1991 AIME Problems/Problem 15
Problem
For positive integer , define to be the minimum value of the sum where are positive real numbers whose sum is 17. There is a unique positive integer for which is also an integer. Find this .
Contents
Solution 1 (Geometric Interpretation)
Consider right triangles joined at their vertices, with bases and heights . The sum of their hypotenuses is the value of . The minimum value of , then, is the length of the straight line connecting the bottom vertex of the first right triangle and the top vertex of the last right triangle, so Since the sum of the first odd integers is and the sum of is 17, we get If this is an integer, we can write , for an integer . Thus, The only possible value, then, for is , in which case , and .
Solution 2
The inequality is a direct result of the Minkowski Inequality. Continue as above.
Solution 3
Let for and . We then have that Note that that . Note that for any angle , it is true that and are reciprocals. We thus have that . By the AM-HM inequality on these values, we have that: This is thus the minimum value, with equality when all the tangents are equal. The only value for which is an integer is (see above solutions for details).
Solution 4 (Vector)
Observe the sum and apply it to the Cartesian plane, we can see that it is basically calculating distance between the origin and two points. Therefore we can assume every to be a vector from the origin to point . Now we can do sum inside the vector so we get , with to be and to be . Then we calculate the length of the vector to be . Since the sum needs to be an integer, we assume equals to . Applying difference between squares, we get that and . Therfore is and to be the final answer.
JINZHENQIAN
See also
1991 AIME (Problems • Answer Key • Resources) | ||
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