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 non perfect squares, 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 greater than the distance between and . Therefore, the segment from to is the longest attainable (the other possible coordinates for and are and , and , and . 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 and vice versa. Using the distance formula, we get that is and that is
Solution 2
We can look at the option choices. Since we are aiming for the highest possible ratio, let's try using (though actually is the highest ratio.) 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 : . 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 ? 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 .
Quality Control by fasterthanlight
(Note by Carrot_Karen: We tried , but some might be confused why we concluded that it was the answer after verifying without trying the others, like why wasn't option tried? This is because the problem can only have one correct answer, so if we have an option that already works, we can conclude that none of the others work and is the answer.
Solution 3
By inspection, when is at and is at it makes as small as possible with a distance of . The greatest possible length of arises when is at and is at Using the distance formula, we find that has a length of The requested fraction is then .
Video Solution
https://youtu.be/umr2Aj9ViOA?t=162
See Also
2017 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.