1984 AHSME Problems/Problem 29
Problem
Find the largest value for for pairs of real numbers which satisfy .
Solution
Let , so that . Substituting this into the given equation yields . Multiplying this out and forming it into a quadratic yields .
We want to be a real number, so we must have the discriminant . The discriminant is . Therefore, we must have , or . The roots of this quadratic, using the quadratic formula, are , so the quadratic can be factored as . We can now separate this into cases:
Case 1: Then, both terms in the factored quadratic are negative, so the inequality doesn't hold.
Case 2: Then, the first term is positive and the second is negative and the second is positive, so the inequality holds.
Case 3: Then, both terms are positive, so the inequality doesn't hold.
Also, when or , the equality holds.
Therefore, we must have , and the maximum value of is .
Solution 2
The equation represents a circle of radius centered at . To find the maximal with is equivalent to finding the maximum slope of a line passing through the origin and intersecting the circle. The steepest such line is tangent to the circle at some point . We have , , because the line is tangent to the circle. Using the pythagorean theorem, we have .
The slope we are looking for is equivalent to where . Using tangent addition,
So is the answer
Solution 3
Following the steps of Solution 2, we get . Let . Based on the distance to the origin and to the centre of the circle
Expanding the latter and substituting the former into the equation gives . . We can square this and remove the 1st equation to get , or .
Substitute to get
Since and are symmetric, we want to be the lower number to maximize , so and .
See Also
1984 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 28 |
Followed by Problem 30 | |
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