2009 AMC 12A Problems/Problem 23
Contents
Problem
Functions and are quadratic, , and the graph of contains the vertex of the graph of . The four -intercepts on the two graphs have -coordinates , , , and , in increasing order, and . Then , where , , and are positive integers, and is not divisible by the square of any prime. What is ?
Solution (Alcumus)
The two quadratics are rotations of each other about . Since we are only dealing with differences of roots, we can translate them to be symmetric about . Now and . Say our translated versions of and are and , respectively, so that . Let be a root of and a root of by symmetry. Note that since they each contain each other's vertex, , , , and must be roots of alternating polynomials, so is a root of and a root of
The vertex of is half the sum of its roots, or . We are told that the vertex of one quadratic lies on the other, so
Let and divide through by , since it will drastically simplify computations. We know and that , or
So . Since , .
The answer is , and .
Note
Actually it is not necessary to solve any quadratic equations, if one utilizes the two facts about the quadratic () that (i) the difference of the two quadratic roots equals to , and (ii) that the minimum value of a quadratic equals to , where . Here is a possible adjustment to the solution:
Without loss of generality we may "shift" ,, units to the left, then the differences of remain the same, and are symmetrical about , so , . The relationship of , becomes . So we may write:
Again without loss of generality, we can assume and (Short argument is needed here instead of the lazy "wlog"). Also, the vertex of is , so , or .
Since are roots of , we have the following relationship of the roots:
So , or .
Therefore .
See also
2009 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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