each other. Another possible pair could be and . To find out which segment is longer, we have to compare the distances from their endpoints to a diameter (which must be the longest possible segment). The closest diameter would be from to . The distance between and is shorter than the distance between and . Therefore, the segment from to is the longest attainable. The least value of is when the two endpoints are in the same quadrant and are very close to each other. This can occur when, for example, is and is They are in the same quadrant and no other point on the circle with integer coordinates is closer to the point than Using the distance formula, we get that is and that is
Solution 2
So what we can do is look at the option choices. Since we are aiming for the highest possible ratio, let's try using . Now, looking at the problem alone, we know that to have the largest ratio possible, we have to let be the minimum possible value while at the same time using integer coordinates. Thus, the smallest possible value of is . Assuming that , we plug in and solve for PQ: . Remember, we don't know if this is possible yet, we are only trying to figure out if it is. But for what values of and does ? Aha! We see that this can easily be made into a triangle. But, instead of substituting into the equation and then using a whole lot of algebra, we can save time and use the little trick, that if in a triangle, the two degree sides have side length , then the hypotenuse is . Using this, we can see that , and since our equation does in fact yield a sensible solution, we can be assured that our answer is .
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See Also
2017 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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