2011 AIME I Problems/Problem 6
Problem
Suppose that a parabola has vertex and equation , where and is an integer. The minimum possible value of can be written in the form , where and are relatively prime positive integers. Find .
Solution
If the vertex is at , the equation of the parabola can be expressed in the form . Expanding, we find that , and . From the problem, we know that the parabola can be expressed in the form , where is an integer. From the above equation, we can conclude that , , and . Adding up all of these gives us . We know that is an integer, so must be divisible by 16. Let . If , then . Therefore, if , . Adding up gives us
Solution 2
Complete the square. Since , the parabola must be facing upwards. means that must be an integer. The function can be recasted into because the vertex determines the axis of symmetry and the critical value of the parabola. The least integer greater than is . So the -coordinate must change by and the -coordinate must change by . Thus, . So .
Solution 3
To do this, we caN use the formula for the minimum (or maximum) value of the coordinate at a vertex of a parabola, and equate this to . Solving, we get . Enter to get so . This means that so the minimum of is when the fraction equals -1, so . Therefore, . -Gideontz
Video Solution
https://www.youtube.com/watch?v=vkniYGN45F4
See also
2011 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
- AIME Problems and Solutions
- American Invitational Mathematics Examination
- Mathematics competition resources
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.