2011 AIME I Problems/Problem 6
Contents
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 . 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
Solution 4
Write this as . Since is equal to the value of this expression when you plug in, we just need to be an integer. Since , we also have which means . The least possible value of is when this is equal to , or , which gives answer .
-bobthegod78, krwang, Simplest14
Solution 5 (You don't remember conic section formulae)
Take the derivative to get that the vertex is at and note that this implies and proceed with any of the solutions above.
~Dhillonr25
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 | |
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