2011 AIME I Problems/Problem 6
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 .
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
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 .
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
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